All India Medical & Engineering Entrance Examination
NEET/JEE (Main)
CHEMISTRY
Previous Years
Chapterwise Objective
Solved Papers
VOLUME-II
Useful for : NEET/AIPMT, AIIMS, JEE (Main), AMU, AP EAMCET (Medical), AP EAMCET (Engg.),
ASSAM CEE, BCECE, BITSAT, Chhattisgarh-PET, COMEDK, Gujarat Common Entrance Test
(GUJCET), Himanchal Pradesh-CET, J & K CET, JCECE, Kerala-CEE, Karnataka-CET(KCET),
MP-PET, MANIPAL, JIPMER, MHT-CET, Odisha-JEE, SCRA, SRM-JEE, TS-EAMCET(Medical),
TS-EAMCET (Engg.), UPCPMT, UPTU, UPSEE, UPSC NDA/NA, VITEEE, WEST BENGAL JEE.
Chief Editor
A.K. Mahajan
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INDEX
NEET & JEE Main Chemistry Syllabus ........................................................................................................4-9
All India Medical & Engineering Entrance Exam, AIIMS, NEET, & JEE Main
Chemistry Previous Years Exam Papers Analysis Chart .........................................................................10-19
Trend Analysis of NEET/JEE (Main) Chemistry Questions Through Pie Chart & Bar Graph ................ 20
SOLUTIONS ................................................................................................................................................21-143
Methods of Expressing Concentration of Solutions ...................................................................................... 21
Colligative Properties and Determination of molar mass .............................................................................. 72
Lowering of Vapour Pressure ........................................................................................................................ 95
Osmosis and Osmotic Pressure of the Solution ........................................................................................... 121
Solubility .................................................................................................................................................... 129
Isotonic Solution ......................................................................................................................................... 133
Coagulation ................................................................................................................................................. 138
Types of Solution ........................................................................................................................................ 143
ELECTROCHEMISTRY ........................................................................................................................144-239
Electrode Potential ...................................................................................................................................... 144
Faraday's Law.............................................................................................................................................. 187
Cell Constant ............................................................................................................................................... 204
Nernst Equation .......................................................................................................................................... 212
Conductance and Conductor ....................................................................................................................... 215
Type of Cell ................................................................................................................................................. 231
Molar conductance ...................................................................................................................................... 239
CHEMICAL KINETICS ..........................................................................................................................240-353
Rate of Chemical Reaction .......................................................................................................................... 240
Rate Law and Rate Constant ....................................................................................................................... 258
Order of Reaction, Molecularity ................................................................................................................. 280
Arrhenius Equation ..................................................................................................................................... 323
Half-life Time of Chemical Reaction .......................................................................................................... 339
Factor Affecting the Rate of Reaction ......................................................................................................... 350
Temperature Quotient.................................................................................................................................. 353
NUCLEAR CHEMISTRY ........................................................................................................................354-379
α, β and γ-rays decay .................................................................................................................................. 354
Rate of Decay and Half-life......................................................................................................................... 362
Isotopes-Isotones and Nuclear Isomer......................................................................................................... 377
Decay Series ................................................................................................................................................ 379
SURFACE CHEMISTRY .........................................................................................................................380-416
Homogenous and Heterogenous Catalysis .................................................................................................. 380
Freundlick's Isotherm .................................................................................................................................. 384
Langmuir's Isotherm .................................................................................................................................... 390
Adsorption and Adsorption Isotherm .......................................................................................................... 391
Collidal Solution ......................................................................................................................................... 401
GENERAL PRINCIPLES AND PROCESS OF ISOLATION OF ELEMENTS ................................417-459
Occurrence of Metals .................................................................................................................................. 417
Process of Extraction of Ore ....................................................................................................................... 425
Refining ....................................................................................................................................................... 441
Electrochemical Principles of Metallurgy ................................................................................................... 446
Alloy ............................................................................................................................................................ 455
Thermodynamic Principles of Metallurgy ................................................................................................... 459
THE d-AND f-BLOCK ELEMENTS ......................................................................................................460-517
Electronic Configuration of d-block and f-block elements ......................................................................... 460
The Lanthanoids .......................................................................................................................................... 471
The Actinoids ............................................................................................................................................. 477
Oxidation State of Transition Element ........................................................................................................ 479
Oxidation State of Inner Transition Element ............................................................................................... 485
d-and f-block Metal ..................................................................................................................................... 486
Magnetic properties of d-and f-block Elements ......................................................................................... 517
Colour of d-block Elements ........................................................................................................................ 517
COORDINATION COMPOUNDS ..........................................................................................................518-628
Isomerism of Coordination Compound ..................................................................................................... 518
Nomenclature .............................................................................................................................................. 536
Magnetic Nature and Coordination Number ............................................................................................... 550
Hybridisation ............................................................................................................................................... 579
2
Spectrochemical Series, Complex Stability................................................................................................. 586
Crystal Field Theory.................................................................................................................................... 604
Oxidation State of Coordination Compounds ............................................................................................. 614
Spectra and Colour of Co-ordination Compounds ...................................................................................... 619
Werner's Theory .......................................................................................................................................... 627
Bonding in Metal carbonyl ......................................................................................................................... 628
Organometallic compound ......................................................................................................................... 628
GENERAL ORGANIC CHEMISTRY ...................................................................................................629-836
Electronic effect and Its Applications ......................................................................................................... 629
Resonance .................................................................................................................................................. 699
Nomenclature of Organic compound ......................................................................................................... 723
Isomerism .................................................................................................................................................... 752
Dipole Moment ........................................................................................................................................... 814
Aromaticity ................................................................................................................................................. 816
Stereochemistry ........................................................................................................................................... 836
Reaction Intermediate.................................................................................................................................. 836
HALOALKANES AND HALOARENES ...............................................................................................837-891
Properties of Alkyl Halides ......................................................................................................................... 837
Preparation of Alkyl Halides ....................................................................................................................... 860
Haloarenes ................................................................................................................................................... 882
Polyhalogen Compounds ............................................................................................................................. 888
ALCOHOLS, PHENOLS AND ETHER .................................................................................................892-969
Identification of Alcohol, Properties of Alcohol ......................................................................................... 892
Preparation of Alcohols, Phenols and Ethers .............................................................................................. 930
Cyclic and Aromatic Ethers ........................................................................................................................ 962
Reaction's of Alcohols, Phenols and Ether .................................................................................................. 964
ALDEHYDES, KETONES AND CARBOXYLIC ACID ....................................................................970-1103
Identification of Aldehydes and Ketones .................................................................................................... 970
Derivatives of Carboxylic Acid ................................................................................................................... 993
Esters ........................................................................................................................................................ 1030
Reactions of Aldehyde and ketone ........................................................................................................... 1046
Properties of Carboxylic acid .................................................................................................................... 1053
Reactions of Carboxylic acids ................................................................................................................... 1062
Aliphatic and Aromatic Carbonyl Group ................................................................................................. 1070
Reagent ..................................................................................................................................................... 1098
AMINES ................................................................................................................................................1104-1163
Aromatic Nitro-Compound ....................................................................................................................... 1104
Diazo-tization Reaction ............................................................................................................................. 1121
Different Nitrogen Containing Compound ................................................................................................ 1125
Nature of Amines ...................................................................................................................................... 1162
Properties of Amines ................................................................................................................................ 1163
BIOMOLECULES ................................................................................................................................1164-1243
Carbohydrates............................................................................................................................................ 1164
Vitamins .................................................................................................................................................... 1200
Proteins ...................................................................................................................................................... 1205
Enzymes .................................................................................................................................................... 1213
Nucleic Acids ............................................................................................................................................ 1219
Amino Acids ............................................................................................................................................. 1226
POLYMERS ..........................................................................................................................................1244-1294
Classification of Polymers......................................................................................................................... 1244
Biodegradable Polymers............................................................................................................................ 1260
Properties of Polymer ................................................................................................................................ 1263
Type of Polymerisation Reaction .............................................................................................................. 1279
CHEMISTRY IN EVERYDAY LIFE .................................................................................................1295-1326
Drugs and Medicines ................................................................................................................................. 1295
Dyes and Pigment, Detergent .................................................................................................................... 1317
Type of Fuel .............................................................................................................................................. 1325
Chemicals in Food ..................................................................................................................................... 1326
REACTION MECHANISM .................................................................................................................1327-1472
3
....
SYLLABUS
UNIT : VII Equilibrium
NEET
CLASS 11
Equilibrium in physical and chemical processes, dynamic
nature of equilibrium, law of chemical equilibrium,
equilibrium constant, factors affecting equilibrium-Le
Chatelier’s principle, ionic equilibrium- ionization of
acids and bases, strong and weak electrolytes, degree of
ionization, ionization of polybasic acids, acid strength,
concept of pH, Hydrolysis of salts (elementary idea),
buffer solutions, Henderson equation, solubility product,
common ion effect (with illustrative examples).
th
UNIT : I Some Basic Concepts of Chemistry
General Introduction Importance and scope of chemistry.
Laws of chemical combination, Dalton’s atomic theory
concept of elements, atoms and molecules. Atomic and
molecular masses. Mole concept and molar mass,
percentage composition and empirical and molecular
formula, chemical reactions, stoichiometry and UNIT : VIII Redox Reactions
calculations based on stoichiometry.
Concept of oxidation and reduction, redox reaction
oxidation number, balancing redox in terms of loss and
UNIT : II Structure of Atom
Atomic number, isotopes and isobars. Concept of shells gain of electron and change in oxidation numbers.
and subshells, dual nature of matter and light, de- UNIT : IX Hydrogen
Broglie’s
relationship,
Heisenberg’s
uncertainty Occurrence, isotopes, preparation, properties and uses of
principle, concept of orbital, quantum numbers, shapes hydrogen, hydrides–ionic, covalent and interstitial,
of s, p and d orbitals, rules for filling electrons in orbitals physical and chemical properties of water, heavy water,
– Aufbau principle, Pauli exclusion principles and hydrogen peroxide-preparation, reaction, uses and
Hund’s rule, electronic configuration of atoms, stability structure.
of half–filled and completely filled orbitals.
UNIT : X s-Block Elements (Alkali and Alkaline
Earth Metals)
UNIT : III Classification of Elements and
Group 1 and group 2 elements General introduction,
Periodicity in Properties
electronic
configuration,
occurrence,
anomalous
properties of the first element of each group, diagonal
relationship, trends in the variation of properties (such as
ionization enthalpy, atomic and ionic radii), trends in
chemical reactive with oxygen, water hydrogen and
halogens, uses. Preparation and Properties of Some
important Compounds. Sodium carbonate, sodium
chloride,
sodium
hydroxide
and
sodium
hydrogencarbonate, biological importance of sodium and
potassium. Industrial use of lime an limestone, biological
importance of Mg and Ca.
Modern periodic law and long form of periodic table,
periodic trends in properties of elements–atomic radii,
ionic radii, ionization enthalpy, electron gain enthalpy,
electronegativity, valence.
UNIT : IV Chemical Bonding and Molecular
Structure
Valence electrons, ionic bond, covalent bond, bond
parameters, Lewis structure, polar character of covalent
bond, valence bond theory, resonance, geometry of
molecules, VSEPR theory, concept of hybridization
involving s, p and d orbitals and shapes of some simple UNIT : XI Some p–Block Elements
molecules, molecular orbital theory of homonuclear General Introduction to p-Block Elements.
diatomic molecules (qualitative idea only). Hydrogen bond.
Group 13 elements General Introduction, electronic
UNIT : V States of Matter : Gases and Liquids
configuration, occurrence, variation of properties,
The states of matter, intermolecular interactions, types of oxidation states, trends in chemical reactivity, anomalous
bonding, melting and boiling points, role of gas laws of properties of first element of the group; Boron, some
elucidating the concept of the molecule, Boyle’s law, important compounds borax, boric acids, boron hydrides.
Charles’ law, Gay Lussac’s law, Avogadro’s law, ideal Aluminium, uses reactions with acids and alkalies.
behavior of gases, empirical derivation of gas equation. General 14 elements General introduction, electronic
Avogadro number, ideal gas equation. Kinetic energy configuration, occurrence, variation of properties,
and molecular speeds (elementary idea), deviation from oxidation states, trends in chemical reactivity, anomalous
ideal behavior, liquefaction of gases, critical temperature. behaviour of first element. Carbon, allotropic forms,
Liquid State –Vapour pressure, viscosity and surface physical and chemical properties, uses of some important
tension (qualitative idea only, no mathematical compounds, oxides.
derivations).
Important compounds of silicon and a few uses, silicon
tetrachloride, silicones, silicates and zeolites, their uses.
UNIT : VI Thermodynamics
First law of thermodynamics internal energy and UNIT : XII Organic Chemistry – Some Basic
enthalpy, heat capacity and specific heat, measurement
Principles and Techniques
of U and H, Hess’s law of constant heat summation, General introduction, methods of purification qualitative
enthalpy of : bond dissociation, combustion, formation, and quantitative analysis, classification and IUPAC
atomization, sublimation, phase transition, ionization, nomenclature of organic compounds. Electronic
solution and dilution. Introduction of entropy as state displacements in a covalent bond: inductive effect,
function, Second law of thermodynamics Gibbs’ energy electromeric effect, resonance and hyper conjugation.
change for spontaneous and non–spontaneous process, Homolytic and heterolytic fission of a covalent bond free
criteria for equilibrium and spontaneity. Third law of radials, carbocations, carbanions, electrophiles and
thermodynamics Brief introduction.
nucleophiles, types of organic reactions.
4
UNIT : XIII Hydrocarbons
UNIT : IV Chemical Kinetics
Alkanes Nomenclature, isomerism, conformations
(ethane only), physical properties, chemical reactions
including free radical mechanism of halogenations,
combustion and pyrolysis. Alkenes Nomenclature,
structure of double bond (ethene), geometrical
isomerism, physical properties, methods of preparation,
chemical reactions, addition of hydrogen, halogen, water,
hydrogen halides (Markovnikov’s addition and peroxide
effect),
ozonolysis,
oxidation,
mechanism
of
electrophilic addition.
Alkynes Nomenclature, structure of triple bond (ethyne),
physical properties, methods of preparation, chemical
reactions, acidic character of alkynes, addition reaction
of – hydrogen, halogens, hydrogen halides and water.
Aromatic
hydrocarbons
Introduction,
IUPAC
nomenclature,
Benzene,
resonance,
aromaticity,
chemical properties, mechanism of electrophilic
substitution – Nitration sulphonation, halogenations,
Friedel Craft’s alkylation and acylation, directive
influence of functional group in mono–substituted
benzene, carcinogenicity and toxicity.
Rate of a reaction (average and instantaneous), factors
affecting rates of reaction, concentration, temperature,
catalyst, order and molecularity of a reaction, rate law
and specific rate constant, integrated rate equations and
half-life (only for zero and first order reactions), concept
of collision theory (elementary idea, no mathematical
treatment). Activation energy, Arrhenious equation.
UNIT : V Surface Chemistry
Adsorption physisorption and chemisorptions, factors
affecting adsorption of gases on solids, catalysis
homogeneous and heterogeneous, activity and
selectivity, enzyme catalysis, colloidal state, distinction
between true solutions, colloids and suspensions,
lyophillic,
lyophobic
multimolecular
and
macromolecular colloids, properties of colloids, Tyndall
effect,
Brownian
movement,
electrophoresis,
coagulation, emulsions – types of emulsions.
UNIT : VI General Principles and Processes of
Isolation of Elements
Principles and methods of extraction concentration,
oxidation, reduction electrolytic method and refining,
occurrence and principles of extraction of aluminium,
copper, zinc and iron,
UNIT : XIV Environmental Chemistry
UNIT : VII p-Block Elements
Environmental pollution Air, water and soil pollution,
chemical reactions in atmosphere, smogs, major
atmospheric pollutants, acid rain ozone and its reactions,
effects of depletion of ozone layer, greenhouse effect and
global warming–pollution due to industrial wastes, green
chemistry as an alternative tool for reducing pollution,
strategy for control of environmental pollution.
Group 15 elements General introduction, electronic
configuration, occurrence, oxidation states, trends in
physical and chemical properties, preparation and
properties of ammonia and nitric acid, oxides of nitrogen
(structure only), phosphorous allotropic forms,
compounds of phosphorous, preparation and properties
of phosphine, halides (PCl3, PCl5) and oxoacids
(elementary idea only).
Group 16 elements General introduction, electronic
configuration, oxidation states, occurrence, trends in
physical and chemical properties, dioxygen, preparation,
properties and uses, classification of oxides, ozone.
Sulphur allotropic forms, compounds of sulphur,
preparation, properties and uses of sulphur dioxide,
sulphuric acid, industrial process of manufacture,
properties and uses, oxoacids of sulphur (structures
only).
Group 17 elements General introduction, electronic
configuration, oxidation states occurrence, trends in
physical and chemical properties, compounds of
halogens, preparation, properties and uses of chlorine
and hydrochloric acid, interhalogen compounds oxoacids
of halogens (structures only).
Group 18 elements General introduction, electronic
configuration, occurrence, trends in physical and
chemical properties, uses.
CLASS 12th
UNIT : I Solid State
Classification of solids based on different binding forces,
molecular, ionic covalent and metallic solids, amorphous
and crystalline solids (elementary idea), unit cell in two
dimensional and three dimensional lattices, calculation of
density of unit cell, packing in solids, packing efficiency,
voids, number of atoms per unit cell in a cubic unit cell,
point defects, electrical and magnetic properties, Band
theory of metals, conductors, semiconductors and
insulators.
UNIT : II Solutions
Types of solutions, expression of concentration of
solutions of solids in liquids, solubility of gases in
liquids, solid solutions, colligative properties- relative
lowering of vapour pressure, Raoult's law, elevation of
boiling point, depression of freezing point, osmotic
pressure, determination of molecular masses using
colligative properties abnormal molecular mass. Van UNIT : VIII d-and f-Block Elements
General
introduction,
electronic
configuration,
Hoff factor.
characteristics of transition metals, general trends in
UNIT : III Electrochemistry
properties of the first row transition metals metallic
Redox reactions, conductance in electrolytic solutions, character, ionization enthalpy, oxidation states, ionic
specific and molar conductivity variation of conductivity radii, colour, catalytic property, magnetic properties,
with concentration, Kohlrausch's Law, electrolysis and interstitial compounds, alloy formation. Preparation and
Laws of electrolysis (elementary idea), dry cell- properties of K2Cr2O7 and KMnO4. Lanthanoids
electrolytic cells and Galvanic cells; lead accumulator, electronic configuration, oxidation states, chemical
EMF of a cell, standard electrode potential, Relation reactivity and lanthanoid contraction and its
between Gibb's energy change and EMF of a cell, fuel consequences. Actinoids Electronic configuration,
oxidation states and comparison with lanthanoids.
cells, corrosion.
5
UNIT : XI Coordination Compounds
UNIT : XVI Chemistry in Everyday Life
Coordination compounds Introduction, ligands,
coordination number, colour, magnetic properties and
shapes, IUPAC nomenclature of mononuclear
coordination compounds, isomerism (structural and
stereo) bonding, Werner's theory VBT, CFT, importance
of coordination compounds (in qualitative analysis,
biological systems).
Chemicals in medicines analgesics, tranquilizers,
antiseptics, disinfectants, antimicrobials, antifertility
drugs, antibiotics, antacids, antihistamines. Chemicals in
food preservatives, artificial sweetening agents,
elementary idea of antioxidants. Cleansing agents soaps
and detergents, cleansing action.
JEE (Main)
UNIT : X Haloalkanes and Haloarenes
Section-A (Physical Chemistry)
Haloalkanes Nomenclature, nature of C–X bond,
physical and chemical properties, mechanism of
substitution reactions. Optical rotation. Haloarenes
Nature of C–X bond, substitution reactions (directive
influence of halogen for mono substituted compounds
only). Uses and environment effects of –
dichloromethane, trichloromethane, tetra chloromethane,
iodoform, freons, DDT.
Unit : I Some Basic Concepts in Chemistry
Matter and its nature, Dalton's atomic theory; Concept of
atom, molecule, element and compound; Physical
quantities and their measurements in Chemistry,
precision and accuracy, significant figures, S.I. Units,
dimensional analysis; Laws of chemical combination;
Atomic and molecular masses, mole concept, molar
UNIT : XI Alcohols, Phenols and Ethers
mass, percentage composition, empirical and molecular
Alcohols Nomenclature, methods of preparation, formulae; Chemical equations and stoichiometry.
physical and chemical properties (of primary alcohols
only), identification of primary, secondary and tertiary Unit : II States of Matter
alcohols, mechanism of dehydration, uses with special Classification of matter into solid, liquid and gaseous
reference to methanol and ethanol. Phenols, states.
Nomenclature, methods of preparation, physical and Gaseous State Measurable properties of gases; Gas laws
chemical properties, acidic nature of phenol, - Boyle's law, Charle's law, Graham's law of diffusion,
electrophillic substitution reactions, uses of phenols. Avogadro's law, Dalton's law of partial pressure;
Ethers, Nomenclature, methods of preparation, physical Concept of Absolute scale of temperature; Ideal gas
equation, Kinetic theory of gases (only postulates);
and chemical properties uses.
UNIT : XII Aldehydes, Ketones and Carboxylic Concept of average, root mean square and most probable
velocities; Real gases, deviation from Ideal behaviour,
Acids
compressibility factor, Van der Waal's Equation,
Aldehydes and Ketones Nomenclature, nature of liquefaction of gases, critical constants.
carbonyl group, methods of preparation, physical and Liquid State Properties of liquids - vapour pressure,
chemical properties, and mechanism of nucleophilic viscosity and surface tension and effect of temperature
addition, reactivity of alpha hydrogen in aldehydes, uses. on them (qualitative treatment only).
Carboxylic Acids Nomenclature, acidic nature, methods
Solid State Classification of solids: molecular, ionic,
of preparation, physical and chemical properties, uses.
covalent and metallic solids, amorphous and crystalline
UNIT : XIII Organic compounds Containing solids (elementary idea); Bragg's Law and its
Nitrogen
applications, Unit cell and lattices, packing in solids (fcc,
Amines Nomenclature, classification, structure, methods bcc and hcp lattices), voids, calculations involving unit
of preparation, physical and chemical properties, uses, cell parameters, imperfection in solids; electrical,
identification of primary secondary and tertiary amines. magnetic and dielectric properties.
Cyanides and Isocyanides will be mentioned at relevant Unit : III Atomic Structure
places. Diazonium salts Preparation, chemical reactions Discovery of sub-atomic particles (electron, proton and
and importance in synthetic organic chemistry.
neutron); Thomson and Rutherford atomic models and
UNIT : XIV Biomolecules
their limitations; Nature of electromagnetic radiation,
photoelectric effect; spectrum of hydrogen atom, Bohr
model of hydrogen atom - its postulates, derivation of the
relations for energy of the electron and radii of the
different orbits, limitations of Bohr's model; dual nature
of matter, de-Broglie's relationship, Heisenberg
uncertainty principle.
Elementary ideas of quantum mechanics, quantum
mechanical model of atom, its important features, ψ and
ψ2, concept of atomic orbitals as one electron wave
functions; Variation of ψ and ψ2 with r for 1s and 2s
orbitals; various quantum numbers (principal, angular
momentum and magnetic quantum numbers) and their
significance; shapes of s, p and d - orbitals, electron spin
and spin quantum number; rules for filling electrons in
orbitals - aufbau principle, Pauli's exclusion principle and
Hund's rule, electronic configuration of elements, extra
stability of half-filled and completely filled orbitals.
Carbohydrates Classification (aldoses and ketoses),
monosaccharide
(glucose
and
fructose),
D.L.
configuration, oligosaccharides (sucrose, lactose,
maltose), polysaccharides (starch, cellulose, glycogen):
importance. Proteins Elementary idea of – amino acids,
peptide bonds, polypeptides, proteins, primary structure,
secondary structure, tertiary structure and quaternary
structure (qualitative idea only), denaturation of proteins,
enzymes. Hormones Elementary idea (excluding
structure). Vitamins Classification and function. Nucleic
Acids DNA and RNA
UNIT : XV Polymers
Classification Natural and synthetic, methods of
polymerization
(addition
and
condensation),
copolymerization. Some important polymers natural
and synthetic like polyesters, Bakelite, rubber,
Biodegradable and non-biodegradable polymers.
6
Unit : IV Chemical Bonding and Molecular Ionic equilibrium Weak and strong electrolytes,
ionization of electrolytes, various concepts of acids and
Structure
bases (Arrhenius, Bronsted-Lowry and Lewis) and their
ionization, acid-base equilibria (including multistage
ionization) and ionization constants, ionization of water,
pH scale, common ion effect, hydrolysis of salts and pH
of their solution, solubility of sparingly soluble salts and
solubility products, buffer solutions.
Kossel Lewis approach to chemical bond formation,
concept of ionic and covalent bonds.
Ionic Bonding Formation of ionic bonds, factors
affecting the formation of ionic bonds; calculation of
lattice enthalpy.
Covalent Bonding Concept of electronegativity, Fajan's
rule, dipole moment; Valence Shell Electron Pair
Repulsion (VSEPR) theory and shapes of simple
molecules.
Quantum mechanical approach to covalent bonding
Valence bond theory - Its important features, concept of
hybridization involving s, p and d orbitals; Resonance.
Molecular Orbital Theory its important features, LCAOs,
types of molecular orbitals (bonding, antibonding),
sigma and pi-bonds, molecular orbital electronic
configurations of homonuclear diatomic molecules,
concept of bond order, bond length and bond energy.
Elementary idea of metallic bonding, Hydrogen bonding
and its applications.
UNIT : VIII Redox
Reactions
Electrochemistry
and
Electronic concepts of oxidation and reduction, redox
reactions, oxidation number, rules for assigning
oxidation number, balancing of redox reactions.
Eectrolytic and metallic conduction,
conductance in electrolytic solutions, specific and molar
conductivities and their variation with concentration:
Kohlrausch's law and its applications.
Electrochemical cells-Electrolytic and Galvanic cells,
different types of electrodes, electrode potentials
including standard electrode potential, half-cell and cell
reaction, emf of a Galvanic cell and its measurement;
UNIT : V Chemical Thermodynamics
Fundamentals of thermodynamics System and Nernst equation and its application; Relationship
surrounding, extensive and intensive properties, state between cell potential and Gibbs' energy change; Dry
cell and lead accumulator; Fuel cells; Corrosion and its
functions, types of processes.
First law of thermodynamics Concept of work, heat prevention.
internal energy and enthalpy, heat capacity, molar heat UNIT : IX Chemical Kinetics
capacity, Hess's law of constant heat summation; Rate of a chemical reaction, factors affecting the rate of
Enthalpies of bond dissociation, combustion, formation, reactions concentration, temperature, pressure and
atomization, sublimation, phase transition, hydration, catalyst; elementary and complex reactions, order and
ionization and solution.
molecularity of reactions, rate law, rate constant and its
Second law of thermodynamics Spontaneity of units, differential and integral forms of zero and first
processes; ∆S of the universe and ∆G of the system as order reactions, their characteristics and half-lives, effect
criteria for spontaneity, ∆G° (Standard Gibb's energy of temperature on rate of reactions-Arrhenius theory,
change) and equilibrium constant.
activation energy and its calculation, collision theory of
UNIT : VI Solution
bimolecular gaseous reactions (no derivation).
Different methods for expressing concentration of UNIT : X Surface Chemistry
solution-molality, molarity, mole fraction, percentage Adsorption physisorption and chemisorptions and their
(by volume and mass both), vapour pressure of solutions
characteristics, factors affecting adsorption of gases on
and Raoult's Law-Ideal and non-ideal solutions, vapour
pressure-compositions plots for ideal and non-ideal solids-Freundlich and Langmuir adsorption isotherms,
adsorption from solutions.
solutions.
Colligative properties of dilute solutions-relative Catalysis Homogeneous and heterogeneous, activity and
lowering of vapour pressure, depression of freezing selectivity of solid catalysts, enzyme catalysis and its
point, elevation of boiling point and osmotic pressure; mechanism.
Determination of molecular mass using colligative Colloidal state distinction among true solutions, colloids
properties; Abnormal value of molar mass, Van't Hoff and suspensions, classification of colloids-lyophilic,
factor and its significance.
lyophobic; multi molecular, macro-molecular and
associated colloids (micelles), preparation and properties
UNIT : VII Equilibrium
Meaning of equilibrium, concept of dynamic of colloids Tyndall effect, Browninan movement,
electrophoresis, dialysis, coagulation and flocculation;
equilibrium.
Equilibria involving physical processes Solid-liquid, Emulsions and their characteristics.
liquid-gas and solid-gas equilibria, Henry's law, general
SECTION-B (Inorganic Chemistry)
characteristics of equilibrium involving physical
UNIT
: XI Classification of Elements and
processes.
Periodicity in Properties
Equilibria involving chemical processes Law of chemical
Periodic
Law
and present Form of the Periodic Table,
equilibrium, equilibrium constants (K and K) and their
significance, significance of ∆G and ∆G° in chemical s,p,d and f block Elements, Periodic Trends in Properties
equilibria, factors affecting equilibrium concentration, of elements, atomic and ionic Radii, ionization Enthalpy,
pressure, temperature, effect of catalyst; Le-Chatelier's Electron Gain Enthalpy, Valence, Oxidation States and
Chemical Reactivity.
principle.
7
UNIT : XII General Principles and Processes of UNIT : XVI d – and f –b Block Elements
Transition Elements General introduction, electronic
Isolation of Metals
configuration, occurrence and characteristics, general
trends in properties of the first row transition elements physical properties, ionization enthalpy, oxidation states.
Atomic radii, colour, catalytic behaviour, magnetic
properties, complex formation, interstitial compounds,
alloy formation; Preparation, properties and uses of K2
Cr2 O7 and KMnO4.
Inner Transition Elements Lanthanoids electronic
UNIT : XIII Hydrogen
configuration, oxidation states, chemical reactivity and
Position of hydrogen in periodic table, isotopes, lanthanoid
contraction.
Actinoids
Electronic
preparation, properties and uses of hydrogen; physical configuration and oxidation states
and chemical properties of water and heavy water;
Structure, preparation, reactions and uses of hydrogen UNIT : XVII Coordination Compounds
peroxide; Classification of hydrides ionic, covalent and Introduction to coordination compounds, Werner's
theory; ligands, coordination number, denticity,
interstitial; Hydrogen as a fuel.
chelation; IUPAC nomenclature of mononuclear
UNIT : XIV s-Block Elements
coordination compounds, isomerism; Bonding Valence
(Alkali and Alkaline Earth Metals)
bond approach and basis ideas of Crystal field theory,
Group 1 and 2 Elements
colour and magnetic properties; importance of
General introduction, electronic configuration and coordination compounds (in qualitative analysis,
general trends in physical and chemical properties of extraction of metal sand in biological systems).
elements, anomalous properties of the first element of UNIT : XVIII Environmental Chemistry
each group, diagonal relationships.
Environmental pollution Atmospheric, water and soil.
Preparation and properties of some important Atmospheric pollution Tropospheric and stratospheric
compounds- sodium carbonate, sodium chloride, sodium Tropospheric pollutants: Gaseous pollutants Oxides of
hydroxide and sodium hydrogen carbonate; industrial carbon, nitrogen and sulphur, hydrocarbons; their
uses of lime, limestone, Plaster of Paris and cement; sources, harmful effects and prevention; Green house
Biological significance of Na, K, Mg and Ca.
effect and Global warming; Acid rain;
Particulate pollutants Smoke, dust, smog, fumes, mist;
UNIT : XV p-Block Elements
Group 13 to Group 18 Elements, General Introduction their sources, harmful effects and prevention. and effects.
Electronic configuration and general trends in physical Water pollution Major pollutants such as, pathogens,
and chemical properties of elements across the periods organic wastes and chemical pollutants their harmful
and down the groups; unique behaviour of the first effects and prevention. Strategies to control
environmental pollution.
element in each group.
Group wise study of the p-Block Elements
SECTION - C (Organic Chemistry)
Group 13 Preparation, properties and uses of boron and UNIT : XIX Purification & Characterization of
aluminium; structure, properties and uses of borax, boric Organic Compounds
acid, diborane, boron tifluoride, aluminium chloride and
Purification
Crystallization,
sublimation,
alums.
distillation,
differential
extraction
and
Group 14 Tendency for catenation; Structure, properties
and uses of allotropes and oxides of carbon, silicon chromatograph and their applications.
tetrachloride, silicates, zeolites and silicones.
Qualitative analysis Detection of nitrogen,
Group 15 Properties and uses of nitrogen and sulphur, phosphorus and halogens.
phosphorus; Allotrophic forms of phosphorus; Quantitative analysis (basic principles only)
Preparation, properties, structure and uses of ammonia
Estimation of carbon, hydrogen, nitrogen,
nitric acid, phosphine and phosphorus halides, (PCl3,
PCl5; Structures of oxides and oxoacids of nitrogen and halogens, sulphur phosphorus. Calculations of
empirical formulae and molecular formulae;
phosphorus.
Numerical
problems in organic quantitative
Group – 16 Preparation, properties, structures an uses of
dioxygen and ozone; Allotropic forms of sulphur; analysis.
preparation, properties, structures and uses of sulphur UNIT : XX Some Basic Principles of Organic
dioxide, sulphuric acid (including its industrial
Chemistry
preparation), Structures of oxoacids of sulphur.
Tetravalency of carbon; Shapes of simple molecules
Group -17 Preparation, properties and uses of chlorine hybridization (s and p)/ Classification of organic
and hydrochloric acid; Trends in the acidic nature of compounds based on functional groups: –– C = C ––, ––
hydrogen halides; Structures of Interhalogen compounds C = C –– and those containing halogens, oxygen,
and oxides and oxoacids of halogens.
nitrogen and sulphur, Homologous series; Isomerism Group - 18 Occurrence and uses of noble gases; structural and stereoisomerism. Nomenclature (Trivial
Structures of fluorides and oxides of xenon.
and IUPAC)
Modes of occurrence of elements in nature, minerals,
ores; steps involved in the extraction of metalsconcentration, reduction (Chemical and electrolytic
methods) and refining with special reference to the
extraction of Al, Cu, Zn and Fe; Thermodynamic and
electrochemical principles involved in the extraction of
metals.
8
Covalent bond fission Homolytic and heterolytic free
radicals, carbocations and carbanions; stability of
carbocations and free radicals, electrophiles and
nucleophiles.
electronic displacement in a covalent bond inductive
effect,
electromeric
effect,
resonance
and
hyperconjugation.
Common types of organic reactions Substitution,
elimination and rearrangement.
Unit : XXV Polymers
General introduction and classification of polymers,
general methods of polymerization-addition and
condensation, copolymerization; Natural and synthetic
rubber and vulcanization; some important polymers with
emphasis on their monomers and uses- polythene, nylon
polyester and Bakelite.
UNIT : XXVI Biomolecules
General introduction and importance of biomolecules.
Carbohydrates Classification: aldoses and ketoses;
monosaccharides (glucose and fructose), constituent
monosaccharides of oligosaccharides (sucrose, lactose,
maltose) and polysaccharides (starch, cellulose,
glycogen).
Proteins Elementary idea of α-amino acids, peptide
bond,. polypeptides; proteins: primary, secondary,
tertiary and quaternary structure (qualitative idea only),
denaturation of proteins, enzymes.
Vitamins Classification and functions.
Nucleic Acids Chemical constitution of DNA and RNA.
Biological functions of Nucleic acids.
UNIT : XXI Hydrocarbons
Classification, isomerism, IUPAC nomenclature, general
methods of preparation, properties and reactions.
Alkanes Conformations: Sawhorse and Newman
projections (of ethane); Mechanism of halogenations of
alkanes.
Alkenes Geometrical isomerism; Mechanism of
electrophilic addition: addition of hydrogen, halogens
water, hydrogen halides (Markownikoff's and peroxide
effect); Ozonolysis, oxidation , and polymerization.
Alkynes acidic character; addition of hydrogen, halogens
water and hydrogen halides; polymerization. Aromatic
hydrocarbons Nomenclature, benzene structure and
aromaticity; Mechanism of electrophilic substitution:
halogenations, nitration, Friedel-Craft's alkylation and UNIT : XXVII Chemistry in Everyday Life
acylation, directive influence of functional group in Chemicals in medicines analgesics, tranquilizers,
mono-substituted benzene.
antiseptics, disinfectants, antimicrobials, antifertility
UNIT : XXII Organic Compounds Containing drugs, antibiotics, antacids, anti-histamin their meaning
and common examples.
Halogens
General methods of preparation, properties and reactions; Chemicals in food preservatives, artificial sweetening
Nature of C–X bond; Mechanism of substitution agents-common examples.
reactions. Uses/environmental effects of chloroform, Cleansing agents Soaps and detergents, cleansing action.
iodoform
UNIT : XXVIII
Principles Related to
UNIT : XXIII Organic Compounds Containing
Oxygen.
Practical Chemistry
•
General methods of preparation, properties, reactions an
uses.
Alcohols, Phenols and Ethers
Alcohols identification of primary, secondary and
tertiary alcohols; mechanism of dehydration.
Phenols Acidic nature, electrophilic substitution
reactions: halogenations, nitration and sulphonation,
Reimer-Tiemann reaction.
Ethers Structure.
Aldhyde and Ketones Nature of charbonyl group;
Nucleophilic addition to >C=O group , relative ractivities
of aldehydes and ketones; important reactions such asNucleophilic addition reactions (addition of HCN, NH3
and its derivatives), Grignard reagent; oxidation;
reduction (Wolff Kishner and Clemmensen) acidity of αhydrogen, aldol condensation, Cannizzaro reaction,
Haloform reaction; Chemical tests to distinguish between
aldehydes and Ketones. Carboxylic Acid Acidic strength
and factors affecting it.
Unit : XXIV
Nitrogen
•
•
•
•
•
•
Organic Compounds Containing •
General methods of preparation, properties, reactions and
uses.
Amines Nomenclature, classification, structure basic
character and identification of primary, secondary and
tertiary amines and their basic character.
Diazonium Salts Importance in synthetic organic
chemistry.
1.
2.
3.
4.
9
Detection of extra elements (N, S, halogens0 in
organic compounds; Detection of the following
functional group: hydroxyl (alcoholic and phenolic),
carbonyl (aldehyde and ketone), carboxyl and amino
groups in organic compounds.
Chemistry involved in the preparation of the
following
inorganic compounds Mohr's salt, potash alum.
Organic compounds Acetanilide,
p-nitroacetainilide, aniline yellow, iodoform.
Chemistry involved in the titrimetirc excercisesAcids bases and the use of indicators, oxaliacid vs
KMnO4, Mohr's salt vs KMnO4.
Chemical principles involved in the qualitative salt
analysis
Cations- Pb2+, Cu2+, Al3+, Fe3+, Zn2+, Ni2+, Ba2+,
Mg2+, NH4+. Anions– CO 24− , S2–, SO 24− , NO2, NO3
Cl–, Br–, I– (Insoluble salts excluded).
Chemical principles involved in the following
experiments.
Enthalpy of solution of CuSO4
Enthalpy of neutralization of strong acid and strong
base.
Preparation of lyophilic and lyophobic sols.
Kinetic study of reaction of iodide ion with hydrogen
peroxide at room temperature.
.
All India Medical & Engineering Entrance Exam, AIIMS, NEET, &
JEE Main Chemistry Previous Years Exam Papers Analysis Chart
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Exam
Proposed Year
Question Paper
Total Question
All India Pre Medical Test/National Eligibility Cum Entrance Test (AIPMT/NEET)
NEET
17.07.2022
50
NEET
12.09.2021
50
NEET
13.09.2020
50
NEET
05.06.2019
50
NEET
06.05.2018
50
NEET
07.05.2017
50
NEET
01.05.2016
Phase-I
50
NEET
24.06.2016
Phase-II
50
NEET/AIPMT
25.07.2015
50
NEET
04.05.2014
50
NEET
05.05.2013
50
AIPMT
2012
50
AIPMT
2011
50
AIPMT
2010
50
AIPMT
2009
50
AIPMT
2008
50
AIPMT
2007
50
AIPMT
2006
50
AIPMT
2005
50
AIPMT
2004
50
AIPMT
2003
50
AIPMT
2002
50
AIPMT
2001
50
AIPMT
2000
50
AIPMT
1999, 98, 97, 96, 95,
600
94, 93, 92, 91, 90,
89, 88
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AIEEE
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All India Institute of Medical Sciences (AIIMS)
AIIMS
26.05.2019
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26.05.2019
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25.05.2019
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25.05.2019
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2018
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2017
60
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2016
60
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2015
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2014
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2013
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60
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1999, 98, 97, 96, 94
300
Assam Combined Entrance Examination (CEE)
ASSAM-CEE
31.07.2022
40
ASSAM-CEE
2021
40
ASSAM-CEE
2020
40
ASSAM-CEE
2019
40
ASSAM-CEE
2018
40
Andhra Pradesh Engineering, Agriculture and Medical Common Entrance Test (AP EAMCET)
AP EAMCET Medical
2013
50
AP EAMCET Medical
2012
50
AP EAMCET Medical
2010
40
AP EAMCET Medical
2009
40
AP EAMCET Medical
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2001
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AP EAMCET Medical
1999
40
AP EAMCET Medical
1998
50
AP EAMCET Medical
1997
50
Andhra Pradesh Engineering, Agriculture and Medical Common Entrance Test (AP EAMCET)
AP EAMCET Engineering
12.07.2022
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12.07.2022
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AP EAMCET Engineering
11.07.2022
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1999, 98, 97, 96, 95,
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Aligarh Muslim University Engineering Entrance Examination (AMUEEE)
2019
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2016
2015
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2012
2011
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40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
360
242.
243.
244.
245.
246.
247.
248.
249.
250.
251.
252.
253.
254.
255.
256.
AMU
AMU
AMU
AMU
AMU
AMU
AMU
AMU
AMU
AMU
AMU
AMU
AMU
AMU
AMU
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
13
257.
258.
259.
260.
261.
262.
263.
264.
265.
266.
267.
268.
269.
270.
271.
272.
273.
274.
275.
276.
277.
278.
279.
280.
281.
282.
283.
284.
285.
286.
287.
288.
289.
290.
291.
292.
293.
294.
295.
296.
297.
298.
299.
300.
301.
302.
303.
304.
305.
306.
307.
308.
309.
310.
311.
312.
313.
314.
315.
316.
317.
318.
319.
AMU
AMU
AMU
AMU
2004
2003
2002
2001
Bihar Combined Entrance Competitive Examination (BCECE)
BCECE
2018
BCECE
2017
BCECE
2016
BCECE
2015
BCECE
2014
BCECE
2013
BCECE
2012
BCECE
2011
BCECE
2010
BCECE
2009
BCECE
2008
BCECE
2007
BCECE
2006
BCECE
2005
BCECE
2004
BCECE
2003
Birla Institute of Technology and Science Admission Test (BITSAT)
BITSAT
2018
BITSAT
2017
BITSAT
2016
BITSAT
2015
BITSAT
2014
BITSAT
2013
BITSAT
2012
BITSAT
2011
BITSAT
2010
BITSAT
2009
BITSAT
2008
BITSAT
2007
BITSAT
2006
BITSAT
2005
Consortium of Medical, Engineering and Dental Colleges of Karnataka (COMEDK)
COMEDK-JEE
2022
COMEDK-JEE
2021
COMEDK-JEE
2020
COMEDK-JEE
2019
COMEDK-JEE
2018
COMEDK-JEE
2017
COMEDK-JEE
2016
COMEDK-JEE
2015
COMEDK-JEE
2014
COMEDK-JEE
2013
COMEDK-JEE
2012
COMEDK-JEE
2011
Chhattisgarh Pre-Engineering Test (CGPET)
Chhattisgarh-PET
22.05.2022
Chhattisgarh-PET
2019
Chhattisgarh-PET
2018
Chhattisgarh-PET
2017
Chhattisgarh-PET
2016
Chhattisgarh-PET
2015
Chhattisgarh-PET
2014
Chhattisgarh-PET
2013
Chhattisgarh-PET
2012
Chhattisgarh-PET
2011
Chhattisgarh-PET
2010
Chhattisgarh-PET
2009
Chhattisgarh-PET
2008
Chhattisgarh-PET
2007
Chhattisgarh-PET
2006
Chhattisgarh-PET
2005
Chhattisgarh-PET
2004
14
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
40
40
40
40
40
40
40
40
40
40
40
40
40
40
60
60
60
60
60
60
60
60
60
60
60
60
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
320.
321.
322.
323.
324.
325.
326.
327.
328.
329.
330.
331.
332.
333.
334.
335.
336.
337.
338.
339.
340.
341.
342.
343.
344.
345.
346.
347.
348.
349.
350.
351.
352.
353.
354.
355.
356.
357.
358.
359.
360.
361.
362.
363.
364.
365.
366.
367.
368.
369.
370.
371.
372.
373.
374.
375.
376.
377.
378.
379.
Gujarat Common Entrance Test (GUJCET)
18.04.2022
06.08.2021
2020
2019
2018
2017
2016
2015
2014
2011
2008
2007
Himachal Pradesh Common Entrance Test (HPCET)
HPCET
2021
HPCET
2020
HPCET
2019
HPCET
2018
HPCET
2017
HPCET
2016
HPCET
2015
HPCET
2014
HPCET
2013
Jammu and Kashmir Common Entrance Test (JKCET)
JKCET
2019
JKCET
2018
JKCET
2017
JKCET
2016
JKCET
2015
JKCET
2014
JKCET
2013
JKCET
2012
JKCET
2011
JKCET
2010
JKCET
2009
JKCET
2008
JKCET
2007
JKCET
2006
JKCET
2005
JKCET
2004
JKCET
2003
JKCET
2002
JKCET
2001
JKCET
2000
JKCET
1999
JKCET
1998
JKCET
1997
Jawaharlal Institute of Postgraduate Medical Education and Research (JIPMER)
JIPMER
2019
JIPMER
2018
JIPMER
2017
JIPMER
2016
JIPMER
2015
JIPMER
2014
JIPMER
2013
JIPMER
2012
JIPMER
2011
JIPMER
2010
JIPMER
2009
JIPMER
2008
JIPMER
2007
JIPMER
2006
JIPMER
2005
JIPMER
2004
GUJCET
GUJCET
GUJCET
GUJCET
GUJCET
GUJCET
GUJCET
GUJCET
GUJCET
GUJCET
GUJCET
GUJCET
15
40
40
40
40
40
40
40
40
40
40
40
40
60
60
60
60
60
60
60
60
60
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
75
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
380.
381.
382.
383.
384.
385.
386.
387.
388.
389.
390.
391.
392.
393.
394.
395.
396.
397.
398.
399.
400.
401.
402.
403.
404.
405.
406.
407.
408.
409.
410.
411.
412.
413.
414.
415.
416.
417.
418.
419.
420.
421.
422.
423.
424.
425.
426.
427.
428.
429.
430.
431.
432.
433.
434.
435.
436.
437.
438.
439.
440.
Jharkhand Combined Entrance Competitive Examination (JCECE)
2018
2017
2016
2015
2014
2013
2012
2011
2010
2009
2008
2007
2006
2005
2004
2003
Kerala Commissioner for Entrance Examinations (K-CEE)
Kerala CEE
04.07.2022
Kerala CEE
29.08.2021
Kerala CEE
2020
Kerala CEE
2019
Kerala CEE
2018
Kerala CEE
2017
Kerala CEE
2016
Kerala CEE
2015
Kerala CEE
2014
Kerala CEE
2013
Kerala CEE
2012
Kerala CEE
2011
Kerala CEE
2010
Kerala CEE
2009
Kerala CEE
2008
Kerala CEE
2007
Kerala CEE
2006
Kerala CEE
2005
Kerala CEE
2004
Karnataka Common Entrance Test (K-CET)
Karnataka-CET
17.06.2022
Karnataka-CET
2021
Karnataka-CET
2020
Karnataka-CET
2019
Karnataka-CET
2018
Karnataka-CET
2017
Karnataka-CET
2016
Karnataka-CET
2015
Karnataka-CET
2014
Karnataka-CET
2013
Karnataka-CET
2012
Karnataka-CET
2011
Karnataka-CET
2010
Karnataka-CET
2009
Karnataka-CET
2008
Karnataka-CET
2007
Karnataka-CET
2006
Karnataka-CET
2005
Karnataka-CET
2004
Karnataka-CET
2003
Karnataka-CET
2002
Karnataka-CET
2001
Madhya Pradesh Pre Engineering Test (MPPET)
MPPET
2013
MPPET
2012
MPPET
2009
MPPET
2008
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
JCECE
16
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
60
50
50
50
50
441.
442.
443.
444.
445.
446.
447.
448.
449.
450.
451.
452.
453.
454.
455.
456.
457.
458.
459.
460.
461.
462.
463.
464.
465.
466.
467.
468.
469.
470.
471.
472.
473.
474.
475.
476.
477.
478.
479.
480.
481.
482.
483.
484.
485.
486.
487.
488.
489.
490.
491.
492.
493.
494.
495.
496.
497.
498.
499.
500.
501.
502.
503.
504.
Manipal University Under Graduate Entrance Test (M-UGET)
2020
50
2019
50
2018
50
2017
50
2016
50
2015
50
2014
50
2013
50
2012
50
2011
50
2010
50
2009
50
2008
50
Maharashtra Common Entrance Test (MHT-CET)
MHT-CET
02-05-2019
Shift-I
50
MHT-CET
02-05-2019
Shift-II
50
MHT-CET
03-05-2019
50
MHT-CET
2018
50
MHT-CET
2017
50
MHT-CET
2016
50
MHT-CET
2015
50
MHT-CET
2014
50
MHT-CET
2013
50
MHT-CET
2012
50
MHT-CET
2011
50
MHT-CET
2010
50
MHT-CET
2009
50
MHT-CET
2008
50
MHT-CET
2007
50
SRM Joint Engineering Entrance Examination (SRM-JEE)
SRM-JEE
2016
40
SRM-JEE
2015
40
SRM-JEE
2014
40
SRM-JEE
2013
40
SRM-JEE
2012
40
SRM-JEE
2011
40
SRM-JEE
2010
40
SRM-JEE
2009
40
SRM-JEE
2008
40
SRM-JEE
2007
40
Telangana State Engineering, Agriculture & Medical Common Entrance Test (TS EAMCET)
TS EAMCET
31.07.2022
Shift-I
40
TS EAMCET
31.07.2022
Shift-II
40
TS EAMCET
30.07.2022
Shift-I
40
TS EAMCET
30.07.2022
Shift-II
40
TS EAMCET
20.07.2022
Shift-I
40
TS EAMCET
20.07.2022
Shift-II
40
TS EAMCET
19.07.2022
Shift-I
40
TS EAMCET
19.07.2022
Shift-II
40
TS EAMCET
18.07.2022
Shift-I
40
TS EAMCET
18.07.2022
Shift-II
40
TS EAMCET
10.08.2021
Shift-I
40
TS EAMCET
10.08.2021
Shift-II
40
TS EAMCET
09.08.2021
40
TS EAMCET
08.08.2021
Shift-I
40
TS EAMCET
07.08.2021
Shift-II
40
TS EAMCET
06.08.2021
40
TS EAMCET
05.08.2021
Shift-I
40
TS EAMCET
05.08.2021
Shift-II
40
TS EAMCET
29.09.2020
Shift-I
40
TS EAMCET
29.09.2020
Shift-II
40
TS EAMCET
28.09.2020
Shift-I
40
TS EAMCET
28.09.2020
Shift-II
40
TS EAMCET
14.09.2020
Shift-I
40
TS EAMCET
14.09.2020
Shift-II
40
TS EAMCET
11.09.2020
Shift-I
40
TS EAMCET
11.09.2020
Shift-II
40
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
M-UGET
17
505.
506.
507.
508.
509.
510.
511.
512.
513.
514.
515.
516.
517.
518.
519.
520.
521.
522.
523.
524.
525.
526.
527.
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
TS EAMCET
528.
529.
530.
531.
Tripura JEE
Tripura JEE
Tripura JEE
Tripura JEE
532.
533.
534.
535.
536.
537.
538.
539.
540.
541.
542.
543.
544.
545.
546.
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
UPSEE
547.
548.
549.
550.
551.
552.
553.
554.
555.
556.
557.
558.
559.
560.
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
UPCPMT
561.
562.
563.
564.
565.
566.
567.
SCRA
SCRA
SCRA
SCRA
SCRA
SCRA
NDA (II)
10.09.2020
Shift-I
10.09.2020
Shift-II
09.09.2020
Shift-I
09.09.2020
Shift-II
09.05.2019
Shift-I
08.05.2019
Shift-I
08.05.2019
Shift-II
09.05.2019
Shift-II
06.05.2019
Shift-I
04.05.2019
Shift-I
04.05.2019
Shift-II
03.05.2019
Shift-I
03.05.2019
Shift-II
07.05.2018
Shift-I
05.05.2018
Shift-I
05.05.2018
Shift-II
04.05.2018
Shift-I
04.05.2018
Shift-II
03.05.2018
Shift-I
02.05.2018
Shift-I
02.05.2018
Shift-II
2017
2016
Tripura Joint Entrance Examination (TJEE)
27.04.2022
2021
2020
2019
Uttar Pradesh State Entrance Examination (UPSEE)
2018
2017
2016
2015
2014
2013
2012
2011
2010
2009
2008
2007
2006
2005
2004
Uttar Pradesh Combined Pre Medical Test (UPCPMT)
2014
2013
2012
2011
2010
2009
2008
2007
2006
2005
2004
2003
2002
2001
UPSC Special Class Railway Apprentice (SCRA)/UPSC NDA
2015
2014
2013
2012
2010
2009
2019
18
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
50
60
60
60
60
60
60
10
568.
569.
570.
571.
572.
573.
574.
575.
576.
577.
578.
NDA (I)
NDA (II)
NDA (I)
NDA (II)
NDA (I)
NDA (II)
NDA (II)
NDA (I)
NDA (II)
NDA (II)
NDA (I)
579.
580.
581.
582.
583.
584.
585.
586.
587.
588.
589.
590.
591.
592.
593.
594.
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
VITEEE
595.
596.
597.
598.
599.
600.
601.
602.
603.
604.
605.
606.
607.
608.
609.
610.
611.
612.
613.
614.
615.
616.
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
WBJEEB
2019
2018
2018
2017
2017
2016
2015
2015
2014
2011
2011
Vellore Institute of Technology Engineering Entrance Examination (VITEEE)
2021
2020
2019
2018
2017
2016
2015
2014
2013
2012
2011
2010
2009
2008
2007
2006
West Bengal Joint Entrance Examination Board (WBJEEB)
30.04.2022
2021
2020
2019
2018
2017
2016
2015
2014
2013
2012
2011
2010
2009
2008
2007
2006
2005
2004
2003
2002
2001
Total
10
10
10
10
10
10
10
10
10
10
10
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
40
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
28705
Note : After detailed analysis of above mentioned papers of NEET/JEE (Main) and State Medical
and Engineering Examination Related to Chemistry 28705 (Volume-II) have been presented
chapterwise. Questions of repeated and similar nature have included so that the technique of
asking question can benefit the competitors.
19
Trend Analysis of NEET/JEE (Main) Chemistry
Questions Through Pie Chart & Bar Graph
20
01.
SOLUTIONS
1.
Methods of Expressing
Concentration of Solutions
1.
In one molal solution that contains 0.5 mole of
a solute there is
(a) 1000 g of solvent
(b) 500 mL of solvent
(c) 500 g of solvent
(d) 100 mL of solvent
NEET-17.06.2022
moles of solute
Ans. (c) : 1 molal solution =
weight of solvent
0.5
weight of solvent =
kg
1
0.5
weight of solvent =
× 1000g
1
weight of solvent = 500g
2.
What mass of 95% pure CaCO3 will be
required to neutralize 50mL of 0.5 M HCl
solution according to the following reaction ?
CaCO3(s) +2HCl(aq)→CaCl2(aq) +CO2(g)+ H2O(I)
[Calculate up to second place of decimal point]
(a) 9.50 g
(b) 1.25 g
(c) 1.32 g
(d) 3.65 g
NEET-17.06.2022
Ans. (c) : Given reaction
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
∴ Number of moles of CaCo3 (pure)
1
= × mole of HCl
2
1
50
= × 0.5 ×
2
1000
= 0.0125
Weight of CaCO3(pure) = mole × molecular weight
= 0.0125 × 100
= 1.25 g
weight of pure substance
% purity =
×100
weight of impure sample
1.25
95 =
×100
weight of impuresample
1.25
∴ weigth of impuresample =
×100
95
= 1.32 g
3.
100 mL of 5% (w/v) solution of NaCl in water
was prepared in 250 mL beaker. Albumin from
the egg was poured into NaCl solution and
stirred well. This resulted in a/an:
(a) Lyophilic sol
(b) Lyophobic sol
(c) Emulsion
(d) Precipitate
JEE Main 29.07.2022, Shift-I
Objective Chemistry Volume-II
Ans. (a) : The discussed method is standard method for
preparation of lyophobic sol.
4.
The normality of H2SO4 in the solution
obtained on mixing 100 mL of 0.1 M H2SO4
with 50 mL of 0.1 M NaOH is ––––×10–1 H.
(Nearest Integer)
JEE Main 27.07.2022, Shift-II
Ans. (1) : Given data–
Volume of H2SO4(V1)=100 mL
Molarity of H2SO4(N1)= 0.1 M
Volume of NaOH (V2)= 50 mL
Molarity of NaOH(N2) = 0.1 m
No. of equivalent of H2SO4= 100×0.1×2=20
No of equivalalent of NaOH = 50×0.1 = 5
N V − N 2 V2
Resultant normality = 1 1
V1 + V2
20 − 5
=
150
15
=
= 0.1
150
5.
While estimating the nitrogen present in an
organic compound by kjeldahl’s method, the
ammonia evolved from 0.25 g of the compound
neutralized 2.5 mL of 2 M H2SO4. The
percentage of nitrogen persent in organic
compound is …….. .
JEE Main 25.07.2022, Shift-I
Ans. (56) : Given data,
Weight of compound = 0.25 g
molarity of H2SO4 = 2
volume of H2SO4 = 2.5 mL
milli equivalent of H2SO4 = 2.5 × 2 × 2
= 10 Meq. of NH3
Milli moles of NH3 = meq. of NH3 [n(−1)] = 10
Milli moles of N = 10, moles of N = 10 × 10−3
weight of N = 10−2 × 14 = 0.14 gm
0.14
% of N =
× 100 = 56%
0.25
6.
20mL of 0.02 M K2CrO7 solution is used for the
titration of 10 mL of Fe2+ solution in the acidic
medium.
The molarity of Fe2+ solution is ––––– × 10–2 M.
(Nearest Integer)
JEE Main-27.07.2022, Shift-I
–2
Ans. (24) : 24 × 10
The redox reaction is a s follows:
Cr2 O 27 − + Fe2+ + 14H+ → 2Cr2+ + Fe3+ + 7H2O
The redox change involved are given
(1) 6e– + Cr2 O 27 − → 2Cr3+ (x = 6)
21
YCT
(2) Fe2+ → Fe3+ + e– (x = 1)
Eq. of K2 Cr2 O7 = Eq. of Fe2+
(M × V × h.f) of K2 Cr2 O7 = (M × V × h.f) of Fe2+
0.02 × 20 × 6 = M × 10 × 1
M = 0.24
Molarity = 24 × 10–2
7.
Boiling point of a 2% aqueous solution of a
non- volatile solute A is equal to the boiling
point of 8% aqueous solution of a non-volatile
solute B. The relation between molecular
weights of A and B is.
(a) MA = 4MB
(b) MB = 4MA
(d) MB = 8MA
(c) MA = 8MB
JEE Main-27.07.2022, Shift-I
Ans. (b): ∆Tb = K b × m
( ∆Tb )A = ( ∆Tb ) B
m A = mB
Molality =
% ( W / W ) ×1000
M A × Wsolvent
2 × 1000 8 × 1000
=
⇒ M B ≈ 4M A
M A × 98 M B × 92
8.
Ans. (25) : Given dataVolume of H2SO4 = 2L
Molarity of H2SO4 = 0.2M
Volume of NaOH = 2L
Molarity of NaOH = 0.1M
Reaction–
→ Na 2SO 4 + 2H 2 O
H 2SO 4 + 2 NaOH
0.2 mol
−
−
0.1mol
0.1
Molarity of Na2SO4 =
= 0.025m
4
= 25 millimolar
10. A gaseous mixture of two substances A and B,
under a total pressure of 0.8 atm is in
equilibrium with an ideal liquid solution. The
mole fraction substance A is 0.5 in the vapour
phase and 0.2 in liquid A is ____ atm. (Nearest
integer)
JEE Main 28.07.2022, Shift-II
Ans. (2) : Given dataTotal pressure = 0.8 atm
Mole fraction of A in vapour phase = 0.5
Mole fraction of A in liquid phase = 0.2
YA = 0.5
YB = 0.5
0.4 mol
0.3mol
When 800 mL of 0.5 M nitric acid is heated in a
= 0.4 atm
beaker, its volume is reduced to half and 11.5 g PA = PB
0
of nitric acid is evaporated. The molarity of the PA = PA × YA
remaining nitric acid solution is x × 10–2 M. P 0 = 2
A
(Nearest Integer)
∆Tf
–1
(Molar mass of nitric acid is 63 g mol )
11. The factor
represents:
Kf
JEE Main 26.07.2022, Shift-I
(a) Molarity
(b) Molality
Ans. (54) : Given that,
(c)
Normality
(d)
Formality
–1 n
molar mass of nitric acid 63 g mol HNO3 = 0.5 × 0.8
CG PET-22.05.2022
= 0.4 mole
Ans. (b) : Freezing point depression is a colligative
property observed in solutions that results from the
11.5
introduction of solute molecule to a solvent.
(nHNO3) = 0.4 –
63
∆Tf =K f .m
or
= 0.2175
∆T
m= f
Kf
Moles solute
Molarity =
or
Volume of solution in(L)
∆T
Molality (m) = f
Kf
0.2175
=
× 1000
Where,
400
m = molality
∆Tf = Freezing point depression
= 0.5437 mol/Lit
Kf = Cryoscopic constant
12. Elevation in boiling point for 1.5 molal
≃ 54 × 10–2 mol/Lit
solutions of glucose in water is 4K. The
9.
2L of 0.2 M H2SO4 is reacted with 2L of 0.1 M
depression in freezing point for 4.5 molal
NaOH solution, the molarity of the resulting
solution of glucose in water is 4K. The ratio of
product Na2SO4 in the solution is _____
molal elevation constant to molal depression
constant (Kb/Kf) is_____.
millimolar. (Nearest integer)
JEE Main-29.06.2022, Shift-II
JEE Main 28.07.2022, Shift-II
Objective Chemistry Volume-II
22
YCT
Ans. (3) : Given, ∆Tb = ∆Tf = 4K, m = 1.5, m = 4.5
Now, from the relation between ∆Tb = and Kb –
∆Tb = mKb
or
∆T
Kb = b
m
4
Kb =
....(i)
1.5
and
∆Tf = mKf
∆T
Kf = f
m
4
Kf =
...(ii)
4.5
From equation (i) and (ii) – we get
K b 4 4.5
= ×
K f 1.5 4
∴
or
i=1+α
0.0198 = (1 + α) × 1.85 × 1.224
α = 1.049 − 1
α = 0.049
or
% α = 4.9%
or
%α≈5%
2 g of a non-volatile non-electrolyte solute is
dissolved in 200g of two different solvents A
and B whose ebullioscopic constants are in the
ratio of 1 : 8. The elevation in boiling points of
A and B are
x
In the ratio
(x : y). The value of y is_____.
y
(Nearest Integer)
JEE Main-27.06.2022, Shift-I
Ans. (y=8×x) : Given, MA = 1, MB = 8
Kb
=3
We know that,
Kf
∆Tb = Kb × m
13. The osmotic pressure exerted by a solution
( ∆Tb )A ( K b × m )A 1
prepared by dissolving 2.0 g of protein of molar
=
=
as m A = m B
mass 60 kg mol–1 in 200 mL of water at 27°C––
( ∆Tb )B ( K b × m )B 8
––– Pa. [Integer value]
x ( ∆Tb ) A
(Use R = 0.083 L bar mol–1 K–1)
(∆Tb )A 1
=
∵ =
JEE Main-26.06.2022, Shift-II ∴
(∆Tb ) B 8
y ( ∆Tb )B
Ans. (415) : Given, amount of solute = 2.0g
x 1
Molar mass = 60 kg mol−1 = 60 × 103 g mol−1
=
y 8
Amount of solution = 200 mL
T = 27oC + 273 K = 300 K
y = 8x
Now,
16. The osmotic pressure of blood is 7.47 bar at
Π = C.R.T
300K. To inject glucose to a patient
2 ×1000
intravenously, it has to be isotonic with blood.
Π=
×
0.083
×
300
The concentration of glucose solution in gL–1
60 ×103 × 200
5
is____. (Molar mass of glucose = 180 g mol–1)
Π = 0.00415bar (∵ 1bar =10 pascal)
R = 0.083 L bar K–1 mol–1) (Nearest integer)
Π = 415 Pa
JEE Main-24.06.2022, Shift-I
14. 1.2 mL of acetic acid is dissolved in water to
–1
make 2.0 L of solution. The depression in Ans. (54 gL ) : Given that,
freezing point observed for this strength of acid Pressure, P = 7.47 bar
−1
−1
is 0.0198ºC. The percentage of dissociation of T = 300 K, R = 0.0832 L bar K mole
We know that,
the acid is –––––.
[Nearest integer]
P = C.R.T
[Given: Density of acetic acid is 1.02 g mL7.47 = C × 0.083 × 300
MOlar mass of acetic acid is 60g/mole
5 92
C= ×
= 0.3M
Kf(H2O) = 1.85 Kg mol–1
92 100
JEE Main-29.06.2022, Shift-I
= 0.3 × 180
Ans. (5) : Given, volume of acetic acid (V) = 1.2 mL,
= 54 gL–1
−1
∆Tf = 0.0198ºC, Density (d) = 1.02 gmL
17. The vapour pressures of two volatile liquids A
M
and B at 25°C are 50 Torr and 100 Torr,
∴
d=
V
respectively. If the liquid mixture, contains 0.3
mole fraction of A, then the mole fraction of
or
M = 1.02 × 1.2 g
x
M = 1.224 g
liquid B in the vapour phase is
. The value
Number of moles of acetic acid = 0.0204 moles in 2L
17
Now, ∆Tf = i × Kf × M
of x is ––––––
∵ for acetic acid,
JEE Main-28.06.2022, Shift-I
Objective Chemistry Volume-II
15.
23
YCT
YB
Po X
= Bo B
1 − YB PA 1 − X B
YB
100 0.7 14
=
=
1 − YB 50 0.3 3
14 x
YB = =
17 17
Hence, the value of x = 14
18. 0.01 M KMnO4 solution was addes to 20.0 mL
of 0.05 M Mohr’s salt solution through a
burette. The initial reading of 50 mL burette is
zero. The volume of KMnO4 solution left in
burette after the end point is ______
ml.[nearest integer]
JEE Main-28.06.2022, Shift-II
Ans. (30) : We know that,
N 1V 1 = N 2V 2
0.01 × 5 × V1 = 0.05 × 1 × 20
V1 = 20 mL used
∴ Volume left = 50 – 20 = 30 mL
19. 2.5 g of protein containing only glycine
(C2H5NO2) is dissolved in water to make 500 ml
of solution. The osmotic pressure of this
solution at 300 K is found to be 5.03x10-3 bar.
The total number of glycine in the protein is
_______.
(Given : R = 0.83 L bar K-1 mol-1)
JEE Main-28.06.2022, Shift-II
Ans. (330) : Given,
π = 5.03 × 10−3 bar, T = 300 K
R = 0.083 L, bar K−1 mol−1
Van’t Hoff equation
π = CRT
5.03 × 10−3 = C × 0.083 × 300
C = 0.202 × 10−3 M
Moles of protein = 0.202 × 10−3 × 0.5
= 1.01 × 10−4
2.5
Molar mass of protein (M) =
1.01×10−4
= 24752
24752
Number of glycine units =
75
= 330.03
20. A 0.5 percent solution of potassium chloride
was found to freeze at –0.24oC. The percentage
dissociation of potassium chloride is ______.
(Nearest integer)
(Molar depression constant for water is 1.80 K
kg mol–1 and molar mass KCl is 74.6 g mol–1)
JEE Main-26.06.2022, Shift-I
0.5 1.8
Ans. (99) : m =
×
= 0.12064
74.6 0.1
We know that,
∆Tf = i × m
0.24 = i × 0.12064
Ans. (14) :
Objective Chemistry Volume-II
21.
i = 1.989
i = 1 + α (n – 1) (∵ n –1 = 1)
1.989 = 1 + α
α = 98.9 % ≃ 99%
A commany dissolves ‘X’ amount of CO2 at 298
K in 1 litre of water to prepare soda water. X =
____×10–3g. (nearest integer)
JEE Main-24.06.2022, Shift-II
Ans. (1221) : P = KH × X CO2
0.835 = 1.67 × 103 ×
0.835 = 1.67 × 103
n CO2
n H 2O
WCO2 / 44
1000
18
1.67 ×18
× WCO2
44
= 1.2222 kg
0.835 =
WCO2
22.
= 1222.2 × 10−3 gm
The rise in boiling point of a solution
containing 1.8 g of glucose in 100g of solvent is
0.1 °C . The molar elevation constant of the
liquid is
(a) 2K kg/mol
(b) 10K kg/mol
(c) 0.1K kg/mol
(d) 1K kg/mol
Karnataka CET-17.06.2022, Shift-II
Ans. (d) :
∆Tb
0.1× 100
=
1.8
m
×1000
180
= 1Kkg / mol
If 3 g of glucose (molar mass=180g) is dissolved
in 60 g of water at 15°C , the osmotic pressure
of the solution will be
(a) 6.57 atm
(b) 5.57 atm
(c) 0.34 atm
(d) 0.65 atm
Kb =
23.
Karnataka CET-17.06.2022, Shift-II
Ans. (a) : T = (15 + 273)K
= 288K
Van’t Hoff equation
3 ×1000
× 0.0821× 288
180 × 60
= 6.56atm
Which of the following colligative properties
can provide molar mass of proteins, polymers,
and colloids with greater precision ?
(a) Depression in freezing point
(b) Osmotic pressure
(c) Relative lowering of vapour pressure
(d) Elevation in boiling point
Karnataka CET-17.06.2022, Shift-II
Π = CRT =
24.
24
YCT
Ans. (b) : Osmotic pressure method is especially
suitable for the determination of molecular masses of
macromolecules such as protein and polymer because
for these substances the value of other colligative
properties such as elevation in bolling.
25. Which property of CO 2 makes it biologically
and geo-chemically important ?
(a) Its low solubility in water
(b) Its high compressibility
(c) Its acidic nature
(d) Its colourless and odourless nature
Karnataka CET-17.06.2022, Shift-II
Ans. (a) : Due to the low solubility of CO2 in water, it
is important in biological and geochemical processes.
26. An aqueous solution of alcohol contains 18g of
water and 414g of ethyl alcohol. The mole
fraction of water is
(a) 0.7
(b) 0.9
(c) 0.1
(d) 0.4
Karnataka CET-17.06.2022, Shift-II
Ans. (c) : Molecular weight of C2H5OH = 24 + 5 + 16 +
1 = 46
Molecular mass of H2O = 18
414
414 g of C2H5OH has
= 9 mole (i.e. n1 = 9 mole)
46
18
18 g of H2O has =
= 1 mole (i.e. n2 = 1 mole)
18
n2
1
mole fraction of water =
=
n1 + n 2 1 + 9
1
= = 0.1
10
27. The neutralization occurs when 10 mL or 0.1M
acid ‘A’ is allowed to react with 30 mL of 0.05
M base M(OH)2. The basicity of the acid ‘A’ is
[M is a metal]
JEE Main-25.06.2022, Shift-II
Ans. (3) :
Acid + Base → salt + H2O
0.1 M
M(OH)2
10 ml
0.05 m
30 ml
At equivalence point equivalent of acid = Equivalent of
base
0.1 ×10 × n = 30 × 0.05 × 2
n=3
28. A solution of Fe2(SO4)3 is electrolyzed for ‘x’
min with a current of 1.5 A to deposit 0.3482 g
of Fe. The value of x is–––.[nearest integer]
–1
Given: 1F = 96500 C mol
–1
Atomic mass of Fe = 56 mol
3 × 96500
56
3 × 96500
For 0.3482 g →
× 0.3482
56
= 1800.06
We know that,
Q = it
1800.06 = 1.5 t
t = 1200 sec. or t = 20 min
29. Solute A associates in water. When 0.7 g of
solute A is dissolved in 42.0 g of water, it
depresses the freezing point by 0.2oC. The
percentage association of solute A in water is:
[Given: Molar mass of A = 93 g mol–1 Molal
depression constant of water is 1.86 K kgmol–1]
(a) 50%
(b) 60%
(c) 70%
(d) 80%
JEE Main-25.06.2022, Shift-II
Ans. (d) : Given that, ∆T = 0.2 K, Kf = 1.86 K
∆T = iKf × m
0.7 1000
0.2 = i × 1.86 ×
×
93
42
0.2 × 93 × 6
i=
1.86 ×100
i = 0.60
1g →
2A ↽ ⇀ A 2
1−α
α
2
i = 1− α +
i = 1−
α
2
α
2
α
= 0.60
2
α
1 − 0.60 =
2
α = 0.80 × 100
α = 80%
30. At 298 K, vapour pressure of two pure liquids
A and B are 200 and 400 mm Hg respectively,
If mole fractions of A and B in solution are 0.7
and 0.3 respectively. What is the mole fraction
of B in vapour phase?
(a) 0.279
(b) 0.721
(c) 0.538
(d) 0.462
AP-EAMCET-06.07.2022, Shift-II
Ans. (d) : Given that,
Vapour pressure of liquids PA0 = 200 mm Hg
1−
Vapour pressure of liquids PB0 = 400 mm Hg
Mole fractions of A (XA) = 0.7
Mole fractions of B (XB) = 0.3
JEE Main-25.06.2022, Shift-II
We, know that,
+3
−
Ans. (20) : Fe + 3e
→ Fe
P = PA0 . XA + PB0 . XB
3F
→1mole Fe is deposited
P = 200 × 0.7 + 400 × 0.3
P = 260
for 56 g → 3 × 96500
Objective Chemistry Volume-II
25
YCT
So, mole fraction of B isP
P 0 .X
XB = B = B B
P
P
400 × 0.3
XB =
260
XB = 0.460
31. 3.1 g of a compound, ‘X’ (molar mass = 62g
mol–1) is dissolved in 19.5 g of other compound,
Y (molar mass = 78 g mol–1). The ratio of mole
fractions of X and Y in the solution is
(a) 1 : 5
(b) 5 : 1
(c) 4 : 1
(d) 1 : 4
AP-EAMCET-07.07.2022, Shift-II
Ans. (a) : Given,
Mass of solute = 3.1 g
Molar mass of solute (X) = 62 g/mol
Mass of solvent = 19.5 g
Molar mass of solvent (Y) = 78 g/mol
Mass
Mole =
Molar Mass
3.1
= 0.05 mole
62
19.5
= 0.25 mole
Mole of solvent (Y) =
78
Total number of Mole = nA + nB = 0.05+ 0.25 = 0.30
Mole of solute
Mole traction of solute (X) =
Total mole
Mole of solute (X) =
N Solute
V solution
Molarity depends on volume, and as well know if the
temperature increases then volume also increases.
Similarly if we decreases.
33. The molality and molarity of a solution of a
glucose in water which is labeled as 10%
(w/w) are respectively (density of solution =
1.2 g mL-1)
(a) 0.57m, 0.517M
(b) 0.67m, 0.617M
(c) 0.617m, 0.67M
(d) 0.517m, 0.57M
AP-EAMCET-07.07.2022, Shift-I
Ans. (c) : According to question
100g of water solution 10% solution of glucose.
w
glucose concentration is 10%
w
Mass of solution = 100g
Mass of solution (glucose) = 10g
Mass of solution = 90g
Molar Mass of glucose is 180g/mol
Mass of solute
1
Molality =
×
Molar Mass of solute Mass of solute
C=
10
1
×
180 .090
= 0.617m
The Molality of the solution is 0.617m
Volume of the solution [density of solution 1.2 g mL–1]
Mass 100
=
=
density 1.2
= 83.33 mL
Mass of solute
1
Molarity =
×
Molar Mass of solute Volume of solution
Molality =
0.05
0.3
= 0.167
0.25
Mole fractional solvent (Y) =
= 0.833
0.3
Then Ratio
0.05 0.25
X:Y=
:
0.3
0.3
X : Y = 5 : 25
=1:5
32. Assertion (A) : Molality of solution increases
with temperature.
Reason (R) : Molality expression does not
involve any volume term.
(a) Both A and R correct and R is the correction
explanation of A.
(b) Both A and R are correct, R is not the
correction explanation of A.
(c) A is correct, R is not correct.
(d) A is not correct, R is correct.
AP-EAMCET-08.07.2022, Shift-II
Ans. (d) : Molarity of a solution depends upon
temperature where as molality does not depend upon
the temperature. Molarity and molality both depend
only on the number of moles of solute particles.
Molarity is number of moles of solute which are
dissolved per liter of solution i.e.
=
Objective Chemistry Volume-II
10 1000
×
180 83.33
= 0.67M
Molarity of the solution is 0.67M
Hence option (c) is correct.
34. Which of the following are correct for an ideal
solution?
(a) ∆Vmix = 0
(b) Vsolvent + Vsolute = Vsolution
(c) ∆Hmix = 0
(d) H2O+CO2→H2CO3 is an example of ideal
solution
(a) a, b only
(b) b, c only
(c) a, b, c only
(d) a, b, c, d
TS-EAMCET-18.07.2022, Shift-I
Ans. (c) : The solutions which obey result's law over
the entire range of concentration are in own as ideal
solutions. The enthalpy of solution is zero.
If the enthalpy of the solution gets closer to zero it is
more likely to show an ideal behavior.
∆ mix H = 0
26
Molarity =
YCT
Thus, the volume of mixing is
Also zero.
∆ mix H = 0
Thus, option (a), (b) and (c) are correct.
35. If 2 g of NaOH is dissolved to make 200 ml
solution at 25 oC, the molarity (M) at 90 oC is
(a) M < 0.25
(b) 0.5 > M > 0.25
(c) M = 0.25
(d) 0.5 < M < 1.0
TS-EAMCET-20.07.2022, Shift-II
Ans. (a) : Given that,
Mass = 2 g
Molecular mass of NaOH = 40
V = 200 ml
We know that
mole
Molarity =
volume in (Liter)
Mole of NaOH =
Molarity =
mass of NaOH
2
=
molecular mass of NaOH 40
15
× 1000
119.5 × (1000, 000 − 15)
= 1.25 × 10–4 mol Kg–1
38. The molarity of 10% (w/w) aqueous NaOH
solution (density 1.11 g mL-1)
(a) 2.50 M
(b) 3.25 M
(c) 2.78 M
(d) 1.52 M
AP-EAPCET-11.07.2022, Shift-I
Ans. (c) :
Molarity(M)=
wt.of solute × density of solution × 10
molecular mass of solute × wt.of solution
∴ Molality =
R Tf 2
1000 ×Lf
8.314 × 290 ×290
1000 × 180
Kf = 3.88 k. kg mol–1
A sample of drinking water has 15 ppm (by
mass) of a carcinogen (molar mass 120 g mol-1).
The molality of carcinogen in water sample in
mol kg-1 is
(a) 2.50×10-4
(b) 2.50×10-3
-4
(c) 1.25×10
(d) 1.25×10-3
AP-EAPCET-11.07.2022, Shift-II
Objective Chemistry Volume-II
10 × 1.11×10
40
= 2.78
50%
of
the
reagent
is
used
for
dehydrohalogenation of 6.45 gm CH3CH2Cl.
What will be the weight of the main product
obtained?
(Atomic Mass of H, C and Cl are 1, 12 & 35.5)
(gm mol-1 respectively)
(a) 0.7 gm
(b) 1.4 gm
(c) 2.8 gm
(d) 5.6 gm
GUJCET-2016, 2015
=
=
37.
Massof chloroform
×1000
MolarMof chloroform×(Mof solution − Mof chloroform)
2 1000
×
= 0.25M
40 200
Molarity is dependent on volume and volume rise as the
temperature does, molarity is inversely proportional to
temperature. So, If temperature increases molarity
decreases.
Thus at 90oC molarity is M < 0.25
36. A solvent freezes at 17 oC and it has latent heat
of fusion 180 J g−1. The molal depression
constant of the solvent is [units of Kf = K kg
mol−1]
(a) 3.88
(b) 3.55
(c) 3.70
(d) 4.77
TS-EAMCET-20.07.2022, Shift-II
Ans. (a) : Given that,
Tf = 17o C
= 17 + 273 = 290k
Latent heat of fusion (Lf) = 180 Jg–1
R = 8.314 J/k/ mol
Molal depression constant (Kf) = ?
We know that,
Kf =
Ans. (c) : 15ppm corresponds to 15g chloroform in
1000,000g of solution.
∴ Molality =
39.
HCl
Ans. (b) : CH 3CH 2 Cl
→ CH 2 = CH 2
Number of moles of CH3CH2Cl
given mass
=
molar mass CH 3CH 2 Cl
6.45
64.5
= 0.1 mol.
50% of product will be formed because of reagent used
50%, So 0.05 mole of CH3CH2Cl will give 0.05 mole
Ethene.
Mass of ethane = moles × molar mass of ethene
= 0.05 × 28
= 1.4 g
40. The boiling points for aqueous solutions of
sucrose and urea are same at constant
temperature. If 3 gm of urea is dissolved in its 1
litre solution, what is the weight of sucrose
dissolved in its 1 litre solution?
[Urea – 60 gm/mole, sucrose = 342 gm/mole]
(a) 34.2 gram
(b) 17.1 gram
(c) 6.0 gram
(d) 3.0 gram
GUJCET-2015, 2016
27
=
YCT
Ans. (b) : From formula
Wurea
Wsucrose
=
(molar mass) urea (molar mass)sucrose
3 Wsucrose
3 × 342
=
=
60
342
60
Wsucrose = 17.1g
41. Which one of the following concentration units
is independent of temperature?
(a) Normality
(b) molarity
(c) molality
(d) ppm
J & K CET-2000, 2008
Ans. (c) : Molality involves only with mass and It does
not depend upon volume. So, we can say Molality is
independent of temperature.
42. 250 mL of a sodium carbonate solution
contains 2.65 g of Na2CO3. If 10 mL of this
solution is diluted to 1 L, what is the
concentration of the resultant solution? (Mol.
wt. of Na2CO3 = 106)
(a) 0.1 M
(b) 0.001 M
(c) 0.01 M
(d) 10–4 M
JCECE-2011
AP EAMCET (Engg.)-2001
Ans. (b) : Molarity of Na2CO3 solution
No.of moleof solute
Molarity =
volumeof solvent ( L )
2.65 1
×
× 1000 = 0.1 M
106 250
Molarity of mixtureM 1V 1 = M 2V 2
10 × 0.1 = 1000 × M2
10 × 0.1
∴
M2 =
= 0.001 M
1000
43. What amount of Cl2 gas liberated at anode, if 1
ampere current is passed for 30 minute from
NaCl solution?
(a) 0.66 mol
(b) 0.33 mol
(c) 0.66 g
(d) 0.33 g
JIPMER-2006, 2005
Ans. (c) : Given,
i = 1 ampere
t = 30 minute = 30×60 sec
We know that,
Q = it
= 1 × 30 × 60
= 1800 C
1× 1800 × 71
Anode =
(∴ 2Cl– → Cl2 + 2e–)
2 × 96500
= 0.66g
44. Molarity is
(a) the number of moles of solute present in 1
dm3 volume of solution
(b) the number o moles of solute dissolved in 1
kg of solvent
=
Objective Chemistry Volume-II
(c) the number of moles of solute dissolved in 1
kg of solution
(d) the number of moles of solute dissolved in
100 dm3 volume of solution
MHT CET-2018, 2015
Ans. (a) : Molarity is the total number of moles of
solute per litre of solution.
45. What is the mole fraction of the solute in a 1.00
m aqueous solution?
(a) 1.770
(b) 0.0354
(c) 0.0177
(d) 0.177
(NEET-2015, 2011)
Ans. (c) : 1 molal solution means 1 mole of solute
dissolved in 1000 gm solvent
∴ nsolute =1, Wsolvent =1000 g
1000
∴ nsolvent =
=55.56
18
1
Xsolute =
= 0.0177
1 + 55.56
46. Which one of the following is independent of
temperature?
(a) Normality
(b) Molality
(c) Molarity
(d) Weight-volume percentage
UPTU/UPSEE-2005
UP CPMT-2001
AIPMT -1995, 1992
Ans. (b) : Volume depends on temperature. So, the
terms which have volume in them will be temperature
dependent. Write the formula of each of them to find
term which is independent of volume.
No.of moles of solute
(a) Molarity =
volume of solution in litre
gram − equivalent of solute
volume of solution in litre
(c) Weight-volume percentage
Weight of solute
=
× 100
volume of solvent
(b) Normality =
(d) Molality =
moles of solute
mass of solvent in kg
∴ Molality does not involve volume
∴ Molality is independent to temperature.
47.
28
15
O 2 ( g ) → 6CO 2 ( g ) + 3H 2 O(liq)
2
Benzene burns is oxygen according to the
above equation. What is the volume of oxygen
(at STP) needed for complete combustion of 39
gram of liquid benzene?
(a) 11.2 litre
(b) 22.4 litre
(c) 84 litre
(d) 168 litre
WB-JEE-30.04.2022
C6 H 6 (liq) +
YCT
2x + 7(−2) = −2
x =6
Ans. (c) : Required Reaction,
2C6 H 6 + 15O 2
→12CO 2 + 6H 2O
Molar mass of 2C6 H 6 = 2 × 78
= 156
Molar mass of 15O2 = 15 × 32
= 480
Now,
39g of benzene required with the respect of
156g of benzene in oxygen (S.T.P.)
15 × 22.4 × 39
=
156
= 84L
48. How much solid oxalic acid (Molecular weight
126) has to be weight to prepare 100 mL
exactly 0.1 (N) oxalic solution in water?
(a) 1.26 g
(b) 0.126 g
(c) 0.63 g
(d) 0.063 g
WB-JEE-30.04.2022
Ans. (c) : Given :
Molecular weight = 126g/mol
'n' factor = 2
126
eq.w.t =
= 63g / mol.
2
W ×1000
M=
molecular weight × Vol
15 × M Fe 2+
20 × 0.03
×1 =
×6
1000
1000
MFe2+ ×15 ×1 = 0.03 × 6 × 20
MFe2+ = 0.24
MFe2+ = 24 × 10–2 M.
51. The mole fraction of a solute in a 100 molal
aqueous solution……×10-2
(Round off to the nearest integer)
[Given, atomic masses H : 1.0u 0 : 16.0u].
[JEE Main 2021, 17 March Shift-I]
Ans. (64): Given,
Moles of solute = 100
1000
Number moles of H2O =
∴ (WH O = 1000g)
18
1800
Mole fraction of solute =
2800
= 64 × 10–2
52. CO2 gas is bubbled through water during a soft
drink manufacturing process at 298 K. If CO2
exerts a partial pressure of 0.835 bar then x m
mol of CO2 would dissolve in 0.9 L of water.
The value of x is ....... . (Nearest integer)
(Henry's law constant for CO2 at 298 K is 1.67
× 103 bar)
W × 100
[AIEEE 2021 Shift-I]
0.1 =
Ans. (25) : Given that, Partial pressure = 0.835 bar
63 × 100
V = 0.9 L = 900 mL
W = 0.63g
Applying Henry laws
49. If the concentration of glucose (C6H12O6) in
-1
blood is 0.72 gL the morality of glucose in PCO2 =K H X gas
blood is….. ×10-3 M. (Nearest integer)
0.835 =1.67 × 103 X CO2
[Given : Atomic mass of C = 12, H = 1,O = 16u]
−3
[JEE Main 2021, 22 July Shift-II] X CO2 =0.5 × 10
Ans. (4) : Given,
900
n CO2
Concentration of glucose = 0.72 gL−1
nH 2 O = 18
X CO2 =
Molecular mass of glucose C6H12O6
n CO2 + nH 2 O
= 50
= 12×6+1×12+96=180 g/mol
n CO2
Concentration
X CO2 =
n CO2 << n H2O
Molarity of Glucose =
n H 2O
molar mass
0.72
n CO2
=
0.5 × 10−3 =
180
50
= 4 ×10–3 M
n CO2 = 25 × 10−3 moles
1 L solution contains 0.04 mole glucose or the molarity
of glucose 0.004 M.
Millimmoles of CO 2 = 25 × 10−3 × 1000 = 25
2+
50. 15 mL of aqueous solution of Fe in acidic 53. The vapour pressures of A and B at 25°C are
medium completely reacted with 20 mL of 0.03
90 mm Hg and 15 mm Hg respectively. If A
and B are mixed such that the mole-fraction of
M aqueous Cr2 O72- . The morality of the Fe2+
A in the mixture is 0.6, then the mole fraction
solution is…..×10-2 M
of B in the vapour phase is x × 10–1. The value
(Round off to the nearest integer).
of X is .......... (Nearest integer)
[JEE Main 2021, 17 March Shift-I]
[AIEEE 2021 Shift-II]
Ans. (24): By the law of equivalence,
Ans. (1) : Given,
neq Fe2+ = n eq Cr2O 72–
PAo = 90 mm Hg XA + XB = 1
n factor Cr2 O 27 −
PBo = 15 mm Hg 0.6 + XB = 1
2
(
Objective Chemistry Volume-II
29
)
YCT
56.
XA = 0.6
XB = 0.4
We know that,
PT = PAo XA + PBo XB
PT = 90 × 0.6 + 15 × 0.4
= 60 mm Hg
Now
Mole factor of B in vapour phase
YB =
PB
P o .X
= B B
PT
PT
∴ PB = PBo .X B
15 × 0.4
= 0.1 or 1 × 10−1
60
Therefore, X × 10–1 = 1 × 10–1
So, X = 1
54. At 20°C, the vapour pressure of benzene is 70
torr and that of methyl benzene is 20 torr. The
mole fraction of benzene in the vapour phase at
20°C above an equimolar mixture of benzene
and methyl benzene is ........ ×10–2 (Nearest
integer)
[AIEEE 2021 Shift-I]
Ans. (78): Given formula,
PTotal = PCo6 H6 × X C6H6 + PCo6 H5CH3 + X C6 H5CH3
YB =
1
1
+ 20 ×
2
2
= 35 + 10
= 45
= 70 ×
70 ×
Mole fraction of Y benzene =
1
2
45
= 0.777
= 0.78 ≃ 78 × 10–2
Sodium oxide reacts with water to produce
sodium hydroxide. 20.0 g of sodium oxide is
dissolved in 500 mL of water. Neglecting the
change in volume, the concentration of the
resulting NaOH solution is……×10--1 M.
(nearest integer)
[Atomic mass : Na = 23.0, O =16.0 H=1.0]
[JEE Main 2021, 31 Aug Shift-II]
Ans. (13) : Required equation
Na2O + H2O → 2NaOH
Number moles of NaOH formed as product to
20
=
×2
62
= 0.6451
Now,
40 / 62
Concentration of resulting NaOH =
500 /1000
= 1.29 M ≃ 13 × 10–1
57. 100 g of propane is completely reacted with
1000 g of oxygen. The mole fraction of carbon
dioxide in the resulting mixture is x ×10-2 The
value of x is……..(Nearest integer) [Atomic
weight H=1.008, C=12.00, O=16.00]
[JEE Main 2021, 27 Aug Shift-II]
Ans. (19): Required reaction–
C3H8(g)+5O2(g) → 3CO2(g) + 4H2O(I)
100 g of propen = 2.27 mol
1000 g of oxygen = 31.25 mol
2.27 mol of propen will react with 11.35 mol oxygen to
give 6.81 mol CO2 and 9.08 mol H2O.
C3H8(g) + SO2(g) → 3CO2(g)+ 4H2O(I)
t = 0 2.27
t=∞ 0
Here,
X × 10−2
So,
X = 78 × 10–2
55. 10.0 mL of Na2CO3 solution is titrated against
0.2 M HCI solution. The following titre values
were obtained in 5 readings. 4.8 mL, 4.9 mL,
5.0 mL, 5.0 mL and 5.0 mL based on these
readings and convention of titrimetirc
estimation of concentration of Na2CO3 solution
is…….mM
(Round off to the nearest integer).
[JEE Main 2021, 18 March Shift-II]
Ans. (50) : Required equation;
Na2CO3 + 2HCl → 2NaCl + H2CO3
Equivalents of Na2CO3 (N × V) = equivalents of HCl
(N × V)
[n1 × M1 × V1] = [n2 × M2 × V2]
2 × M × 10 = 1 × 0.2 × 5
0.2 × 5 1.0
M=
=
2 × 10 20
M Na 2CO3
= 0.05 mol / L
= 0.05 × 1000
= 50 mM
Objective Chemistry Volume-II
31.25
19.9
0
6.81
0
9.08
6.81
19.91 + 6.81 + 9.08
= 0.1902 = 19.02×10–2
So, volue of x is 19.
58. If 80g of copper sulphate CuSO4 ⋅ 5H 2O is
dissolved in deionised water to make 5L of
solution. The concentration of the copper
sulphate solution is x × 10–3 mol L–1. The value
of x is..........
[Atomic masses Cu = 63.54 u, S=32 u, O=16 u,
H=1 u]
[AIEEE 2021 Shift-II]
Ans. (64) : Given, CuSO4.5H2O with mass = 80 g
Molar mass of CuSO4.5H2O = 63.54+32+16×4+5×18
= 249.54 g/mole
The concentration of copper sulphate solution x×10–3
mol/L.
weight of solute
Number of mole of solute =
molecular mass of solute
80
=
= 0.32 mol
249.5
30
X CO2 =
YCT
61.
Now,
Molarity =
Number of moles of solute
Volume(litre)
0.32
= 64.11× 10−3 mol/L
5
Hence, = 64.11 ≈ 64
59. The molarity of the solution prepared by
dissolving 6.3 g of oxalic acid (H2C2 O4 . 2 H2O)
in 250 mL of water in mol L–1 is x × 10–2. The
value of x is .......... (Nearest integer) [Atomic
mass H = 1.0, C = 12.0, O = 16.0]
[AIEEE 2021 Shift-I]
Ans. (20) : Given, Oxalic acid C2H2O4.2H2O of mass
6.3 g
Then, molar mass = 126 g/mole
weight
Mole =
molar mass
= =
=
6.3
126
Then,
Molarity=
weight of solute
molecular mass of solute
100 mL of Na3PO4 solution contains 3.45 g of
sodium. The molarity of the solution is.......×10–
2
mol L–1. (Nearest integer)
[Atomic masses – Na = 23.0 u,
O=16.0 u, P = 31.0 u]
[AIEEE 2021 Shift-II]
Ans. (50) : Required equation:
Na3PO4 → 3Na+ + PO 4–
3.45
So, no of moles Na+ →
23
3.45
No. of moles of Na3PO4=
= 0.05
23 × 3
0.05
Molarity (M)=
× 1000 = 0.5 = 50 ×10−2 mol L−1
100
62. The exact volumes of 1 M NaOH solution
required to neutralise 50 mL of 1M H3PO3
solution and 100 mL of 2M H3PO2 solution,
respectively, are
(a) 100 mL and 100 mL (b) 100 mL and 50 mL
(c) 100 mL and 200 mL (d) 50 mL and 50 mL
[AIEEE 2021 Shift-II]
Ans. (c) : Required reaction,
2NaOH + H3PO3 → Na2HPO3 + 2H2O
50 ml 1M
1M
V=?
n NaOH 2
[n=m×v]
=
n H3PO3 1
1× V 2
= ⇒ VNaOH = 100ml
50 × 1 1
NaOH + H3PO2 → NaH2PO2 + H2O
100 ml
1M
2M
V=?
6.3 /126
Molarity =
250 /1000
6.3 × 1000
126 × 250
= 0.02 M
= 20×10–2
n ×10−2 =
Hence,
n NaOH 1
=
n = 20
n H3PO3 1
–3
60. The density of NaOH solution is 1.2 g cm . The
1× V 1
molality of this solution is ...........m.
= ⇒ VNaOH = 200ml
2
× 100 1
(Round off to the nearest integer)
of KOH (aq) has a
[Use : Atomic masses : Na = 23.0 u, O=16.0 u, 63. A 6.50 molal solution
–3
–3
.
The molarity of the
density
of
1.89
g
cm
H=1.0 u, density of H2O : 1.0g cm ]
solution is ........ mol dm–3. (Round off to the
[AIEEE 2021 Shift-I]
nearest integer).
Ans. (5) : Given, Density = 1.2 g/cm3
[Atomic masses: K : 39.0 u,
Then, Mass of Solution = 1.2 × 1000
O : 16.0 u, H : 1.0 u]
[AIEEE 2021 Shift-I]
= 1200 gm
Mass of NaOH
= (1200 – 1000) = 200 gm.
Ans. (9) : Wt. of solute= 6.5×56= 364 g
Wt. of solution = 1000+364= 1364 g
weight
Then, mole of NaOH=
1364
molar mass
Volume of solution =
ml
1.89
Moler mass of NaOH= 23+16+1 = 40 g/mole
Moleof solute
6.5 × 1.89 × 1000
200g
Molarity =
=
=9
Mole of NaOH =
= 5 mol
Volumeof solution
1364
40g / mol
64. 4.5g of compound A (MW = 90) was used to
mole of solute
make 250 mL of its aqueous solution. The
Now, Molality =
weight of solvent
molarity of the solution in M is x × 10–1. The
value of x is ......... (Rounded off to the nearest
5
Molality = = 5m.
integer).
1
[AIEEE 2021 Shift-I]
Objective Chemistry Volume-II
31
YCT
Ans. (2) : Given, Mass of compound = 4.5 g and molar
mass = 90 g/mole.
Then,
weight
4.5
Moles =
=
Mol
molar mass
90
Then,
Number of moles of solute
Molarity =
Volume
4.5
= 90
250
1000
= 0.2
= 2 × 10–1 m.
Here, M= x × 10–1 m
So, x = 2
65. If a compound AB dissociates to the extent of
75% in an aqueous solution, the molality of the
solution which shows a 2.5 K rise in the boiling
point of the solution is ........ molal. (Rounded
off to the nearest integer)
[Kb = 0.52 K kg mol–1]
[JEE Main 2021, 25 Feb Shift-II]
Ans. (3) : Given:
α = 0.75
n=2
i = 1– α + nα =1– 0.75 + 2 × 0.75
= 1.75
∆Tb = iKbm
2.5
m=
= 2.74 ≃ 3 (nearest integer value)
1.75 × 0.52
66. 1 molal aqueous solution of an electrolyte A2B3
is 60% ionised. The boiling point of the solution
at 1 atm is ....... K (Rounded off to the nearest
integer). [Given, Kb for (H2O)= 0.52 K kg mol–
1
]
[JEE Main 2021, 25 Feb Shift-I]
Ans. (375) : From formula
∆Tb = iKbm
= (1 + 4α) × 0.52 × 1
(α=0.60)
= 3.4 × 0.52 × 1
= 1.768
Tb
= 1.768 + 373.15
= 374. 918K
= 375 K
67. Liquids A and B form an ideal solution in the
entire composition range. At 350 K, the vapor
pressures of pure A and pure B are 7 × 103 Pa
and 12 × 103 Pa, respectively. The composition
of the vapour is in equilibrium with solution
containing 40 mole percent of A at this
temperature is:
(a) xA = 0.37; xB = 0.63
(b) xA = 0.28; xB = 0.72
(c) xA = 0.4; xB = 0.6
(d) xA = 0.76; xB = 0.24
BITSAT – 2021
Objective Chemistry Volume-II
Ans. (b) : Given that
Vapour pressure (PA° ) = 7 × 103 Pa
Moles of A (xA)= 0.4
Vapour pressure (PB) = 12 × 103
Moles of B(xB) = 0.6
From formula,
Moles fraction of A (YA) =
PA° x A
P x A + PB° x B
°
A
7 × 103 × 0.4
7 × 10 × 0.4 + 12 × 103 × 0.6
2.8
YA =
10
YA = 0.28
And, moles fraction of B (YB) = 1–0.28
= 0.72
68. The OH– concentration in a mixture of 5.0 mL
of 0.0504 M NH4Cl and 2 mL of 0.0210 M NH3
solution is X × 10–6 M. The value of X is ..........
(Nearest integer)
[Given, Kw = 1×10–14 and Kb = 1.8 × 10–5]
[JEE Main 2021, 26 Aug Shift-I]
Ans. 3
Required equation.
–
NH3 + H2O → NH +4 + OH
From formula
YA =
3
Kb =
Kb =
NH 4+ OH –
[ NH3 ]
0.0504 × 5 OH –1
0.0210 × 2
2 0.0210
OH = 1.8 × 10 × ×
5 0.0504
756 × 10 –3
=
= 0.3 × 10 –5
2520
So,
OH – = 3 × 10 –6
–1
–5
69.
32
3.0 gram ethanoic acid in 50 gram benzene
having –––––– molality?
(Atomic weights : H = 1, C = 12, O = 16).
(a) 0.1
(b) 1.0
(c) 0.6
(d) 0.06
GUJCET-2021
Ans. (b) : Given,
Molecular weight of ethanoic acid (CH3COOH)
= 60 g/mol
We know that,
moles of solute
Molality =
Amount of solvent (in kg)
=
3gm / 60g mol −1
50 × 10 −3 ( kg )
=
3 ×1000
=1
60 × 50
YCT
In basic medium CrO 42– oxidizes S 2O 32– to form We know that, molality = number of moles of solute
–
present per kg of the solvent
SO42– and itself changes into Cr ( OH )4 . The
∵ 6.5 mole KOH solvent present in 1000g solvent
volume of 0.154 M CrO 42– required to react
∴ Weight of solute (KOH) = 6.5×56= 364 g
with 40 mL of 0.25 M S 2O 32– is _____ mL. ∴ Weight of solution = 1000+364= 1364g
(Rounded –off to the nearest integer)
Mass ( m )
JEE Main 25-02-2021, Shift-I Density ( ρ ) = Volume v
( )
Ans. : (173.16 mL) Reaction of CrO 2–
4 ion with
m
or Vol. =
S2 O32– in basic medium is given as follows –
ρ
–
2–
2–
CrO 2–
+
S
O
→
SO
+
Cr
OH
(
)
1364
4
2 3
4
4
Volume =
= 721.69mL = 721.69 × 10–3L
Balance this equation using oxidation number method,
1.89
we get –
no.of moles of solute
∴ Molarity ( M ) =
+6
+2
+6
+3
–
2–
2–
2–
Volume
of solution ( in litre )
8Cr O + 3S 2 O → 6 S O + 8C r ( OH )
70.
4
Now,
3
0.154 M CrO
4
2–
4
4
= 0.154 × 3 N CrO 2–
4
6.5 × 1000
721.69
Molarity (M) = 9.00 mol dm−3
73. If a compound AB dissociated to the extent of
75% in an aqueous solution, the molality of the
solution which shows a 2.5 K rise in the boiling
point of the solution is
molal.
(Rounded-off to the nearest integer)
[Kb = 0.52 K kg mol–1]
JEE Main 25.02.2021, Shift-II
−1
Ans.: (3mol Kg ) Reaction is given by,
AB ⇌ A + + B−
1–α
α
α
∴
α = 3/4 and n = 2
i = 1+(n–1) α
3
⇒
i = 1+ (2 − 1)
4
3 7
⇒
i = 1+ =
4 4
∆Tb = iKbm
∆Tb
m=
i × Kb
∴ Molarity =
Similarly, the normality of 0.25 M S2 O32– solution is
0.25 × 8 N
gm equivalent of CrO42− = S2O32−
⇒ N 1V 1 = N 2V 2
⇒ V × 0.154 × 3 = 0.25 × 40 × 8
0.25 × 40 × 8
⇒V=
0.462
80
⇒V=
= 173.16 mL .
0.462
71. If the density of a 2 M solution of ethylene
glycol in water is 1.11 g/ml, the molality (in 'm')
of the solution is approximately
(a) 1.92
(b) 1.57
(c) 2.05
(d) 2.15
TS EAMCET 04.08.2021, Shift-I
Ans. (c) : Mole of ethylene glycol = 2
Mass of solution = Volume × density
= 1000 × 1.11
= 1110 gm
Mass of ethylene glycol = 2 × 62.07 = 124.14 gm
Where, 62.07 = molecular mass of ethylene glycol
Mass of water = 1110 – 124.14 = 985.86 gm
Moles of solute ( ethyleneglycol )
∴ Molality(m) =
Mass of solution ( water ) in kg
2
2000
=
≃ 2.05
985.86 × 10−3 985.86
72. A 6.50 molal solution of KOH (aq.) has a
density of 1.89 g cm-3. The molarity of the 74.
solution is __________ mol dm-3. (Round off to
the Nearest Integer).
[Atomic masses: K = 39.0 u; O = 16.0 u; H = 1.0
u]
JEE Main 16.03.2021, Shift-I
Ans. (9.00) :
Given :- Density ( ρ ) = 1.89g cm-3
=
Molarity (M) =?
Objective Chemistry Volume-II
33
2.5
7
× 0.52
4
2.5 × 4
10
=
=
= 2.747 ≈ 3 mol kg −1
7 × 0.52 3.64
Consider titration of NaOH solution versus
1.25 M oxalic acid solution. At the end point
following burette readings were obtained.
(i) 4.5 mL
(ii) 4.5 mL
(iii) 4.4 mL
(iv) 4.4 mL
(v) 4.4 mL
If the volume of oxalic acid taken was 10.0 mL
then the molarity of the NaOH solution is
M. (Rounded-off to the nearest integer)
JEE Main 25.02.2021, Shift-II
=
YCT
Ans. (6M) : Average burette reading = volume of
NaOH solution (V1)
4.5 + 4.5 + 4.4 + 4.4 + 4.4
=
5
= 4.44 mL
Strength of NaOH solution = S1(M) = S1(N)
Volume of oxalic acid solution (V2) = 10mL
Strength of oxalic acid solution (S2) = 1.25 M
= 1.25×2 N
So,
V1S1 = V2S2 (∵ Law of eqivalence)
V S 10 × (1.25 × 2)
⇒ S1 = 2 2 =
V1
4.44
S1 = 5.63 M
S1 ≈ 6 M
75. 4.5 g of compound A (MW = 90) was used to
make 250 mL of its aqueous solution. The
molarity of the solution in M is x×10–1. The
value of x is ______.(Rounded off to the nearst
integer)
JEE Main 24.02.2021, Shift-I
Ans. : (2) Given that,
W = 4.5g
Molecular weight (m) = 90
V = 250 mL
Molarity (M) = x×10–1
x=?
Now, we use the following molarity equation –
W ×1000
Molarity =
m×V
4.5×1000
–1
x ×10 =
= 2 × 10–1
90× 250
or x = 2
Hence, the value of x is 2.
76. 50 mL of 0.2 N HCl is titrated against 0.1 N
NaOH solution. The titration is discontinued
after adding 50 mL of NaOH solution. The
remaining titration is completed by adding 0.5
N KOH solution. What is the volume of KOH
solution required for completing the titration?
(a) 10 mL
(b) 11 mL
(c) 12 mL
(d) 10.5 mL
AP-EAPCET-6 Sep. 2021, Shift-II
Ans. (a) : Amount of HCl left after completing 1st
titration.
HCl= 0.2 N, NaOH = 0.1N
VHCl= 50 ml
N1×V1=N2×V2
0.5×VKOH= 0.2×25
0.2 × 25
⇒
VKOH=
= 10 ml
0.5
Therefore, 10 ml of KOH solution required for
completing the titration.
77. Concentrated HNO3 is 63% HNO3 by mass and
has a density of 1.4 g/ml. How many milliliters
of this solution are required to prepare 250 ml
of a 1.2M HNO3 solution?
(a) 18.0
(b) 21.42
(c) 20.0
(d) 14.21
AP EAPCET-6 Sep. 2021, Shift-II
Ans. (b): Concentration of a solution is the amount of
moles of the solute present in 1 liter volume of solution.
No.of moles of solute
Conc. of solutions =
volume of solution (in litre)
Given:– Concentration of HNO3 is 63% by weight
density of HNO3= 1.4 gram/ml
Molar mass of HNO3 = 63 gram/mol
63 gram of HNO3 is present in 100 gram of solution
∴ No, of moles of HNO3 in solution
63gram
=
= 1 mol
63gram / mol
⇒
Volume of solution=
100gram
m
∵ ρ =
1.4gram / ml
V
= 71.428 ml
No.of moles of solute
Molarity of HNO3 solution=
Volume of solution in L
=
1
0.071428
= 14 m
Therefore, volume of HNO3 solution needed to prepare
1.2m of 250 ml HNO3 solution,
=
M1V1 = M 2 V2
V1 =
M 2 V2
M1
1.2 × 250
14
= 21.4 ml
The vapour pressure of a solvent decreased by
20 mm of Hg when a non-volatile solute was
added to the solvent. The mole-fraction of the
solute in the solution is 0.5. What should be the
mole-fraction of the solvent for the decrease in
the vapour pressure needs to be 10 mm of Hg?
3
2
(a)
(b)
4
3
1
3
(c)
(d)
4
2
AP EAPCET 20.08.2021 Shift-II
=
78.
We Know that, N1 × V1 = N 2 × V2
⇒ 0.1×50 = 0.2×V
50
⇒ V=
= 25 ml
2
i.e, 25 ml of 0.2 N HCl has been consumed and
therefore 0.2 N HCl left 50 – 25 = 25ml
Calculation of KOH used in 2nd titrationObjective Chemistry Volume-II
Mass
density
34
YCT
Ans. (a): Ist condition:
Vapour pressure of solvent decreased = 20 mm of Hg
Mole fraction of solute = 0.5
Applying the formula∆P1 X a1
Po − P
=
= Xa
∆P2 X a 2
Po
∆P
= Xa
Po
⇒
81.
The equivalent weight of KMnO4 in acidic and
strongly alkaline medium, respectively, are,
(a) 158; 79
(b) 31.6; 158
(c) 158 ; 158
(d) 31.6; 79
TS EAMCET 10.08.2021, Shift-II
Ans. (b) : Molecular weight of KMnO 4 = 158g/mol.
In neutral medium:-
20 0.5
=
10 X a 2
{Mn
0.5
2
1
Xa2 =
4
Mole fraction of solvent = 1– 1/4
= 3/4
79. 0.63 g of Oxalic acid as dissolved in order to
obtain 250 cm3 of its solution. Find the
normality of this solution [Oxalic acid
(COOH)2 2H2O]
(a) 0.05 N
(b) 0.01 N
(c) 0.04 N
(d) 0.02 N
AP EAPCET 20.08.2021 Shift-II
Ans. (c): Given that,
weight of oxalic acid = 0.63 gm
volume of solution = 250 cm3 = 0.25 liter
∴Normality
gram weight of sulute
=
........( i )
volume of solution × equi.volume weight
Equivalence weight of oxalic acid = 63 gm
Putting these value in eqn. (i) we get
0.63
N=
0.25 × 63
1
N=
25
Normality N = 0.04N
80. A 100 mL of aqueous solution contains 10 gm
of urea. If the solvent is hypertonic with respect
to a 100 mL glucose solution containing W gm
of the compound. Which of the following is
correct?
(a) W glucose = 10 gm
(b) W glucose < 30 gm
(c) W glucose > 30 gm
(d) W glucose < 10 gm
TS EAMCET 10.08.2021, Shift-II
Ans. (b) : From,
N1 N 2
=
V1 V2
10
W
=
60×100 180×100
180×10
W=
60
W = 30gm
If solvent is hypertonic than Wglucose < 30 gm
+ 3e −
→ Mn +4 }
Vf → 3
mol.weight 158
∴ Eq. weight =
=
= 52.6
v.f .
3
In acidic medium :-
Xa2 =
Objective Chemistry Volume-II
+7
{Mn
+7
+ 5e −
→ Mn +2 }
Vf →5
158
= 31.6
5
In alkaline medium :∴ Eq. weight =
{Mn
+7
+ e −
→ Mn +6 }
Vf →1
158
= 158
1
So, the equivalent weight of KMnO4 in acidic and
strongly alkaline medium are :31.6; 158
82.The exact volumes of 1 M NaOH solution required
to neutralise 50 mL of 1 M H3PO3 solution and
100 mL of 2 M H3PO2 solution, respectively,
are:
(a) 50 mL and 50 mL
(b) 100 mL and 50 mL
(c) 100 mL and 200 mL
(d) 100 mL and 100 mL
JEE Main-16.03.2021, Shift-II
Ans. (c) : During the process, the following reaction
occurred-
35
∴ Eq. weight =
H3PO3 + 2NaOH
→ Na2HPO3 + 2H2O
1M
1M
50mL VNaOH = ?
n NaOH 2
=
n H3PO3 1
or
1× V 2
=
50 × 1 1
VNaOH = 100mL
H3PO2 + 2NaOH
→ NaH2PO2 + H2O
2M
1M
100mL vNaOH = ?
n NaOH 1
=
n H3PO2 1
1× V
1
=
2 ×100 1
VNaOH = 200mL
YCT
If 500 ml of CaCl2 solution contains 3.01 × 1022
chloride ions, molarity of the solution will be
(a) 0.05 M
(b) 0.01 M
(c) 0.1 M
(d) 0.02 M
AP EAPCET 20.08.2021 Shift-I
Ans. (a) : Given that,
Volume = 500ml = 0.5 litre
Since CaCl2 has 2 chloride ions is
3.01×1022
=
moles
2
moles of solute
Molarity =
Volumeof solution
83.
3.01× 1022
2 × 6.023 ×1023 × 0.5
3.01
=
6.023 ×10
= 0.0499
= 0.05 M
84. A solution of 122 g of benzoic-acid in 1000g of
benzene shows a boiling point elevation of 1.4°.
If the solute is dimerized to an extent of 80%
the boiling of pure benzene will be ______
(Molar enthalpy of vaporization of benzene =
7.8 kcal. mol-1)
(a) 420 K
(b) 370 K
(c) 540 K
(d) 460 K
AP- EAPCET- 07-09-2021, Shift-I
Ans. (a) : Given data –
Weight of solute (benzoic acid) = 122 gm
Weight of solute (benzene) = 1000 gm
Boiling point elevation = 1.4°
Tb − Ti ( ∆Tb ) = iK b m
[Where, I = van’t Hoff factor, m = molality]
Ans. (d) : Given that,
K sp ( SrCO3 ) = 7.0 ×10−10
K sp ( SrF2 ) = 7.9 ×10−10
It is clear that It is AB2 type salt so,
From formula.
K sp = 4s3
(
= 7.9 × 10−10 = 1.2 × 10−3 × F−
2+
Because Sr Show equality with the Co
So,
F− = 3.7 × 10−2 M
=
2
)
2−
3
86.
The mole fraction of ethanol in water is 0.08.
Its molality is
(a) 6.32 mol kg–1
(b) 4.83 mol kg–1
–1
(c) 3.82 mol kg
(d) 2.84 mol kg–1
WB-JEE-2020
Ans. (b) : Given:
Mole fraction of ethanol = 0.08
Molar mass of solvent = 18
Mole fraction of H 2 O = 1 − 0.08
= 0.92
Now,
0.08 1000
Molality =
×
0.92 18
= 4.83 mol kg −1
A solution is prepared by dissolving 20g NaOH
in 1250 mL of a solvent of density 0.8 g/mL.
Then the molality of the solution is
(a) 0.2 mol kg–1
(b) 0.08 mol kg–1
–1
(c) 0.25 mol kg
(d) 0.0064 mol kg–1
–1
122 1000
1
(e) 0.5 mol kg
Tb − Ti =
×
× K b 1 + − 1 α
122 1000
n
Kerala-CEE-2020
Ans.
(e)
:
Weight
of
NaOH
=
20
gram
1
Tb − Ti = K b 1 − × 0.8
Molar mass of NaOH = 23+16+1= 40 gram
2
20 1
Tb – Ti = 0.6 Kb
So, no. of moles =
= mole
40 2
2
RTb × M solvent
Now, volume of solvent = 1250 ml and density = 0.8
Kb =
1000 × L vap
g/ml
So, mass of solvent = 0.8×1250
2
2 × Tb × 78
= 1000 gram
Kb =
1000 × 7.8 × 1000
= 1 kg.
0.4 × 2 2
moles of solute
1.4 =
T
b
So, molality =
105
mass of solvent in kg
Tb = 418.33K ≈ 420K
1/ 2 1
85. A solution is saturated with SrCO3 and SrF2
=
=
1
2
-3
–1
2
−
The [CO3 ] is found to be 1.2×10 M. The
= 0.5 mol kg
88. In qualitative analysis, NH4Cl is added before
concentration of F- in the solution would be
NH4OH
Given: Ksp (SrCO3) = 7.0×10–10
(a) to increase [OH–] concentration
Ksp(SrF2) = 7.9×10–10
(b) for making HCl
(a) 3.7×10–6 M
(b) 3.2×10–3 M
(c) to decrease [OH–] concentration
–7
–2
(c) 5.1×10 M
(d) 3.7×10 M
(d) statement is wrong
WB-JEE-2020
Manipal-2020
Objective Chemistry Volume-II
87.
36
YCT
Ans. (c) : NH4Cl is added before NH4OH to decrease
[OH–] concentration by the common ion effect.
89. Certain electric current for half an hour can
collect 11.2 L of hydrogen at NTP. Same
current when passed through an electrolytic
solution for one hour, can deposit how much
silver?
(a) 216 g
(b) 108 g
(c) 47 g
(d) 60 g
Manipal-2020
Ans. (a) : Given that :
11.2L H2 gas at N.T.P. = 1g of H2
from formula,
W1 E1
1
1
=
=
=
W2 E 2 W2 108
W2 = 108 g
Now,
Silver will be deposited in
One hour = 108 × 2
= 216 g
90. The morality of HNO3 in a sample which has
density 1.4 g/mL and mass percentage of 63%
is……(Molecular weight of HNO3 = 63)
[JEE Main 2020, 9 Jan Shift-I]
Ans. (14): Given that,
mass percentage
= 63% or 63 gm in weight
d = 1.4 g / ml
100
v=
= 71.428ml
1.4
So,
No.of moles of solute
Molarity =
Volume (L)
63 × 1000
Molarity =
63 × 71.428
63/ 63
=
0.0714
= 14 M.
91. A 100 mL solution was made by adding 1.43 g
of Na2CO3. xH2O. The normality of the solution
is 0.1 N. The value of x is…… .
[JEE Main 2020, 4 Sep Shift-II]
Ans. (10): Molecular mass of Na2CO3 × H2O
= 23 × 2 + 12 + 48 + 18 x
= (106 + 18x)
Charge present in dissolution = 2
M
So,
Eqwt =
= ( 53 + 9x )
2
Number equction of solute
Normality =
Volume(L)
1.43
53 + 9x
0.1
⇒ 53 + 9x = 143
9x = 90
x = 10
=
{normality = 0.1N is given}
Objective Chemistry Volume-II
92.
The mole fraction of glucose (C6H12O6) in an
aqueous binary solution is 0.1 The mass
percentage of water in it, to the nearest integer,
is…… .
[JEE Main 2020, 3 Sep Shift-I]
Ans. (47): Given that,
Molecular mass of C6H12O6 = 180g.
Molecular mass of H2O = 18g
Mole fraction = 0.1
0.1 × 180
Mass % C6H12O6 =
× 100
0.1 × 180 + 0.9 × 18
1800
=
%
34.2
= 52 .63 %
= 53 %
∴ mass % of H2O = 100–53%
= 47 %
22
93. 6.023×10 molecules are present in 10 g of a
substance 'X'. The morality of a solution
containing 5 g of substance 'X' in 2 L solution is
…….×10-3.
[JEE Main 2020, 3 Sep Shift-II]
Ans.(25):
10 × 6.0 23 × 1023
= 100g / mol
Molar mass =
6. 023 × 1022
Molarity
5
100
2
= 0.025 Or 25 × 10–3
=
From equation
X × 10–3 = 25×10–3
So,
X = 25
94. At 300 K, the vapour pressure of a solution
containing 1 mole of n-hexane and 3 moles of nheptane is 550 mm of Hg. At the same
temperature, if one more mole of n-heptane is
added to this solution, the vapour pressure of
the solution increases by 10 mm of Hg. What is
the vapour pressure in mm Hg of n-heptane in
its pure state............. ?
[JEE Main 2020, 4 Sep Shift-I]
Ans. (600) : Given, Vapour pressure PT = 550
We know that- PT = PA × A + PB × B
For first condition;
1
3
550 = PA + PB
4
4
PA + 3PB = 2200
….(1)
Second condition;
1
4
560 = PA + PB
5
5
PA + 4PB = 2800
…..(2)
From first and second equation,
PA = 400 mm Hg
PB = 600 mmHg
37
YCT
95.
Two open beakers one containing a solvent and
the other containing a mixture of that solvent
with a non-volatile solute are together sealed in
a container. Over time
(a) The volume of the solution decreases and the
volume of the solvent increases
(b) The volume of the solution does not change
and the volume of the solvent decreases
(c) The volume of the solution increases and the
volume of the solvent decreases
(d) The volume of the solution and the solvent
does not change
[JEE Main 2020, 7 Jan Shift-I]
Ans. (c) : If a non–volatile solute is added to a solvent
to form a solution, the vapour pressure gets decreased.
According to Raoult’s law, If vapour pressure of pure
solvent is P1o , vapour pressure of solvent in solution (p).
So,
X has lower intermolecular interaction as compared to
Y.
Because at a particular tempurature as the
intermolecular force of attraction increases vapour
pressure decreases
97. A solution of two components containing n1
moles of the 1st component and n2 moles of the
2nd component is prepared. M1 and M2 are the
molecular weights of component 1 and 2
respectively. If d is the density of the solution in
g mL–1, C2 is the molarity and χ2 is the molefraction of the 2nd component, then C2 can be
expressed as
1000χ2
(a) C2 =
M1 + χ2 (M 2 − M1 )
p = P1o × x1
(b) C2 =
o
1
dχ 2
M 2 + χ2 (M 2 − M1 )
p< P
1000dχ2
Since, vapour pressure of solution is less, there will be
(c) C2 =
net backward reaction [i.e. vapour → liquid] in that
M1 + χ2 (M 2 − M1 )
beaker.
dχ 2
Thus, its volume increases.
(d) C2 =
In another beaker containing only net reaction is
M 2 + χ2 (M 2 − M1 )
forward.
[AIEEE 2020 Shift-I]
Thus, volume decreases.
Ans.
(c)
:
Component
‘A’
molecular
weight = m1
96. A graph of vapour pressure and temperature
Component ‘B’ molecular weight = m2
for three different liquids X, Y and Z is shown
below :
Mole fraction of (B) = χ2
Molarity of component (B) = C2
So,
Total moles of two component = 1
χ2 = moles of component (2)
(1– χ2) = moles of component (1)
χ2M2 = mass of component (2)
(1–χ2)M1= mass of component (1)
The following inference are made :
Total
mass
= (χ2 M2 + (1– χ2) M1) gm
(A) X has higher intermolecular interactions compared
to Y.
χ 2 M 2 + (1 − χ 2 ) M1
(B) X has lower intermolecular interactions compared Total volume =
ml
d
to Y.
(C) Z has lower intermolecular interactions compared
Moles
Molarity =
to Y.
Volume
The correct inference(s) is/are:
χ 2 .d
(a) (C)
(b) (B)
×1000
Molarity (C2) =
χ 2 M 2 + (1 − χ 2 ) M1
(c) (A)
(d) (A) and (C)
d
[JEE Main 2020, 8 Jan Shift-I]
d
Ans. (b) :
1000.d.χ 2
C2 =
M1 + χ 2 ( M 2 − M1 )
98.
Objective Chemistry Volume-II
38
The hardness of a water sample containing 10–3
M MgSO4 expressed as CaCO3 equivalents (in
ppm) is........
(molar mass of MgSO4 is 120.37 g/mol)
[AIEEE 2020 Shift-II]
YCT
Ans. (100):
Hardness of water is measured in PPM (CaCO3)
10−3 ×100
×106
=
1000
100
(∴ Grams of CaCO3 in 1M =
× 10−3 grams)
1000
= 100PPM
99. 10.30 mg of O2 is dissolved into a litre of sea
water of density 1.03 g/mL. The concentration
of O2 in ppm is
[AIEEE 2020 Shift-II]
Ans. (10.00): Given,
WO2 =10.30 mg = 10.30×10–3 g
Vwater = 1 litre = 1000 mL
Density of solution (ρ) = 1.03g/mL
M=ρ×V = 1.03×1000 = 1030 g
mass of solute
×106
PPM =
Total mass of solution
10.30 ×10−3
× 106 = 10 PPM of O 2 conc.
1030
100. A cylinder containing an ideal gas (0.1 mol of
1.0 dm3) is in thermal equilibrium with a large
volume of 0.5 molal aqueous solution of
ethylene glycol at its freezing point. If the
stoppers S1 and S2 (as shown in the figure) are
suddenly withdrawn, the volume of the gas in
litres after equilibrium is achieved will be
..............
(Given, Kf (water) = 2.0 K kg mol–1, R =0.08
dm3 atm K–1mol–1)
=
P = 2.176 atm
Applying Boyle’s law, P1V1 = P2V2
2.176 × 1 = 1atm × V
V2 = 2.176 ≃ 2.18 dm3.
101. Which of the following is a reversible sol?
(a) Gelatin solution
(b) As2S3 solution
(c) Fe(OH)3 solution
(d) Gold solution
GUJCET-2020
Ans. (a) : Reversible solutions are those, where
components can be separated again from the solution.
ex – Gelatin solution
102. Maximum amount of a solid solute that can be
dissolved in a specified amount of given liquid
solvent does not depend upon_____
i. Temperature
ii. Nature of solute
iii. Pressure
iv. Nature of Solvent
(a) i. and iii.
(b) Only ii.
(c) ii. & iv.
(d) Only iii.
GUJCET-2020
Ans. (d) : Maximum amount of a solid solute that can
be dissolved in a specified amount of given liquid
solvent does not depend upon pressure. Solid and liquid
are highly incompressible so, they are unaffected by
change in pressure so we can say that given solvent
does not depend upon pressure.
103. The molality of aqueous solution of any solute
having mole fraction 0.25 is_____.
(a) 18.52 m
(b) 16.67 m
(c) 33.33 m
(d) 9.26 m
GUJCET-2020
Ans. (a) : Given that:
Mole fraction = 0.25
Mole fraction of solute in solution = 1– 0.25
= 0.75
Is moles of solutions 0.75 moles of
dissolved 0.75 mole = 18 × 0.75
= 13.5 g
= 0.0135 kg
So,
0.25
molality =
0.0135
= 18.52 m
104. The vapour pressure of two pure liquids A and
B that form an ideal solution, are 400 and 800
mm of Hg respectively at a temperature toC.
The mole fraction of A in a solution of A and B
whose boiling point is toC will be
(a) 0.4
(b) 0.8
(c) 0.1
(d) 0.2
BITSAT – 2020
Ans. (c) : From formula,
P = PAº .x A + PBº .(1 − x A )
[JEE Main 2020, 9 Jan Shift-II]
Ans. (2.18) : Given, Kf = 2.0
m = 0.5
As we know,
∆Tf = Kf.m
∆Tf = 2.0 × 0.5 = 1
Now,
∆Tf = 273 – Ti
1 = 273 – Ti
Ti = 272 K
PV
= nRT
nRT
P
=
V
0.1× 0.08 × 272
P=
⇒
1
Objective Chemistry Volume-II
39
)
(∵ forx binarysolution
+ x =1
A
B
760 = 400xA + 800 (1 – xA)
YCT
⇒
–40 = –400xA
xA = 0.1
The mole fraction of A is 0.1
105. The molority of solution containing 8 g of
NaOH in 250 mL water would in
(a) 0.2 m
(b) 0.4 m
(c) 0.8 m
(d) 1.0 m
Assam CEE-2020
Ans. (c) :
Volume of Solution = 250 mL
Mass of NaOH = 8 gm
Molar mass of NaOH = 40 g/mol
WNaOH × 1000
Molarity =
M NaOH × Vsolution (L)
8 × 1000
40g / mol × 250(L)
= 0.8 mol/L
106. If active mass of a 6% solution of a compound
is 2, its molecular weight will be
(a) 30
(b) 15
(c) 60
(d) 22
COMEDK-2020
Ans. (a) : Given,
The percentage of solute in the solution = 6%
Mass of the compound = 2moles/ liter
As 100 ml solution contains 6 gm of solute,
6 × 1000
∴ 1000 ml Solution contains =
= 60 gm of solute.
100
60
Thus, molecular weight of the compound =
= 30
2
107. Calculate the molarity of NaOH solution
prepared by dissolving 0.4 g of NaOH in
enough water to form 500 mL of the solution.
(a) 0.02
(b) 0.05
(c) 0.04
(d) 0.03
AP EAMCET (Engg.) 18.9.2020 Shift-I
0.4
× 100
WB gL−1
500
Ans. (a) : Molarity =
=
= 0.02 M
M Bg mol−1
40
1 kg toothpaste will contain 200 × 2 = 400 mg fluoride
ions.
Hence, concentration of fluoride ion is 400 ppm or 4 ×
102 ppm.
Hence, option (b) is correct.
109. 10 g of NaOH is dissolved in 500 mL of aqueous
solution. Calculate the molarity of this solution
? (Given, formula weight of NaOH = 40)
(a) 0.5×10–3 M
(b) 0.4 M
(c) 0.25×10–3 M
(d) 0.5 M
AP EAMCET (Engg.) 17.09.2020 Shift-I
Ans. (d) :
Moles of solute
Molarity =
Volume of solution ( in litre )
=
=
Weight of solute × 1000
Molecular weight solute × Volume ( mL )
10 ×1000
= 0.5M
40 × 500
110. A 40% HCl solution has density 1.2 g mL–1.
The molarity of the solution is nearly ........... .
(a) 11 M
(b) 12 M
(c) 13 M
(d) 14 M
AP EAMCET (Engg.) 17.09.2020 Shift-I
=
Ans. (c) : As, density (d) of HCℓ solution is given (pc)
(1.2 g mL-1), the percentage strength of the solution
(40%) is of (w/w) type.
⇒ Molarity =
pc ( w / w ) × d ( gmL−1 ) × 10
M HCℓ
40 ×1.2 × 10
= 13.15 ≃ 13M
36.5
111. The density of oxygen gas at 5 atm and 127ºC
will be
(a) 2.80 g/L
(b) 4.88 g/L
(c) 1.49 g/L
(d) 5.60 g/L
AP EAMCET (Engg.) 17.09.2020, Shift-II
Ans. (b) : Density of a gas,
pM
[∴ WB = mass of solute (NaOH) = 0.4 g/mL
d=
–1
RT
MB = molar mass of NaOH = 40 g mol ]
∵
p
= pressure = 5 atm
108. Calculate the concentration of fluorine in ppm
M
= molar mass of O2 gas = 32 g mol–1
in 500 g of toothpaste containing 0.2 g of
R = 0.082L atm mol–1 K–1
fluorine.
2
2
(a) 2 × 10
(b) 4 × 10
T = 400 K
(c) 4 × 100
(d) 4 × 103
5 × 32
d=
AP EAMCET (Engg.) 21.09.2020, Shift-II
0.082 × ( 273 + 127 )
Ans. (b) : PPM or Parts Per Million is expressed as 1
= 4.878 ≈ 4.88 gL–1
milligram of substance per kilogram of parent
112. The molarity of 0.2 N Na2CO3 solution will be
substance.
(a) 0.05 M
(b) 0.2 M
Since, mass of toothpaste = 500 g = 0.5 kg,
and mass of fluoride ion = 0.2 g = 200 mg,
(c) 0.1 M
(d) 0.4 M
0.5 kg toothpaste contains 200 mg fluoride ions,
AP EAMCET (Engg.) 21.09.2020, Shift-I
Objective Chemistry Volume-II
40
=
YCT
Ans. (c) : n-factor for Na2CO3 = 2
The n-factor of such salts is defined as the number of
moles of electrons exchanged (lost or gained) by one
mole of the salt and there is the exchange of 2 electrons,
so its n-factor is 2.
Therefore, normality = n-factor × molarity ⇒ Molarity
= 0.2/2 = 0.1 M
113. The density of 2 M aqueous solution of NaOH
is 1.28 g/cm3. The molality of the solution is
[Given that molecular mass of NaOH = 40 g
mol–1]
(a) 1.20 m
(b) 1.56 m
(c) 1.67 m
(d) 1.32 m
NEET-Odisha 2019
Ans. (c) : given
Molarity = 2M
d = 1.28g / cm3
From formula,
1000 × molarity
Molality =
( density ) − ( molarity × molecular mass )
=
10000 × 2
(1.28 × 1000 ) − ( 2 × 40 )
2000
1200
= 1.67m
114. In water saturated air, the mole fraction of
water vapour is 0.02, If the total pressure of the
saturated air is 1.2 atm, the partial pressure of
dry air is
(a) 1.18 atm
(b) 1.76 atm
(c) 1.176 atm
(d) 0.98 atm
(Odisha NEET-2019)
Ans. (c) : Given, mole fraction of water vapour
( X H2O ) = 0.02
=
116. Calculate molarity of one litre solution of 22.2
gm CaCl2.
(a) 0.4 M
(b) 0.2 M
(c) 0.8 M
(d) 0.6 M
JIPMER-2019
Ans. (b) : Applying molarity formula.
Number of molessolute
Molarity =
Volumeof solution (L)
22.2
(∴ CaCl2 molar mass =111g )
111×1
= 0.2M
117. If molarity of Cu+2 ions is 3×10-4 express this in
ppm
(a) 0.3
(b) 0.2
(c) 0.1
(d) 0.6
JIPMER-2019
Ans. (a) : Given,
Molarity of Cu+2 ions = 3 × 10–4 in 1000 ml
3 ×10−4
Now, changing in ppm =
× 106
1000
= 0.3 ppm
118. Liquid M and liquid N form an ideal solution.
The vapour pressures of pure liquids M and N
are 450 and 700 mmHg, respectively, at the
same temperature. Then correct statement is
xM = mole fraction of M in solution;
xN = mole fraction of N in solution;
yM = mole fraction of M in vapour phase;
yN = mole fraction of N in vapour phase;
x
y
x M yM
(a) M > M
(b)
=
x N yN
x N yN
Total pressure of saturated air (PT) = 1.2 atm
PO2 = PT × H 2 O
=
(c)
x M yM
<
x N yN
(d)
( x M − yM ) < ( x N − y N )
[JEE Main 2019, 9 April Shift-I]
Ans. (a) :
Given,
PM° = 450 mm Hg
So,
Po2 = 1.2 × 0.02
= 0.024
Then partial pressure of dry air = 1.2 – 0.024
PN° = 700 mm Hg
x = 1.176 atm
115. A mixture of NaCl and K2Cr2O7 is heated with We know
conc. H2SO4, deep red vapours are formed.
PMo x M
y
=
M
Which of the following statements is false?
Ps
(a) The vapours give a yellow solution with NaOH
o
P X
(b) The vapours contain CrO2Cl2 only
yN = N N
(c) The vapours contain CrO2Cl2 and Cl2
Ps
(d) The vapours when passed into lead acetate in
o
y
P X /P
acetic give a yellow precipitate
Then, M = Mo M s
y N PN X N / Ps
Karnataka-CET-2019
Ans. (c) : Reaction will be,
y M PMo x M
y
450 x M
= o×
⇒ M=
×
K2Cr2O7 + 4NaCl + 6H2SO4 → 2CrO2Cl2 + 4NaHSO4 +
y N 700 x N
y N PN x N
2KHSO4 + 3H2O
y M 450 x M
x
y
If the salt is NaCl on heating deep red vapoure of
×
=
⇒ M> M
y N 700 x N
x N yN
chyroml chloride are evolved.
Objective Chemistry Volume-II
41
YCT
119. What would be the molality of 20%
92
=
(mass/mass) aqueous solution of KI?
23
Molar mass of KI = 166 g mol–1)
=4
(a) 1.48
(b) 1.51
4
Molality =
(c) 1.35
(d) 1.08
1
[AIEEE 2019 Shift-I]
Hence, Molality = 4
122. The amount of sugar (C12H22O11) required to
Ans. (b) : Given,
prepare 2L of its 0.1M aqueous solution is
100g solution contain 20g KI
(a) 17.1 g
(b) 68.4 g
Then solvent = 100 – 20 = 80 g
(c) 136.8 g
(d) 34.2 g
Molar mass of KI = 40 +126 = 166g/mole
[AIEEE 2019 Shift-II]
weight
20
Ans. (b) : We know
Then, mole =
=
molar mass 166
Number of mole
Molarity =
mole of solute
Volume of solution
So, Molality =
weight of solvent
Given, Molarity = 0.1 M
Then,
20
n
0.1 =
molality = 166
80
2
Moles (n) = 0.2
1000
Now,
m = 1.506 mol / kg
Weight of solute
120. The mole fraction of a solvent in aqueous
Moles (n) =
Molar
mass of suger
solution of a solute is 0.8. The molality (in mol
kg–1) of the aqueous solution is
(a) 13.88×10–2
(b) 13.88×10–1
(c) 13.88
(d) 13.88×10–3
[AIEEE 2019 Shift-I]
Ans. (c) : Given, η solvent = 0.8
We know,
ηTotal = ηsolute + ηsolvent
1 = 0.8 + η solute
ηsolute = 0.2
mole of solute
Molality =
weight of solvent
Molality =
0.2
×1000 = 13.88
0.8 × 18
121. A solution of sodium sulphate contains 92g of
Na+ ions per kilogram of water. The molality of
Na+ ions in that solution in mol kg–1 is
(a) 16
(b) 4
(c) 132
(d) 8
[AIEEE 2019 Shift-I]
Ans. (b) : Molality is define as the number of moles of
solute present in per kg of solvent.
Hence,
Number of moles of solute
Molality =
weight of solvent
Then,
Number of moles of solute =
Objective Chemistry Volume-II
weight
molar mass
W
342 g
W = 68. 4g
123. 8g of NaOH is dissolved in 18g of H2O. Mole
fraction of NaOH in solution and molality (in
mol kg–1) of the solution respectively are
(a) 0.2, 11.11
(b) 0.167,22.20
(c) 0.2, 22.20
(d) 0.167, 11.11
[AIEEE 2019 Shift-II]
Ans.(d): Given, Weight of NaOH = 8 g
Then, molar mass NaOH = 23+16+1 = 40 g/mol
8
Number of moles of 8g NaOH =
= 0.2 mol
40
18g H2O mole has 1 mol
0.2
Mole fraction of NaOH =
1.2
= 0.167
So,
0.2 ×1000
Molality =
18
= 11.11m
124. A mixture of 100 mmol of Ca(OH)2 and 2 g of
sodium sulphate was dissolved in water and the
volume was made up to 100 mL. The mass of
calcium sulphate formed and the concentration
of OH- in resulting solution, respectively, are:
(Molar mass of Ca(OH)2 Na2SO4 and CaSO4
are74, 143 and 136 g mol-1 respectively; Ksp of
Ca(OH)2 is 5.5×10-6
(a) 13.6 g, 0.28 mol L-1 (b) 1.9 g, 0.28 mol L-1
(c) 13.6 g, 0.28 mol L-14 (d) 1.9 g, 0.14 mol L-1
[JEE Main 2019, 10 Jan Shift-I]
42
0.2 =
YCT
Ans. (b): Required equation,
Ca(OH)2 + Na2SO4 → CaSO4 + 2NaOH
1 mol
1 mol
1 mol
2 mol
Now,
Mass of CasO4 = 14 × 10–3 × 136
= 1.9g
–
28
[OH ]
=
100
= 0.28 mol/L
125. The strength of 11.2 volume solution of H2O2 is
[Given that molar mass of H = 1g mol-1 and
O=16g mol-1]
(a) 1.7%
(b) 34%
(c) 13.6%
(d) 3.4%
[JEE Main 2019, 8 April Shift-II]
Ans. (d) : Required equation
At S.T.P.
2H2O2 → 2H2O + O2
68g
22.4L. at STP
Now,
11.2 L of O2 at STP will be produced by H2O2
68
=
× 11.2 = 34 g/L
22.4
Now
w
34
Percentage
=
× 100
w
1000
= 3.4 %
126. On mixing 20 mL of acetone with 60 mL of
chloroform, the total volume of the solution is
(a) less than 80 mL
(b) more than 80 mL
(c) equal to 80 mL
(d) unpredictable.
J & K CET-2019
Ans. (a) : A mixture of chloroform (CHCl3) and
acetone (CH3)2 CO slows negative deviations from
Raoult’s law due to formation of hydrogen bonding
between the two molecules. Hence, a slight decrease in
volume takes place.
127. A solution contain 62g of ethylene glycol in
250g of water is cooled upto –10°C. If K, for
water is 1.86 K kg mol–1, then amount of water
(in g) separated as ice is
(a) 32
(b) 48
(c) 64
(d) 16
[JEE Main 2019, 9 Jan Shift-II]
Ans. (c) : Considering the expression of the depression
in freezing point of a solution,
∆Tf = Kf × mi
w ×1000
Tfo − Tf = K f × B
×i
..…..(i)
MB × w A
Here, Tfo = 0oC, Tf = –10oC
mass of ethylene glycol (wB)= 62g
molar mass of ethylene glycol(MB)=32 g mol–1
mass of water in g as liquid solvent (wA),
van’t-Hoff factore (i)= 1 (for ethylene glycol in water)
Kf = 1.86 K kg mol–1
Objective Chemistry Volume-II
On substituting in Eq. (i), we get
62 ×1000
0 − (−10) = 1.86 ×
×1
62 × w A
1.86 × 62 ×1000
= 186g
10 × 62
So, amount of water separated as ice (solid solvent)
= 250 – wA= 250 – 186 = 64 g
128. Liquids A and B form an ideal solution in the
entire composition range. At 350 K, the vapour
pressures of pure A and pure B are 7×103 Pa
and 12×103 Pa, respectively. The composition of
the vapour in equilibrium with a solution
containing 40 mole percent of A at this
temperature is
(a) XA = 0.76; XB = 0.24
(b) XA = 0.28; XB = 0.72
(c) XA = 0.4; XB = 0.6
(d) XA = 0.37; XB = 0.63
[JEE Main 2019, 10 Jan Shift-I]
Ans. (b) : Given that;
PA° = 7 × 103 Pa
wA =
PB° = 12 × 103 Pa
Ptotal = YA PAo + YB PBo
Ptotal = 0.4 × 7 × 103 + 0.6 × 12 × 103
Mole fraction of A (vapour phase) is
XA × Pt = YA × PA°
0.4 × 7 × 103
0.4 × 7 × 103 + 0.6 + 12 × 103
∴ mole fraction of A = 0.4
molefraction
of
Bin
Solution
=
0.6
2.8
⇒XA =
10
= 0.28
∴XB = 1 – 0.28
= 0.72
129. Mole fraction of the solute in a 1.00 molal
aqueous solution is
(a) 0.1770
(b) 0.0177
(c) 0.0344
(d) 1.7700
BITSAT – 2019
Ans. (b) : 1.00 molal solution means 1 mole of solute in
1 kg of water.
Number of moles of solute = 1
1000
Number of moles of solvent =
18
= 55.55
mole fraction of solute
moleof solute
=
Total number of molesin solution
1
=
56.6
≈ 0.0177
43
XA =
YCT
130. 25 mL of the given HCl solution requires 30
mL of 0.1 M sodium carbonate solution. What
is the volume of this HCl solution required to
titrate 30 mL of 0.2 M aqueous NaOH
solution?
(a) 75 mL
(b) 25 mL
(c) 12.5 mL
(d) 50 mL
[JEE Main 2019, 11 Jan Shift-II]
Ans. (b) : From Normality equation,
2HCl+Na2CO3→2NaCl+H2O+CO2
m. eq of HCl solution = m.eq of Na2Co3 Solution
25 × N1 = 30 × N2
25 × n – factor × M1 = 30 × n – factor × M2
25 × 1 M1 = 30 × 2 × 0.1
60 × 0.1
M1 =
25
6
=
25
Now,
m. eq of HCl = m. eq. of NaOH
N1 × V1 = N2 × V2
⇒ n-factor × M1 × V1 = n – factor × M1 × V2
6
⇒ 1× × V1 = 1× 0.2 × 30
25
0.2 × 30 × 25
⇒ V1 =
6
V1 = 25 mL
131. Which one of the following is an example of
ideal solution?
(a) Carbon tetrachloride + Chloroform
(b) Carbon tetrachloride + Toluene
(c) Toluene + Benzene
(d) Benzene + Acetone
Assam CEE-2019
Ans. (c) : Benzene and toluene are non-polar, operating
intermolecular forces are almost similar, So, they form
an ideal solution.
132. The mole fraction of gas in the solution is
(a) Proportional to the partial pressure of the gas
over the solution
(b) Proportional to the partial pressure of the gas
in the solution
(c) Proportional to the square of the partial
pressure of the gas over the solution
(d) Proportional to the square of the partial
pressure of the gas in the solution
Assam CEE-2019
Ans. (a) : The mole fraction of gas in the solution is
proportional to the partial pressure of the gas over the
solution.
133. The molarity of urea (molar mass 60 g mol–1)
solution by dissolving 15 g of urea in 500 cm3 of
water is
(a) 2 mol dm–3
(b) 0.5 mol dm–3
–3
(c) 0.125 mol dm
(d) 0.0005 mol dm–3
MHT CET-2018
Objective Chemistry Volume-II
Ans. (b) : Given :
V = 500 cm3
= 0.500 L
mole
Molarity
volume(L)
15 / 60mol
0.500ml
= 0.5 mol dm–3
134. If 1 mol of NaCl solute is dissolved into the 1 kg
of water, at what temperature will water boil at
1.013 bar ?
(Kb of water is 0.52 K kg mol-1)
(a) 373.15 K
(b) 373.67 K
(c) 374.19 K
(d) 373.19 K
(e) 375 K
Kerala-CEE-2018
Ans. (b) : Given that,
Kb = 0.52k kg mol–1
n = 1, WA = 1kg
now,
we know that
n
∆Tb = Kb
WA
=
0.52 ×1
= 0.52K
1
So, the boiling point of solution (at 1atm)
= (373.15 + 0.52)
= 373.67K
135. When attraction between A–B is more than
that of A–A and B–B, the solution will show:
(a) positive deviation from Raoult's law
(b) negative deviation from Raoult's law
(c) no deviation from Raoult's law
(d) Cannot be predicted
Manipal-2018
Ans. (b) : When attraction b/w A–B is more than that of
A–A and B–B, the solution will show negative
deviation from Raoult’s law because, the forces of
attraction b/w components A and B is greater.
136. What is the density of solution of sulphuric acid
used as an electrolyte in lead accumulator?
(a) 1.5 gmL–1
(b) 1.2 gmL–1
–1
(c) 1.8 gmL
(d) 2.0 gmL–1
MHT CET-2018
Ans. (b) : H2SO4 used as an electrolyte in lead
accumulator with the quantity of 1.2 gm–1, which is
38% of its mass.
137. Calculate the molarity of a solution of 30g of
Co (NO3)2. 6H2O in 4.3 L of lsolution?
Consider atomic mass of Co = 59 u, N = 14 u,
O = 16 u, H = 1u
(a) 0.023 M
(b) 0.23 M
(c) 0.046 M
(d) 0.46 M
J & K CET-2018
44
=
YCT
Ans. (a) : Given :
Volume of solution = 4.3L
Weight of CO(NO3)2 6H2O in solution = 30 g
Molar mass = 59 + 14 × 2 + 12 × 16 + 12 × 1 = 291
30
Molarity =
291
4.3
= 0.023M.
138. A 100 mL flask contained H2 at 200 torr, and a
200 mL flask contained He at 100 torr. The two
flask were then connected so that each gas
filled their combined volume. Assume, no
change in temperature, total pressure is
(a) 104 torr
(b) 163.33 torr
(c) 279 torr
(d) 133.33 torr
JIPMER-2018
Ans. (d) : We can take T as constant.
Now, from equation
P 1V 1 + P 2V 2 = P f V f
100 × 200 + 200 × 100 = Pf × 300
40000
Pf =
torr
300
= 133.33 torr
139. Chloroform, CHCl3, boils at 61.7oC. If the Kb
for chloroform is 3.63oC/molal, what is the
boiling point of a solution of 15.0kg of CHCl3
and 0.616 kg of acenaphthalene, C12 H10?
(a) 61.9
(b) 62.0
(c) 52.2
(d) 62.67
BITSAT – 2018
Ans. (d) : Given:
Kb = 3.63ºC/ Molal
WA = 15kg
= 15000 gm
WB = 0.616 kg
= 616 gm
Now,
MB=Acenaphthalene (C12H10)
Molecular mass of C12H10
MB = 12 × 12 + 10 × 1
= 154 g/mol
From formula,
∆Tb = Kb.m
= 3.63 ×
Ans. (c) : Molality =
720 × 1000
18 × 720
= 55.5M
141. Which one of the following is the correct
statement?
(a) Surface tension of a liquid decreases with
increase in temperature
(b) Vapour pressure of a liquid decreases with
increase in temperature
(c) Viscosity of a liquid decreases with decreases
in temperature
(d) In gravity free environment droplets of a
liquid on flat surface are slightly flattened.
BCECE-2018
Ans. (a) : Kinetic energy of molecules increases due to
increasing temperature and then the intermolecular
force decreases, because of this surface tension
decreases.
142. Equimolar solution of urea and KCl are
separated by a semi permeable membrane.
Which one of the following will take place?
(a) No net flow of solvent in either directions
(b) Solvent will flow from KCl to urea solution
(c) Solvent will flow from urea to KCl solution
(d) Nothing can be predicted
BCECE-2018
Ans. (c) : Osmosis is the movement of solvent, which
allows solvent to move through a barrier from a less
concentrated solution to more concentrated solution.
The molecules of solvent can pass through semi
permeable membrane.
143. What will be the normality of the salt solution
obtained by neutralising x mL y (N) HCI with y
mL x (N) NaOH, and finally adding (x + y) mL
distilled water
2(x + y)
xy
(a)
N
(b)
N
xy
2(x + y)
2xy
(c)
N
x+y
Ans. (b) : Required Reaction.
+
HCl
Initial = xy ×10
616 1
W 1000
× ∵ m = B ×
154 15
M B WA
x+y
(d)
N
xy
WB-JEE-2017
Final = 0
−3
NaOH
xy ×10
→ NaCl +
−3
0
0
xy × 10 −3
H2O
0
xy × 10 −3
Now,
∆Tb = 0.968
Volume of water added = ( x + y ) mL
Now,
VT of solution = ( x + y ) + ( x + y ) mL
Tbs = Tbº + ∆Tb = 61.7 + 0.968 = 62.668
xy ×10 −3
= 62.67
Normality =
× 1000
( x + y) + ( x + y)
140. The molality of 720g of pure water is:
(a) 40 M
(b) 4 M
xy
=
N
(c) 55.5 M
(d) None
2( x + y)
HP CET-2018
Objective Chemistry Volume-II
45
YCT
144. Which of the following solutions will turn violet So,
when a drop of lime juice is added to it?
20ml
From 15ml, H2O2 oxygen liberated =
×15
(a) A solution of NaI
1000ml
(b) A solution mixture of KI and NaIO3
= 0.3 litre
(c) A solution mixture of NaI and KI
= 300 ml
(d) A solution mixture of KIO3 and NaIO3
148. The molality of the 3M solution of methanol if
WB-JEE-2017
the density of the solution is 0.9 g cm–3 is
(a) 3.73
(b) 3.0
Ans. (b) : Reaction will be,
+
+
+
(c)
3.33
(d) 3.1
NaIO3 + 5KI + 6H
→ 3H 2 O + 3I 2 + 5K + Na
(e) 3.2
From reaction, It is clear that A solution mixture of KI
Kerala-CEE-2017
and NaIO3 was the solution which turn violet.
Ans. (a) : We know that,
145. The molality of solution containing 15.20 g of
WB (%) × 1000
…….(i)
urea, (molar mass = 60) dissolved in 150 g of Molality(m) =
100 − WB (%) × M B
water is
C × MB
3 × 32
(a) 1.689 mol kg–1
(b) 0.1689 mol kg–1
WB(%) =
=
–1
–1
(c) 0.5922 mol kg
(d) 0.2533 mol kg
d × 10 0.9 ×10
MHT CET-2017
= 10.66 gm (%)
Where, WB = Mass of solute
Ans. (a) : Given:
MB = Molecular mass
W = 15.20 g
Putting
the
value in eq. (1)
Molar mass of urea (NH2CONH2) = 60g
10.66 × 1000
Mass of water = 150 g
m
=
= 0.150 kg
(100 − 10.66) × 32
moleof sloute
= 3.728 ≅ 3.73
Molality =
mass of solven(kg)
149. The volume of water to be added to 100 cm3 of
0.5 NH2SO4 to get decinormal concentraction is
Mass of urea
=
(a) 100 cm3
(b) 450 cm3
Molar mass of urea × Mass of solvent (kg)
3
(c) 500 cm
(d) 400 cm3
Mass of solute
JIPMER-2017
∵ Mole of solute =
Molar mass of solute
Ans. (d) : Given, N1 = 0.5 N, V1 = 100 cm2, N2=0.1N
From Normality equation,
15.20g
=
N 1V 1 = N 2V 2
60 g / mol × 0.150 kg
0.5 × 100 = 0.1 × V2
= 1.689 mol/kg
V2 = 500 cm3 (Total volume)
146. Which of the following is dependent on So,
temperature
Volume Should be added
(a) Molarity
(b) Mole fraction
= 500 - 100
(c) Weight percentage (d) Molality
= 400cm3
(NEET-2017) 150. The normality of 10%(w/v) of acetic acid is
(a) 1 N
(b) 1.3 N
Ans. (a) : Molarity depends on the volume of a solution
(c) 1.7 N
(d) 1.9 N
which can be changed with change in temperature.
JIPMER-2017
Molarity is defined as moles present in per unit volume
of solution. On changing the temperature the volume of Ans. (c) : Given:
the solution changes due to which molarity will also be W = 10g.
affected. So as far the temperature changes, the volume Molar mass of acetic acid = 60g
will also change due to which molarity changes.
We know that,
147. The volume of oxygen liberated at STP from
gram eq.of solute
Normality =
15mL of 20 volume H2O2is
volumeof solution in ( L )
(a) 100 mL
(b) 150 mL
W × 1000
(c) 200 mL
(d) 250mL
N=
M × V(ml)
(e) 300 mL
10
×
1000
Kerala-CEE-2017
=
60 × 100
Ans. (e) : Given,
= 1.66 ≈ 1.7N
amount of H2O2 = 15ml and volume = 20
Objective Chemistry Volume-II
46
YCT
151. When a mixture of 10 moles of SO2 and 16
moles of O2 were passed over a catalyst, 8 mole
of SO3 were formed at equilibrium. The
number of moles of SO2 and O2 remaining
unreacted were
(a) 2, 12
(b) 12, 2
(c) 3, 10
(d) 10, 3
AMU-2017
Ans. (a) : Required equation:
2SO2 +
O2 ⇌ 2SO3
Initial t = 0
10
16
0
Final teq
(10 – 2x) (16 – x)
2x
∵ 8 mol of SO3 were formed at equilibrium
So,
2x = 8
x=4
Remaining SO2 = 10 – 8
= 2mol.
Remaining O2 = 16 – 4
= 12mol.
152. Water + CH3OH mixture shows +ve deviation
from ideal solution behaviour. 100 mL of water
is mixed with 100 mL of CH3OH. Then total
volume of the mixture will be.
(a) 200 mL
(b) less than 200 mL because of additional Hbonding between H2O and CH3OH
(c) more than 200 mL because H-bonding within
H2O molecules vanishes
(d) more than 200 mL because H-bond between
H2O and CH3OH is weaker than that between
H2O and H2O
AMU-2017
Ans. (d) : Total volume of mixture (water + CH3OH)
V = Vwater + VCH3OH
Putting the values,
i = 1+ (2 – 1) 0.9
= 1.9
Now
Π = iCRT
Where, C = Molar concentration
R = Ideal gas constant
T = (27° + 273 = 300 K)
Π = 1.9 × 0.002 × 0.082 × 300
= 0.0934 or 0.094 bar.
154. Commercially available H2SO4 is 98 g by
weight of H2SO4 and 2 g by weight of water.
It's density is 1.38 g cm–3. Calculate the
molality (m) of H2SO4 (molar mass of H2SO4 is
98 g mol–1)
(a) 500 m
(b) 20 molal
(c) 50 m
(d) 200 m
TS EAMCET-2017
Ans. (a) : We know that,
Mass of solute
Moles of solute
Molar mass of solute
Molality =
=
Weight of solvent Weight of solvent(kg)
98
× 1000
= 98
2
Molality = 500 m
155. How many millilitres of 20 volume H2O2
solution is needed to react completely with 500
mL of acidified 1N KMnO4 solution?
(a) 224
(b) 280
(c) 140
(d) 56
AP EAMCET-2017
Ans.
(c)
:
Given
that
:= 100 + 100
Normality of KMnO4 (N1) = 1N
= 200 mL
Volume of KMnO4 = 500 mL
Here, solution does not follow the Roult’s law i.e.
Volume of H2O2 solution = 20
positive deviation from ideal solution behavior. Which
means solution is non-ideal.
vol.of O 2 at NTP
So, answer will be ‘more than 200mL because H– bond Normality of the H2O2 (N2) = 5.6 ( eq.mass )
between H2O and CH3OH is weaker than that between
20
H2O and H2O.
N2 =
153. 0.002 molar solution of NaCl having degree of
5.6
N2 = 3.5 > N
dissociation of 90% at 27o C has osmotic
pressure equal to
∴ N 1V 1 = N 2V 2
(a) 0.94 bar
(b) 9.4 bar
1 × 500 = 3.57 × V2
(c) 0.094 bar
(d) 9.4 × 10–4 bar
500
V2 =
= 140 mL
AMU-2017
3.57
Ans. (c) : We know thatHence, the volume of H2O2 used = 140 mL.
i = 1 + ( n – 1) α
156. 5.0 g of sodium hydroxide (molar mass 40 g
mol–1) is dissolved in little quantity of water
Required equation,
and the solution is diluted upto 100 mL. What
NaCl ↽ ⇀ Na + + Cl−
is the molarity of the resulting solution?
Where,
(a) 0.1 mol dm–3
(b) 1.0 mol dm–3
–3
n=2
(c) 0.125 mol dm
(d) 1.25 mol dm–3
α = 0.9
MHT CET-2016
Objective Chemistry Volume-II
47
YCT
Ans. (d) : Suspension is a heterogeneous mixture. The
particles in a suspension are so much greater than the
Wsolute = 5.0g
Msolute = 40 g mol
solution. The dispersed particle diameter in a
suspension is 1000 times greater than sand in the
V = 100 ml
solution. In a suspension, the diameter of dispersed
mole
molarity =
particle is of the order of 2000Å.
volume(L)
161. 40 mL of xM KMnO4 solution is required to
5.0 / 40
react completely with 200 mL of 0.02 M oxalic
molarity =
100 /1000
acid solution in acidic medium. The value of x
is
= 1.25 mol dm–3
(a) 0.04
(b) 0.01
157. Milk is an emulsion in which
(c)
0.03
(d) 0.02
(a) a gas is dispersed in water
AP-EAMCET – 2016
(b) lactose is dispersed in water
Ans.
(a)
:
Given
data,
(c) milk fat is dispersed in water
Volume of KMnO4 (V1) = 40 mL
(d) a solid is dispersed in water
UPTU/UPSEE-2016 Molarity of KMnO4 (M1) = x M
Volume of oxalic acid (V2) = 200 mL
Ans. (c) : Milk is an emulsion, which are colloids and
Molarity of oxalic acid (M2) = 0.02 M
they get dispersed and dispersion medium are liquids.
2KMnO 4 + 3H 2SO 4 + 5H 2 C 2 O 4 → K 2SO 4 + 2MnSO 4 + 8H 2 O + 10CO 2
158. The normality of 26% (wt/vol) solution of
M1V1 M 2 V2
ammonia (density = 0.855) is approximately :
∴
=
n1
n2
(a) 1.5
(b) 0.4
Ans. (d) : Given ,
−1
(c) 15.3
(d) 4
BITSAT 2016
26
Ans. (c) : Weight of NH3 = 26 g = g = 1.53 g
17
Volume of solution = 100 mL = 0.1 L
1.53
∴ Normality =
= 15.3 N
0.1
159. Calculate the molality of a solution that
contains 51.2 g of naphthalene (C10H8) in 500
mL of carbon tetrachloride. The density of
CCl4 is 1.60 g/mL.
(a) 0.250 m
(b) 0.500 m
(c) 0.750 m
(d) 0.840 m
(d) 1.69 m
Kerala-CEE-2016
Ans. (b) : Given,
Mass of naphthalene WB=51.2g
Molar mass of naphthalene MB = C10H8
= 12×10+8×1
= 128 g
Mass (WA) = density(d) ×Volume (v)
WA= 1.60 ×500 =800 g
Since MA=
WB 1000
×
M B WA
51.2 × 1000
= 0.500 m
128 × 800
160. In a suspension, the diameter of the dispersed
particles is of the order
MA=
o
o
(a) 10 A
(b) 100 A
o
(c) 1000 A
o
(d) 2000 A
AMU-2016
Objective Chemistry Volume-II
x × 40 0.02 × 200
=
2
5
x = 0.04 M
162. The molality of an aqueous dilute solution
containing non-volatile solute is 0.1 m. What is
the boiling temperature (in °C) of solution?
(Boiling point elevation constant, Kb = 0.52 kg
mol–1 K; boiling temperature of water =
100°C).
(a) 100.0052
(b) 100.052
(c) 100.0
(d) 100.52
TS-EAMCET-2016
Ans. (b) : Given that,
Kb = 0.52 kg mol–1 K
Boiling temperature of water = 100°C
Molality = 0.1 m
∴
∆Tb = Kb.m
= 0.52 × 0.1
∆Tb = 0.052
Thus, the boiling point of solution–
(Tb) = 100 + ∆Tb
= 100 + 0.052
Tb = 100.052oC
163. For preparing 3.00L of 1M NaOH by mixing
portions of two stock solutions (A and B) of
2.50 M NaOH and 0.40 M NaOH respectively.
Find out the amount of B stock solution (in L)
added.
(a) 8.57 L
(b) 2.14 L
(c) 1.28 L
(d) 7.51 L
JIPMER-2015
Ans. (b) : Moles needed = 3.00L×1.00 M
= 3 mol
Let xL of 2.50 M NaOH is added to (3–x) L of .40 M
NaOH is added ……..(I)
48
YCT
The number of moles of solute from more concentrated
solutions 2.50 is (0.40) (3.00–x)
Total moles = 3.00
∴ 2.50 x+0.40(3.00–x) = 3.00
2.10x = 1.8
x = 0.857 L of 2.50 M NaOH…….(II)
Putting value of x form (II) in (I)
3–0.857 = 2.14 L of 0.40 M NaOH the amount of B
stock solution to be added is 2.14 L
164. Which one of the following binary liquid
mixture exhibit positive deviation from
Raoult's law?
(a) Carbon disulphide -acetone
(b) Chloroform-acetone
(c) Bromobenzene-chlorobenzene
(d) Benzene-toluene
(e) Phenol-aniline
Kerala-CEE-2015
Ans. (a) : Carbon disulphide – acetone (CS2 – CH3C
CH3) shows positive
deviation from Raoults law.
165. Vapour pressure of a solution at 100ºC having
3.42 g of cane sugar in 180 g water is
(a) 759.2mm
(b) 760 mm
(c) 740 mm
(d) 748.5 mm
CG PET- 2015
Ans. (a) : Raoult's law,
So,
32
0.8
32 ×10
=
8
= 40 mL.
167. The vapour pressure of pure water is 23.5mm
Hg. Then, the vapour pressure of an aqueous
solution which contains 5 mass percent of urea
is (Molar mass of urea is 60).
(a) 23 mm Hg
(b) 18 mm Hg
(c) 31 mm Hg
(d) 35 mm Hg
BCECE-2015
Ans. (a) : Mole fraction of solvent =
Number of moles of water
Total Number of moles
32g CH3OH =
=
Wmassof water / M molar mass of water
Number of moles of water + Number of moles of urea.
∵ Mass of urea = 5 g
∴ Mass of water = 100 – 5 = 95gm
95
No. of moles of water =
= 5.278
18
5
No. of moles of urea =
= 0.083
60
95 /18
=
5.278 + 0.083
PA° – Ps
nB
95 /18
=
(relative lowering in pressure)
Ps =
× 23.5
PA°
nA + nB
5.278 + 0.083
= 23.14mm Hg
180
nA =
≃ 23mm Hg.
18
168. What is the mole fraction of the solute in 2.5
= 10
molal aqueous solution?
3.42
(a) 0.043
(b) 0.053
nB =
= 0.01
342
(c) 0.063
(d) 0.073
Now,
AMU-2015
760 – Ps
0.01
Ans. (a) : Given; moles of solute = 2.5
=
Molecular weight of solvent (water) = 18g
760
10 + 0.01
1000
760 – Ps = 0.75924
Number of moles of solvent =
Ps = 760 – 0.75924
18
= 55.56 mol.
= 759.2mm
166. If the density of methanol is 0.8 kg L–1, what is So,
nA
its volume needed for making 4 L of its 0.25 M
Mole fraction of solute =
solution?
nA + nB
(a) 4 mL
(b) 8 mL
2.5
=
(c) 40 mL
(d) 80 mL
2.5 + 55.56
J & K CET-2015
= 0.043
Ans. (c) : Given
169. A solution containing 2.44 g of a solute
density of methanol = 0.8 kg L-1
dissolved in 75 g of water boiled at 100.413oC.
Molarity = 0.25M
What will be the molar mass of the solute?
(Kb for water = 0.52 K kg mol–1)
1ML → 0.8g CH3OH
(a) 40.96 g mol−1
(b) 20.48 g mol−1
No. of moles = 4×0.25
−1
(c) 81.96 g mol
(d) None of these
= 1 Mole
AMU-2015
molecular mass of CH3OH (12+4+16) g = 32 g
Objective Chemistry Volume-II
49
YCT
Ans. (a) : Given:
2.44g of a solute dissolved in 75g of water boiled at
100.413°C.
Wsolute = 2.44g
Wsolvent = 75g
Tb = 100.413oC
So,
∆Tb = 100.413 – 100°C
= 0.413°C
From formula,
k × Wsolute × 1000
Msolute = b
∆Tb × Wsolvent
0.52 × 2.44 ×1000
=
75 × 0.413
= 40.96g mol–1.
170. Density of 3 M NaCl solution is 1.25 g/cc. The
molality of the solution is
(a) 2.79 Molal
(b) 0.279 Molal
(c) 1.279 Molal
(d) 3.85 Molal
AMU-2015
Ans. (a) : Given:
Msolution = 3
Density = 1000 × 1.25
= 1250
3M NaCl solution molecular weight = 2 × 58.5
= 175.32gms
So, kg of solvent = 1250 – 175.32 = 1074.7
3
m=
× 1000
1074.7
= 2.79 mol/kg
171. The number of Cl− ions in 100 mL of 0.001 M
HCl solution is
(a) 6.022 × 1023
(b) 6.022 × 1020
19
(c) 6.022 × 10
(d) 6.022 × 1024
AMU-2015
173. Gaseous benzene reacts with hydrogen gas in
presence of a nickel catalyst to form gaseous
cyclohexane according to the reaction,
C6 H 6 ( g ) + 3H 2 ( g ) → C6 H12 ( g )
A mixture of C6H6 and excess H2 has a pressure
of 60 mm of Hg in an unknown volume. After
the gas had been passed over a nickel catalyst
and all the benzene converted to cyclohexane,
the pressure of the gas was 30 mm of Hg in the
same volume at the same temperature. The
fraction of C6H6 (by volume) present in the
original volume is
(a) 1/3
(b) 1/4
(c) 1/5
(d) 1/6
VITEEE-2015
Ans. (d) :
C6 H 6 (g) + 3H 2 (g)
→ C6 H12 (g)
Intial
P1
P2
0
At, t = 0
P1 + P2 = 60 mm Hg
....(i)
C6 H 6 (g) + 3H 2 (g)
→ C6 H12 (g)
After
P1 − P1
P2 − 3P1
P1
time,(t)
So, total pressure = P2 – 3P1 + P1 = 30 mm Hg
= P2 – 2P1 = 30 mm Hg
….(ii)
On solving equation (i) and (ii)
P1 = 10 mm Hg, P2 = 50 mm Hg
Fraction of C6H6 by volume = moles fraction of
10 1
pressure
=
=
60 6
174. Which one of the following is the wrong
statement about the liquid?
(a) It has intermolecular force of attraction
(b)
Evaporation of liquids increase with the
N A × 10–3
decrease
of surface area
Ans. (c) : Number of molecules =
× 100
1000
(c) It resembles a gas near the critical
temperature
6.022 ×1023 × 10−3
=
× 100
(d) It is an intermediate state between gaseous
1000
and solid state
= 6.022 × 1019
–
AP-EAMCET (Engg.) 2015
172. The molarity of NO3 in the solution after 2L of
Ans. (b) : Evaporation occurs when molecules get
3M AgNO3 is mixed with 3L of 1 M BaCl2 is
enough energy from heat to escape the liquid. An
(a) 1.2 M
(b) 1.8 M
increased surface area means more liquid will be
(c) 0.5 M
(d) 0.4 M
VITEEE-2015 exposed to air at one time.
175. The volume of oxygen evolved at STP by
Ans. (a) : 2L of 3M AgNO3 will contain 6 moles of
decomposition of 0.68 g '20 volume' hydrogen
AgNO3.
peroxide solution is
3L of 1M BaCl2 will contain 3 moles of BaCl2.
(a) 2.24 mL
(b) 22.4 mL
2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2
(c)
224
mL
(d)
2240 mL
So, 6 moles of AgNO3 will react with 3 moles of BaCl2
MHT CET-2014
it means, two solution will react completely to form 3
−
→ 2H 2O + O 2
moles of Ba(NO3)2 = 6 moles of NO3 ions in 2+3 = 5L Ans. (c) : 2H 2 O 2
1mol
2mol
solution.
=2×34=
68g
22400
ml
6
Hence, the molarity of NO3− = = 1.2 M
∴ At STP, 68g H2O2 produced O2 = 22400 ml
5
Objective Chemistry Volume-II
50
YCT
∴ At STP, 0.68g H2O2 will produce
22400 × 0.68
68
= 224 mL.
176. What is the molality of a
200 mg of urea (molar
dissolved in 40 g of water?
(a) 0.0825
(b)
(c) 0.498
(d)
Ans. (a) :
O2 =
solution containing
mass 60 g mol-1)
0.825
0.0013
MHT CET-2014
Ans. (a) : Given:
W1 = 200 mg = 0.2g, W2 = 40 g = 0.04 kg of water
Molar mass = 60 g mol–1
0.2g
No. of moles of urea =
= 0.0033mol
60g mol−1
40
= 0.04 kg
1000
0.0033
Molality of solution =
= 0.0825mol kg −1
0.04
177. Molality of 2.5 g of ethanoic acid (CH3COOH)
in 75 g of benzene is
(a) 0.565 mol kg–1
(b) 0.656 mol kg–1
–1
(c) 0.556 mol kg
(d) 0.665 mol kg–1
J & K CET-2014
Ans. (c): Given that,
Molecular mass of CH3COOH = 60g
w
Number of moles of CH3COOH =
m
2.5
=
60
= 0.0417 mol.
75g
Mass of benzene =
kg −1
10000
= 7.5 × 10–3 kg.
2.5g × 1000
Molality =
60g mol−1 × 7.5
Mass of water in kg =
= 0.556 mol kg–1
178. Consider the separate solutions of 0.500 M
C2H5OH (aq), 0.100 M Mg3 (PO4)2 (aq), 0.250M
KBr (aq) and 0.125 M Na3PO4 (aq) at 25°C.
Which statement is true about these solutions,
assuming all salts to be strong electrolytes?
(a) They all have same osmotic pressure
(b) 0.100 M Mg3 (PO4)2 (aq) has the highest
osmotic pressure
(c) 0.125 M Na3PO4 (aq) has the highest osmotic
pressure
(d) 0.5000 M C2H5OH (aq) has the highest
osmotic pressure
JEE Main-2014
Objective Chemistry Volume-II
π = iCRT
Electrolyte
i
C
i×C
KBr
2
0.25
0.5
C2H5OH
1
0.5M 0.5
Mg3(PO4)2
5
0.1M 0.5
Na3PO4
4
0.125 0.5
So, option (a) is correct.
179. If 10 ml of 0.1 M aqueous solution of NaCl is
divided in to 1000 drops of equal volume. What
will be the concentration of one drop?
(a) 0.01 M
(b) 0.10 M
(c) 0.001 M
(d) 0.0001 M
GUJCET-2014
Ans. (b) : Given.
Volume (V) = 10mL
Now, n = M × V = 0.1 × 10 mL = 1 m moL
1m moL
n=
1000
10mL
1
V=
=
1000 100
1
1
M=
=
= 0.1M
1000 10
1
100
180. What will be the value of molality for an
aqueous solution of 10% w/w NaOH (Na = 23,
O = 16, H = 1)
(a) 2.778
(b) 5
(c) 10
(d) 2.5
GUJCET-2014
Ans. (a) : Molality =
given weight of NaOH
Molar Mass of NaOH
Mass of solvent
Volume
10 / 40
100
=
=
90 /1000
36
= 2.778 mole/kg
181. 2.5 mL of 2/5 M weak monoacidic base
(Kb = 1× 10–12 at 250C) is titrated with 2/15 M
HCl in water at 250C. The concentration of H+
at equivalence point is (Kω = 1 × 10–14 at 250C)
(a) 3.7 × 10–13 M
(b) 3.2 × 10–7 M
–2
(c) 3.2 × 10 M
(d) 2.7 × 10–2 M
BCECE-2014
Ans. (d) : Given:
2
N1 = 2.5, V1 = 2/5 ,
N2 =
V2 =?
15
From Normality equation
2 2
2.5 × = × V2 (∵ N1V1 = N2V2)
5 15
V2 = 7.5mL
51
YCT
Required equation,
BOH + HCl → BCl + H 2 O
Salt of BCl formed = 2.5 ×
2
5
= 1mol.
Volume of total solution = (2.5 + 7.5)ml = 10ml
Now,
1
For salt [BCl]. =
= 0.1
10
Concentration of H+ at equivalent point
kx =
Ch 2 k w 10−14
=
=
= 10–2
1 − h k b 10−12
0.1× h 2
1− h
Here, h=0.27 (significant will be suitable, not negligible)
So, [H+] = 0.1× 0.27
= 2.7 × 10 –2 M.
182. A solution containing 10g per dm3 of urea
(molecular mass = 60 gmol-1) is isotonic with a
5% solution of a non-volatile solute. The
molecular mass of this non volatile solute is
(a) 300 g mol-1
(b) 350 g mol-1
-1
(c) 200 g mol
(d) 250 g mol-1
AIIMS-2014
Ans. (a): Given,
Molecular mass of urea = 60g/mol.
Let m = molecular mass of non-volatile solute
Concentration of the both compound is same
So,
C1 = C2
(∵ Isotonic compound)
Now,
5 1000
Concentration of non- volatile solute =
×
100
m
According to question,
10 5 × 1000
=
60 m × 100
m = 5 × 60
= 300g/mol or 300g mol-1.
183. A one molal solution of sodium chloride in
water has the same boiling point as
(a) 1 m solution of magnesium sulphate
(b) 1 m solution of magnesium chloride
(c) 1 m solution of aluminium sulphate
(d) 1 m solution of aluminium chloride.
COMEDK-2014
Ans. (a) : We know that,
∆Tb = iK b m
For NaCl, i=2
MgSO4, i=2
MgCl2, i=3
Al2(SO4)3, i=5
AlCl3, i=4
Two solutions having same molality and same
Vanthoff's factors will have same elevation in boiling
points (∆Tb ) and thus, have same boiling points as one
molal solution of MgSO4.
10−2 =
Objective Chemistry Volume-II
184. When the concentration is expressed as the
number of moles of solute per kilogram of the
solvent, it is known as
(a) molarity
(b) molality
(c) normality
(d) mole fraction
COMEDK-2014
Ans. (b) : Molality is defined as concentration is
expressed as the number of moles of solute per
kilogram of the solvent.
185. KMnO4 reacts with KI, in basic medium to
form I2 and MnO2. When 250 mL of 0.1 M KI
solution is mixed with 250 mL of 0.02 M
KMnO4 in basic medium, what is the number
of moles of I2 formed?
(a) 0.015
(b) 0.0075
(c) 0.005
(d) 0.01
AP-EAMCET (Engg.) - 2014
Ans. (b) :
Chemical equivalents on left hand side is equal to
Chemical equivalent on right hand side.
No. of equivalent = No. of equivalent of I2
of KMnO4
M×n-factor × V = Moles of I2 × n-factor of I2
250
0.02 × 3 ×
= Moles of I 2 × 2
1000
0.02 × 3 × 250
Moles of I2 =
2 × 1000
Moles of I2 = 0.0075
186. 19.85 mL of 0.1 N NaOH reacts with 20 mL of
HCl solution for complete neutralisation. The
molarity of HCl solution is
(a) 9.9
(b) 0.99
(c) 0.099
(d) 0.0099
VITEEE-2014
Ans. (c) : Given data –
Volume of NaOH = 19.8 mL
Volume of HCl = 20 mL.
Normality of NaOH = 0.1 N
Normality 0.1
Molarity of base =
=
= 0.1
Acidity
1
∴
M 1 V1 = M 2 × V2
0.1 × 19.85 = M2 × 20
M2 = 0.09925 ≈ 0.099
187. The volume strength of 1.5 N H2O2 solution is
(a) 16.8L
(b) 8.4L
(c) 4.2L
(d) 5.2L
VITEEE-2014
Ans. (b) : Given that,
Normality = 1.5 N
Volume strength of H2O2 = 5.6 × normality
= 5.6 × 1.5
= 8.4 litre.
52
YCT
188. The molarity of a solution in which 5.3 g
Na2CO3 is dissolved in 500 mL will be
(a) 1.0M
(b) 0.1 M
(c) 0.20 M
(d) 0.2 M
UPTU/UPSEE-2013
Ans. (b) : Molecular mass of Na2CO3= 106 g mol–1
5.3
∴ Number ofmoles =
= 0.05mol
106
As 0.05 mol are dissolved in 500 mL;
Number of moles dissolved in 1000 mL (1L) or
molarity of the solution
0.05
=
× 1000 = 0.1M
500
189. A gas X is passed through water to form a
saturated solution. The aqueous solution on
treatment with silver nitrate gives a white
precipitate. The saturated aqueous solution
also dissolves magnesium ribbon with evolution
of colourless gas Y. Identify 'X' and 'Y' are
respectively
(a) CO2, Cl2
(b) O2, CO2
(c) Cl2, H2
(d) N2, H2
UP CPMT-2013
Ans. (c) : Chlorine reacts with water forming HCl and
HClO (hypo-chlorous acid). HClO further decomposes
to give HCl and nescent oxygen.
Cl2 + H2O → HCl + HClO
HClO → HCl + O
Thus, saturated aqueous solution of Cl2 has HCl acid
and HCl when reacts with AgNO3 solution, gives white
precipitate of AgCl.
AgNO3 + HCl → AgCl ↓+ HNO3
White ppt
HCl solution also reacts with magnesium ribbon quite
rapidly forming magnesium chloride and H2 gas.
Mg(s) + 2HCl(aq) → MgCl2 (aq) + H 2 ↑
Magnesium
Hydrogen
190. The molar concentration of chloride ions in the
resulting solution of 300 mL of 3.0M NaCl and
200 mL of 4.0 M BaCl2 will be
(a) 1.7 M
(b) 1.8 M
(c) 5.0 M
(d) 3.6 M
UP CPMT-2013
Ans. (c) : The number of moles of chlorides ion in 300
mL of 3.0 M NaCl =
3
× 300 = 0.9mol. The number of
1000
moles of chloride ion in 200 mL of 4.0 M BaCl2
solution =
As
4
× 200 = 0.8 mol.
1000
BaCl2 ⇌ Ba2+ + 2Cl–
∴ Moles of Cl– ions = 0.8 × 2 = 1.6 mol.
Total volume of solution = 200 + 300 = 500 mL.
As 1.6 + 0.9 = 2.5 moles of chloride ions are present in
500 mL solution.
∴ Molar concentration of Cl– ions in the resulting
solution =
2.5
× 1000 = 5.0M .
500
Objective Chemistry Volume-II
191. How many grams of concentrated nitric acid
solution should be used to prepare 250mL of
2.0 M HNO3? The concentrated acid is 70%
HNO3.
(a) 70.0g con. HNO3
(b) 54.0g con. HNO3
(c) 45.0g con. HNO3
(d) 90.0g con. HNO3
(NEET-2013)
Ans. (c) : given
M = 2.0 Vsol = 250ml
M w = 63
Molarity =
W × 1000
M w × Sol( m1)
W 1000
×
63 250
1
W = 2 × 63 ×
4
= 31.5gm
31.5gm is the weight of 100% HNO3 for 70%,
2=
Weight = 31.5 ×
= 45g
100
70
192. A current strength of 9.65 A is passed through
excess fused AlCl3 for 5 h. How many litres of
chlorine will be liberated at STP? (F = 96500
C)
(a) 2.016
(b) 1.008
(c) 11.2
(d) 20.16
(e) 10.08
Kerala-CEE-2013
Ans. (d) : Given,
i = 9.65A
t = (5 × 60 × 60) sec.
1
Now, C1– → Cl 2 + ie −
(Faraday electrolysis)
2
35.5 × 9.65 × 5 × 60 × 60
w=
(∴w = z it)
1× 96500
= 63.9g
Now,
Chlorine will be librated at STP
22.4 × 63.9
=
71
= 20.16 L
193. Which of the following aqueous solutions will
exhibit highest boiling point?
(a) 0.01 M urea
(b) 0.01 M KNO3
(c) 0.01 M Na2SO4
(d) 0.015 M C6H12O6
Karnataka-CET-2013
Ans. (c) : Urea and Glucose having covalent molecules
and they do not dissociate.
As we know that elevation in boiling point is directly
proportional to solute particle concentration.
The highest number of solute particles in 0.01M
Na2SO4 solution.
Therefore, 0.01 M Na2SO4 solution will show the
highest boiling point.
53
YCT
194. A beaker contains a solution of substance 'A'.
On dissolving substance 'A' in small amount in
this solution, precipitation of substance 'A'
takes place. The solution is
(a) concentrated
(b) saturated
(c) unsaturated
(d) super saturated
JCECE-2013
Ans. (d) : A saturated solution cannot dissolve any
more solute at that temperature. If precipitation occurs,
It is supersaturated solution.
195. 4L of 0.02 M aqueous solution of NaCl was
diluted by adding 1L of water. The molality of
the resultant solution is
(a) 0.004
(b) 0.008
(c) 0.012
(d) 0.016
JCECE-2013
Ans. (d) : Given,
Molarity of NaCl (M1) = 0.02M
Volume (V1) of NaCl solution = 4 L
We know that,
Molality of mixture isM 1V 1 = M 2V 2
or
0.02 × 4 = M2 × 5
or
M2 = 0.016
196. At equilibrium the rate of dissolution of a solid
solute in a volatile liquid solvent is
(a) less than the rate of crystallization
(b) greater than the rate of crystallisation
(c) equal to the rate of crytallisation
(d) zero
JIPMER-2013
Ans. (c) : At equilibrium, the rate of dissolution of a
solid solute in a volatile liquid solvent is equal to the
rate of crystallization.
197. Which of the following solutions will have the
highest boiling point?
(a) 1 M glucose solution
(b) 1 M sodium nitrate solution
(c) 1 M barium chloride solution
(d) 1 M aluminium chloride solution
J & K CET-2013
Ans. (d) : Elevation of boiling point is a colligative
property and it depends upon the number of solute
particles.
So, 1M aluminium chloride solution have highest
boiling point.
198. The molarity of a solution obtained by mixing
750 mL of 0.5 M HCI with 250 mL of 2M HCI
will be
(a) 0.875 M
(b) 1.00 M
(c) 1.75 M
(d) 0.0975 M
JEE Main-2013
Ans. (a): Given,
750 mL of 0.5 M HCl
M1= 750 mL
V1= 0.5 M
Objective Chemistry Volume-II
250 mL of 2 M HCl
M2= 250 mL
V2= 2M
Total molarity(M f ) =
M1V1 + M 2 V2
V1 + V2
750 × 0.5 + 250 × 2
750 + 250
875
=
1000
Mf = 0.875 M
199. Equal weights of NaCl and KCl are dissolved
separately in equal volumes of solutions.
Molarity of the two solutions will be:
(a) Equal
(b) That of NaCl will be less than that of KCl
(c) That of NaCl will be more than that of KCl
solution
(d) That of NaCl will be about half of that of KCl
solution
BITSAT – 2013
Ans. (c) : Molecular mass of KCl = (39+35.5)
= 74.5
Molecular mass of NaCl= 23+35.5
= 58.5
Weight
n=
Atomic weight
=
74.5
mol = 1 mol
74.5
74.5
n NaCl =
= 1.27 mol
58.5
number of moles 1.27
Molality =
=
volume
1
n NaCl 1.27
(Molarity) NaCl =
=
1
1
n KCl =
(Molality) KCl =
1
=1
1
(Molality)NaCl > (Molality)KCl
Hence, the molality of NaCl is greater than molality of
KCl.
200. Volume of 3% solution of sodium carbonate
necessary to neutralize a litre of 0.1 N
sulphuric acid
(a) 176.66 ml
(b) 156.6 ml
(c) 116.0 ml
(d) 196.1ml
BITSAT-2013
Ans. (a) : Given:
N1 = 0.1,
V1 = 100mL
Normality of Volume of 3% Na2CO3 solution,
3 × 100
N2 =
53 × 100
= 0.0566 N.
54
YCT
Ans. (b) : Given that,
So, from normality equation
V = 200mL, Molarity = 2M
0.1 × 100 = 0.0566 × V2
Molecular
weight (w) of urea = 60g
0.1×100
V2 =
Mole of solute
0.0566
Molarity(M) =
Volumeof solution in litre
= 176.66mL
201. Which condition is not satisfied by an ideal
w ×1000
Molarity =
solution ?
M×V
(a) ∆ mix V = 0
M × V × Molarity
w=
(b) ∆ mix S = 0
1000
(c) Obeyance to Raoult’s Law
60 × 200 × 2
=
(d) ∆ mix H = 0
1000
w = 24 gm of Urea
Karnataka NEET-2013
204. The volume of water to be added to 100 mL of
Ans. (b) : For ideal solution is follow as0.5N H2SO4 acid solution to get solution
(i) Volume Change (∆V) of mixing should be zero.
decinormal concentration is :
(ii) Obey Raoult's law at every range of concentration.
(a) 400 mL
(b) 450 mL
(iii) Heat change (∆H) on mixing should be zero.
(c) 500 mL
(d) 100 mL
Hence option (b) not follow for ideal condition of
MPPET-2013
solution.
Ans. (a) : Given that, N1 = 0.5 N, N2 = 1/10N
202. Solution "X" contains Na2CO3 and NaHCO3.
V1 = 100 mL, V2 = ?
20 mL of X when titrated using methyl orange ∴
According to law of dilution,
indicator consumed 60 mL of 0.1 M HCl
N 1V 1 = N 2V 2
solution. In another experiment, 20 mL of X
1
solution when titrated using phenolphthalein
0.5 × 100 = × V2
10
consumed 20 mL of 0.1 M HCl solution. The
–1
V
=
0.5
×
10
× 100
2
concentrations (in molL ) of Na2CO3 and
= 500 mL
NaHCO3 in X are respectively
Therefore, volume to be added
(a) 0.01,0.02
(b) 0.1,0.1
= 500 – 100 = 400mL
(c) 0.01,0.01
(d) 0.1,0.01
AP-EAMCET (Engg.) 2013 205. A 5 molar solution of H2SO4 acid is diluted
from 1 litre to 10 litre. What is normality of
Ans. (b) :
solution?
Na 2 CO3 + HCl
→ NaHCO3 + NaCl
(a) 0.25 N
(b) 1N
(c) 2N
(d) 7N
∴ Volume of Na 2 CO3 = Volume of consumed HCl
20 mL of 0.1M = 20 mL of 0.1M
Conc. of Na2CO3 = 0.1M.
∴ The concentration of Na2CO3 in solution X = 0.1M
MPPET-2013
Ans. (b) : Given that,
Molarity of solution = 5 M
V1 = 1 litre, V2 = 10 litre
NaHCO3 + HCl
→ NaCl + CO 2 + H 2 O
Normality = Molarity × Valency factor of H2SO4
Given
= 5 × 2 = 10N
Na 2 CO3 = 2 × 20mL = 40mL
N1 = 10N, N2 = ?
If methyl orange is used, the end point is indicated
Therefore, by law of dilutionwhen all the alkali is neutralized.
N 1V 1 = N 2V 2
NaHCO3 + HCl
→ NaCl + CO 2 + H 2 O
10 × 1 = N2 × 10
N2 = 1N
As 40 mL of 0.1M HCl is consumed in complete
neutralizing of Na2CO3 at methyl orange end point
206. Which of the following statement is false?
remaining HCl solution = 60–40 = 20 mL of 0.1 M
(a) Raoult's law states that the vapour pressure of
a component over a solution is proportional to
As the equation = 1 mole of NaHCO3 = 1 moL of HCl
its mole fraction
∴ 0.1 mol of NaHCO3 = 0.1mol of HCl.
(b) The osomotic pressure (π) of a solution is
203. Weight of Urea required to prepare 200 ml of
given by the equation π = MRT, where M is
2M solution will be :
the molarity of the solution
(a) 12 gm
(b) 24 gm
(c) The correct order of osmotic pressure for 0.01
(c) 20 gm
(d) 60 gm
M aqueous solution of each compound is
BaCl2 > KCl > CH3COOH > sucrose
MPPET-2013
Objective Chemistry Volume-II
55
YCT
(d) Two sucrose solution of same molality 210. The density of a solution prepared by
dissolving 120 g of urea (mol. mass=60u) in
prepared in different solvents will have the
1000 g of water is 1.15 g/ml. The molarity of
same freezing point depression
this
UPTU/UPSEE-2012
(a) 0.50 M
(b) 1.78 M
Ans. (d) : Two sucrose solutions of the same molality
(c)
1.02
M
(d) 2.05 M
prepared in different solvents will have the same
AlEEE-2012
freezing point depression. Different solvents have
Given
that,
Ans.
(d)
:
different values of kf. Thus ∆Tf will be different even if
The density of a solution = 120 g
molality is same.
Molecular make of urea = 60 u
207. Aluminum chloride exists as dimer, Al2Cl6, in
Total mass of solution = 1000+120
solid state as well as in solution of non-polar
= 1120 g
solvents such as benzene. When dissolved in
water, it gives
1120
Volume of solution =
mL
1.15
(a) Al3+ + 3Cl−
3+
mass of solution
(b) [ Al(H 2 O6 )] + 3Cl−
molecular
mass of solution
3−
Molarity =
(c) [ Al(HO)6 ] + 3HCl
Volumeof solution
(d) Al2 O3 + 6HCl
120 / 60
2 × 1000 × 1.15
=
× 1000 =
UPTU/UPSEE-2012
1120 /1.15
1120
= 2.05 M
Ans. (b) : Required reaction
3+
211. What will be the volume of O2 at NTP liberated
AlCl3 + 6H 2 O → Al ( H 2O )6 + 3Cl −
by 5 A current flowing for 193 and through
Hence option (b) is correct
acidulated water?
(a) 56 mL
(b) 112mL
208. The molar concentration of chloride ions in the
(c)
158mL
(d)
965mL
resulting solution of 300 mL of 3.0 M NaCl and
BITSAT – 2012
200 mL of 4.0 M BaCl2 will be
(a) 1.7 M
(b) 1.8 M
Ans. (a) : Volume of 32g of O2 = 22400 mL
And,
(c) 5.0 M
(d) 3.5 M
MHT CET-2012
I× t × E
W=
Ans. (c) : Given, M1 = 3.0, M2 = 4.0 × 2
F
Reaction :
5 × 193 × 8
W=
NaCl ↽ ⇀ Na+ + Cl −
96500
=
0.08g
2+
−
BaCl 2 ↽ ⇀ Ba + 2Cl
Now,
Now,
At NTP, volume of 32 g O2 = 22400 mL
Molar Concentration of Cl–
22400 × 0.08
Volume of 0.08g of O2 =
M1V1 + M 2 V2
32
=
= 56 mL
V1 + V2
212. The normality of 0.2 M H3PO2 is
3.0 × 300 + 2 × 4.0 × 200
=
(a) 0.2 N
(b) 0.4 N
300 + 200
(c) 0.6 N
(d) 0.06 N
= 5.0 M
CG PET- 2012
209. The molarity of a solution obtained by mixing
Ans. (b) : : Given that:
800 mL of 0.5 M HCl with 200 mL of 1 M HCl
M = 0.2
will be
Normality
=
molarity×basicity
(a) 0.8 M
(b) 0.6 M
(basicity of base is equal to two)
(c) 0.4 M
(d) 0.2 M
Normality = M × basicity of H3PO2
JCECE-2012
= 0.2 × 2
Ans. (b) : We know that,
= 0.4 N
Molarity of mixture is213. The molarity of pure water is
M1V1 + M2 V2 = M3 V3
(a) 18
(b) 5.56
Final
Initial
(c) 55.6
(d) 100
V3 = V1 + V2 = 800 + 200 = 1000 mL
CG PET- 2012
= 0.5 × 800 + 1 × 200 = M3 × 1000
Ans. (c) : Mass of 1L of water = Density × volume
600
= 1kg
M3 =
= 0.6M
1000
= 1000g
Objective Chemistry Volume-II
56
YCT
Molecular mass of H2O = 18
Molarity of pure water
number of moles of water (in kg)
=
volumeof solution in litre
1000
18
1L
= 55.56M.
214. Two solutions of HCl, A and B, have
concentrations of 0.5 N and 0.1 M respectively.
The volume of solutions A and B required to
make 2 litres of 0.2 N HCl are
(a) 0.5 L of A+1.5 L of B
(b) 1.5 L of A+0.5 L of B
(c) 1.0 L of A+1.0 L of B
(d) 0.75 L of A+1.25 L of B
AMU-2012
Ans. (a) : let x L of A and (2−x) L of B are mixed.
From normality Equation:M1 V1+ M2 V2 = MR (V1+V2)
Substituting the values, we get :0.5 × x + 0.1 (2 − x) = 0.2 × 2
(0.5−0.1) x = 0.4 − 0.2
0.4x = 0.2
x = 0.5 L
∴0.5 L of A and 1.5 L of B should be mixed.
215. The mole fraction of benzene in a solution
containing 39% by mass in an organic solvent
of molecular mass 122 is
(a) 0.5
(b) 0.6
(c) 0.4
(d) 0.35
COMEDK-2012
Ans. (a) : Given, Molecular Mass of solvent (M1)=122
g/mol
Molecular mass of benzene (M2)= 78 g mol–1
Mass of benzene ( W2 ) = 39gm
=
Mass of solvent ( W1 ) = 61gm
39
78
39 61
+
78 122
= 0.5
216. The experimentally determined molar mass of
a non-volatile solute, BaCl2 in water by
Cottrell's method, is
(a) equal to the calculated molar mass
(b) more than the calculated molar mass
(c) less than the calculated molar mass
(d) double of the calculated molar mass
AP-EAMCET (Engg.) - 2012
Ans. (c) : Required Reaction,
BaCl2 → Ba2+ + 2Cl–
Now,
(Normal molar mass )
i=
Abnormal molar mass
∴Mole fraction of benzene =
Objective Chemistry Volume-II
Mc 3
=
Mo 1
Mc = 3Mo
Hence, Mo < Mc
217. At a certain temperature vapour pressure of
pure water is 3000 Nm–2. 5 g of non-electrolyte
and non-volatile solute is added to 100 g of
water. Vapour pressure of the solution is 2985
Nm–2. Assume that it is a dilute solution, find
the molar mass of the solute.
(a) 90
(b) 180
(c) 200
(d) 270
SRMJEEE – 2012
Ans. (b) : Given :Mass of non-electrolyte, non-volatile solute (W1) = 5g ;
Mass of water (W2) = 100g
Vapour pressure of solution (Ps) = 2985 Nm−2
Vapour pressure of pure water (Po) = 3000 Nm−2
Molecular weight of water (H2O) = 18
Po − Ps W1 × M 2
Now,
=
Po
W2 × M1
3000 − 2985
5 ×18
=
3000
100 × M1
15
90
=
3000 100 × M1
90 × 30 2700
M1 =
=
15
15
M1 = 180
218. Normality of 0.25 M phosphorous acid H3PO3
is
(a) 0.125 N
(b) 0.75 N
(c) 0.50 N
(d) 0.25 N
SRMJEEE – 2012
Ans. (c) : Given that,
Normality of H3PO3 is 0.25 M.
∴ Normality = Molarity × no. of equivalents
N = M×n
Since phosphorous acid (H3PO3) is dibasic, thus the
valency is 2.
Therefore Normality = 0.25 × 2
= 0.50 N
219. What is the concentration of a solution
prepared by dissolving 4.20 g of NaF in 500 g of
water ?
(a) 0.00840 Molal
(b) 0.00840 Molar
(c) 0.200 Molar
(d) 0.200 Molal
SCRA-2012
Ans. (d) : Given that,
Weight of NaF (w) = 4.20g,
Weight of water (W) = 500g
Molecular weight of NaF = 42 g/mol
Concentration = ?
w ×1000
Now, Molality =
M×W
57
YCT
4.20 × 1000
42 × 500
Molality = 0.200 molal
220. In which mode of expression of concentration
of a solution remains independent of
temperature?
(a) Molarity
(b) Molality
(c) Formality
(d) Normality
MPPET-2012
Moles of solute
Ans. (b) : Molality =
Mass of solvent (in kg)
The molality of a solution remains independent of
temperature because it involves only mass, which is
independent of temperature.
221. 100 mL of an acid solution is neutralized by 50
mL of NaOH solution containing 0.2 g NaOH.
The concentration of acid solution is:
(a) 0.1 N
(b) 0.05 N
(c) 0.5 N
(d) 0.25 N
MPPET-2012
Ans. (a) : Given that,
V1 = 50 mL, N1 = 0.2 N, V2 = 100 mL, N2 = ?
N1 V1 = N2 V2
0.2 × 50
N2 =
100
N2 = 0.1 N
222. The normality of '30 volume H2O2' is
(a) 2.678N
(b) 5.336N
(c) 8.034N
(d) 6.685N
WB-JEE-2011
Ans. (b) : Given,
Volume = 30
Now,
Volume
Normality =
5.6
30
=
5.6
= 5.3N
223. One mole of P4O10 is allowed to react fully with
dust and salt-free doubly distilled water and
the volume is made up to 1L. What is the
normality of the so-generated orthophosphoric
acid?
P4O10+6H2O→4H3PO4
(a) 1.0 N
(b) 8.0 N
(c) 12.0 N
(d) 4.0 N
UPTU/UPSEE-2011
Ans. (c) : Required reaction.
P4 O10 + 6H 2 O → 4H 3 PO 4
Molality =
From reaction
Molality = 4M
Normality = M × basicity
= 4×3
= 12N
Objective Chemistry Volume-II
224. An aqueous solution of urea containing 18 g
urea in 1500 cm3 of the solution has a density
equal to 1.052. If the molecular weight of urea
is 60, the molality of the solution is
(a) 0.200
(b) 0.192
(c) 0.100
(d) 1.200
MHT CET-2011
Ans. (b) : Given :
d = 1.052 g/ml
According to question,
1500 cm3 of the solution has a density equal to 1.052,
So,
It will correspond = 1.052 × 1500
= 1578 g
18g
No. of mole of urea =
= 0.3mol
60 g (mol)
Mass of H2O in solution = (1578–18)g
= 1560 g
= 1.560 kg
Now,
0.3
molality =
1.560
= 0.192 m.
225. What volume of 2 M H2SO4 is required to form
0.2 N of 100 mL of solution?
(a) 5 mL
(b) 20 mL
(c) 10 mL
(d) 50 mL
MHT CET-2011
Ans. (a) : Given :
N1 = 2 M correspond 4NH2SO4
N2 = 0.2, V1 = 9, V2 = 100 ml
From formula
N 1V 1 = N 2V 2
N × V2
V1 = 2
N1
0.2 × 100
4
= 5 ml
226. A solution made by dissolving 40 g NaOH in
1000 g of H2O is
(a) 1 molar
(b) 1 normal
(c) 1 molal
(d) None of these
JIPMER-2011
Ans. (c) : Given:
Weight of NaOH in Solution = 40g
Amount of Solvent
= 1000gm
W
no. of moles of NaOH =
M
40
no. of moles of NaOH is n =
=1
40
1000
Mass of solvent in kg ism =
= 1kg
1000
n 1
So, molality of solution is
= = 1molal
m 1
58
=
YCT
1 molal:– 1 molal aqueous solution means 1 mole of
solute in 1 kg of water.
Hence option (c) is correct.
227. In a volumetric experiment, it was found that a
solution of KMnO4 is reduced to MnSO4 If the
normality of the solution is 1.0N, then molarity
of the solution will be
(a) 0.5 M
(b) 0.2 M
(c) 1.0 M
(d) 0.4 M
J & K CET-2011
Ans. (b) : Required equation;
+7
+2
KMnO4 → MnSO4
molecular weight
Equivalent weight =
5
Normality
Molarity =
5
1
= = 0.2 N
5
228. The molality of a urea solution in which 0.0100
g of urea, [(NH2)2 CO] is added to 0.3000 dm3
of water at STP is
(a) 5.55×10-4 M
(b) 33.3 M
(c) 3.33×10-2 M
(d) 0.555 M
AlEEE-2011
Ans. (a): Given that,
0.0110
Number of moles of urea =
mol
60
3
water at STP = 0.3 dm = 0.3kg [∵ 1 dm3=1000 ml]
moles of solute
molality =
kg of water
=
0.010
= 5.55 × 10−4 M.
60 × 0.3
229. A 5.2 molal aqueous solution of methyl alcohol,
CH3OH, is supplied. What is the mole fraction
of methyl alcohol in the solution?
(a) 0.100
(b) 0.190
(c) 0.086
(d) 0.050
[AIEEE 2011]
Ans. (c): Given of moles of CH3OH= 5.2 and mass of
water = 1Kg
weight
Mole of water =
molar mass
1000g
= 55.56 mol
18g / mol
Mole fraction of solute methyl alcohol
m
=
m + 55.56
=
5.2
5.2 + 55.56
= 0.086
=
Objective Chemistry Volume-II
230. Consider the following statements.
I. The colour of the hydrophobic sol depends
on the wavelength of the light scattered by
the dispersed particle.
II. The smaller the gold number value of a
hydrophilic colloid, the greater is its
protective power.
III. The movement of sol particle under an
applied electric potential is called
electroosmosis.
Which of the above statements are correct?
(a) I and II
(b) I and III
(c) II and III
(d) I, II and III
BCECE-2011
Ans. (a) : The movement of sol particles under the
influence of electric potential is called electrophoresis.
All other given statement are correct.
231. Choose the correct statement:
When concentration of a salt solution is
increased........
(a) Boiling point increases while vapour pressure
decreases.
(b) Boiling point decreases while vapour pressure
increases.
(c) Freezing point decreases while vapour
pressure increases.
(d) Freezing point increases while vapour
pressure increases.
GUJCET-2011
Ans. (a): On increasing the concentration of salt
solution, the boiling point of salt solution increases
while vapour pressure of the solution decreases.
232. To prepare 2 dm3 of decinormal solution of
oxalic acid, the mass of oxalic acid required is
(a) 6.3 g
(b) 0.63 g
(c) 12.6 g
(d) 1.26 g
COMEDK-2011
Ans. (c) :Given,
Volume = 2 dm3
1
Normality =
N
10
Weight of Oxalic acid
∴N =
Eq.Wt × Volumein L
W = N × Eq.Wt × volume in L
1
× 63 × 2
10
W = 12.6 gm
233. 25 mL of 0.08 N HCl neutralizes with 20 mL of
NaOH. The mass of NaOH present in 5 dm3 is
(a) 20 g
(b) 40 g
(c) 10 g
(d) 30 g
COMEDK-2011
Ans. (a) : Given that, N1 = 25, V1 = 0.08
V2 = 20, N2 = ?
59
W=
YCT
Ans. (a) : Given, W = 4g
250
V = 250mL =
L
1000
Now,
Moles
4
Molarity =
Moles =
Volume
40
4.0
40g molL−1
= 0.4M
Molarity =
250
1000
W
0.1 =
237.
Density
of
a 2.05 M solution of acetic acid in
40 × 5
water
is
1.02
g / mL. The molality of the
W = 0.1 × 40 × 5 = 20 gm
solution is
234. 19g of a mixture containing NaHCO3 and
(a) 2.28 mol kg-1
(b) 3.28 mol kg-1
-1
Na2CO3 on complete heating liberated 1.12 L of
(c) 0.44 mol kg
(d) 1.14 mol kg-1
CO2 at STP. The weight of the remaining solid
UPTU/UPSEE-2010
was 15.9 g. What is the weight (in g) of Na2CO3
Ans. (a) : Given : M = 2.05, m = 60; d = 1.02
in the mixture before heating?
M
(a) 8.4
(b) 15.9
Molality ( m ) =
× 1000
1000d − Mm
(c) 4.0
(d) 10.6
2.05
AP-EAMCET- (Engg.)-2011
=
× 1000
Ans. (d) : Molecular weight of NaHCO3 =23+1+12+48
(1000 ×1.02 ) − ( 2.05 × 60 )
= 84
= 2.28mol kg −1
Molecular weight of Na2CO3 = 46+12+48 = 106
238. The normality of 10 L volume H2O2 is
Total weight = 84+106 = 190
(a) 0.176
(b) 0.88
∵ In 190 gm of a mixture, weight of Na2CO3 is = 106
(c)
1.78
(d)
3.52
19 gm of a mixture weight contains
UPTU/UPSEE-2010
106 × 19
Na 2 CO3 =
=10.6gm.
Ans. (c) : Hydrogen peroxide dissociates in the
190
following manner
235. The volume in mL of 0.1 M solution of NaOH
1
H 2O2 → H 2O + O2
required to completely neutralise 100 mL of 0.3
2 1 mol
1mol
M solution of H 3 PO 3 is
2
(a) 60
(b) 600
34g
11.2L
∵
1M H2O2 solution = 2N
(c) 300
(d) 30
AP-EAMCET- (Engg.)-2011 Normality = Molarity × n factor
2 × 10
Ans. (b) : Phosphorus acid ( H 3 PO3 ) is a dibasic acid.
∴ Normality =
= 1.78N
11.2
Normality = molarity × basicity
239. An aqueous solution is 1.00 molal in KI. Which
For H3PO3 , Normality = 0.3M × 2 = 0.6N
change will cause the vapour pressure of the
Given that, N1 = 0.1 N, N2 = 0.6 N, V2 = 100 mL
solution to increase?
We know that,
(a) Addition of NaCl
N1V1 = N 2 V2
(b) Addition of Na2So4
(c) Addition of 1.00 molal KI
0.1× V1 = 0.6 ×100
(d)
Addition of water
0.6 ×100
V1 =
(AIPMT -2010)
0.1
Ans. (d) : Addition of water to the aqueous solution of
V1 = 600ml
KI, will cause the vapour pressure of the solution to
Hence, 600mL of volume is required to completely increase. Vapour pressure depends upon the surface
neutralize 0.3 M of H3PO3.
area of the solution.
236. The molarity of a NaOH solution by dissolving 240. 2.5 cm3 of 0.2 M H2SO4 solution is diluted to 0.5
4 g of it in 250 mL water is.
dm3. Find normality of the diluted solution.
(a) 0.4M
(b) 0.8M
(a) 0.2 N
(b) 0.02 N
(c) 0.2M
(d) 0.1M
(c) 0.002 N
(d) 0.04 N
WB-JEE-2010
MHT CET-2010
We know,
N 1V 1 = N 2V 2
25×0.08= N2×20
25 × 0.08
N 2=
20
N 2 = 0.1
Therefore,
W
Normality =
Eq.wt × Volumein dm3
Objective Chemistry Volume-II
60
YCT
from Raoult's law,
p1o − p1
= X2 = X
p1
3000 − 2985
9
=
3000
10x
x = 180
180g/mol is the molecular formula of glucose.
244. What is the molarity of 0.2 N Na2CO3 solution?
(a) 0.1 M
(b) 0 M
−3
(c)
0.4
M
(d)
0.2 M
2.5 × 10 × 0.4
=
JCECE-2010
0.5
Ans.
(a)
:
Given,
= 0.002 N
Normality of Na2CO3 is solution = 0.2 N
241. 450 mg of glucose is dissolved in 100 g of
Molecular weight = 2M
solvent. What is the molality of solution?
(∵ Na2CO3 is dipositive.)
(a) 0.0025 m
(b) 0.025 m
equivalent weight
(c) 0.25 m
(d) 2.5 m
Molarity = normality ×
MHT CET-2010
molecular weight
Ans. (b) : Given :
M
∴
Molarity = 0.2 ×
Wsolute = 450 mg
2M
= 0.1 M
molar mass of glucose = 180
245. 36 g of glucose (molar mass=180g/mol) is
Wsolvent = 100
present in 500 g of water, the molarity of the
Now,
solution is
450 × 10−3 × 1000
(a) 0.2
(b) 0.4
Molality =
(c) 0.8
(d) 1.0
180 × 100
J & K CET-2010
= 0.025 m
242. 200 mL of water is added to a 500 mL of 0.2M Ans. (b) : Given,
solution. What is the molarity of this diluted W= mass of glucose = 36g
Volume of water = 500 mL
solution?
Mass of glucose = 180g
(a) 0.5010 M
(b) 0.2897 M
Now.
(c) 0.7093 M
(d) 0.1428 M
W ×1000
(e) 0.4005 M
Molarity =
M×V
Kerala-CEE-2010
36 ×1000
Ans. (d) : Given that, M1 = 0.2, V1 = 500, V2 = 700
Molarity =
M 1V 1 = M 2V 2
180 × 500
= 0.4M.
0.2 × 500 = M2 (500 + 200)
246. On mixing, heptane and octane form an ideal
M2 = 0.14M
solution. At 373 K, the vapour pressures of the
243. The vapour pressure of 100g water reduces
two liquid components (heptane and octane)
from 3000 Nm–2 to 2985 Nm–2 when 5g of
are 105 kPa and 45 kPa, respectively. Vapour
substance 'X' is dissolved in it. Substance 'X' is
pressure of the solution obtained by mixing 25
(a) methanol
(b) glucose
g of the heptane and 35 g of octane will be
(c) carbon dioxide
(d) cannot predict
(molar mass of heptane = 100 g mol–1 and of
JIPMER-2010
octane = 114 g mol–1).
(a) 72.0 kPa
(b) 36.1 kPa
Ans. (b) : Given:
o
−2
(c)
96.2
kPa
(d) 144.5 kPa
p1 = 3000Nm
AIEEE-2010
-2
p1= 2985Nm
Ans. (a) : Given,
X2 = mole - fraction of Solute
PHo = 105kPa,POo = 45kPa
Let molecular whight of substance X is x.
Ans. (c) : Given:
V1 = 2.5 cm3 V2 = 0.5 dm3
Concentration of H2SO4 before dilution = 0.2M
and as we know that,
1N = 2M
So, 0.2 M = 0.4N (N1)
From Normality equation
NV
N2 = 1 1
V2
PT = X H .PHo + X o Poo
So, number of moles X will be 5/x
100
No. of moles of solvent is
18
5/ x
9
Mole fraction of X=
=
100 /18 10x
Objective Chemistry Volume-II
25 /100
= 0.45
25 /100 + 35 /114
Xo = 1–XH =1– 0.45= 0.55
PT = 0.45×105+0.55×45= 72.0 kPa
XH =
∴
61
YCT
247. Volume of 0.6M NaOH required to neutralize
30cm3 of 0.4M HCl is
(a) 30 cm3
(b) 45 cm3
3
(c) 20 cm
(d) 50 cm3
BCECE-2010
Ans. (c) : We know that,
N 1V 1 = N 2V 2
So,
n × M 1V 1 = n × M 2 V 2
1 × 0.4 × 30 = 1 × 0.6 × V
V = 20cm3
248. What is the volume of 0.1 M H2SO4 required in
litres to neutralize completely 1 L of a solution
containing 20 g of NaOH ?
(a) 5.0
(b) 0.5
(c) 2.5
(d) 10.0
AP- EAMCET(Medical) -2010
Ans. (c): Given data,
Molarity (M1) of H2SO4 = 0.1 M
Volume of NaOH solution (V2) = 1 L
Mass of NaOH (M2) = 20 g
20 1
Molarity of NaOH =
× = 0.5M
40 1
H 2SO 4 + 2NaOH → Na 2SO 4 + 2H 2 O
1mole
⇒
2 mole
M1V1 M 2 V2
=
n1
n2
0.1× V1 0.5 × 1
=
1
2
⇒ Volume of H2SO4 (V1) = 2.5 litre.
249. Equal weight of methane and oxygen are mixed
in an empty container at 25oC. The fraction of
the total pressure exerted by oxygen is
(a) 1/2
(b) 2/3
(c) 1/4
(d) 1/3
AP-EAMCET(Medical) -2010
Ans. (d): GivenWeights of methane (CH4) = weights of oxygen (O2) =
x
x
32
∴ Mole fraction of O 2 X O =
2
x x
+
32 16
x
32
XO =
2
x 1
+ 1
16 2
x
32
XO =
2
x 3
×
16 2
⇒
( )
Objective Chemistry Volume-II
x 16 × 2
×
32 x × 3
1
XO =
2
3
XO =
2
1
3
250. What is the normality of a solution containing
20 g acetic acid in 2 L of solution?
(a) 0.20 N
(b) 1.06 N
(c) 4.00 N
(d) 0.166 N
SCRA-2010
Ans. (d) :
No. of gram equivalent
Normality (N) =
volume of solution (L)
weight
Gram equivalent weight =
equivalent weight
Molecular weight
Equivalent weight =
n
Where, n = Valence factor
20
So,
N=
60 × 2
{Molecular mass of CH3COOH (M) = 60}
1
N=
6
or
N = 0.166 N
251. What is the total number of moles of H2SO4
needed to prepare 5.0L of a 2.0M solution of
H2SO4?
(a) 2.5
(b) 5.0
(c) 10
(d) 20
UPTU/UPSEE-2009
Ans. (c) : Moles of solute in 5 litres of 2.0M Solution
No. of moles = Molarity × Volume
= 5 × 2.0 = 10.0
252. 2.5 L of NaCl solution contain 5 moles of the
solute. What is the molarity?
(a) 5 M
(b) 2 M
(c) 2.5 M
(d) 12.5
UPTU/UPSEE-2009
Moles of solute
Ans. (b) : Molarity =
volumeof solution
5
Molarity =
2.5
=2M
N
N
253. 60 mL. of
H 2SO4 , 10 mL of
HNO 3 and
5
2
N
30 mL. of
HCI, aremixed together. The
10
Strength of the resulting mixture is
(a) 0.1 N
(b) 0.2 N
(c) 0.3 N
(d) 0.4 N
UP CPMT-2009
62
Thus, fraction of pressure exerted by oxygen =
YCT
257. 0.01 mole of a non-electrolyte is dissolved in
10g of water. The molality of the solution is
(a) 0.1 m
(b) 0.5 m
(c) 1.0 m
(d) 0.18 m
N
N
N
JCECE-2009
60 × + 10 × + 30 ×
5
2
10
Ans.
(c)
:
Given,
=
60 + 10 + 30
10 g water contain 0.01 mol
N = 0.2N
∵
1000g water contain = 1 mol
254. When 10 A current is passed for 80 min, the
no.of mole of Solute
volume of hydrogen gas liberated is
Molality of Solution (m) =
Mass
of Solvent in (kg)
(a) 11.14 L
(b) 5.57 L
(c) 22.4 L
(d) 2.78 L
0.01
=
× 1000
MHT CET-2009
10
Ans. (b) : Given :
= 1.0 m
1
−
+
258.
25
g
of
a
solute
of
molar mass 250 g mol–1 is
→ H2
H + e
2
dissolved in 100 mL of water to obtain a
IF
10 A current is passed for 80 min is
solution whose density is 1.25 g mL–1. The
molarity and molality of the solution are
equal to 10 × 80 × 60 C
respectively
It will liberated hydrogen
(a) 0.75 and 1
(b) 0.8 and 1
11.2 × 10 × 80 × 60
gas volume =
(c)
1
and
0.8
(d)
1 and 0.75
96500
J & K CET-2009
(∴ 96500 C = 11.2 H2)
= 5.57 L H2 gas.
Ans. (c) : Given.
255. The volume of 2 N H2SO4 solution is 0.1 dm3. W = 25 g
The volume of its decinormal solution (in dm3) Molar mass = 250
will be
density = 1.25g mL–1
(a) 0.1
(b) 0.2
Now,
(c) 2
(d) 1.7
No.of moles of solute
MHT CET-2009 Molarity =
Volume
N
Ans. (c) : Given, N1 = 2N, N2 = , V1 = 0.1, V2 = ?
25
0.1
Molarity =
=
= 1M.
10
250
100
For solution of same substance,
100 1000
N 1V 1= N 2V 2
1000
N
2N×0.1= ×V2
And
10
No.of moles of solute
V2 = 2 dm3.
Molality =
mass
256. In an oxidation-reduction reaction, MnO 4− ion
0.1
is converted to Mn2+. What is the number of Molality =
= 0.8M
(∵ m = d × v)
125 /1000
equivalents of KMnO4 (mol. wt. = 158) present
in 250 mL of 0.04 M KMnO4 solution?
259. Two liquids X and Y form an ideal solution at
(a) 0.02
(b) 0.05
300 K, vapour pressure of the solution
(c) 0.04
(d) 0.07
containing 1 mol of X and 3 mol of Y is 550
mmHg. At the same temperature, if 1 mol of Y
JIPMER-2009
is further added to this solution, vapour
UPCPMT-2008
pressure of the solution increases by 10 mm
Ans. (b) : Required equation.
Hg. Vapour pressure (in mmHg) of X and Y in
+7
Mn O 4− + 8H + + 5e −
→ Mn 2+ + 4H 2 O
their pure states will be, respectively
(a) 200 and 300
(b) 300 and 400
Normality = Molarity × Change in oxidation number.
= 0.04 × 5
(c) 400 and 600
(d) 500 and 600
= 0.2N
AIEEE-2009
Gram eq.of Solute
Ans. (c) : Given.
Normality =
1
3
Volume of Solution ( litre )
P = 550mm Hg; X =
,Y=
1
+
3
1
+
3
0.2 × 250
no. of equivalents =
From Raoult’s law
1000
P = Px° X + Py° Y
= 0.05
Ans. (b): Strength of resulting mixture,
N V + N 2 V2 + N 3 V3
N= 1 1
V1 + V2 + V3
Objective Chemistry Volume-II
63
YCT
Ans. (c) : Given that,
∆Tb = 1.00 K
Kb = 0.512 K kgmol–1
According to the Raoult's law–
∆Tb = Kb × m
Where, m = molality
∆T
or
m= b
Kb
1
3
+ Py° ×
4
4
o
o
2200= Px + 3Py ......(i)
550 = Px° ×
Adding 1mole of Y then,
1
4
560 = Px° × + Py° ×
5
5
2800= Pxo + 4Pyo
......(ii)
From (i) and (ii)
Py° = 600 mm Hg and Px° = 400 mm Hg
260. A binary liquid solution is prepared by mixing
n-heptane and ethanol. Which one of the
following statements is correct regarding the
behaviour of the solution?
(a) The solution formed is an ideal solution
(b) The solution is non-ideal, showing positive
deviatioin from Raoult’s law.
(c) The solution is non-ideal, showing negative
deviation from Raoult's law
(d) n-heptane shows positive deviation while
ethanol show negative deviation from
Raoult's law
AIEEE-2009
Ans. (b) : n- heptanes and ethanol forms a non-ideal
solution that shows positive deviation from Raoult’s
law.
In solution, heptane– ethanol interactions are weaker
than ethanol– ethanol interactions.
261. Sodium bicarbonate on heating decomposes to
sodium carbonate, CO2 and H2O. If 0.2 moles
of sodium bicarbonate are completely
decomposed, how many moles of sodium
carbonate are formed ?
(a) 0.2
(b) 0.1
(c) 0.05
(d) 0.025
SCRA - 2009
Ans. (b): When sodium bicarbonate (NaHCO3)
decomposed on heating it gives, sodium carbonate
(Na2CO3), carbon dioxide (CO2) and water (H2O).
∆
2NaHCO3 ( s )
→ Na 2CO3 ( s ) + Ca 2 ( g ) + H 2 O ( l )
1
0.512
m = 1.95 m
263. 20 mL of 10N HCl and mixed with 10 ml of 36
N H2SO4 and the mixture is made one litre.
Normality of the mixture will be
(a) 0.56 N
(b) 0.50 N
(c) 0.40 N
(d) 0.35 N
MPPET- 2009
Ans. (a) : Given,
HCl (1)
H2SO4(2)
Mixture
V1 = 20mL
V2 = 10mL
VM = 1L
N1 = 10N
N2 = 36N
NM = ?
N 1V l + N 2V 2 = N M V M
10 × 20 + 36 × 10 = NM × 1000
200 + 360 = NM × 1000
560
NM =
= 0.56 N.
1000
264. The molarity of 98% H2SO4 (d = 1.8g/ml) by
weight is
(a) 6 M
(b) 18 M
(c) 10M
(d) 4 M
MPPET- 2009
Ans. (b) : Molar mass of solute (H2SO4) = 98 gm
Mass of solution = 100 gm
d = 1.8 gm/mL = 1800 gm/litre
98
∴
No. of moles =
= 1 mole.
98
Mass of solution M
and
Density (d) =
=
Volume
V
M 100
1
V=
=
= litre
d 1800 18
No.of moles of solute
1
∴
Molarity =
=
Volume of solution
1/18
m=
Form the above reaction, we can say that 2 moles of
sodium bicarbonate decompose into 1 mole of sodium
carbonate.
Then,
0.2 mole of sodium bicarbonate will decompose into =
1
= 18 M
× 0.2 = 0.1 mole of sodium carbonate
2
265. The vapour pressure of pure liquid A is 0.80
262. Molality of an aqueous solution that produces
atm. When a non volatile B is added to A its
an elevation in boiling point of 1.00 K at 1 atm
vapour pressure drops to 0.60 atm. The mole
pressure is (Kb for water = 0.512 K kg mol–1)
fraction of B in the solution is
(a) 0.512 m
(b) 0.195 m
(a) 0.125
(b) 0.25
(c) 1.95 m
(d) 5.12 m
(c) 0.5
(d) 2.5
UPTU/UPSEE-2008
AP-EAMCET(MEDICAL) - 2009
Objective Chemistry Volume-II
64
YCT
(a) 400 cm3
(c) 500 cm3
Ans. (b) : Given that,
P o = 0.80, PS = 0.60
Applying Raoults law–
(b) 450 cm3
(d) 100 cm3
BCECE-2008
Ans. (a) : From formula.
P o − PS
N 1V 1 = N 2 V 2
XB =
Po
0.5 × 100 = 0.1 × V2
0.80 − 0.60
V2 = 500 cm3
So,
XB =
Volume of water to be added to 100 cm3
0.80
= 500 – 100
= 0.25
= 400 cm3
266. In transforming 0.01 mole of PbS to PbSO4 the
volume of ‘10 volume’ H2O2 required will be 270. The volume of CO2 formed at STP on burning
A mixture of 0.5 mole of methane and 24 gram
(a) 11.2 mL
(b) 22.4mL
of oxygen is
(c) 33.6 mL
(d) 44.8 mL
(a) 84 litre
(b) 8.4 litre
WB-JEE-2008
(c)
22.4
litre
(d) 0.84 litre
Ans. (d) : Required Reaction.
AMU – 2008
PbS + 4H 2 O 2
→ PbSO 4 + 4H 2 O
Ans. (b) : CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
∵ 1 mole of PbS requires 4 moles of H2O2
16gm 64gm
44gm
36gm
1 mol 2 mol
1 mol 2 mol
∴ 0.01 moles of PbS requires 0.04 moles of H2O2
Volume Strength = M ×11.2
24
0.5 mol =
= 0.375 mol
10 = M × 11.2
64
M = 0.892
O2 is a limiting reactant.
So, if 64 g of O2 gives 22.4 L of CO2
MV
No. of moles =
22.4 × 24
1000
then 24 g of O2 will give =
= 8.4 L of CO2.
0.892 × V
64
0.04 =
271. How many grams of NaOH will be required to
1000
prepare 500 gram solution containing 10%
V = 44.8mL
w/w NaOH solution?
267. Which of the following concentration factors is
(a) 5.0 grams
(b) 0.5 grams
affected by change in temperature?
(c)
50
grams
(d)
100 grams
(a) Molarity
(b) Molality
GUJCET-2008
(c) Mole fraction
(d) Weight fraction
Ans.
(c)
:
Given
that
,
MHT CET-2008
Ans. (a) : Molarity is dependent on temperature. The Mass of solution = 500gm.
w
definition of molarity is, the number of moles per unit
% of NaOH solution = 10%
volume. When volume changes the temperature is also
w
changes.
Let x = Mass of NaOH (solute)
268. The concentration of H2O2 solution of ’10 Then,
volume’ is
w
Mass of solute
%=
× 100
(a) 30%
(b) 3%
w
Total mass of solution
(c) 1%
(d) 10%
x
JCECE-2008
10 =
× 100
500
Ans. (b) : Required equation:
10 × 500
2H 2 O 2
→ O2 + H 2O
x=
100
Molar mass of H 2 O 2 = 2 × 34
x = 50gram
∴10L of O2 will produced from (S.T.P)
272. Molarity is expressed as
68 × 10
=
(a) Litre/mole
(b) Moles/litre
22.7
(c)
Moles/1000gms
(d) Grams/litre
= 29.9 ≈ 30g H2O2
MPPET-2008
So,
Ans.
(b)
:
Molarity
is
expressed
as
the
moles
of solute
30
Concentration of H2O2 Solution of 10 Volume =
per litres of solution. Molarity is also known as molar
10
concentration of a solution.
= 3%
moles of solute
269. The volume of water to be added to 100 cm3 of Molarity =
Volume (litre)
0.5N H2SO4 to get decinormal concentration is–
Objective Chemistry Volume-II
65
YCT
273. 10.6 g of a substance of molecular weight 106 276. If 20 mL of 0.4 N NaOH solution completely
was dissolved in 100 mL, 10 mL of this solution
neutralised 40 mL of a dibasic acid, the
was pipetted out into a 1000 mL flask and
molarity of the acid solution is
made up to the mark with distilled water. The
(a) 0.1M
(b) 0.2M
molarity of the resulting solution is
(c)
0.3M
(d) 0.4M
(a) 1.0M
(b) 10–2M
AP-EAMCET-2008
(d) 10–4M
(c) 10–3M
Ans. (a) : Given that,
AP-EAMCET-2008
N1 = 0.4N
Ans. (b) : Given dataV1 = 20mL
Weight of substance = 10.6 g
V2 = 40mL
Molecular weight of substance = 106 g
N2 = ?
Volume of solution =100 mL
Now, from normality equation–
Volume of flask = 1000 mL
N 1 V 1 = N 2V 2
Pipetted volume of solution = 10mL
0.4 × 20 = N2×40
Then,
0.4 × 20
N2 =
= 0.2N
moles of solute
40
Initial molarity =
volume of solution ( L )
N
0.2
For dibasic acid, molarity (M) = 2 ⇒
= 0.1 M
10.6 × 1000
2
2
M1 =
=1 M
106 × 100
277. 9.8 g of H2SO4 is present in two litres of a
We know thatsolution. The molarity of the solution is
M 1V 1 = M 2V 2
(a) 0.1 M
(b) 0.05 M
1×10 = M2×1000
(c) 0.2 M
(d) 0.01 M
AP-EAMCET-2008
10
M2 =
= 10−2 M
Ans.
(b)
:
Given
that,
1000
274. The weight, in gram, of KCl (mol. wt. = 74.5) in Weight of H2SO4 = 9.8 gm
Volume of solution (V) = 2 litre
100 mL of a 0.1 M KCl solution is
We know that(a) 74.5
(b) 7.45
Molecular weight of H2SO4 = 98 g
(c) 0.745
(d) 0.0745
AP-EAMCET-2008
W
Ans. (c) : Given that,
m = 74.5
∴
M=
V = 100 mL
m×V
M = 0.1
9.8
=
Weight 1000
98 × 2
∴
Molarity =
×
= 0.05 M
m
V(mL)
278. The weight of oxalic acid crystals,
w 1000
0.1 =
×
H2C2O4⋅2H2O required to prepare 500 mL of
74.5 100
0.2 N solution is
7.45
(a) 3.4 g
(b) 63 g
w=
10
(c) 6.3 g
(d) 126 g
w = 0.745 gm
AP-EAMCET-2008
275. 5.85 g of NaCl (mol. wt. 58.5) is dissolved in Ans. (c) : Given that,
water and the solution is made up to 500ml. Volume (V) = 500 mL
The molarity of the solution will be
N = 0.2
(a) 0.1
(b) 0.2
Molecular weight
(c) 1.0
(d) 0.117
Gram equivalent weight =
Basicity
AP-EAMCET-2008
Equivalent weight of oxalic acid
Ans. (b) : Given that,
Molecular mass of H 2 C 2 O 4 ⋅ 2H 2 O
Weight of NaCl (w) = 5.85 gm
=
Molecular weight of NaCl (Mol.wt.) = 58.5
2
Volume of solution (V) = 500 mL
126
=
= 63 g/mol
We know that –
2
w ×1000
N × E × V 0.2 × 63 × 500
Molarity (M) =
∴
W=
=
Mol.wt. × V
1000
1000
5.85 × 1000
W = 6.3 gm
or
M=
The mass of oxalic acid crystals required to prepare
58.5 × 500
500mL of a 0.2 N solution = 6.3gm.
M = 0.2 M
Objective Chemistry Volume-II
66
YCT
279. 138 g of ethyl alcohol is mixed with 72 g of Ans. (a) : Let V litre of 10 N HCl be mixed with (1 –
water. The ratio of mole fraction of alcohol to V) litre of 4 N HCl to give (V + 1 – V) = 1L of 7 N
water is
HCl.
N1V1 + N2V2 = NV
(a) 3 : 4
(b) 1 : 2
10 V + 4 (1 – V) = 7 × 1
(c) 1 : 4
(d) 1 : 1
6V = 7 – 4
AP-EAMCET-2008
3
Ans. (a) : Given that,
V = = 0.05 L
6
Mass of H2O, m H2O = 72g
∵
Volume
of
10
N
HCl
= 0.50 L
Mass of C2H5OH, m C2 H5OH = 138g
∴ Volume of 4 N HCl = 1 – 0.50 = 0.50 L
72
282.
When a sulphur sol is evaporated sulphur is
No. of moles of H2O =
=4
18
obtained. On mixing with water sulphur sol is
not found. The sol is
138
No. of moles of C2H5OH =
=3
(a) lyophilic
(b) reversible
46
(c) hydrophobic
(d) hydrophilic
3
3
Karnataka-CET-2007
∴ Mole fraction of C2H5OH =
=
3+ 4 7
Ans. (c) : Hydrophobic sol are irreversible in nature.
4
4
They have a affinity between the dispersed phase and
and mole fraction of H2O =
=
the dispersion medium (H2O). Further once precipitated,
4+3 7
they do not form the colloidal sol by simple addition of
Therefore, the ratio of mole fraction of ethyl alcohol to
water.
3
283. 4.5 g of aluminium (Atomic mass 27 amu) is
X C2 H5OH 7 3
H2O is
= = ⇒ 3: 4
deposited at cathode from Al3+ solution by a
4 4
X H 2O
certain quantity of electric charge. The volume
7
of hydrogen produced at STP from H+ ions in
solution by the same quantity of electric charge
280. Equal masses of hydrogen gas and oxygen gas
will be
are placed in a closed container at a pressure of
(a) 22.4 L
(b) 44.8 L
3.4 atm. The contribution of hydrogen gas of
(c) 5.6 L
(d) 11.2 L
the total pressure is
JIPMER-2007
(a) 1.7 atm
(b) 0.2 atm
(c) 3.2 atm
(d) 3.02 atm
Ans. (c) : Required equation.
SRMJEEE – 2008
2Al → 2Al3+ + 6e–
6H+ + 6e– → 3H2
Ans. (c) : Given that,
2Al + 6H+ → 3H2 + 2Al3+
WO2 = WH2 = M
From faraday second law;
Total pressure (P) = 3.4 atm
m Al E Al
=
Mole fraction of X H2 –
mH EH
M
4.5 27 / 3
n H2
16
=
2
X H2 =
=
=
m
1
H
M
M
n H 2 + n O2
17
+
mH = 0.5g
2 32
∵ Volume of 2g H2 at S.T.P = 22.4L
∴ Partial pressure PH2 = Ptotal × X H2
Volume of 0.5g H2 at S.T.P
16
22.4 × 0.5
PH2 = 3.4 ×
=
L
17
2
= 5.6 L
PH2 = 3.2 atm
284. The mass of carbon anode consumed (giving
281. The volume of 10 N and 4 N HCl required to
only carbondioxide) in the production of 270 kg
make 1 L of 7 N HCl are
of aluminium metal from bauxite by the Hall
(a) 0.50 L of 10 N HCl and 0.50 L of 4N HCl
process is
(b) 0.60 L of 10 N HCl and 0.40 L of 4N HCl
( Atomic mass Al = 27 )
(c) 0.80 L of 10 N HCl and 0.20 L of 4N HCl
(a) 180 kg
(b) 270 kg
(d) 0.75 L of 10 N HCl and 0.25 L of 4N HCl
(c) 540 kg
(d) 90 kg
Karnataka-CET-2007
JIPMER-2007
( )
Objective Chemistry Volume-II
67
YCT
288. The density (in g mL–1) of a 3.60M sulphuric
acid solution that is 29% H2SO4 (molar mass =
98 g mol–1) by mass will be
equivalent w.t of
27
(a) 1.64
(b) 1.88
Al =
(c)
1.22
(d) 1.45
3
[AIEEE 2007]
No. of gram of equivalent of AlAns. (c) : Let the density of the solution is 'd'
270 × 103
1L of solution contains = 3.6 ×98gm of H2SO4
Al =
27 / 3
Now,
= 30 × 103
100/d ml solution contains 29gm of H2SO4
No. of gram equivalent of CThen,
Mass
mass
1000 mL solution contains 3.6 × 98g of H2SO4
C=
=
Gram equivalent weight
3
So,
(∵gram equivalent w.t of Al 2 O3 = Gram equivalent of C )
29 × d
3.6 × 98 =
× 1000
100
Now,
d = 1.22
Mass
30 × 103 =
289.
Equal
masses of methane and oxygen are
3
mixed in an empty container at 25°C. The
fraction of the total pressure exerted by oxygen
mass
= 90 × 103g
is
= 90kg
2
1 273
285. The molarity of the solution obtained by
(a)
(b) ×
dissolving 2.5g of NaCl in 100 mL of water is
3
3 298
(a) 0.00427 moles
(b) 427 moles
1
1
(c)
(d)
(c) 0.427 moles
(d) 0.027 moles
3
2
J & K CET-2007
[AIEEE 2007]
Ans. (c) : Given:
Ans. (c):
V= 2.5g
Mass of methane = wm
Molar mass of NaCl = = 58.5
Mass of oxygen = wo
Now,
Mole fraction (x) of oxygen =
N 2.5 ×1000
Number of moles of oxygen
Molarity = =
V 58.5 × 100
Number of moles of oxygen + Number of moles of methane
= 0.427 mol
1
286. One part of solute in one million parts of
1
solvent is expressed as
= 32 =
1 1 3
(a) ppm
(b) milligrams/100 cc
+
32 16
(c) grams/litre
(d) grams/100 cc
J & K CET-2007 Partial pressure of oxygen = X O2 × PTotal
Ans. (a) : One parts per million as ppm is
1
Concentration of solution and can be numerically
=P×
3
expressed as.
Hence,
option
(c)
is
correct
answer.
No.of parts of thecomponent
=
× 106
290. A mixture of ethyl alcohol and propyl alcohol
Tota l number of parts of the solution
has a vapour pressure of 290 mm at 300 K. The
287. On mixing ethyl acetate with aqueous sodium
vapour pressure of propyl alcohol is 200 mm. If
chloride, the composition of the resultant
the mole fraction of ethyl alcohol is 0.6, its
solution is
vapour pressure (in mm) at the same
(a) CH3COOC2H5 + NaCl
temperature will be
(b) CH3COONa + C2H5OH
(a) 350
(b) 300
(c) CH3COCl + C2H5OH + NaOH
(c) 700
(d) 360
(d) CH3Cl+C2H5COONa
AIEEE-2007
AIEEE-2007
Ans. (a) : Given;
Ans. (a) : CH3COOC2H5 + NaCl → no reaction.
PT = 290mm
The resultant solution contains ethyl acetate and sodium
Vapour pressure of propyl alcohol = 200mm
chloride.
Aqueous NaCl is in neutral state so there will be no Mole fraction of C2H5OH = 0.6
such a reaction between ethyl acetate and aqueous Mole faction of propyl Alcohol = 1– 0.6
= 0.4
NaCl.
Ans. (d) : Required equation:
2Al2O3 + 3C → 4Al + 3CO2
Objective Chemistry Volume-II
68
YCT
From formula,
PT = PA° × XA + PB° × XB
290 = 200 × 0.4 + PB° × 0.6
PB° = 350mm.
294. The volume of water to be added to 100 cm2 of
0.5 N H2SO4 to get decinormal concentration is
(a) 400 cm3
(b) 450 cm3
3
(c) 500 cm
(d) 100 cm3
UP CPMT-2006
Ans. (a) : Given, N1= 0.5N, N2 = 0.1
V1= 100 cm2
N1V1 = N2V2 ⇒ 0.5 × 100 = 0.1 × V2
V2 = 500 cm3
∴ Water to be added to 100 cm3 solution = 500 − 100 =
400 cm3
295. Which of the following concentration terms
is/are independent of temperature?
(a) Molarity
(b) Molarity and mole fraction
(c) Mole fraction and molality
(d) Molality and normality
(e) Only molality
Kerala-CEE-2006
Ans. (c) : Molality is the ratio of moles and mass and
mass does not change at any temperature so, we can say
molality is independent of temperature.
Mole fraction is the number of moles of solute divided
by the total number of solute moles and solvent moles.
It will also not change with the respect of temperature
and it does not depend on volume.
296. 5 litres of a solution contains 25 mg of CaCO3.
What is its concentration in ppm? (mol. wt. of
CaCO3 is 100)
(a) 25
(b) 1
(c) 5
(d) 2500
J & K CET-2006
Ans. (c) : Given:
mass of solute = 25 × 10-3g
dilute Solution volume = 5 kg
= 5000 g.
5000 gm contains 25 × 10−3 gm of CaCO3
291. The volume strength of 1.5 N H2O2 solution is
(a) 8.4
(b) 4.8
(c) 5.2
(d) 8.8
CG PET -2007
molar mass
Ans. (a) : Equivalent weight of H2O2 =
2
1× 2 + 16 × 2
=
2
= 17
So,
Strength = Normality × Equivalent weight
= 1.5 × 17
= 25.5
Now, required equation2H2O2 → 2H2O + O2
(2 × 34 = 68 g) (22.4L)
Since, 68 grams H2O2 produces 22.4 litres oxygen at
NTP, therefore 25.5 grams of H2O2 will produce.
22.4
=
× 25.5
68
= 8.4
Thus, volume strength of given H2O2 solution is 8.4 L.
292. Disperse phase and dispersion medium in
butter are respectively
(a) solid and liquid
(b) liquid and solid
(c) liquid and liquid
(d) solid and solid
AP-EAMCET (Medical) - 2007
Ans. (b) : Butter is type of dairy product which contains
buffer fat upto 80% which is solid in state when preserved
in a cooled environment or at room temperature. It is an
example of solid emulsion type colloid. The continuous
medium in which the dispersed phase is distributed
throughout is called dispersion medium. Water in milk is
an example of dispersion medium. The phase that is
composed of particles that are distributed through another
25 × 10−3 g
phase is called the dispersed phase.
106 gm contains =
×106
5000g
293. An aqueous solution of glucose is 20% in
strength. The volume in which 1 g-mole of it is
= 5 ppm.
dissolved will be:
297. Density of a 2.05 M solution of acetic acid in
(a) 9 L
(b) 1.8 L
water is 1.02 g/mL. The molality of the solution
(c) 8 L
(d) 0.9 L
is
UPTU/UPSEE-2006
(a) 1.14 mol kg–1
(b) 3.28 mol kg–1
Ans. (d) : Molar Mass of Glucose = 180g
–1
(c) 2.28 mol kg
(d) 0.44 mol kg–1
20g glucose
→100mL
[AIEEE 2006]
Solution
Ans. (c) : Given that,
So,
Density D = 2.05, Molar mass = 60
180g glucose will be disolved
Weight of acetic acid = d × molar mass
100 × 180
=
= 2.05 × 60
20
= 123 g
= 900mL
Weight of solution
= 1000 × 1.02 = 1020g
= 0.9L
Objective Chemistry Volume-II
69
YCT
∴ Weight of water
300. The mole fraction of the solute in one molal
aqueous solution is
(a) 0.009
(b) 0.018
(c) 0.027
(d) 0.036
(AIPMT -2005)
Ans. (b) : nA = solute 1kg
nB = number of moles of water (solvent)
nA
mole fraction of solute nA =
nA + nB
= Weight of solution – weight of
acetic acid
= 1020 – 123
= 897g
Now,
2.05 ×1000
897
= 2.285 mol kg–1
298. Acidified KMnO4 oxidises oxalic acid to CO2.
1mol
What is the volume (in liters) of 10-4 M KMnO4
=
required to completely oxidise 0.5 L of 10-2 M
1mol + 55.56 mol
oxalic acid in acid medium?
= 0.018
(a) 125
(b) 1250
301. Identify the incorrect statement:
(c) 200
(d) 20
(a) The molarity of a solution is independent of
AP-EAMCET (Medical), 2006
temperature
Ans. (d) : Following reaction occurs during the process
(b) The tendency for catenation is much higher
for carbon than for silicon
2MnO4– + 5C2O 24 − + 16H+→2Mn2+ + 10CO2 + 8H2O
(c) Nitriles and iso nitriles constitute metamers
–4
N KMnO4 = 5 × molarity = 5 × 10 N
(d) t-buty 1 carbocation has planar carbons and is
very reactive
N C O 2− = 2 × molarity = 2 × 10−2 N
2 4
JCECE-2005
Now, from the normality equationAns. (a) : Molarity is dependent on temperature, but
N 1 V 1 = N 2V 2
molality is independent on temperature because its
or N KMnO4 × V1 = N C O2− × V2
value does not change with changes in temperature.
2 4
−2
302. If 117g NaCl is dissolved in 1000g of water the
2 ×10 × 0.5
or V1 =
concentration of the solution is said to be:
−4
5 ×10
(a) 2 molar
(b) 2 molal
2
V1 = 0.2 ×10 L
(c) 1 normal
(d) 1 molal
or V1 = 20 L
JCECE-2005
299. A solution has a 1:4 mole ratio of pentane to
Ans. (b) : Given.
hexane. The vapour pressures of the pure
Weight of solute = 117g
hydrocarbons at 20°C are 440 mm Hg for
pentane and 120 mm Hg for hexane. The mole Molar mass (NaCl) = 58.5
fraction of pentane in the vapour phase would Molality = no.of mole of solute
mass of solvent in ( kg )
be
(a) 0.200
(b) 0.549
117 / 58.5
Molality =
(c) 0.786
(d) 0.478
1kg
(AIPMT -2005)
= 2 molar
Ans. (d) : Given, mole ratio = 1:4
303.
Benzene
and toluene form nearly ideal
PCo5 H12 = 440 mmHg
solutions. At 20°C, the vapour pressure of
benzene is 75 torr and that of toluene is 22 torr.
PCo6 H14 = 120 mmHg
The partial vapour pressure of benzene at 20°C
for a solution containing 78g of benzene and 46
Totel pressure (Pt) = PCo5 H12 × X C5H12 + PCo6 H14 × X C6H14
g of toluene in torr is
1
4
(a) 53.5
(b) 37.5
(Pt ) = 440 × + 120 ×
5
5
(c) 25
(d) 50
920
AIEEE-2005
=
= 184
2
Ans. (d) : Given,
PC5H12
Vapour pressure of benzene (PBo ) = 75 torr
YC5H12 =
Pt
Vapour pressure of toluene (P o ) = 22 torr
molality
=
T
Now,
YC5H12
88
=
= 0.478
184
Objective Chemistry Volume-II
Molar mass of benzene = 78 g
Molar mass of toluene = 92 g
70
YCT
Mole fraction of benzene (χ) =
Data: MA = 0.1, VA = 20
MB = 0.1 VB =?
Required equation,
H 3 PO3 + 2KOH
→ K 2 HPO3 + 2H 2 O
78 / 78
1
=
78 46 1 + 0.5
+
78 92
= 0.667
Partial pressure of benzene = Po×χ
0.1 × 20 0.1× VB
=
= 75 × 0.667 = 50 torr
1
2
304. Two solutions of a substance (non-electrolyte) So ,
are mixed in the following manner. 480 mL of
0.1 × VB = 2(0.1 × 20)
1.5 M first solution + 520 mL of 1.2 M second
⇒
0.1 × VB = 4
solution. What is the molarity of the final
4
mixture?
VB =
= 40
0.1
(a) 2.70 M
(b) 1.34 M
VB = 40mL.
(c) 1.50 M
(d) 1.20 M
[AIEEE 2005] 307. Assertion: The molecular weight of acetic acid
determined by depression in freezing point
Ans. (b) : Given,
method in benzene and water was found to be
m1 = 1.5, v1 = 480
different.
m2 = 1.2, v1 = 520
Reason:
Water is polar and benzene is nonM V + M 2 V2
Now, total molarity = 1 1
polar
V1 + V2
(a) If both Assertion and Reason are correct and
1.5 × 480 + 1.2 × 520
the Reason is a correct explanation of the
=
480 + 520
Assertion.
(b) If both Assertion and Reason are correct but
= 1.34 M
Reason is not a correct explanation of the
305. Which of the following solutions will have the
Assertion.
highest boiling point?
(c)
If the Assertion is correct but Reason is
(a) 0.5 molal BaCl2
incorrect
(b) 1.0 molal KBr
(d) If both the Assertion and Reason are
(c) 1.8 × 1024 glucose molecules per litre
incorrect.
(d) 100 g powdered glucose in one litre water
(e)
If the Assertion is incorrect but the Reason is
BITSAT – 2005
correct.
Ans. (c) : Boiling points depends on:
AIIMS-2005
More no of ions → more elevation in boiling point.
Ans.
(a):
The
freezing
point
depression
is directly
On increasing the no. of moles → molality also
proportional
to
the
molal
concentration
of
the
solute.
increases and it is directly proportional to boiling point.
In
this
case,
the
number
of
particles
are
different
for
For electrolytes multiply the concentration by van’t
benzene
and
water
because
acetic
acid
associated
with
Hoff’s factor ‘i’.
benzene whether dissociated in water.
(a) 0. 5×3=1.5 for BaCl2;
So, the depression in freezing point method is different.
(b) 1.0×2=2 for KBr,
308.
In an oxidation-reduction reaction, MnO4 ion is
1.8 × 1024
converted to Mn2+. What is the number of
(c)
= 2.98for Glucose
6.023 × 1023
equivalents of KMnO4 (mol. Wt. = 158) present
100
in 250 mL of 0.04 N KMnO4 solution?
(d)
= 0.55 for glucose
(a) 0.02
(b) 0.05
180
Hence, the highest boiling point will be observed in
(c) 0.04
(d) 0.07
case of option (c).
AP-EAMCET (Engg.)-2005
306. To neutralise completely 20 mL of 0.1 M Ans. (b) : For the given oxidation reduction reactionaqueous solution of phosphorus acid (H3PO3),
the value of 0.1 M aqueous KOH solution
required is
Change in oxidation number = (+7) – (+2)
(a) 40mL
(b) 20mL
=5
(c) 10mL
(d) 60mL
The normality of solution will be 5×0.04 = 0.20 V
BITSAT – 2005 Given:- V = 250 mL
Ans. (a) : From formula:
250
∴ Number of equivalents of KMnO4 = 0.20 ×
M A VA M B VB
1000
=
nA
nB
= 0.05
Objective Chemistry Volume-II
71
YCT
2.
Colligative Properties and
Determination of Molar Mass
(0.5 mL × 1.05 g ml−1)HCOOH in 1L
0.525 g HCOOH in 1L
0.525
Molarity =
moles/lit
46
∆Tf = i × Kf × Molarity
0.525
0.0405 = i × 1.86 ×
46
46 × 0.0405
i=
1.86 × 0.525
i = 1.9
312. The elevation in boiling point for 1 molal
solution of non-volatile solute A is 3K. The
depression in freezing point for 2 molal
solution of A in the same solvent is 6 K. The
ratio of Kb and Kf i.e., Kb/Kf is 1: X. The value
of X is [nearest integer]
JEE Main-26.07.2022, Shift-II
Ans. (1) : Given ∆Tb = 3k
∆Tf = 6k
∆Tb = ikb ml , ∆Tf = i k f m 2
309. 1.80 g of solute A was dissolved in 62.5 cm2- of
ethanol and freezing point of the solution was
found to be 155.1 K. The molar mass of solute
A is ___g mol-1.
[Given: Freezing point of ethanol is 156.0K.
Density of ethanol is 0.80 g cm-3.
Freezing point depression constant of ethanol is
2.00 K kg mol-1]
JEE Main 29.07.2022, Shift-II
Ans. (80) : Mass of C2H5OH = 62.5× 0.8=50g
∆Tf = Kf × m
∆Tf = 156 –155.1= 0.9
2 × 1.8 ×1000
0.9 =
M w × 50
2 ×1.8 × 1000
Mw =
= 80
0.9 × 50
310. Two solution A and B are prepared by
∆Tb k b ×1 3 1 k b 1
dissolving 1 g of non-volatile solutes X and Y.
=
= = = ×
respectively in 1 kg of water. The ratio of
∆Tf k f × 2 6 2 k f 2
depression in freezing points for A and B is
kb 1
found to be 1 : 4. The ratio of molar masses of
= ⇒ x =1
X and Y is:
kf 1
(a) 1 : 4
(b) 1 : 0.25
313. The freezing point of equimolal aqueous
(c) 1 : 0.20
(d) 1 : 5
solution will be highest for
JEE Main 25.07.2022, Shift-II
+
−
(a) C6 H 5 − NH N H 3 Cl (b) Ba(NO3)2
Ans. (b) :
∵
∆Tf = Kf. m
(c) LaCl3
(d) C6H12O6
TS-EAMCET-20.07.2022, Shift-I
Ans. (d) : Higher the value of i, leaser will be the
∴
freezing point. Glucose has not least value of i, or it is
non-electrolyte so has the highest freezing point.
∆Tfx 1/ M x /1
=
∴
314. The freezing point depression of a solution
∆Tfy 1/ M y /1
containing 0.6 g of urea (molar mass = 60 g
Given :mol-1) in 100g of benzene (in K) is (Kf of
∆Tfx 1
CH3COOH = 4.0 K Kg mol-1)
=
(a) 0.30
(b) 0.58
∆Tfy 4
(c) 0.40
(d) 0.24
1 My
AP-EAMCET-06.07.2022, Shift-II
∴
=
4 Mx
Ans. (c) : Given that, Mass of urea = 6 g
∴
Mx : My = 1: 0.25
Molar mass of urea = 60 g
311. The depression in freezing point observed for a
Mass of benzene = 100 g
formic acid solution of concentration 0.5 mL
Kf = 4.0 kgmol–1
L−1 is 0.0405oC. density of formic acid is 1.05 g We know that
mL−1. The Van’t Hoff factor of the formic acid
∆Tf = Kf × m
solution is nearly: (Given for water kf = 1.86 K Where,
−1
kg mol )
m = molality
(a) 0.8
(b) 1.1
Mole solute
(c) 1.9
(d) 2.4
=
kg solvent
JEE Main 25.07.2022, Shift-I
Ans. (c) : Given that,
kf × mole of soluble
∆Tf =
Conc. of formic acid [HCOOH] = 0.5 mL
kg of solvent
−1
L ∆Tf = 0.0405
.01
.6
Density of formic acid = 1.05 gm/mL
4×
= 0.40 ∴Mole of soluble =
= 0.01
.1
60
Kf = 1.86 K kg mol−1
Objective Chemistry Volume-II
72
YCT
315. 0.05 mole of a non-volatile solute is dissolved in
500 g of water. What is the depression in
freezing point of resultant solution? (Kf (H2O) =
1.86 k kg mol-1)
(a) 0.047 K
(b) 0.372 K
(c) 0.093 K
(d) 0.186 K
AP-EAMCET-05.07.2022, Shift-I
Ans. (d) : Given data, mole = 0.05, W (solvent) = 500 g
Tf = ?, Kf = 1.86 Kg mo–1
Now, Tf = Kf × m
0.05
Tf = 1.86 ×
×1000
500
Tf = 0.186 K
316. At T (K) x g of a non-volatile solid (molar mass
78 g mol–1) when added to 0.5 kg water,
lowered its freezing point by 1.0°C. What is x
(in g)? (Kf of water at T(K) = 1.86 K Kg mol–1)
(a) 10.48
(b) 20.96
(c) 41.92
(d) 5.24
AP-EAMCET-04.07.2022, Shift-I
Ans. (b) : Given that, W1= w g
W2 = 0.5 kg = 500g
M1 = 78g mol–1
kf = 1.86 k kg mol–1
∆Tf = 1°C
we know that
w × 1000
∆ T f = Kf × l
W2 M 2
x × 1000
1 = 1.86 ×
500 × 78
x = 20.96 g
317. 1.8 g of glucose (molar mass 180 g mol–1) is
dissolved in 0.1 kg of water. The freezing point
of the solution (in ºC) is
(Kf for water = 1.86 K kg mol–1)
(a) +0.186
(b) –0.372
(c) –0.186
(d) +0.372
AP-EAMCET-07.07.2022, Shift-I
Ans. (c) : Given, Mass of solute (glucose) = 1.8 g
Molar Mass of solute = 180 mol–1
Mass of solvent = 0.1 kg
Kf for water = 1.86 K kg mol–1
∆Tf = Kf × m
Where m is modality
Mass of solute
1
Molality =
×
Molar Mass of solute Mass of solution in Kg
1.8
1
×
= 0.1m
180 0.1
∆Tf = 1.86 × 0.1
= 0.186 K
∆Tf = To – Tf [To = 273.15K
0.186 = 273.15K–Tf
Tf = 273.15 –0.186
= 272.964K
Tf = 272.964 K
Tf = – 0.186 °C
=
Objective Chemistry Volume-II
318. Which of the following solution has the highest
freezing point?
(a) 0.1 mol KCℓ in 1 kg water
(b) 0.1 mol K 2SO 4 in 1 kg water
(c) 0.1 mol Urea in 1 kg water
(d) 30 g of glucose in 1 kg water
AP-EAMCET-06.07.2022, Shift-I
Ans. (c) : Freezing point will be given by formula ∆Tf
= iKf m
(a) 0.1 mol KCl in 1 kg water
KCl → K+ + Cl–
Thus, i = 2
(b) 0.1 mol K2SO4 in 1 kg water
Na2SO4 → 2Na+ + SO 24−
Thus, i = 3
(c) 0.1 mol urea in 1 kg water
urea is a non -electrolyte
Thus, i = 1
∆Tf = i × Kf × m = I × Kf × 0.1 = 0.1Kf
(d) 30g glucose in 1 kg water
∆Tf = i × Kf × m
glucose is non-electrolyte i = 1
30 /180
∆Tf = i × Kf ×
= 0.166 K f
1
The highest freezing point will be of the solution having
the lowest ∆Tf value. Thus, the highest freezing point of
option (c) because it, give lowest ∆Tf vlue.
319. Which of the following is not a colligative
property?
(a) Osmotic pressure
(b) Optical activity
(c) Depression in freezing point
(d) Elevation in boiling point
Karnataka-CET-2016, 2013
MHT CET-2009, UPTU/UPSEE-2009
AMU-2007, J & K CET-2003
AIIMS-2001, UP CPMT-2001
Ans. (b) : Certain properties of an solution depend only
on the number of particles of the solute in a definite
amount of the solvent and do not depend on the nature
of solute. Such properties are called colligative
properties. Osmotic pressure, lowering of vapour
pressure, elevation in boiling point and depression in
freezing point are some colligative properties. Hence,
the one which is not colligative property is optical
activity.
320. An aqueous solution freezes at – 0.186ºC, then
elevation in boiling point is :
(Kb = 0.512, Kf = 1.86)
(a) 0.0512ºC
(b) 100.0512ºC
(c) – 0.0512ºC
(d) None of these
Manipal-2019
AIEEE-2002, BCECE-2018
Ans. (a) : Given that–
Tf (freezing point of an aqueous solution = –0.186°C)
∴ ∆Tf = Tf° − Tf = 0 − (−0.186°C)
∆Tf = 0.186°C
73
YCT
Since, ∆Tf = Kf × molality
∆Tf
molality (m) =
Kf
it is known that , ∆Tb = Kb × m
So,
∆Tf
Kf
0.816
= 0.512 ×
1.86
= 0.0512°C
321. If the elevation in boiling point of a solution of
10g of solute (mol. wt.=100) in 100g of water is
∆Tb, the ebullioscopic constant of water is
(a) 10
(b) 100 Tb
∆Tb
(c) ∆Tb
(d)
10
(e) 10Tb
Kerala-CEE-2007, BITSAT-2012
Ans. (c) : We know that–
∆Tb = m.K b
1000× w
∆Tb =
×K b
W×M
where, w = weight of solute
W = weight of solvent
Kb = ebullioscopic constant of water
∆Tb = elevation in boiling point
M = molar mass of solute
1000×10
∆Tb =
×Kb
100×100
∆Tb = K b
∆Tb = 0.512 ×
322. Tyndall effect shown by colloids is due to:
(a) scattering of light by the particles
(b) movements of particles
(c) reflection of light by the particles
(d) coagulation of particles
UPTU/UPSEE-2005, UP CPMT-2010
Ans. (a) : Tyndall effect is due to scattering of light by
colloidal particles. This effect can be observed in
theaters when a beam of light is thrown on the screen
we observe the path of light clearly visible because dust
particles in air scatter light.
323. A solution of urea (mol. Mass 56g mol-1) boils
at 100.18oC at the atmospheric pressure. If kf
and kb for water are 1.86 and 0.512 k kg mol-1
respectively, the above solution will freeze at
(a) – 6.54o C
(b) 6.54o C
(c) 0.654o C
(d) – 0.654o C
AIIMS-2015, JIPMER-2007
Ans. (d) : Given that–
Molecular mass of urea = 56 g mol–1
Boiling point = 100.18°C
Kf for water = 1.86 K kg mol–1
Kb for water = 0.512 K kg mol–1
From formula,
∆Tf = Kf × molality of solution and
∆Tf = Kb × molality of solution
Objective Chemistry Volume-II
∆Tf K f
=
∆Tb K b
∆Tf
1.86
=
0.18 0.512
1.86 × 0.18
or ∆Tf =
0.512
= 0.654
Now,
∆Tf = T1 – T2
0.654 = 0° C – T2
∴T2 = – 0.654°C
324. Gold sol is not a
(a) lyophobic sol
(b) negatively charged sol
(c) macromolecular sol
(d) multimolecular colloid
Karnataka-CET-2018, 2014
Ans. (c) : Gold sol is not a macromolecular sol colloid.
Gold sol contains a large number of particles so it is
multimolecular, lyphobic and negatively charged sol.
325. During the depression in freezing point
experiment, an equilibrium is established
between the molecules of
(a) liquid solvent and solid solvent
(b) liquid solute and solid solvent
(c) liquid solute and solid solute
(d) liquid solvent and solid solute
AP EAMCET (Engg.)-2009
VITEEE- 2009
Ans. (a) : Freezing point of a substance is the
temperature at which the solid and the liquid forms of
the substance are in equilibrium. During the depression
in freezing point experiment, an equilibrium is
established between the molecules of liquid solvent and
solid solvent.
326. Which among the following is a colligative
property?
(a) Surface tension
(b) Osmotic pressure
(c) Optical rotation
(d) Viscosity
AP EAPCET 24.08.2021 Shift-II
CG PET-2006
Ans. (b) : Colligative properties of solutions are
properties that depend upon the concentration of solute
molecules or ions. Colligative properties include vapour
pressure lowering, boiling point elevation, freezing
point depression and osmotic pressure.
327. Assertion: If one component of a solution obeys
Raoult’s law over a certain range of
composition, the other component will not obey
Henry’s law in the range.
Reason: Raoult’s law is a special case of
Henry’s law
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
74
YCT
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are
incorrect.
[AIIMS-2017, 2013, 2011]
Ans. (b): If one compound obeys Henry's law, then the
other component must obey Raoult's law over the same
composition range. Conversely if one component obeys
Raoult's law over a given composition range, the other
component must be obey Henry's law over the same
range. Raoult's law is in general only obeyed by
component A in the limiting case when the solution
consists practically entirely of component A, so that
component B, being present in very small amount,
obeys Henry's law.
Yes the Raout's law special case of Henry's law.
328. In a 0.2 molal aqueous solution of a weak acid
HX the degree of ionization is 0.3. Taking Kf
for water as 1.85, the freezing point of the
solution will be nearest to
(a) –0.360oC
(b) –0.260oC
o
(c) +0.480 C
(d) –0.480oC
BITSAT-2008, AIEEE-2003
Ans. (d) : Given,
α = 0.3, m = 0.2, Kf = 1.85
HX ↽ ⇀ H+ +X–
i = 1 + α = 1.3
∆Tf = molality × Kf × i
= 0.2 × 1.85 × 1.3
= 0.481º
∴Freezing point = –0.481ºC ≈ –0.480 °C
329. Which colligative property is more useful to
determine the molecular weight of the
substances like proteins and polymers?
(a) Lowering of vapour pressure
(b) Elevation in boiling point
(c) Depression of freezing point
(d) Osmotic pressure
GUJCET-2016, 2015
Ans. (d) : Osmotic pressure is the best method for the
determination of molar mass of proteins and polymers.
w RT
M2 = 2
ΠV
Where,
Π = osmotic pressure
R = gas constant
V = Volume of the solution in (l)
M2 = Molar mass
w2 = weight of solute
330. C6H6 freezes at 5.5°C. The temperature at
which a solution 10 g of C4H10 in 200 g of C6H6
freeze is .........°C. (The molal freezing point
depression constant of C6H6 is 5.12°C/m.)
[JEE Main 2021, 24 Feb Shift-II]
Ans. (1) : Given data –
Freezing point of C6H6 = 5.5 °C
Mass of C4H10 = 10 g
Molecular mass of C6H6 = 200 g
Tk(solvent) = 5.5°C
Objective Chemistry Volume-II
Tf(solution) = ?
From formula –
∆Tf = i × Kf × m
1× 5.12 × 10
∆Tf =
58200 × 1000
5.12 × 50
∆Tf =
58
= 4.414
We know that–
Tf = TK – ∆Tf
= 5.5 – 4.414
= 1.086°C ≃ 1°C
331. When 3.00 g of a substance 'X' is dissolved in
100 g of CCl4, it raises the boiling point by 0.60
K. the molar mass of the substance 'X' is ........ g
mol–1. (Nearest integer)
[Given, Kb for CCl4 is 5.0 K kg mol–1]
[JEE Main 2021, 25 July Shift-II]
Ans. (250) : Given that–
Mass of substance 'X' (solute) = 3 g
Mass of CCl4 (solvent) = 100 g
Difference in boiling point (∆Tb) = 0.60 K
Kb for CCl4 = 5.0 K kg mol–1
From formula–
∆Tb = iKb.m where i = 1
n (solute ) × 1000
0.6 = 1 × 5 ×
w (solvent )
0.6 = 5 ×
3 × 1000
M x × 100
150
0.6
= 250g/mol
332. Which one of the following 0.06 M aqueous
solutions has lowest freezing point?
(a) Al2(SO4)3
(b) C6H12O6
(c) KI
(d) K2SO4
[JEE Main 2021, 22 July Shift-II]
Ans. (a) : ∆Tf = i × Kf × m or ∆Tf ∝ i
Van't hoff factor 'i' for the given compound areKCl: 2; C6H12O6: 1; Al2 (SO4)3: 5; K2SO4:3
Hence, Al2(SO4)3 will show the largest depression in
freezing point.
333. A solute a dimerises in water. The boiling point
of a 2 molar solution of A is 100.52°C.
The percentage association of A is ...... (Round
off to the nearest integer)
[Use : Kb for water = 0.52 K kg mol–1, boiling
point of water = 100°C]
[JEE Main 2021, 18 March Shift-II]
Ans. (100) :
∆Tb = i × Kb × m
0.52 = i × 0.52 × 2
1
i=
2
75
Mx =
YCT
We know that–
1
i = 1 + − 1 α
n
n = 2 for dimerization
1
1
∴
= 1 + − 1 α
2
2
α =1
Now
α = 1 × 100 = 100
334. 2 molal solution of a weak acid HA has a
freezing point of 3.885°C. The degree of
dissociation of this acid is ......... × 10–3. (Round
off to the nearest integer).
[Given : Molal depression constant of water =
1.85 K kg mol–1, freezing point of pure water =
0°C]
[JEE Main 2021, 18 March Shift-II]
Ans. (50) : Given that–
Molality of solution = 2
Freezing point of weak acid HA = 3.885 °C
Kf for water = 1.85 K kg mol–1
∆Tf = i × Kf × m
3.885 = i × 1.85 × 2
3.885
i=
1.85 × 2
i = 1.05
Now, 1.05 = 1 + (2 – 1) α
1.05 = 1 + α
α = 1.05 – 1
α = 0.05
α = 50 × 10–3
335. 40g of glucose (Molar mass = 180) is mixed
with 200 mL of water. The freezing point of
solution is............. K. (Nearest integer)
[Given, Kf = 1.86K kg mol–1
density of water = 1.00 g cm–3
freezing point of water = 273.15 K]
[JEE Main 2021, 27 Aug Shift-II]
Ans. (271) : Given data–
Mass of solute = 40 g
Molecular mass of solute = 180 g
Mass of solvent = 200 mL
Tf = 273.15k
Tf' = ?
From formula
∆Tf = K f m
40 × 1000
∆Tf = Tf – Tf' = 1.86 ×
180 × 200
Tf' = 273.15 – 2.06
= 271.09K.
336. 1 kg of 0.75 molal aqueous solution of sucrose
can be cooled up to –4°C before freezing. The
amount of ice (in g) that will be separated out is
......... (Nearest integer)
[Given, Kf (H2O) = 1.86 K kg mol–1]
[JEE Main 2021, 27 Aug Shift-I]
Objective Chemistry Volume-II
Ans. (518) : Molar mass of sucrose = 342 g mol–1
So, Mass of sucrose per kg of solvent = 0.75× 342
= 256.5g
Mass of sucrose in 1 kg of solution
256.5
=
×1000
1000 + 256.5
= 204.14g
Mass of solvent = 1000 – 204.14
= 795.86g
204.14 1000
∆Tf = i × Kf × m = 1× 1.86 ×
×
342
W
1.86× 204.14×1000
4=
342× W
W = 277.55 g
Amount of ice will be separated = 795.86 – 277.55
= 518.31g
337. The following solutions were prepared by
dissolving 10 g of glucose (C6H12O6) in 250 ml
of water (P1). 10 g of urea (CH4N2O) in 250 ml
of water (P2) and 10g of sucrose (C12H22O11) in
250 ml of water (P3)
The right option for the decreasing order of
osmotic pressure of these solutions is.
(a) P3> P1> P2
(b) P2> P1> P3
(c) P1> P2> P3
(d) P2> P3> P1
(NEET-2021)
Ans. (b) : Osmotic pressure = i CRT Higher will be the
molar mass of solute, smaller will be molar
concentration and smaller will be osmotic pressure.
Decreasing order = sucrose > Glucose >urea.
338. 1.22 g of an organic acid is separately dissolved in
100 g of benzene (Kb = 2.6 K kg mol–1) and 100 g
of acetone (Kb = 1.7 K kg mol–1). The acid is
known to dimerise in benzene but remain as a
monomer in acetone. The boiling point of the
solution in acetone increases by 0.17°C. The
increase in boiling point of solution in benzene in
°C is x × 10–2. The value of x is ....... (Nearest
integer) [Atomic mass : C = 12.0, H = 1.0, 0=16.0]
[JEE Main 2021, 31 Aug Shift-II]
Ans. (13) : Benzene as a solvent–
From formula–
∆Tb = iKbm
1
1.22 Mw
= × 2.6 ×
×
.....(i)
2
100 1000
Now,
Acetone as solvent–
1.22 × Mw
∆Tb = 1 × 1.77 ×
100 × 1000
1.22 × Mw
0.17 = 1× 1.7 ×
…….(ii)
100 × 1000
Dividing equation (i) by (ii) we get–
0.26
∆Tb =
= 0.13
2
∆Tb = 13×10–2
So,
x = 13
76
YCT
339. Which one of the following 0.10 M aqueous 342. 1 molal aqueous solution of an electrolyte A2B3
solutions will exhibit the largest freezing point
is 60% ionised. The boiling point of the solution
depression?
at 1 atm is _______K. (Rounded-off to the
(a) Hydrazine
(b) Glucose
nearest integer)
(c) Glycine
(d) KHSO4
[Given Kb for (H2O) = 0.52 kg mol–1]
[JEE Main 2021, 31 Aug Shift-II]
JEE Main 25-02-2021, Shift-I
Ans. (d) : We know that –
∆Tf = i × K f × m
∆Tf ∝ i
Here –
∆Tf = Depression in freezing point
Kf = Molar depression constant
m = molality
i = van't hoff factor
van't of hoff factor ' i ' for the given compounds are –
Hydrazine (NH2–NH2) : 1; Glucose (C6H12O6) : 1 ;
Glycine (H2N–CH2–COOH) : 1 ; KHSO4 : 2
Hence, KHSO4 will so the largest depression in freezing
point.
340. When 12.2 g of benzoic acid is dissolved in 100g
of water, the freezing point of solution was
found to be –0.93°C (Kf (H2O) = 1.86 K kg mol–
1
). The number (n) of benzoic acid molecules
associated (assuming 100% association) is
[JEE Main 2021, 26 Feb Shift-II]
Ans. (2) : Given that,
Mass of benzoic acid = 12.2g
mass of water = 100 g
freezing point of solution = – 0.93 °C
Kf (H2O) 1.8 K kg mol–1
According to freezing point depression is –
∆Tf = i × Kf × m
12.2
0 – (– 0.93) = i × 1.86 ×
× 1000
122 × 100
0.93
i=
= 0.5
1.86
Now,
1
i = 1 + – 1 α
x
1
1
= 1 + – 1 × 1
2
x
x=2
341. 83 g of ethylene glycol dissolved in 625 g of
water. The freezing point of the solution is ........
K. (Nearest integer)
[Use, molal freezing point depression constant
of water = 1.86 K kg mol–1, Freezing point of
water = 273 K and Atomic masses : C = 12.0 u,
O = 16.0 u, H = 1.0 u]
[JEE Main 2021, 26 Aug Shift-II]
Ans. (269) : We know that–
∆Tf = Tf° – Tf (s) = i × K f × m.
83 × 1000
273.15 – Tf(s) = 1 × 1.86 ×
62 × 625g
273.15 – Tf(s) = 3.984
Tf(s) = 269.166 K
≃ 269 K
Objective Chemistry Volume-II
Ans. : Reaction is given by, A 2 B3 ⇌ 2A 3+ + 3B2–
No. of ions = 5
⇒ i = 1 + (n–1) α
⇒ i = 1 + (5–1) × 0.6
⇒ i = 1 + 4 × 0.6
⇒ i = 3.4
We know that, ∆Tb = i Kbm
where, Kb = boiling point elevation constant or
molal elevation constant/Ebullioscopic constant.
m = molality of solution
∆Tb = Tb - Tb° is elevation of boiling point.
Tb° = boiling point of pure solvent
Tb = boiling point of solution.
∴ ∆Tb = i × m × Kb
= 3.4 × 1 × 0.52
= 1.768 ºC
( )
⇒ ∆Tb = (Tb) solution – Tb°
puresolvent
1.768 = (Tb)solution – 100
(Tb)solution = 101.768 ºC
= 374.768K ≈ 375 K
343. Which one of the following graphs correctly
represents change in freezing point as a
function of solute concentration?
(a)
(b)
(c)
(d)
TS-EAMCET 09.08.2021, Shift-I
Ans. (b) :
Graph represents change in freezing point as a function
of solute concentration.
344. On mixing urea, the boiling point of H2O
changed to 100.5oC. Calculated the freezing
point of the solution, if Kf of water is 1.87 K.
Kg.mol–1 and Kb of water is 0.52 K.kg.mol–1.
(a) –1oC
(b) –0.5oC
o
(c) –1.8 C
(d) 0oC
TS EAMCET 05.08.2021, Shift-I
77
YCT
Ans. (c):
Given that, Boiling point of solution = 100.5oC
Kb = 0.52K kg mol–1, Kf = 1.87K kg mol–1
∆Tb = 100.5 – 100oC = 0.5
∆T = Kb × molality
0.5
Molality =
0.52
∆TFreezing = Kf × molality
0.5
= 1.87 ×
0.52
= 1.8 oC
Freezing point of solution = 0 °C – 1.8 oC
= –1.8 oC
345. Match of following
(A) Ebullioscopic
(I) Depression of freezing
constant
point
(B) Cryosopic
(II) Total pressure is the
sum of partial
constant
pressures
of the
components
(C) Henry's law
(III) Elevation of boiling
point
(D) Dalton's law
(IV) Solubility of a gas in
liquid
The correct match is
A
B
C
D
(a) III I
II
IV
(b) I
III
II
IV
(c) III I
IV II
(d) I
III
IV II
TS EAMCET 05.08.2021, Shift-I
Ans. (c) :
Ebullioscopic constant
Elevation of boiling point
Cryosopic constant
Depression of freezing
point
Henry's law
Solubility of a gas in liquid
Dalton's law
Total pressure is the sum
of partial pressures of the
components.
346. AB2 is 10% dissociated in water to A2+ and B–.
The boiling point of a 10.0 molal aqueous
solution of AB2 is_______ 0C. (Round off to the
Nearest Integer).
[Given : Molal elevation constant of water
Kb= 0.5 K kg mol-1, boiling point of pure water
= 1000C]
JEE Main 16.03.2021, Shift-I
Ans. (106) : Given that,
∆Tb = i × m × K b
α =0.1, m=10, K b = 0.05K kg mol−1
The dissociation of AB2 is :
AB2 ↽ ⇀ A 2+ + 2B−
t=0
t=T
a
a − aα
0
aα
0
2aα
Total mole ( n ) = a − aα + aα + 2aα = a (1 + 2α )
Objective Chemistry Volume-II
∴ i = (1+2α )
Now,
∆Tb = i × m × K b
∆Tb = (1 + 2 × 0.1) × 10 × 0.5
∆Tb = 6 °C
∵∆Tb = Tb − Tbo
or Tb = ∆Tb + Tbo
Tb = 6 + 100 = 106°C
So, boiling point = 106 °C
347. During which of the following processes, does
entropy decrease?
(a) Freezing of water to ice at 0°C
(b) Freezing of water to ice at –10°C
(c) N 2 (g) + 3H 2 (g) → 2NH 3 (g)
(d) Adsorption of CO(g) on lead surface.
(e) Dissolution of NaCl in water
Choose the correct answer from the options
given below:
(a) (a),(b),(c) and (d) only
(b) (a), (c) and (e) only
(c) (a) and (e) only
(d) (b) and (c) only
JEE Main 17.03.2021, Shift-II
Ans. (a) :The entropy decreases during the formation of
ice from water at 0°C i.e.
0° C
Water →
ice, ∆S = –ve
(b) At –10°C, the conversion of water into ice.
∆S having negative value.
–10° C
Water
→ ice, ∆S = –ve
(c) The formation of ammonia by Haber's process
which is given asN 2 (g) + 3N 2 (g) ↽ ⇀ 2NH 3 (g), ∆S = –ve
(d) Adsorption of CO(g) at lead surface decrease the
entropy of the system.
(e) Dissolution of NaCl in water increases the entropy.
So, in the process a, b, c, d entropy decrease.
348. At 298 K, the enthalpy of fusion of a solid (X) is
2.8 kJ mol–1 and the enthalpy of vaporisation of
the liquid (X) is 98.2 kJ mol–1. The enthalpy of
sublimation of the substance (X) in kJ mol–1 is
............ . (Nearest integer)
[JEE Main 2021, 25 July Shift-I]
Ans. 101
Given that,
∆Hfus = 2.8 kJ mol–1
∆Hvap = 98.2 KJ mol–1
∆Hsub = ∆Hfus + ∆Hvap
= 2.8 + 98.2
= 101 KJ/mol
349. 2 g of a non-electrolyte solute (molar mass is
500 g mol-1) was dissolved in 57.3 g of xylene. If
the freezing point depression constant Kf of
Xylene is 4.3 K kg mol-1 . Then the depression
in freezing point of Xylene is ____
(a) 57.3 k
(b) 0.3 k
(c) 4.3 k
(d) 0.002 k
AP EAPCET 24.08.2021 Shift-II
78
YCT
Ans. (a) : We have formula of elevation constant (Kb)
and elevation of boiling point (∆Tb) –
∆Tb= i Kb m
For binary electrolyte – i.e.
i=2
∆Tb= 2 Kb × m
or
∆Tb
10 × 1000
Kb =
∵ m=
2m
100 ×100
m=1
Ans. (b): Given that,
Weight of solute (W1) = 2 gm
Weight of solvent (W2) = 57.3 gm
Molar mass of solute (M1) = 500 g mol-1)
Freezing point depression const (Kf ) = 4.3 Kkg mol-1
∆Tf = ?
∆Tf = Kf ×
W1 × 1000
W2 × M1
2 × 1000
57.3 × 500
∆Tb
2
∆Tf = 0.3k
353. Which of the following aqueous solution has
highest boiling point? (Assume identical
350. Which of the following will form an ideal
conditions)
solution?
(a) 1% Glucose
(b) 1% Sucrose
(a) C2H5OH & H2O
(c)
1%
Urea
(d)
1% Phenol
(b) HNO3 & H2O
TS EAMCET 10.08.2021, Shift-I
(c) CHCl3 & CH3COCH3
Ans. (c) : ∆Tb = mk b
(d) C6H6 & C6H5CH3
AP EAPCET 19-08-2021 Shift-I Elevation in boiling point is directly proportional to
Ans. (d) : An ideal solution or ideal mixture is a molality.
Mole
Molality (m)
solution that exhibits thermodynamic properties Species
1
1/ 80
analogous to those of a mixture of ideal gases. Benzene
= 0.126 m
(C6H6) and C6H5CH3 (Toluene) will form an ideal 1% glucose
180
99 × 10−3
solution over the entire range of composition. The
1
1/ 342
bonding interactions among the molecules will be same 1% sucrose
= 0.088m
342
33
× 10−3
before and after mixing. Other species given in the
option are not form ideal solution because they deviate
1
1/ 60
1% Urea
= 0.168m
from the Raoult's Law.
60
99 ×10−3
351. If a solution of maltose (molecular mass = 342.3
1/ 94
1/ 94
= 0.107 m
g.mol-1) was prepared by dissolving 72.4 g of 1% Phenol
99 ×10−3
maltose in 1000 g of ethanol then depression in
So, urea will have more boiling point.
freezing point of ethanol is ______ K.
-1
354.
Calculate the molal depression constant of a
Given for ethanol. Kf = 1.23 K kg mol and
solution, which freezes at 15°C. The latent heat
molecular mass = 46.07g.mol-1
of fusion is 180.7 Jg–1
(a) 0.26
(b) 272.74
(a) 3.81 K molal–1
(b) 0.381 K molal–1
(c) 46.07
(d) 72.40
–1
(c) 1.90 K molal
(d) 0.19 K molal–1
AP EAPCET-6 Sep. 2021, Shift-II
TS-EAMCET (Engg.), 07.08.2021 Shift-II
Ans. (a) : The depression in freezing point
Ans. (a) : Given data,
∆Tf = m × K f
Tf = 15°C + 273 = 288K
Lf = 180.7 J/g, R = 8.314 JK–1 mol–1
K f × w B × 1000
Molal depression constant–
∆Tf =
M B × WA
RTf2
Kf =
1.23 × 72.4 × 1000
1000 × Lf
=
342.3 × 1000
Where, Kf = Molal depression constant
R = Gas constant
89052
=
Tf = Freezing point of the solution
342300
And
Lf = Latent heat of fusion.
∆Tf = 0.26 K
8.314 × (288) 2
Kf =
352. The elevation in boiling point of a solution of 10 ∴
1000 ×180.7
g of a binary electrolyte (of molecular mass
Kf = 3.81K molal–1
100) in 100 g of water is ∆Tb. Then the value of
355. Which condition is not satisfied by an ideal
Kb water is ______.
solution?
∆Tb
(a) Both ∆mix H = 0 and ∆mix S = 0
(a)
(b) 10
2
(b) Obeyance on Raoult's law
∆Tb
(c) Both ∆mix H = 0 and ∆mix V = 0
(c) 10∆Tb
(d)
(d) ∆mix H = 0
10
AP-EAPCET 19.08.2021, Shift-II
AP EAPCET 24.08.2021, Shift-I
∆Tf = 4.3 ×
Objective Chemistry Volume-II
∴ Kb =
79
YCT
358. The freezing point depression constant (Kf) of
benzene is 5.12 k kg mol–1. The freezing point
depression for the solution of molality 0.078 m
containing a non-electrolyte solute in benzene is
(rounded off upto two decimal places)
(a) 0.20k
(b) 0.80k
(c) 0.40k
(d) 0.60k
(NEET-2020)
Ans. (c) : We know that,
∆Tf = i Kf m
Where,
Kf = freezing point depression constant
m = molality
∆Tf = 1 × 5.12 × 0.078 (because i = 1)
= 0.399
≈ 0.40K
359. Function of potassium ethyl xanthate in froth
floatation process is to make the ore
(a) lighter
(b) hydrophobic
K b = 0.512 K kg mol−1
(c) hydrophilic
(d) heavier
Karnataka-CET-2020
∆Tb = K b × molality
...( i )
Ans. (b) : Hydrophobic compounds do not dissolve
∆Tf = K f × molality
...( ii )
easily in water, and they are usually non-polar. The
function of potassium ethyl xanthate in froth floatation
°
Tb – Tb = Kb m = ∆Tb
process is to make the are hydrophobic.
100.20 °C – 100 °C = 0.20 °C
360. The elevation of boiling point of 0.10 m
From (i) and (ii) we get–
aqueous CrCl3 . xNH3 solution is two times that
of 0.05 m aqueous CaCl2 solution. The value of
∆Tb K b
=
x is.......
∆Tf K f
[Assume 100% ionisation of the complex and
CaCl2, coordination number of Cr as 6, and
K
1.86
∆Tf = ∆Tb f = 0.20 ×
or
that all NH3 molecules are present inside the
0.512
Kb
coordination sphere]
[JEE Main 2020, 6 Sep Shift-II]
∆Tf = 0.726º C
Ans.
(5.00)
:
Given,
Freezing point (Tf)= 0 – 0.726ºC = – 0.726ºC.
Molality = 0.05m (CaCl2 solution)
357. Kf (water)=1.86 K kg mol-1 The temperature at From formula
which ice begins to separate from a mixture of
∆Tb = iKbm = 3 × Kb × 0.05
10 mass % ethylene glycol is
= 0.15 Kb…..(i)
o
o
(a) –1.86 C
(b) – 3.72 C
Molality of CrCl3.xNH3 = 0.10m from formula,
o
o
(c) –3.3 C
(d) –3 C
∆Tb = i Kb × 0.10
WB-JEE-2020 Now, It is given that
Ans. (c) : Given that–
∆Tb = 2∆Tb
Kf = 1.86 k kg mol–1
iKb × 0.10 = 0.15 Kb
We know that–
i=3
Coordination Number of Cr = 6
∆Tf = Kf × m
Now,
mole
molality (m) =
from the reaction.
solvent(in kg)
–
[Cr(NH3)5 Cl] Cl2 → [Cr(NH3)5Cl]2+ + Cl
10 1000
x=5
∆Tf = 1.86 ×
62 90
361. How much amount of NaCl should be added to
600g of water (ρ = 1.00 g/mL) to decrease the
= 3.3oC
freezing point of water to –0.2°C? ...... (The
Since the freezing point of ice is 0°C.
freezing point depression constant for water =
freezing point at which ice becomes to seprate
2 K kg mol–1)
∆Tf = 3.3°C
[JEE Main 2020, 9 Jan Shift-I]
Tf° – Tf = 3.3°C
Ans. (1.76) : Given that,
0 – Tf = 3.3°C
∆Tf = 0.2 i = 2
Tf = –3.3°C
∆Tf = i (Kf . m)
Ans. (a) : The condition,
∆mix H = 0 and ∆mix S = 0 is not satisfy the ideal solution
because an ideal solution follows;
• Volume change (∆V) of mixing should be zero. i.e.
∆mix V = 0
• Heat change (∆H) of mixing should be zero. i.e. ∆mix
H = 0.
• Obeys Raoult's law at every range of concentration.
356. A solution of urea (molar mass 60 g mol-1) boils
at 100.20ºC at the atmospheric pressure. if Kf
and Kb for water are 1.86 and 0.512 K kg mol-1
respectively. The freezing point of the solution
will be ____.
(a) – 0.654º C
(b) + 0.654º C
(c) – 0.726º C
(d) + 0.726º C
AP EAPCET 19.08.2021, Shift-II
Ans. (c) : Given - Kf = 1.86 K kg mol-1, Tf = ?
Objective Chemistry Volume-II
80
YCT
So,
NaCl ↽ ⇀ Na+ +Cl–
0.2 = 2 × 2 ×
w
58.5
600
1000
2 × 585 × 6
4000
w = 1.76 gm
362. 160 g of non-volatile solute 'A' is disolved in 54
mL of water at 373 K. What is the vapour
pressure of aqueous solution of A.
(Given, molecular weight of A = 160 g mol-1)
(a) 760 Torr
(b) 720 Torr
(c) 570 Torr
(d) 450 Torr
AP EAMCET (Engg.) 17.09.2020 Shift-I
w=
Ans. (c) : Raoult's law of vapour pressure =
( pº -p )
pº
[ ∵ pº = vapour pressure of H2O(g) at 373 K = 760 bar
Let, d of water = 1 g/mL
⇒ 54 mL = 54 g water
then,
nA
mole fraction ( χ A ) =
n A + n H 2O
160
760 − p
1
= 160 =
160
54
760
4
+
160 18
p
1
1−
=
760 4
3
⇒ p = × 760 = 570 torr.
4
363. Assertion : Standard boiling point of a liquid is
slightly higher than the normal boiling point.
Reason : 1 bar pressure is slightly less than 1
atm pressure.
(a) Assertion and Reason are correct statements
and reason is the correct explanation for
Assertion.
(b) Assertion and Reason are correct statements
and Reason is not the explanation for
Assertion.
(c) Assertion is correct, Reason is incorrect.
(d) Assertion is incorrect, Reason is correct.
AP EAMCET (Engg.) 18.09.2020, Shift-I
Ans. (d) :
• Assertion is incorrect because standard boiling point
(b.p. at 1 bar pressure) of a liquid is slightly lower
than the normal boiling point (b.p. at 1 atm pressure).
It is because 1 bar pressure is slightly lower than 1
atm pressure.
• Reason is correct because 1 atm = 1.01325 bar.
Objective Chemistry Volume-II
364. Match the item given in Column-I (method
used for determining colligative property) and
Column-II (for the corresponding colligative
property) and find the correct order.
Column I
Column II
A. Beckmann
1. Osmotic pressure
method
B. Ostwald-Walker 2. Elevation in B.P
method
C. Berkeley-Hartley 3. Depression in F.P.
method
D. Landsberger
4. Relative lowering of
method
vapour pressure
Code:
A
B
C
D
(a) 2
4
3
1
(b) 1
4
2
3
(c) 2
3
4
1
(d) 3
4
1
2
AP EAMCET (Engg.) 18.9.2020 Shift-I
Ans. (d) : The correct matching between Column I and
Column II are–
Method
Colligative property
Backmann method
Depression in freezing point
Ostwald-Walker method
Relative lowering of vapour
pressure
Berkeley-Hartley method Osmotic pressure
Landsberger method
Elevation in boiling point
365. What is the density of water vapour at boiling
point of water?
(a) 1×10–4 g cm–3
(b) 1 g cm–3
–4
–3
(c) 6×10 g cm
(d) 4×10–4 g cm–3
MHT CET-02.05.2019, SHIFT-II
Ans. (c) : Water has never absolute density, because
density of water directly belongs to temperature.
So, The density of water vapour at 100°C is 6 × 10–4g
cm–3.
366. A non-volatile solute, 'A' tetramerises in water
to the extent of 80%. 2.5 g of 'A' in 100 g of
water, lower the freezing point by 0.3oC. The
moler mass of 'A' in g is
(Kf for water = 1.86 K kg mol–1)
(a) 62
(b) 221
(c) 155
(d) 354
Karnataka-CET-2019
Ans. (a) We know that–
i × K f × w 2 × 1000
∆Tf =
m 2 × 100
81
α=
1– i
1
1–
4
1– i
1
1–
4
i = 0.4
⇒ 0.8 =
YCT
Given–
w2 = 2.5g
∆Tf = 0.3°C
Kf = 1.86 k.kg mol–1
0.4 × 1.86 × 2.5 ×10
m2 =
0.3
m2 = 62 g mol–1
367. The vapour pressures of pure liquids A and B
are 400 and 600 mmHg, respectively at 298 K.
On mixing the two liquids, the sum of their
initial volumes is equal to the volume of the
final mixture. The mole fraction of liquid B is
0.5 in the mixture. The vapour pressure of the
final solution, the mole fractions of components
A and B in vapour phase, respectively are
(a) 450 mmHg, 0.4, 0.6 (b) 500 mmHg, 0.5, 0.5
(c) 450 mmHg, 0.5, 0.5 (d) 500 mmHg, 0.4,0.6
[JEE Main 2019, 8 April Shift-I]
Ans. (d) : Given that–
Vopour pressure of pure liquid A = 400 mmHg
Vopour pressure of pure liquid B = 600 mmHg
From Partial pressure formula–
PT = X A PA° + X B PB°
PT = 0.5 × 400 + 0.5 × 600 = 500 mmHg
Now,
Mole fraction of A in vapour–
P
0.5 × 400
YA = A =
PT
500
= 0.4
Then, mole fraction of B
YB = 1 – 0.4 = 0.6
368. Elevation in the boiling point for 1 molal
solution of glucose is 2 K. The depression in the
freezing point for 2 molal solution of glucose in
the same solvent is 2 K. The relation between
Kb and Kf is
(a) Kb = 1.5 Kf
(b) Kb = 0.5 Kf
(c) Kb = Kf
(d) Kb = 2 Kf
[JEE Main 2019, 10 Jan Shift-II]
Ans. (d) : Given that –
∆Tb = 2, ∆Tf = 2
m1 = 1, m2 = 2
∆Tb = iK b m1 , ∆Tf = iK f m 2
∆Tb iK b m1
=
∆Tf iK f m 2
2 1× K b
=
2 2 × Kf
iKf × 2 = iKb × 1
Kb = 2Kf (∴ i of glucose = 1)
369. Freezing point of a 4% aqueous solution of X is
equal to freezing point of 12% aqueous solution
of Y. If molecular weight of X is A, then
molecular weight of Y is
(a) 4A
(b) 2A
(c) 3A
(d) A
[JEE Main 2019, 12 Jan Shift-I]
Objective Chemistry Volume-II
Ans. (c) : For same freezing point, molality of both
solution should be same.
Mx = My
4 ×1000 12 ×1000
=
96 × M x
88 × M y
4×1000×88× My = 12 × 1000 × 96 × Mx
96 × M x × 12
or, My =
88 × 4
My = 3.27A ≃ 3A
370. 1 g of a non-volatile, non-electrolyte solute is
dissolved in 100g of two different solvents A
and B, whose ebullioscopic constants are in the
ratio of 1:5. The ratio of the elevation in their
∆T (A)
, is
boiling points, b
∆Tb (B)
(a) 5 : 1
(b) 10 : 1
(c) 1 : 5
(d) 1 : 0.2
[JEE Main 2019, 10 April Shift-II]
Ans. (c) : Given that –
Mass of non-volatile non-electrolyte solute = 1 g
Mass of solvents A & B = 100 g
Ebullioscopic constants ratio = 1 : 5
From formula,
∆Tb = Kb × m
Here, molality of A and B is same.
∆Tb (A) K b (A)
=
as mA = mB
∆Tb (B) K b (B)
=
∆Tb (A) 1
=
∆Tb (B) 5
371. The freezing point of a diluted milk sample is
found to be –0.2°C, while it should have been –
0.5°C for pure milk. How much water has been
added to pure milk to make the diluted
sample?
(a) 2 cups of water to 3 cups of pure milk
(b) 1 cup of water to 3 cups of pure milk
(c) 3 cups of water to 2 cups of pure milk
(d) 1 cup of water to 2 cups of pure milk
[JEE Main 2019, 11 Jan Shift-I]
Ans. (c) : Applying formula for both diluted milk and
pure milk
∆Tf = i × Kf × m
For diluted milk.
W ×1000
0 – (– 0.2) = Kf ×
× 1......(i)
M × W1 (H 2O)
For pure milk.
W × 1000
0 – (0.5) = 1 × Kf ×
......(ii)
M × W2 (H 2 O)
From eq. (i) and (ii) we get–
0.2 W2 (H 2 O)
=
,
0.5 W1 (H 2O)
2 W2 (H 2 O)
=
5 W1 (H 2 O)
82
YCT
372. Molecules of benzoic acid (C6H6COOH)
dimerise in benzene 'w' g of the acid dissolved
in 30 g of benzene shows a depression in
freezing point equal to 2K. If the percentage
association of the acid to form dimer in the
solution is 80, then w is (Given that Kf = 5K kg
mol–1, molar mass of benzoic acid = 122g mol–1)
(a) 1.8 g
(b) 1.0 g
(c) 2.4 g
(d) 1.5 g
[JEE Main 2019, 12 Jan Shift-II]
Ans. (c) : Given that–
i = 1 – 0.8 + 0.4 = 0.6
∆Tf = 2, Kf = 5K kg mol–1
Molar mass of benzoic acid = 122g mol–1
∆Tf = i × Kf × m
W ×1000
2 = 0.6 × 5 ×
122 × 30
W = 2.44g
373. If morality of solution is 0.05 and elevation in
boiling point is 0.16 K then, what is the molal
elevation constant of the solvent?
(a) 3.2
(b) 1.6
(c) 2.2
(d) 2.3
GUJCET-2019
Ans. (a) : Given that–
∆Tb = 0.16
m = 0.05m
∆Tb = Kbm
0.16 = Kb × 0.05
0.16
Kb =
0.05
Kb = 3.2
374. The depression in freezing point of 0.1 M
aqueous solution of HCI. CuSO4 and K2SO4 are
in the ratio
(a) 1 : 1 : 1
(b) 1 : 2 : 3
(c) 1 : 1 : 1.5
(d) 2 : 4 : 3
Assam CEE-2019
Ans. (c) : ∆Tf = i × Kf × m
Thus, ∆Tf ∝ i (as concentration is same for all solution)
i for HCl = 2, i for CuSO4 = 2 , is for K2SO4 = 3
Thus, ratio of depression in tracing point =
2:2:3=1:1:1.5
[Note: If 1 molecule of solute dissociates to given ions,
then i = n]
375. w2 gram of a solute having molar mass M2
dissolve in w1 gram of solvent produces the
depression of freezing point ∆Tf of the solvent if
Kf is the cryoscopic constant of the solvent then
K × w 2 × 1000
(a) ∆Tf = f
M 2 × w1
(b) ∆Tf =
K f × w 2 × 100
M 2 × w1
(c) ∆Tf =
K f w1 ×1000
M2 × w 2
Objective Chemistry Volume-II
(d) ∆Tf =
K f × w1 × 100
M2 × w 2
Assam CEE-2019
Ans. (a) : ∆Tf = Kf M
Molar of solvent = W2/M2
Weights of solvent = W1×10-3 Kg
W2
Morality (m) =
M 2 W1 × 10−3
∆Tf =
K f × W2 ×1000
M 2 W1
376. Assertion: Ideal solutions obey Raoult's law
Reason: ∆Hmix and ∆Vmix for an ideal solution are
less than zero.
(a) If both Assertion and Reason are correct and
Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.
AIIMS 25 May 2019 (Evening)
Ans. (c) : If any solution followed by Raoult's law then
it will be a ideal solution. In ideal solution, SoluteSolute behave same interaction with molecules as well
as solvent shows with solvent.
377. Vapour pressure of CCl4 at 25º C is 143 mm
Hg. 0.5 g of a non-volatile solute (mol. wt. 65) is
dissolved in 100 mL of CCl4. Find the vapour
pressure of the solution. (Density of CCl4 = 1.58
g/cm3)
(a) 141.93 mm Hg
(b) 94.39 mm Hg
(c) 199.34 mm Hg
(d) 143.99 mm Hg
AIIMS 25 May 2019 (Evening)
Ans. (a) : Given that,
Vapour pressure of pure solvent P° = 143 mm
Ps = vapour pressure of solute
WB = weight of solute = 0.5 g
WA = weight of solvent = (100 × 1.58) = 158g
MB = molar mass of solute = 65
MA = molar mass of solvent = 154g/mol
P° − Ps WB M A
=
×
P°
M B WA
Ps = P0 –
WB M A
×
× P0
M B WA
0.5 ×154
× 143
65 × 158
Ps = 141.93mm Hg .
Ps =143 –
378. 0.5 molal aqueous solution of a weak acid (HX)
is 20% ionised. If Kf for water is 1.86 K kg
mol–1, the lowering in freezing point of the
solution is
(a) –1.12 K
(b) 0.56 K
(c) 1.12 K
(d) –0.56 K
UPTU/UPSEE-2018
83
YCT
Ans. (c) : HX →H++ X–
+
HX → H + X
Lasser is the Van't Hoff factor greater will be the
freezing point of solution.
(a) [Co(H2O)6] Cl3
i=4
(b) [Co(H2O)5] Cl2 . H2O
i=3
(c) [Co(H2O)4Cl2] Cl.2H2O i = 2
(d) [Co(H2O)3 Cl3].3H2O i = 1
[Co(H2O)3 Cl3].3H2O this complex will show highest
freezing point.
382. Ethylene glycol is used as an antifreeze to
reduce freezing point of water to –2.4 ºC. What
mass of antifreeze is required for 2 L water?
(Kf water = 1.86 K. kg/mol)
(a) 16 kg
(b) 160 kg
(c) 1.60 kg
(d) 16 g
[AIIMS-27 May, 2018 (E)]
Ans. (b): Given that,
∆Tf = 2.4
−
initial
1
0
0
final
1− α
α
α
From formula
(i − 1)
α=
(dissociation of electrolyte)
(n − 1)
20
α = 20% =
= 0.2
100
(i − 1)
0.2 =
(2 − 1)
i = 1+0.2= 1.2
Given, Kf = 1.86, m = 0.5
∆Tf = i × Kf × m
= 1.2 × 1.86 × 0.5
w water = 2
= 1.12 K
Now,
379. Which one of the following solutions exhibits
∆T × w water × msolute
the maximum elevation in boiling point?
w solute = f
(a) 0.1 m NaCl
(b) 0.1 m FeCl3
Kf
(c) 0.1 m CaCl2
(d) 0.1 m CaCl2
2.4 × 2 × 0.062
=
Manipal-2018
1.86
Ans. (b) : Greater the number of ions greater will be
=
160g
elevation in boiling point. FeCl3 furnishes more number
383. Freezing point of 0.4 m solution in a weak
of ions (i = 3).
monoprotic acid is – 0.1 ºC what is its van’t
380. The correct relation between elevation of
Hoff factor i?
boiling point and molar mass of solute is
(Kf = 1.86 ºC/m)
k .w
K .w
(a) 1.5
(b) 1.6
(a) M 2 = b 2
(b) M 2 = b 1
∆Tb .w1
∆Tb .w 2
(c) 1.34
(d) 1.1
[AIIMS-27 May, 2018 (M)]
∆Tb .K b
∆Tb .w1
(c) M 2 =
(d) M 2 =
Ans.
(c):
The
freezing
point
depression can be seen as –
w1.w 2
K b .w 2
°
∆
T
=
T
−
T
f
f
f
MHT CET-2018
Tf° is the freezing point of the pure solvent.
Tf is the freezing point of the solution.
∆Tf = 0°C − ( −0.1°C ) = 0.1°C
Now by putting all the values in formula we obtain
∆Tf = i.K f .m
Ans. (a) : For dilute solution,
∆Tb ∝ m
∆Tb = Kb m
(where Kb is molal boiling point elevation constant
Now,
K b .w 2
∆Tb =
M 2 × w1
So,
K .w
M2 = b 2
∆Tb .w1
i=
Objective Chemistry Volume-II
0.1°C
1.86°C m × 0.4 molkg −1
i = 1.34
384. When 45 g solute is dissolved in 600 g water,
freezing point is lowered by 2.2K, calculate
molar mass of solute (Kf =1.86 K kg mol-1)
(a) 63.4 g/mol
(b) 80 g/mol
(c) 90 g/mol
(d) 21 g/mol
[AIIMS-26 May, 2018 (M)]
Ans. (a): We know that–
∆Tf = Kf × m
.....(i)
m = molality
Given that, ∆Tf = 2.2, Kf = 1.86
i=
381. For 1 molal aqueous solution of the following
compounds, which one will show the highest
freezing point?
(a) [Co(H2O)6] Cl3
(b) [Co(H2O)5] Cl2 . H2O
(c) [Co(H2O)4Cl2] Cl.2H2O
(d) [Co(H2O)3 Cl3].3H2O
[JEE Main-2018]
Ans. (d) : We know, ∆Tf = i Kf × m
At same molality , the depression in freezing point will
depend on Van't Hoff factor.
∆Tf
K f .m
84
−1
YCT
45 × 1000
Molar mass of solute × 600
Putting the value of m in the equation (i) we get–
∆Tf = 1.86 × m
75
2.2 = 1.86 ×
Molar mass
1.86 × 75
Molar mass of solute =
= 63.4 gm
2.2
385. Assertion: A non volatile solute is added in
liquid solvent the freezing point of mixture
decreases.
Reason: Vapour pressure decreases by addition
of non volatile solute, so equilibrium point
where V.P. of solid and V.P. of liquid are equal
can reach at lower temp.
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.
[AIIMS-27 May, 2018 (M)]
Ans. (a): When a non-volatile solute is added in solvent
liquid then freezing point of mixture decreases.
On adding a non volatile solute in a liquid the vapour
pressure decreases because some molecules of the
solvent on the surface are replaced by the molecules of
the solute.
386. If molality of the dilute solution is doubled, the
value of molal depression constant (Kf)will be
(a) halved
(b) tripled
(c) unchanged
(d) doubled
(NEET-2017)
Ans. (c) : Kf (molal depression constant) is independent
of molality. It is a characteristic of solvent. Hence
molality of the dilute solutionis doubled, the value of
molal depression constant (Kf) unchanged.
387. The freezing point of benzene decreases by
0.45°C when 0.2 g of acetic acid is added to 20 g
of benzene. If acetic acid associates to form a
dimer in benzene, percentage association of
acetic acid in benzene will be (Kf for benzene =
5.12 K kg mol–1)
(a) 64.6%
(b) 80.4%
(c) 74.6%
(d) 94.6%
JEE Main-2017
Ans. (d) : Given that,
∆Tf = 0.45ºC
Kf = 5.12 K kg mol–1
∆Tf = Kf.m.i
w 1000
0.45 = 5.12 × 2 ×
×i
m2
w1
m=
2CH3COOH → (CH3COOH)2
1
0
1–α
α/2
α
i = 1– = 0.527
2
2 (1 – 0.527) = α
α = 0.946
% dissociation = 94.6%
388. The colligative properties of a dilute solution
depend upon
(a) The nature of solute
(b) The diffusion of solvent
(c) The number of particles of solute
(d) The number of particles of solvent.
J & K CET-2017
Ans. (c) : Colligative properties depends only on the
number of the solute particles in a definite amount of
solution not on the nature of the solute.
389. When the pure solvent diffuses out of the
solution
through
the
semi-permeable
membrane then the process is called
(a) sorption
(b) dialysis
(c) reverse osmosis
(d) osmosis
Karnataka-CET-2017
Ans. (c) : When the pure solvent diffuses out of the
solution through the semi-permeable membrane then the
process is called reverse osmosis. In Reverse osmosis
excess pressure which must be applie to a solution to
prevent the passage of solvent in to it through a
semipermeabele membrane.
390. The freezing point of equimolal solution will be
highest for
(a) C6H5NH3Cl
(b) AgNO3
(c) Ca (NO3)2
(d) La(NO3)3
(e) D-fructose
Kerala-CEE-2017
Ans. (e) : Lower the value of i, smaller will be
depression in freezing point, higher will be the freezing
temperature, if molalities are equal.
Thus D–fructose is lower the value of i = 1 hence
freezing point will be highest.
391. Raoult’s law becomes a special case of Henry’s
law when
(a) K H = P1o
(b) K H > P1o
0.45 × 60 × 20
0.2 × 1000 × 5.12
i = 0.527
i=
Objective Chemistry Volume-II
Now,
85
initial
Final
(c) K H < P1o
(d) K H > P1o
BITSAT-2017
Ans. (a) : Due to Raoult’s law;
Vapour pressure of a volatile component
P = P°x …….(i)
By Henry’s law,
Partial solubility of a component,
P = KH x ……(ii)
On compairing equation (i) and (ii)
P o = KH
When KH = Po henry law becomes special case of
raoults law.
YCT
392. Which of the following aqueous solution will
have the boiling point 102.2oC? The molal
elevation constant for water is 2.2 K kg mol–1.
(a) 1m CH3COOH
(b) 1m NaCl
(c) 1M NaCl
(d) 1m glucose
GUJCET-2017
Ans. (d) : Given,
Molal elevation constant Kb = 2.2 K kg mol–1
We know that elevation in boiling point given by
∆Tb = iKbm
= i × 2.2 × m ……(i)
Here,
∆Tb = 102.2 – 100…..(i)
= 2.2° C
Comparing both- equation we get,
i = 1, m = 1
So, option (d) is correct.
393. The depression in freezing point for 0.01 m
aqueous solution of Kx[Fe(CN)6] is 0.0744 K.
the molal depression constant for solvent is 1.86
K kg mol-1. If the solute undergoes complete
dissociation, what is the correct molecular
formula for the solute?
(a) K2[Fe(CN)6]
(b) K3[Fe(CN)6]
(c) K[Fe(CN)6]
(d) K4[(CN)6]
GUJCET-2017
Ans. (b) Depression in freezing point (∆Tf) = i × Kf × m
0.0774 = i× 1.86× 0.01
0.0774
i=
1.86 × 0.01
i = 4.16 ≈ 4
i −1
Degree of dissociation (α) =
n −1
4 −1
1=
(x + 1) − 1
x ≃3
Molecular formula of compound is K3[Fe(CN)6]
394. The solution having lowest freezing point is
(a) 0.1 M potassium chloride
(b) 0.1 M potassium sulphate
(c) 0.1 M potassium nitrate
(d) 0.1 M aluminium sulphate.
COMEDK-2017
Ans. (d) : Depression in freezing point is a colligative
property. Higher the value of 'i' higher will be
depression in freezing point and lower will be the
freezing point of solution.
For KCl,
i=2
KNO3, i = 2
K2SO4 i = 3
Al2(SO4)3, i = 5
Aluminium sulphate has highest value of 'i' among
given option hence (d) is correct.
395. Which observation (s) reflect (s) colligative
properties?
(i) A 0.5 m NaBr solution has a higher vapour
pressure than a 0.5 m BaCl2 solution at the
same temperature.
Objective Chemistry Volume-II
(ii) Pure water freezes at the higher temperature
than pure methanol.
(iii) A 0.1 m NaOH solution freezes a lower
temperature than pure water.
Choose the correct answer from the codes
given below.
(a) (i),(ii) and (iii)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (i) and (iii)
AIIMS-2017
Ans. (d): It is clear that colligative properties depends
upon the no. of particles. Here, methanol shows non
electrolyte property hence cannot be considered.
396. 31g of ethylene glycol (C2H6O2) is dissolved in
600 g of water. The freezing point depression of
the solution is (Kf for water is 1.86 K kg mol–1)
(a) 0.77 K
(b) 1.55 K
(c) 4.65 K
(d) 3.10 K
AP EAMCET-2017
Ans. (b) : Given that, molecular mass of ethylene
glycol = 62
Mass of ethylene glycol (W1) = 31 gm.
Mass of water (W2) = 600 gm.
Kf for water = 1.86 K kg mol–1
We know that
W ×1000
∆Tf = Kf × m = K f × 1
M × W2
1.86 × 31× 1000
∆Tf =
62 × 600
∆Tf = 1.55 K.
397. Which one of the following solution of
compounds shows highest osmotic pressure?
(AB, AB2 and A2B3 are ionic compounds)
(a) 5.0 M urea i = 1.0 and temperature is 67°C
(b) 1.5 M A2B3 type i = 4.1 and temperature is
27°C
(c) 3.0 M AB type i = 1.6 and temperature is
27°C
(d) 2.5 M AB2 type i = 2.5 and temperature is
57°C
AP EAMCET-2017
Ans. (d) : We know that,
Osmotic pressure (π) = iCRT
(i) i = 1, C = 5, T = 67 + 273 = 340 K
so , π = 1 × 5 × R × 340
π = 1700R
(ii) i = 4.1 , C = 1.5, T = 27 + 273 = 300 K
∴ π = 4.1 × 1.5 × 300 R
π = 1845 R
(iii) i = 1.6, C = 3 , T = 27 + 273 = 300 K
∴ π = 1.6 × 3 × 300 R
π = 1440 R
(iv) i = 2.5, C = 2.5, T = 330 K
∴ π = 2.5 × 2.5 × 330 R
π = 2062.5 R.
Hence, the highest osmotic pressure we will get in
option (d)
86
YCT
398. To prepare XeF6. Xe and F2 are mixed at 573 K
and 60-70 bar in the ratio of
(a) 20 : 1
(b) 1 : 5
(c) 5 : 1
(d) 1 : 20
AP EAMCET-2017
Ans. (d):
573K
Xe + 3F2
→ XeF6
60–70bar
1
20
Xenon hexafluoride (XeF6) is formed by taking Xe and
F2 in ratio 1 : 20 under pressure of 60 – 70 bar at 573 K.
399. The experimental depression in freezing point
of a dilute solution is 0.025 K. If the van't Hoff
factor(i) is 2.0, the calculated depression in
freezing point (in K) is
(a) 0.00125
(b) 0.025
(c) 0.0125
(d) 0.05
TS-EAMCET-2016
Ans. (c) : Given that, depression in freezing point of a
dilute solution = 0.025 K, i = 2.0
Observed colligative property
Van't Hoff factor (i) =
Calculated colligative property
0.025
2=
Calculated depression in freezing point
Calculated depression in freezing point = 0.0125 K.
400. Cryolite is
(a) Na3AlF6 and is used in the electrolysis of
alumina for decreasing electrical conductivity
(b) Na3AlF6 and is used in the electrolysis of
alumina for lowering the melting point of
alumina only
(c) Na3AlF6 and is used in the electrolysis of
alumina for lowering the melting point and
increasing the conductivity of alumina
(d) Na3AlF6 and is used in the electrolytic
refining of alumina
Karnataka-CET-2015
Ans. (c) : Cryolite (Na3AlF6) is used in the electrolysis
of alumina for lowering the melting point and
increasing the conductivity of alumina.
401. After adding non-volatile solute freezing point
of water decreases to –0.186oC. Calculate ∆Tb,
if Kf = 1.86 K kg mol–1 and Kb = 0.521 K Kg
mol–1.
(a) 0.521
(b) 0.0521
(c) 1.86
(d) 0.0186
Karnataka-CET-2015
0.186
Ans. (b) : Molality =
1.86
= 0.1
Now,
∆Tb = Kb × m
= 0.521 × 0.1
= 0.0521°C
402. Which of the following aqueous solution has
the highest freezing point?
(a) 0.1 M sucrose
(b) 0.01 M NaCl
(c) 0.1 M NaCl
(d) 0.01 M Na2SO4
Karnataka-CET-2015
Objective Chemistry Volume-II
Ans. (b) : Depression in freezing point, ∆Tf = i Kf m
(i)
∆Tf for 0.1 M NaCl
m = 0.1, i = 2
∆Tf = 0.1 × 2 × Kf = 0.2 Kf
(ii)
∆Tf for 0.1 M sucrose
m = 0.1, i = 1
∆Tf = 0.1 × 1 × Kf
∆Tf = 0.1 Kf
(iii)
∆Tf for 0.01 M Na2SO4
i = 3, m = 0.01
∆Tf = 3 × 0.01 × Kf
∆Tf = 0.03 Kf
(iv)
∆Tf for 0.01 M NaCl
i = 2, m = 0.01
∆Tf = 2 × 0.01 × Kf
∆Tf = 0.02 Kf
∆Tf will follow the order
0.1 M NaCl > 0.1 M sucrose > 0.01M Na 2SO 4 > 0.01 M NaCl
i=2
i=1
i=3
i=2
Depression in freezing point is minimum in case of
0.01M NaCl solution hence, it will have maximum
freezing point.
403. Fog is colloidal solution of
(a) Liquid particles dispersed in a gas
(b) Gaseous particles dispersed in a liquid
(c) Solid particles dispersed in a liquid
(d) Solid particles dispersed in a gas
JCECE-2015
Ans. (a) : Fog is a colloidal solution in which water
(liquid, dispersed phase) is dispersed in air (gas,
dispersion medium).
A collection of liquid water droplets or ice crystals
suspended in the Earth's surface is called fog. We can
say that fog is a colloidal solution in which water is the
dispersed phase and air is the dispersion medium. i.e.
liquid dispersed is gas.
404. If a homogeneous colloid placed in dark is
observed in the direction of light, it appears
clear and if it is observed from a direction at
right angles to the direction of light beam, it
appears perfectly dark. This is known as
(a) Brownian effect (b) Hardy-Schulze effect
(c) Einstein effect
(d) Tyndall effect.
J & K CET-2015
Ans. (d) : When a beam of light is passed through a
colloidal solution, placed in a dark room then beam of
light is scattered by the colloidal particles and the path
of the beam becomes visible and this phenomenon is
called Tyndall effect.
405. By dissolving 68.4g of a compound whose
molecular mass is 342 in 1 kg of water, a
solution is prepared. If Kf for water is 1.86 K kg
mol-l then freezing point of the solution will be
(a) 272.8 K
(b) 273.5 K
(c) 282.3 K
(d) 263.7 K
SCRA-2014
87
YCT
So, It shows that ∆Tf
Ans. (a) : Depression in freezing point (∆Tf) = Kf × m
W × 1000
Where, molality (m) = B
M B × WA
68.4 ×1000 68.4
=
342 ×1000 342
68.4
∴
∆Tf = 1.86 ×
= 0.372K
342
Hence, the freezing point of solution will be–
273 – 0.372 = 272.8K.
406. A solution prepared from 1.25g of oil of
wintergreen in 99.0 g of benzene has a boiling
point of 80.31°C. What is the molar mass of the
compound, given that the normal boiling point of
benzene is 80.10°C and its boiling point elevation
constant Kb is 2.53 K kg mol-1?
(a) 137 gm mol-1
(b) 150 gm mol-1
-1
(c) 560 gm mol
(d) 117 gm mol-1
SCRA-2014
Ans. (b) : Given that,
Weight of solute (W2) = 1.25gm
Weight of solvent (W1) = 99.0gm
Boiling point of solution (Tb) = 80.31 °C
Molar mass (m) = ?
Al2 ( SO 4 )3 because of 'i' is maximum (i.e. = 5) for
Al2(SO4)3.
Hence, Al2 ( SO 4 )3 has lowest freezing point.
=
( )
Boiling point of solvent Tbo = 80.10°C
∴
or
Kb = 2.53 K kg mol–1
∆ Tb = Tb − Tb° = 80.31°C − 80.10°C = 0.21°C
Now,
K × W2 × 1000
M= b
∆Tb × W1
M=
2.53 K kg mol−1 × 1.25 × 1000
0.21× 99.0
is maximum for 0.10m
408. The depression in freezing point of water
observed for the same amount of acetic acid (I),
trichloroacetic acid (II) and trifluoroacetic
acid (III) decreases in the order
(a) I > II > III
(b) II > I > III
(c) III > I > II
(d) III > II > I
AMU-2014
Ans. (d) : Flurine is more electronegative than Cl.
Trifluroacetic acid gives maximum ions in solution so it
is the strongest acid. Depression of freezing point will
be maximum in this case and it shows least for acetic
acid which is the weakest acid.
409. Which among the following is a non-colligative
property?
(a) Elevation in boiling point
(b) Osmotic pressure
(c) Refractive index
(d) Lowering of vapour pressure
J & K CET-2014
Ans. (c) : Colligative properties of solutions are
properties that depend upon the concentration of solute
molecules or ions.
Elevation in boiling point, osmotic pressure, lowering
of vapour pressure are colligative properties but
refractive index is a non-calligative property.
410. Glucose is added to 1 L water to such an extent
that ∆Tf/Kf becomes equal to 10–3 the weight of
glucose (C6H12O6) added is
(a) 180 g
(b) 18 g
(c) 1.8 g
(d) 0.18 g
UPTU/UPSEE-2013
Ans. (d) : Given–
∆Tf
= 10−3
Kf
3162.5
gm mol−1
20.79
or
M = 152.1 gm mol–1 ≈ 150 gm mol–1.
M 2 = 180 gmol−1 (molecular mass of glucose)
407. Which of the following 0.10 m aqueous solution
will have the lowest freezing point?
W1 = 1000gm
(a) Al2(SO4)3
(b) C6H12O6
we know that,
(c) KCl
(d) C12H22O11
1000K f W2
∆Tf= =
AIIMS-2014
W1 × M 2
Ans. (a): Ionization,
Or
Al2 ( So 4 )3
→ 2Al3+ + 3SO 42 −
(Here, i = 5)
∆T W M
W2 = f 1 2
+
−
1000 × K f
KCl ↽ ⇀ K + Cl
(Here, i = 2)
10−3 × 1000 ×180
∆Tf = i × K f × m
=
1000
∆Tf ∝ i
W2 = 0.18g
It is clear that Depression in freezing point is directly
411. The freezing point of equimolal aqueous
proportional to number of particle.
solution will be highest for
So,
(a) C6 H 5 NH 3+ Cl −
(b) Ca(NO3 )2
i C12 H22O11 = 1, i( KCl ) = 2
(c) La(NO3 )2
(d) C6 H12 O 6
i( Al (SO ) ) = 5, i C6 H12O6 = 1
2
4 3
AIIMS-2013
or
M=
Objective Chemistry Volume-II
88
YCT
Ans. (d): Glucose does not dissociate (i.e., i = 1), so it
has the minimum number of particles and therefore it
shows minimum depression in freezing point. Hence,
glucose have highest freezing point.
412. 25 mL of an aqueous solution of KCI was
found to require 20 mL of 1 M AgNO3 solution
when titrated using a K2CrO4 as indicator.
Depression if freezing point of KCl solution
with 100% ionisation will be [KF = 2.00 mol–1 kg,
molarity = molality]
(a) 3.20
(b) 1.60
0
(c) 0.8
(d) 5.00
BCECE-2013
Ans. (a) : Given, V1 = 25 ml, V2 = 20 ml, M1 = 1 M
During neutralization
M 1V 1 = M 2V 2
1× 20
M1 =
= 0.8M
25
i = 1+ α = 1 + 1 = 2
∴
∆Tf = molality × Kf × i [molality ≈ molarity]
= 0.8 × 2 × 2
= 3.2ºC
∴
Tfo = Tf + ∆Tf = 0 + 3.2ºC
= 3.2ºC
413. 2.56 g sulphur in 100 g of CS2 has depression in
freezing point of 0.01ºC. Atomicity of sulphur
in CS2 in–
(Given, Kf = 0.10 molal–1)
(a) 2
(b) 4
(c) 6
(d) 8
BCECE-2013
Ans. (d) : Let atomicity be n, then it exist as (S)n
molar mass of sulphur = 32n
1000w1 k f
∆Tf =
m1 w 2
given,
w1 = 2.56g
w2 = 100g
kf = 0.1º (molal)–1
∆Tf =0.010º
1000 × 256 × 0.1
m1 =
= 256
0.01×100
∴
32n = 256
n=8
414. The unit of ebullioscopic constant is
(a) K or K (molality)–1
(b) mol kg K–1 or K–1 (molality)
(c) kg mol–1 K–1 or K–1 (molality)–1
(d) K mol kg–1 or K (molality)
JCECE-2013
Ans. (a) : The unit of ebullioscopic constant, Kb = K or
K (molality)–1.
415. If the elevation in boiling point of a solution of
non-volatile,
non-electrolytic
and
nonassociating solute in a solvent (Kb = x K kg mol1
) is y K, then the depression in freezing point
of solution of same concentration would be (Kf
of the solvent = z K kg mol-1)
Objective Chemistry Volume-II
2xz
y
xz
(c)
y
xz
(e)
2y
(a)
yz
x
yz
(d)
2x
(b)
Kerala-CEE-2013
Ans. (b) : we know that,
∆Tb = Kb × m
y = x × molality
y
molality =
x
∆Tf = Kf × molality (Kf = z)
∆Tf = z × molality
y
∆Tf = z ×
x
yz
∆Tf =
x
416. The vapour pressure of pure benzene and
toluene at a particular temperature are 100 mm
and 50 mm respectively. Then the mole fraction
of benzene in vapour phase in contact with
equimolar solution of benzene and toluene is
(a) 0.67
(b) 0.75
(c) 0.33
(d) 0.50
(e) 0.20
Kerala-CEE-2013
Ans. (a) : PTotal = P1 + P2
PT = 100 + 50
= 150 mm
P1
Mole fraction (benzene) x =
P1 + P2
100
=
150
= 0.67
417. Colloidal solutions of gold, prepared by
different methods are of different colours
because of
(a) Variable valency of gold
(b) Different concentration of gold particles
(c) Impurities produced by different methods
(d) Different diameters of colloidal gold particles
JCECE-2012
Ans. (d) : Colour of the colloidal solution depends upon
the particle size of colloid.
So, colloidal solutions of gold, prepared by different
methods are of different colours because of different
diameters size of colloidal gold particles.
418. Assertion (A): Molar mass of acetic acid found
by the depression of freezing point method,
separately in the solvents water and benzene
are different.
Reason (R): Water helps in ionization but
benzene brings association of acetic acid.
Identify the correct option.
89
YCT
(a) Both A and R are correct and R is the correct
explanation for 'A'.
(b) Both A and R are correct but R is not the
correct explanation for 'A'.
(c) A is true but R is false.
(d) A is false but R is true.
SRMJEEE – 2012
Ans. (a) : Acetic acid dissociates in water sharply but
the acetic acid exists in dimer form when we dissolve
the acetic acid in benzene due to hydrogen bonding.
(iii) NaCl ↽ ⇀ Na2+ Cl–
(iv) KCN ↽ ⇀ K+ CN–
Because PO34− has highest charge among the given
anions, therefore, Mg3(PO4)2 is the most effective in the
coagulation of Fe(OH)3 solution.
421. Ethylene glycol is used as an antifreeze in a
cold climate. Mass of ethylene glycol which
should be added to 4 kg of water to prevent it
from freezing at –6°C will be (Kf for water =
1.86 K kg mol–1 and molar mass of ethylene
H 2O
–
+
⇀
glycol = 62 g mol–1)
CH3COOH ↽
CH3COO + H
(a) 804.32 g
(b) 204.30 g
C6 H 6
⇀ (CH3COOH)2
2CH3COOH ↽
(c) 400.00 g
(d) 304.60 g
Hence, molar mass of acetic acid is different when
AIEEE-2011
dissolved in benzene and water.
Ans. (a) : Mass of water = 4 kg
419. The correct order of increasing boiling points
Mass of glycol = W
of the following aqueous solutions
Molar mass of glycol = 62 g / mol
0.0001 M NaCl (I), 0.0001 M Urea (II)
Molality of glycol solution =
0.001 M MgCl2 (III), 0.01 M NaCl (IV) is
W / 62g mol
W
=
(a) I < II < III < IV
(b) IV < III < II < I
4 kg
62 g mol × 4 kg
(c) II < I < III < IV
(d) III < II < IV < I
o
∆
T
=
6
C
f
AMU-2012
Kf = 1.86k kg mol–1
Ans. (c) : Applying formula –
Using the formula, ∆Tf = Kf .m
∆T = i ×Kb × m
o
W
For 0.01M NaCl, i × m
6 C = 1.86k kg mol–1 ×
62g mol−1 × 4 kg
= 2 × 0.01
= 0.02
6o C × 62g mol−1 × 4kg
= 800g
W=
since, i = 2 and conc. of ions will be 2 × 10–2
1.86o C kgmole −1
For 0.001MgCl2 = 3×0.001
422. An aqueous solution of NaCl shows the
= 0.003
depression of freezing point of water equal to
i = 3 as it is a electrolyte and conc. of ions will be 3 ×
0.372 K. The boiling point of BaCl2 solution of
–3
10
same molality will be K f (H 2 O) = 1.86 K kg
For 0.0001 NaCl = 2×0.0001
mol-1 ; K b (H 2 O) = 0.52 K kg mol -1 )
= 0.0002
(a) 100.52 ºC
(b) 100.104 ºC
i = 2 and conc. of ions will be 2 × 10–4
(c)
101.56
ºC
(d) 100.156 ºC
For urea = 1×0.0001
AIIMS-2011
= 0.0001
Given
that,
Ans.
(d):
i = 1 as it is a non electrolyte and conc. of ions will be
K f ( H 2 O ) = 1.86K kg mol−1
10–4
Hence, II < I < III < IV
K b ( H 2 O ) = 0.52K kg mol −1
Priority given to concentration, option decide by
∆Tf = 0.372
concentration.
i1 = 2 for NaCl
420. Which of the following will be the most
effective in the coagulation of Fe(OH)3 soil?
i 2 = 3 for BaCl2
(a) Mg3(PO4)2
(b) BaCl2
Now,
(c) NaCl
(d) KCN
From formula–
BITSAT-2012
∆Tf i1 K f m
=
Ans. (a) : According to hardy-schulze rule, coagulation
∆Tb i 2 K b m
power of ions is directly proportional to charge of an
2 × 1.86
0.372
ion.
⇒
=
∆Tb
3 × 0.52
∴ Fe (OH)3 is positively charged colloid.
o
∴ It will coagulated by anion
∆Tb = 0.156 C
(i) Mg3(PO4)2 ↽ ⇀ 3 Mg2+ 2PO34−
i.e.
Tb = 100 + 0.156
2+
–
⇀
(ii) BaCl2 ↽
Ba 2Cl
= 100.156 oC
Objective Chemistry Volume-II
90
YCT
423. 1.00 g of a non-electrolyte solute (molar mass
250g mol-1) was dissolved in 51.2 g of benzene.
If the freezing point depression constant, Kf of
benzene 5.12 K Kg mol-1, the freezing point of
benzene will be lowered by
(a) 0.3 K
(b) 0.5 K
(c) 0.4 K
(d) 0.2
BITSAT-2011
Ans. (c) : Given that,
Kf = 5.12K Kg mol–1, W1 = 1 g, M1 = 250 gmol–1
W ×1000
∆Tf = Kf × 1
W2 × M1
1×1000
= 5.12 ×
51.2 × 250
∆Tf = 0.4k
424. The size of colloidal particles is in between
(a) 10–7 - 10–9 cm
(b) 10–9-10–11 cm
–5
–7
(c) 10 -10 cm
(d) 10–2-10–3 cm
JCECE-2010
Ans. (c) : The size of colloidal particles is in the range
of 100 nm to 1 nm or 10–5 cm to 10–7 cm.
425. If sodium sulphate is considered to be
completely dissociated into cations and anions
in aqueous solution, the change in freezing
point of water ( ∆Tf ) , when 0.01 mole of
sodium sulphate is dissolved in 1 kg of water, is
(Kf = 1.86 K kg mol–1)
(a) 0.0372 K
(b) 0.0558 K
(c) 0.0744 K
(d) 0.0186 K
AIEEE-2010
Ans. (b) : Reaction will be
Na2SO4 → 2Na+ + SO42– (i = 3 ions)
Here,
i=3
m = 0.01
∆Tf = i Kfm
= 3 × 1.86 × 0.01
= 0.0558K
426. Which one of the following 1.0×10-3 molal
aqueous solutions has the highest boiling point?
(a) Aluminium (III) chloride
(b) Lead (II) nitrate
(c) Sodium chloride
(d) Magnesium nitrate
AP- EAMCET(Medical) -2010
Ans. (a): Colligative properties like boiling point
elevation depends only upon the number of particles in
solution but not on their nature.
Now, ∆T = i × K b × m
where- i = Vant Hoff's factor
Kb = boiling point elevation constant
m = molality.
∴ More the number of particles, higher will be the
value of elevation of boiling point.
For AlCl3
→ Al3+ + 3Cl− ( i = 4 ions )
For Pb ( NO3 )2
→ Pb2+ + 2NO3− ( i = 3 ions )
Objective Chemistry Volume-II
For NaCl
→ Na + + Cl− ( i = 2 ions )
For Mg ( NO3 )2
→ Mg 2+ + 2NO3− ( i = 3 ions )
Hence, Aqueous solution of AlCl3 has the highest
boiling point.
427. Exactly 1 g of urea dissolved in 75 g of water
gives a solution that boils at 100.114ºC at 760
torr. The molecular weight of urea is 60.1. The
boiling point elevation constant for water is
(a) 1.02
(b) 0.51
(c) 3.06
(d) 1.51
AIIMS-2010
Ans. (b): Given that,
weight of solute (w) = 1 g
weight of solvent (W) = 75 g
Boling point of solution = 100.114°C
Boling point of solvent = 100°C
∆T = 100.114 – 100 = 0.114°C
Molecular weight of solute (m) = 60.1
Boiling Point elevation Constant (K)
m × ∆T × W
=
100 × w
60.1× 0.114 × 75
=
1000 × 1
= 0.513
428. Gold number indicates
(a) protective action of lyophlic colloid
(b) charge on gold sol
(c) protective action of lyophobic colloid
(d) quantity of gold dissolved in a given sol
AMU-2010
Ans. (a) : Gold number is defined as the number of
milligrams of a lyophilic colloid that will just prevent
the precipitation of 10 cc of a gold solution on the
addition of 1 cc of 10% sodium chloride solution.
Higher the value of gold number, lower will be the
protective power.
429. Maximum depression in freezing point is
caused by
(a) potassium chloride
(b) sodium sulphate
(c) magnesium sulphate
(d) magnesium carbonate
MHT CET-2009
Ans. (b) : Sodium sulphate (Na2SO4) produces
maxinmum number of particles so it show maximum
depression in Freezing point. because depression in the
freezing point is a colligative property, so it depends
upon the number of particles.
430. One component of a solution followos Raoult’s
law over the entire range 0 ≤ x1 ≤ 1. The second
component must follow Raoult’s law in the
range when x2 is
(a) close to zero
(b) close to 1
(c) 0 ≤ x2 ≤ 0.5
(d) 0 ≤ x2 ≤ 1
AMU-2009
91
YCT
Ans. (d) : Ideal solution obeys Raoult's law at every
range of concentration. so, the second component must
follow Raoult's law in the range, when
x2 is 0 ≤ x2 ≤ 1
431. Which one of the following impurities present
in colloidal solution cannot be remove by
electrodialysis?
(a) Sodium chloride
(b) Potassium sulphate
(c) Urea
(d) Calcium chloride
(e) Magnesium chloride
Kerala-CEE-2009
Ans. (c) : Urea, being non-electrolyte and covalent
compound hence, it does not dissociate to give ion so it
can not be removed by electrodialysis method.
Other impurities are electrolyte so it is easily removed
by electodialysis method.
432. Abnormal colligative properties are observed
only when the dissolved non-volatile solute in a
given dilute solution
(a) Is a non-electrolyte
(b) Offers an intense colour
(c) Associates or dissociates
(d) Offers no colour
J & K CET-2008
Ans. (c) : When experimentally measured colligative
property of a solution shows is different in theoretical
form by the law of osmosis, then it means it has
abnormal colligative property.
When non-volatile solute associates or dissociates in a
solution, values of colligative properties deviates i.e.,
abnormal colligative properties are obtained.
433. Which is not a colligative property in the
following
(a) pH of a buffer solution
(b) Boiling point elevation
(c) Freezing point depression
(d) Vapour pressure
MPPET-2008
Ans. (a) : Colligative properties are those properties of
solution that depend on the ratio of the number of solute
particles to the number of solvent particles in a solution
eg- Boiling point of elevation and freezing point of
depression and lowering of vapour are examples of
colligative properties. Whereas the pH of a buffer
solution is not a colligative property bcz pH depends
only upon the molar concentration of H+ ion not in their
number of particles in it.
434. 0.01 M solution of KCl and BaCl2 are prepared
in water. The freezing point of KCl is found to
be –2ºC. What is the freezing point of BaCl2 to
be completely ionised?
(a) –3ºC
(b) +3ºC
(c) –2ºC
(d) –4ºC
AIIMS-2008
Ans. (a): Ionic reaction,
KCl
→ K + + Cl− (Here, i = 2)
BaCl2
→ Ba 2+ + 2Cl− (Here,i = 3)
As, we know that,
∆Tf α i
Objective Chemistry Volume-II
∆Tf is directly proportional to the molality of the
solution i
∆Tf ( KCl ) = −2°C
Now,
∆Tf ( KCl )
i ( KCl )
=
∆Tf ( BaCl2 ) i ( BaCl2 )
⇒
∆Tf ( KCl )
2
=
∆Tf ( BaCl2 ) 3
3 × ( −2 )
2
= −3°C
435. The change in entropy for the fusion of 1 mole
of ice is [m.p. of ice = 273 K, molar enthalpy of
fusion for ice = 6.0 kJ mol–1]
(a) 11.73 JK–1 mol–1
(b) 18.84 JK–1 mol–1
–1
–1
(c) 21.97 JK mol
(d) 24.47 JK–1 mol–1
JCECE-2007
Ans. (c) : Given data,
Molar enthalpy of fusion for ice = 6.0 kJ mol–1
∆Hfusion. = 6.0 × 103 J mol–1
∆H fusion
∴Sfusion =
Tm.p.
⇒ ∆Tf (BaCl 2 ) =
6.0 × 103 Jmol –1
273K
= 21.976JK–1 mol–1
436. 1 mol each of the following compounds is
dissolved in 1L of solution. Which will have the
largest ∆Tb value?
(a) HF
(b) HCl
(c) HBr
(d) HI
AIIMS-2007
Ans. (d): HF is most stable halogen acid and HI is least
stable.
Higher the bond dissociation energy, lower is the degree
of ionization.
Bond dissocation energy increase in the order –
HI < HBr < HCl < HF
As we know ∆Tb = i.Kb. m
Hence, ∆Tb is largest for HI. All colligative properties
depend upon number of particles.
437. If for a sucrose solution, elevation in boiling
point is 0.1°C then what will be the boiling
point of NaCl solution for same molal
concentration?
(a) 0.1°C
(b) 0.2°C
(c) 0.08°C
(d) 0.01°C
JIPMER-2006
Ans. (b) : ∆Tb ∝ n
It is clear that Elevation in a boiling point is a
colligative property as it depends upon the number of
particle.
Number of particle for sucrose, n = 1, ∆Tb = 0.1°C
Number of particle for NaCl , n = 2, ∆Tb = 0.2°C
92
=
YCT
438. Assertion: The water pouch of instant cold
pack for treating athletic injuries breaks when
squeezed and NH4NO3 dissolves lowering the
temperature.
Reason: Addition of non-volatile solute into
solvent results into depression of freezing point
of the solvent.
(a) If both Assertion and Reason are correct and
the Reason is a correct explanation of the
Assertion.
(b) If both Assertion and Reason are correct but
Reason is not a correct explanation of the
Assertion.
(c) If the Assertion is correct but Reason is
incorrect
(d) If both the Assertion and Reason are
incorrect.
(e) If the Assertion is incorrect but the Reason is
correct.
AIIMS-2006
Ans. (a): Ammonium nitrate being a non-volatile solute
when dissolved in the water, lowers the temperature of
the solution and reason is also correct as depression in
freezing point of the solvent occurs due to the addition
of non-volatile solute into solvent.
Note:Freezing point of a substance is defined as the
temperature at which the substance changes its states
from liquid to solid, also the vapour pressure of both
the solid state and liquid state must be equal.
439. Assertion: If red blood cells were removed
from the body and place in pure water,
pressure inside the cells increase.
Reason: The concentration of salt content in
the cells increase.
(a) If both Assertion and Reason are correct and
the Reason is a correct explanation of the
Assertion.
(b) If both Assertion and Reason are correct but
Reason is not a correct explanation of the
Assertion.
(c) If the Assertion is correct but Reason is
incorrect
(d) If both the Assertion and Reason are
incorrect.
(e) If the Assertion is incorrect but the Reason is
correct.
AIIMS-2006
Ans. (c) : If the red blood cells are placed in pure water,
pressure inside the cells increases due to osmosis.
440. A 5% solution (by mass) of cane sugar in
water has freezing point of 271 K and freezing
point of pure water is 273.15 K The freezing
point of a 5% solution (by mass) of glucose in
water is
(a) 271 K
(b) 273.15 K
(c) 269.07 K
(d) 277.23K
AIIMS-2006
Objective Chemistry Volume-II
Ans. (c): For
C12 H 22 O11
WSolute = 5g, ammount of solution = 100g
Molar mass of C12 H 22 O11 = 342g
WSolvent = 100 − 5 = 95g
From formula.
∆Tf = K f ×
( 273 − 271) = K f ×
WSolute ×1000
Molar mass × WSolvent
5 × 1000
342 × 95
K f = 13.97
Molar mass of glucose = 180 g/mol
since water contains 5% of glucose
5g of glucose is present in (100 – 5) g
5
Number of moles of glucose =
= 0.0278 mol
180
0.278mol
Molality of the solution, m =
0.095kg
–1
= 0.2926 mol kg
For C6 H12 O6 (i = 1)
∆Tf = iKf. m
∆Tf = 13.99 K kg mol–1 × 0.2926 mol kg–1
= 4.09 K (approx)
∴ The freezing point of 5% glucose solution
= (273.15 – 4.09) K
= 269.06 K
441. Which of the following solution will have lowest
boiling point?
(a) 0.1M NaCl
(b) 0.1M KCl
(c) 0.1M CaCl2
(d) 0.1M glucose
BITSAT-2006
Ans. (d) : Elevation in boiling point is a colligative
property. Smaller the no. of solute particles lesser will
be elevation in boiling point.
442. Observe the following abbreviations
Π obs = observed colligative property
93
Π cal =
theoretical
colligative
property
assuming normal behavior of solute. Van’t
Hoff factor (i) is given by
(a) i = Πobs × Π cal
(b) i = Π obs + Π cal
(c) i = Π obs – Πcal
(d) i =
Π obs
Π cal
J & K CET-2006
Ans. (d) :
Van't Hoff factor (i) =
So,
i=
Observed colligative property
Theoreticalcolligative property
Π obs
Π cal
YCT
443. Lyophilic sols are more stable than lyophobic
sols because the particles :
(a) are positively charged
(b) are negatively charged
(c) are solvated
(d) repel each other
(e) are heavy
Kerala-CEE-2005
Ans. (b) : Lyophilic sol is more stable than Lyphobic
sol due to the extensive salvation (solvated). Stability of
lyophilic sols is a result of two factors. the presence of
charge and the salvation of colloidal particles. while
lyophobic sols is only because of the presences of
charge.
444. The molal elevation constant for water is 0.52.
What will be the boiling point of 2 molal sugar
solution at 1 atm pressure? (Assume B.P. of
pure water as 100oC)
(a) 101.04oC
(b) 100.26oC
o
(c) 100.52 C
(d) 99.74oC
AMU-2005
Ans. (a) : Given that,
Kb = 0.52 k mol–1, m = 2 m, i = 1
∆Tb = Kb × m × i
∆ Tb = 0.52 × 2 × 1 = 1.04°C
∆Tb = boiling point of soluction – boiling point
of solvent (i.e. H2O)
∴ boiling point of solution = ∆Tb+boiling point of H2O
= 1.04°C + 100°C
= 101.04°C.
445. Molal solution means 1 mole of solute present
in:
(a) 1000g of solvent
(b) 1 L of solvent
(c) 1 L of solution
(d) 1000 g of solution
BCECE-2005
Ans. (c) : Molar solution means 1 mole of solution
present in 1L of solution.
Molality is the ratio of number of moles of solute and
volume of solution in liter present in solution.
446. Which of the following shows maximum
depression in freezing point?
(a) K2SO4
(b) NaCl
(c) Urea
(d) Glucose
BCECE-2005
Ans. (a) : The substance which have maximum number
of particles on dissociation, will have low at freezing
point and involve with the high depression.
447. A solution of 4.5 g of a pure non-electrolyte in
100 g of water was found to freeze at 0.465°C.
The molecular weight of the solute is closest to
(Kf = 1.86):
(a) 135.0
(b) 172.0
(c) 90.0
(d) 180.0
JCECE-2005
Objective Chemistry Volume-II
Ans. (d) : Given data –
Kf = 1.86
∆Tf = 0.465
w(non-electolyte) = 4.5
W = 100g
From formula,
1000 × K f × w (H2O) 1000 ×1.86 × 4.5
=
m=
∆Tf × W
0.465 × 100
= 180g.
448. If the elevation in boiling point of a solution of
10 g of solute (mol. Wt. = 100) in 100 g of water
is ∆ Tb, the ebullioscopic constant of water is
(a) 10
(b) 100 Tb
∆Tb
(c) ∆ Tb
(d)
10
VITEEE-2014
Ans. (c) : Given,
W2 = 100
M= 100
W1 = 10
We know that,
1000 × K b × W1
∆Tb =
W2 × M
1000 × K b × 10
100 × 100
∆Tb = K b .
∆Tb =
449. Equimolar solution in the same solvent have
(a) Different boiling point and different freezing
point
(b) Same boiling point and same freezing point
(c) Same freezing point but different boiling
point
(d) Same boiling point but different freezing
point
AIEEE-2005
Ans. (b) : Equimolar solutions are the solution which
contains the same number of moles of solute dissolved
in a solvent. Hence the concentration of solution or
molarity or molality will be the same. The colligative
properties are the physical change in the solutions
which result from the addition of solute to the solvent.
So, Equimolar Solution in the same solvent have same
boiling point and same freezing point.
450. A pressure cooker reduces cooking time for
food because
(a) Heat is more evenly distributed in the cooking
space
(b) Boiling point of water involved in cooking is
increased
(c) The higher pressure inside the cooker crushes
the food material
(d) Cooking involves chemical changes helped
by a rise in temperature
Rajasthan PMT-2008
Ans. (b) : High temperature causes food to cook faster.
So, we can say that a pressure cooker reduces cooking
time for food because boiling point of water involved in
cooking is increased.
94
YCT
S1 S2
=
P1 P2
Where, S = solubility
451. Calculate the quantity of CO2 required to
P = Pressure
prepare 1 L of soda water when the soda water Henry’s law is valid for
was packed under 2 atm of CO2.
(a) Ammonia gas dissolution in water.
[Henry's law constant for CO2 is 1.67 × 108 Pa] (b) CO dissolution in water.
2
(a) 5.98 g
(b) 1.21 g
454.
The
vapour pressure of two liquids X and Y
(c) 2.9 g
(d) 67.1 g
are 80 and 60 Torr respectively. The vapour
TS-EAMCET-19.07.2022, Shift-II
pressure of the ideal solution obtained by
Ans. (c) : As we know that Henry's law : –
mixing 3 moles of X and 2 moles of Y would be
P = KH × x
(a) 68 Torr
(b) 140 Torr
5
(c)
48
Torr
(d)
72 Torr
2×1.013×10
x=
(e) 54 Torr
8
1.6×10
Kerala-CEE-2009, JCECE - 2008
n
CO 2
2×1.013
UPTU/UPSEE-2007
=
100 /18 1.6×1000
Ans. (d) : Given that,
20 1.013
n
The vapour pressure of two liquids X and Y areCO 2 = ×
16
18
PXo = 80Torr
20 44×1.013
PYo = 60Torr
Mass of CO2 dissolved = ×
16
18
Moles of X = 3
20×11×1.013
Moles of Y = 2
=
4×18
∴ Total vapour pressure = PXo ⋅ X + PYo ⋅ Y
= 3.0
3
2
≈ 2.9 gram
= 80 ×
+ 60 ×
2
+
3
3
+
2
452. The Henry’s law constant for the solubility of
+5
3
2
N2 gas in water at 298 K is 1×10 atm. The
= 80 × + 60 ×
5
5
mole fraction of air is 0.8. The number of moles
of N2 from air dissolved in 10 mole of water at
= 16 × 3 + 12 × 2
298 K and 5 atm pressure is
= 48 + 24
(a) 4×10−5
(b) 4×10−4
= 72 Torr
(c) 5×10−4
(d) 4×10−6
455. Which option is inconsistent for Raoult’s law?
TS-EAMCET-19.07.2022, Shift-I
(a) Volume of liquid solvent + volume of liquid
solute = volume of solution.
Ans. (b) : PN 2 = Total pressure × mole fraction
(b) The change in heat of dilution for solution = 0
= 5 × 0.8 = 4atm
(c) Solute does not undergo association in
From Henry's Law,
solution
PN2 = K H × N 2
(d) Solute undergoes dissociation in solution
GUJCET-2016, 2015
4
X N2 =
= 4 × 10−5
Raoult’s
law
is
not
applicable
to solutes
Ans.
(d)
:
5
1× 10
which
dissociates
in
the
particular
solutions.
-5
∴ 1 mole of solution will contain 4 × 10 moles of N2
and 1 mole of water contain 1 – 4 × 10–5 is 456. 18g of glucose (C6H12O6) is added to 178.2 g
water. The vapour pressure of water (in torr)
approximately equal to 1.
for this aqueous solution is
∴ 10 moles of water contain 4 × 10–4 moles of N2
(a) 76.0
(b) 752.4
453. Henry's law is valid for
(c) 759.0
(d) 7.6
(A) Ammonia gas dissolution in water
[JEE Main 2016], [AIEEE 2006]
(B) O 2 gas dissolution in unsaturated blood
Ans. (b) : Solution:–
(C) O 2 dissolution in water
18
Number of moles of glucose =
= 0.1mol
(D) CO 2 dissolution in water
180
(a) A and B
(b) B and C
178.2
Number of moles of water =
= 9.9 mol
(c) C and D
(d) B and D
18
TS-EAMCET-18.07.2022, Shift-II
Mole fraction of water in solution–
Ans. (c) : Henry’s law states that, at a given
9.9
9.9
temperature the solubility, (S) of gas in a liquid is
X H 2O =
=
= 0.99
9.9 + 0.1 10
directly proportional to the pressure (P)
So, vapour pressure of H2O
S∝P
3.
Lowering of Vapour Pressure
Objective Chemistry Volume-II
95
YCT
PH2O = PH° 2O × X H2O
M2 = molecular mass of benzene (solvent)
P° = vapour pressure of pure solvent.
Ps = vapour pressure of solution
{
PH2O = 760 torr × 0.99
PH2O = 752.4 torr
}
(640 − 600)mmHg 2.175 × 78
=
457. For dilute solutions, Raoult’s law states that
640mmHg
M1 (39.08)
(a) lowering of vapour pressure is equal to the
2.175 × 78 × 640
M1 =
mole fraction of the solute
40 × 39.08
(b) Relative lowering of vapour pressure is equal
= 69.457 mol–1 ∼ 69.60g mol–1
to the mole fraction of the solvent
(c) Relative lowering of vapour pressure of the 460. The vapour pressure of a solvent decreased by
10 mm of mercury when a non-volatile solute
solvent is equal to the mole fraction of the
was added to the solvent. The mol fraction of
solute
the solute in the solution is 0.2. What should be
(d) Vapour pressure of the solution is equal to the
the mole fraction of the solvent if the decrease
vapour pressure of the solvent
in the vapour pressure is to be 20 mm of
JCECE - 2014
mercury ?
J & K CET-(2010),(1999)
(a) 0.4
(b) 0.6
Ans. (c) : For dilute solutions, Raoult’s law states that
(c)
0.8
(d) 0.2
the relative lowering of vapour pressure of the solvent is
(AIIMS
-2015), (AIPMT -1998)
equal to the mole fraction of the solute.
°
Ans.
(b)
:
From
Raoult's
formula,
P −P
i.e, mathematically
= XB
P º − Ps
P°
= XB
Where, P ° − P = lowering in vapour pressure
Pº
10
X B = mole fraction of solute.
= 0.2
Pº
458. The relative lowering of vapour pressure of
Pº = 50
aqueous solution containing non-volatile solute
is 0.0125. The molality of the solution is
P º − Ps 20
=
=
Now,
for
X
(a) 0.70
(b) 0.50
B
Pº
50
(c) 0.60
(d) 0.80
=
0.4
(e) 0.40
XA + XB = 1
Kerala-CEE-2008, 2007
Mole fraction of solvent (XA) = 1– 0.4
Ans. (a): By Raoult’s law X B = 0.0125
= 0.6
m × MA
∴
XB =
461. The mixture which shows positive deviation
1000 + m × M A
from Rault's Law is
m × 18
(a) Ethanol +acetone
0.0125 =
∴
(b) Benzene + toluene
1000 + m ×18
12.5 + 0.225m = 18m
(c) Acetone + chloroform
(d) Chloroform +bromoethane.
17.775m = 12.5
NEET-2020, 2006
12.5
m=
Ans.
(a)
:
Mixture
of
ethanol
and
acetone shows
17.775
positive
deviation
from
Raoult’s
law.
In
pure ethanol,
m = 0.70
molecules are hydrogen bonded. On adding acetone, its
459. The vapour pressure of benzene at a certain molecules get in between the host molecules and break
temperature is 640 mm of Hg. A non-volatile some of the hydrogen bonds between them. Due to
and non-electrolyte solid weighing 2.175 g is weakening of interaction, the solution shows positive
added to 39.08 g of benzene. If the vapour deviation from Raoult’s law.
pressure of the solution is 600 mm of Hg, what
462. Value of Henry's constant KH _____.
is the molecular weight of solid substance?
(a) no effect by changing temperature
(a) 49.50
(b) 59.60
(b) decreases with increase in temperature
(c) 69.60
(d) 79.82
(c) increases with increase in temperature
MHT CET-2008, (AIPMT -1999), AIIMS-1999
(d)
first decreases and then increases by increase
Ans. (c) : From Raoult's second law,
in temperature
°
P − Ps w1 × M 2
GUJCET-2022
=
P°
M1 × w 2
Ans. (c) : Henry’s law constant (KH) increases with
increasing in temperature. Henry’s constant value
Where, w1 = weight of solid substance
depends only on the nature of gas or liquid and
M1 = molecular mass of solid substance
temperature.
w2 = weight of benzene (solvent)
Objective Chemistry Volume-II
96
YCT
463. Pick the correct statement
(a) Relative lowering of vapour pressure
independent of T
(b) Osmotic pressure always depends on the
nature of solute
(c) Elevation of boiling point is independent of
nature of the solvent
(d) Lowering of freezing point is proportional to
the molar concentration of solute
WB-JEE-30.04.2022
Ans. (a) : Colligative properties depends on relative no.
of solute particles and does not depend on nature of
solute particles.
So, we can say that relative lowering of vapour pressure
independent of T is correct statement.
464. When 12.2 g of benzoic acid is dissolved in 100g
of water, the freezing point of solution was
found to be – 0.93°C [Kf(H2O)=1.86 K kg mol–
1
]. The number (n) of benzoic acid molecules
associated (assuming 100% association) is
_____.
JEE Main 26.02.2021,Shift-II
Ans.(2) : Given that Mass of benzoic acid(m) = 12. 2 gm
Molecular mass of benzoic acid (M) = 122
Freezing point of solution ( ∆Tf ) = 0.93°C
466. 224 mL of SO2(g) at 298 K and 1 atm is passed
through 100 mL of 0.1 M NaOH solution. The
non-volatile solute produced is dissolved in 36 g
of water. The lowering of vapour pressure of
solution (assuming the solution is dilute)
( P(H 2O) = 24mm of Hg) is x×10–2 mm of Hg, the
value of x is ______ (Integer answer)
JEE Main 26.02.2021, Shift-I
Ans. (18) : The balanced equation is,
SO2+2 NaOH → Na 2SO3 + H 2 O
∴ Moles of NaOH= molarity ×Volume (litre)
= 0.1×0.1 = 0.01 moles
2mole NaOH will form1moleof Na 2SO3
Then,
1
0.01moleof NaOH willform = × 0.01 moles of Na 2SO3
2
= 0.005moleof Na 2SO3
36
= 2 moles
18
According to relative lowering vapour pressure
Therefore, moles of H2O =
PA° – PA
n
= iX B = i B (∵ n A > n B )
PA°
nA
Na 2SO3 → 2Na + + SO32−
i = 2+1=3
24 – PA 3 × 0.005
=
24
2
24–PA =0.18
PA = 23.82
K f = 1.86K kg.mol –1
We know that ,
1
∴ i = and m = molality
n
1
12.2 × 1000
0 – (–0.93) = × 1.86 ×
n
122 ×100
1.86
n=
×1
0.93
n=2
465. The freezing point of an aqueous solution
containing 25 g of ethanol in 1000g of H2O is
[Kf = 1.86 K kg mol–1]
(a) 0.25°C
(b) 0.5°C
(c) –1.5°C
(d) –1°C
TS-EAMCET (Engg.), 06.08.2021
Ans. (d): Given data,
Mass of ethanol = 25 g
Mass of solvent (H2O) = 1000g = 1kg
We know that,
Molecular mass of ethanol (C2H5OH) = 46 g
W × 1000
∆Tf = K f B
M B × WA
WB = Weight of solute
MB = Molecular weight of solute
WA = Mass of solvent
1.86 × 25 × 1000
∆Tf =
= 1.01
46 × 1000
Freezing point of solution = 0 – 1.01 = – 1.01°C ≈ –1°C
∆Tf = i × K f × m
Objective Chemistry Volume-II
Lowering in pressure = 0.18 mm of Hg = 18×10 –2 mm Hg
x = 18
467. The vapour pressure of pure liquids A and B
are 450 and 700 mm at Hg of 350 K
respectively. If the total vapour pressure of the
mixture is 600 mm of Hg, the composition of
the mixture in the solution is
(a) xA = 0.4 and xB = 0.6
(b) xA = 0.6 and xB = 0.4
(c) xA = 0.3 and xB = 0.7
(d) xA = 0.7 and xB = 0.3
Kerala-CEE-29.08.2021
Ans. (a) : Given that,
97
Vapour pressure of pure liquid A ( PAo ) = 450 mm Hg
Vapour pressure of pure liquid B ( PBo ) = 700 mm Hg
Total vapour pressure = 600 mm Hg
From the Raoult's laws 600 = (450 – 700) xA + 700
– 100 = – 250 xA
100
xA =
= 0.4
250
And
xB = 1- xA
= 1– 0.4 = 0.6
Hence, the composition of the mixture in the solution is
xA = 0.4 and xB = 0.6
YCT
468. Henry's law constant for the solubility of N2 gas
1
2
in water at 298 K is 1.0 × 105 atm. The mole
PT = × 21 + × 18
3
3
fraction of N2 in air is 0.8. The number of moles
PT = 7+12
of N2 from air dissolved in 10 moles of water at
PT = 19kPa
298 K and 5 atm pressure is
(a) 4.0 × 10-4
(b) 4.0 × 10-5
471. A set of solutions is prepared using 180g of
(c) 5.0 × 10-4
(d) 4.0 × 10-6
water as a solvent and 10 g of different nonvolatile solutes A,B and C. The relative
Kerala-CEE-29.08.2021
lowering of vapour pressure in the presence of
IIT-JEE, 2009
these solutes are in the order
Ans. (a) : Given that, X N2 = 0.8
[Given,
molar mass of A = 100g mol–1: B = 200g
-1
Henry's law constant (KH) = 1.0×105atm
mol : C = 10,000g mol-1]
Total pressure (PTotal) = 5 atm
(a) B > C > A
(b) C > B > A
Now, Partial pressure of N2 = PTotal × X N2
(c) A > B > C
(d) A > C > B
[JEE Main 2020, 6 Sep Shift-II]
P = 5 × 0.8 = 4 atm
Ans.
(c)
:
From
formula,
According to the Henry's law –
∆P
P = K HX
Relative lowering of vapour pressure = o = Xsolute
Where, P = Partial pressure of gas
P
KH = Henry law constant
So,
X = Mole fraction of the gas
0.1
1
for PA =
=
4atm
−5
10.1
101
∴
X=
= 4 × 10
1× 105 atm
0.05
1
PB =
=
n
10.05 201
∴
4 × 10−5 = ⇒ n = 4 × 10−4 mol
10
10−3
P
=
= 10−4
C
469. The correct option for the value of vapour
10
pressure of a solution at 45°C with benzene to Hence option (c) A > B > C is correct option.
octane in molar ratio 3:2 is
[At 45°C pressure of benzene is 280 mm Hg 472. Solute 'X' dimerises in water to the extent of
80%. 2.5g of 'X' in 100g of water increases the
and that of octance is 420 mm Hg. Assume
boiling point by 0.3oC. The molar mass of 'X' is
ideal gas]
[Kb = 0.52 K kg mol–1]
(a) 350 mm of Hg
(b) 160 mm of Hg
(a) 13
(b) 52
(c) 168 mm of Hg
(d) 336 mm of Hg
(c)
65
(d)
26
(NEET-2021)
Karnataka-CET-2020
Ans. (d) : According to the Raoult's law
Ans. (d) : Given, ∆Tb = 0.3
PT = PAo .X a + PBo X B
α
i =l–α+
3
2
2
= 280 × + 420 ×
5
5
= 1 – 0.8 + 0.4 = 0.6
From formula.
nC6 H 6 : nC8 H18 = 3 : 2
W 1000
XB = 2 / 5
∆Tb = iKb × B ×
X A = 3/ 5
M WA
= 336 mm of Hg.
2.5 1000
470. At 363 K, the vapour pressure of A is 21 kPa
0.3 = 0.6 × 0.52 ×
×
and that of B is kPa. One mole of A and 2
M 100
moles of B are mixed. Assuming that this
M = 26g
solution is ideal, the vapour pressure of the 473. In comparison to a 0.01 M solution of glucose,
mixture is ---------- kPa. (Round off to the
the depression in freezing point of a 0.01M
Nearest Integer).
MgCℓ2 solution is .......... (Molecular weight of
JEE Main 16.03.2021, Shift-II
MgCℓ2=95, molecular weight of glucose = 180).
(a) the same
(b) about twice
(c) about three times
(d) about six times
AP EAMCET (Engg.) 21.09.2020, Shift-II
Ans. (c) : 0.01 M solution of glucose does not ionise
while 0.01 M MgCℓ2 solution furnishes 3 ions (Mg+2 +
Ans. (19) : Given data : P°A = 21kPa, P°B = 18kPa
An ideal solution is prepared by mixing 1mol A and
2mol B.
1
1
Mole fraction of A(XA) =
=
2 +1 3
2
2
Mole fraction of B(XB) =
=
1+ 2 3
According to the Raoult's law PT = X A PAo + X B PBo
Objective Chemistry Volume-II
2Cℓ–) in the solution.
Hence, the value of colligative property of MgCℓ2 solution
is about 3 times. Hence, the correct option is (c).
98
YCT
474. Which of the following relations is true based
on the partial pressure and mole fraction ?
(a) ρi = xi ×T
(b) ρi = xi × ρtotal
ρ
(d) ρixi = ρtotal
(c) i = V
xi
AP EAMCET (Engg.) 18.9.2020 Shift-I
Ans. (b) : Partial pressure = mole fraction × total
pressure
So, the ith, component of a gas mixture, which obeys
Dalton's law of partial pressure,
Pi = Xi × Ptotal
475. Which of the following does not show positive
deviation from Raoult’s law?
(a) Benzene-Chloroform
(b) Benzene-Acetone
(c) Benzene-Ethanol
(d) Benzene-Carbon tetrachloride
COMEDK-2020
Ans. (a) : Benzene-Chloroform does not show a
positive deviation from Raoult's law instead it shows a
negative deviation.
In the presence of strong Van-der Wall's force of
attraction between benzene and chloroform leads to
decrement in vapour pressure which implies negative
deviation from Rault’s law.
476. Which of the following change decrease the
vapour pressure of water kept in a sealed vessel?
(a) Decreasing the quantity of water
(b) Adding salt to water
(c) Decreasing the volume of the vessel to onehalf
(d) Keeping the temperature of water constant
AP-EAMCET (Engg.) 17.09.2020, Shift-II
Ans. (b) : When salt (non-volatile solute) is added to
water kept in a sealed vessel, boiling point of the
solution will be elevated or increased. So, vapour
pressure of water (solution) will be decreased.
Lowering of vapour pressure (∆P = Pº - P) is also
proportional to mole fraction of the solute (XB) present
in the solution.
∆P ∝ XB [Pº = Vapour pressure of pure water
P = Vapour pressure of solution]
477. The Henry's law constant for O2 dissolved in
water is 4.34 × 104 atm at certain temperature.
If the partial pressure of O2 in a gas mixture
that is in equilibrium with water is 0.434 atm,
what is the mole fraction of O2 in solution?
(a) 1 ×10–5
(b) 1 ×10–4
(c) 2 ×10–5
(d) 1 ×10–6
–6
(e) 2 ×10
Kerala-CEE-2019
Ans. (a) : According to Henry's law, P = KH × x
PO
0.434
xO = 2 =
= 1 × 10–5
2
K H 4.34 × 104
Mole fraction of O2 in water =1×10–5.
478. 'K' is Henry's constant and has the unit
(a) atm mol–1 dm3
(b) mol–1 dm3 atm–1
–3
(c) atm mol dm
(d) mol dm–3 atm–1
MHT CET-02.05.2019, SHIFT-III
Objective Chemistry Volume-II
Ans. (d) : The partial pressure of gas in the vapour
phase (p) is proportional to the mole fraction of the gas
(X), in the solution, "according to Henry".
p ∝ X or p = kHX
Where, kH = Henry's law constant.
479. The elevation in boiling point of 0.25 molal
aqueous solution of a substance is (Kb = 0.52 K
kg mol–1)
(a) 0.15 K
(b) 0.50 K
(c) 0.13 K
(d) 2.08 K
MHT CET-03.05.2019, SHIFT-I
Ans. (c) : From formula,
Given, m=0.25
Kb=0.52K kg mol–1
∆Tb = Kb × m
= 0.52 × 0.25
=0.13K
480. 9 gram anhydrous oxalic acid (mol. wt. = 90)
was dissolved in 9.9 moles of water. If vapour
pressure of pure water is p10 1 the vapour
pressure of solution is
(a) 0.99 p10
(b) 0.1 p10
0
(c) 0.99 p1
(d) 1.1 p10
MHT CET-02.05.2019, SHIFT-II
Ans. (a) : By Raoult law
Pw = Xw Pwº
Anhydrous oxalic acid is a non-volatile chemical.
Vapour pressure of solution = vapour pressure of water
(pw).
9.9 ( moles of water )
No.of moles of oxalic acid =
9.9 + 0.1
= 0.99
⇒ Ps = Pw
= 0.99 × P1º
481. Relative lowering of vapour pressure of a dilute
solution of glucose dissolved in 1 kg of water is
0.002. The molality of the solution is
(a) 0.004
(b) 0.222
(c) 0.111
(d) 0.021
Karnataka-CET-2019
Ans. (c) : Given ,
∆P
W1 = 1000g,
= 0.002, m1 = 18
P°
∆P n 2 W2 m1
=
=
×
P°
n1 m 2 W1
18
0.002 = n 2 ×
⇒ n 2 = 0.111mol
1000
n
0.111
×1000
Molality (m) = 2 ×1000 =
W1
1000
= 0.111mol / kg
482. Solution 'A' contains acetone dissolved in
chloroform and solution 'B' contains acetone
dissolved in carbon disulphide. The type of
deviations from Raoult's law shown by solution
A and B, respectively are
99
YCT
(a)
(b)
(c)
(d)
positive and positive
positive and negative
negative and negative
negative and positive
n = moles of urea
N = moles of water
P ° − Ps
0.60 / 60
=
0.60 360
P°
+
60
18
0.01
°
∆P = P ×
0.01+20
∆P = 35× 0.01 = 0.0175 mm Hg
20.01
Karnataka-CET-2019
Ans. (d) : Acetone in chloroform show negative
deviation and carbon disulphide show positive
deviation.
483. The vapour pressure of pure CHCl3 and
CH2Cl2 are 200 and 41.5 atm respectively. The
weight of CHCl3 and CH2Cl2 are respectively
11.9 g and 17 g. The vapour pressure of 485. Acetic acid dimerizes when dissolved in
benzene. As a result boiling point of the
solution will be?
solution rises by 0.36ºC, when 100g of benzene
(a) 80.5
(b) 79.5
is mixed with "X" g of acetic acid. In this
(c) 94.3
(d) 105.5
solution, if experimentally measured molecular
AIIMS-25 May 2019 (Morning)
weight of acetic acid is 117.8 and molar
Ans. (c):Given,
elevation constant of benzene is 2.57 K kgmol–1,
Vapour pressure of CHCl3 (PV1 ) = 200 atm
what is the weight % and degree of dissociation
(in %) of acetic acid in benzene?
Vapour pressure of CH2Cl2 (PV2 ) = 41.5 atm
(a) 1.62 and 98.3
(b) 0.81 and 98.3
Weight of CHCl3 (W1) = 11.9 gram
(c) 0.5 and 8.6
(d) 1 and 98.3
Weight of CH2Cl2 (W2) = 17 gram
(e) 1.4 and 99
weight
Kerala-CEE-2019
Number of moles =
Ans. (a) : Given that,
molecular weight
∆Tb = 0.36ºC, Kb = 257 Kg/ mol, Ma = 100gm.
W 11.9
n1 = 1 =
= 0.1
Experimental molecular weight of acetic acid = 117.8
M1 119
normal molar mass
60
i=
=
= 0.51
W2 17
n2 =
=
= 0.2
Abnormal molar mass 117.8
M 2 85
We know that, ∆Tb = iK b × m
Vapour pressure of solution (Pv)
x
PV = PV1 X1 +PV2 X 2
× 1000
× 100
0.36 = 0.51 × 257 × 60
Where , X1 = mole fraction of CHCl3
100
x = 1.65 gm
X 2 = mole fraction of CH 2 Cl2
Weight of CH 3COOH
0.1
0.2
% weight of CH3COOH=
× 100
X1 =
,X 2 =
Weight of solution
0.3
0.3
1
2
1.65
=
=
=
×100 = 1.62
3
3
101.65
1
2
PV = 200 × + 41.5 ×
i −1
Degree of dissociation (α) =
3
3
1
PV = 94.33 atm
−1
n
484. At room temperature, a dilute solution of urea
0.51 − 1
is prepared by dissolving 0.60g of urea in 360 g
(α) =
= 0.98 = 98%
of water. If the vapour pressure of pure water
1
−1
at this temperature is 35 mm Hg, lowering of
2
vapour pressure will be (Molar mass of urea =
486. Calculate Van't Hoff-factor for 0.2 m aqueous
60g mol–1)
solution of KCl which freezes at - 0.680°C.
(a) 0.027 mmHg
(b) 0.031 mmHg
–1
(K
f = 1.86 K kg mol )
(c) 0.017 mmHg
(d) 0.028 mmHg
(a) 3.72
(b) 1.83
[JEE Main 2019, 10 April Shift-I]
(c) 6.8
(d) 1.86
Ans. (c) : Given data,
MHT CET-02.05.2019, SHIFT-III
Weight of Urea = 0.60 g
Ans.
(b)
:
Given
that,
Weight of Water = 360 g
Molality = 0.2m
Molar mass of urea = 60 g/mole
Kf = 1.86 K Kg mol–1
o
Vapour pressure of pure water (P ) = 35mm Hg
∆Tf = 0.680oC
°
We know that,
P − Ps
n
=
∆Tf = i × Kf × m
P°
n+N
Objective Chemistry Volume-II
100
YCT
0.680 = i × 1.86 × 0.2
0.680
i=
1.86 × 0.2
i = 1.83
487. A solution of two liquids boils at a temperature
more than the boiling point of either of them.
Hence, the binary solution shows
(a) negative deviation from Raoult’s law
(b) positive deviation from Raoult’s law
(c) no deviation from Raoult’s law
(d) positive or negative deviation from Raoult’s
law depending upon the composition.
COMEDK-2019, KCET-2011
Ans. (a) : A solution of two liquid boils at a high
temperature that means vapour pressure of solution is
less then its shows negative deviation from Raoult's
law.
488. The latest heat of vaporization of a liquid at
227oC and 1 atm pressure is 12 kcal mol–1. The
change in internal energy, if 3 moles of the
liquid changes to vapour at same conditions is
(a) 33 kcal
(b) 39 kcal
(c) 4 kcal
(d) 15 kcal
BCECE-2018
Ans. (a) : Given, ∆ng = 3 – 0 = 3 moles
T = 227 + 273 = 500K
Form formula,
∆E = ∆H –∆ngRT
= 3×12–3×500×0.002
= 36 – 3 = 33 kcal
489. Molal depression constant =
(a)
R × M1 × Tb2
1000 × ∆vap H
(b)
R × M2 × Tf2
1000 × ∆f H
491. If x1 and x2 represent the mole fraction of a
component A in the vapour phase and liquid
mixture respectively and pAo and poB represent
Vapour pressure of pure A and B, then total of
vapour pressure of liquid mixture is
Po x
Po x
(b) A 2
(a) A 1
x2
x1
(c)
PBo x1
x2
(d)
PBo x 2
x1
AMU-2018
Ans. (b) : According to Dalton’s law of partial
pressures of gassesPA
= x1
Ptotal
PA = x1 × Ptotal .......(i)
According to Raoult’s law
PA = PA° x 2 .......(ii)
From equation (i) and equation (ii) we get:–
Ptotal x1 = PA° x 2
Ptotal =
PA° x 2
x1
492. A 0.0020 m aqueous solution of an ionic
compound freezes at -0.00732ºC. Number of
moles of in which one mole of the ionic
compound produces on being dissolved in
water will be (Kf = 1.88K kgm-1)
(a) 3
(b) 4
(c) 1
(d) 2
Assam CEE-2018
AIPMT-2009
R × M1 × Tf2
Ans. (d) : Given, K f = 1.88K kg m −1
1000 × ∆f H
Molalty of solution = 0.0020 m
JCECE - 2018
∆Tf = T –Ts
Ans. (d) : The correct formula for molal depression
= 0 – (–0.00732)
2
R × M1 × Tf
= 0.00732oC
constant (Kf) =
1000 × ∆f H
From formula–
It is defined as the depression in freezing point for 1
∆Tf
i=
molal solution, i.e. a solution containing 1 g mole of
Kf × m
solute dissolved in 1000 g of solvent.
0.00732
490. What happens on increasing pressure at
=
constant temperature?
1.88 × 0.002
(a) Rate of Haber process decreases.
= 1.946 ≃ 2
(b) Solubility of gas increases in liquid.
493. 58.5 g of NaCl and 180 g of glucose were
(c) Solubility of solid increases in liquid.
separately dissolved in 1000 mL of water.
(d) 2C(s) + CO 2 (g) → 2CO(g) reaction
moves
Identify the correct statement regarding the
elevation of boiling point (b.p) of the resulting
forward.
solutions.
[AIIMS-26 May, 2018 (M)]
(a) NaCl solution will show higher elevation of
Ans. (b): According to Henry’s law, “the partial
boiling point
pressure applied by any gas on a liquid surface is
(b) Glucose solution will show higher elevation
directly proportional to its mole fraction present in a
liquid solvent.”
of boiling point
So, on increasing pressure at constant temperature,
(c) Both the solutions will show equal elevation
solubility of gas will increase in liquid.
of boiling point
(c)
R × M2 × Tb2
1000 × ∆vap H
(d)
Objective Chemistry Volume-II
101
YCT
(d) The boiling point elevation will be shown by ∆T = T – T o
b
b
b
neither of the solutions
0.324 = Tb –80.1oC
JCECE - 2017, WBJEE-2012
Tb = 80.42oC
Ans. (a): Elevation in boiling point,
496. If Po and P are the vapour pressure of the pure
∆Tb = i × K b × m
solvent and solution and n1 and n2 are the moles
of solute and solvent respectively in the solution
n
Molality of NaCl solution = ×1000
then the correct relation between P and Po is
W
n1
n2
58.5 / 58.5
1000
(a) P o = P
(b) P o = P
=
×1000 =
WH 2O
WH2O
n1 + n 2
n1 + n 2
180
100 ×1000 = 1000
WH2O
WH2O
Both the solutions have same molality but the values of
(i) for NaCl and glucose are 2 and 1 respectively.
∆TNaCl = 2×∆TC6H12O6
Molality of C6H12O6 solution =
n1
(d) P = P o
n1 + n 2
WB-JEE-2016
Ans. (c) : According to the Raoult’s law –
Hence, NaCl will show higher elevation of boiling.
494. Which of the following aqueous solution has
Or
highest freezing point?
(a) 0.1 molal Al2(SO4)3 (b) 0.1 molal BaCl2
(c) 0.1 molal AlCl3
(d) 0.1 molal NH4Cl
Or
Karnataka-CET-2017
Ans. (d) : Depression in freezing point.
Or
∆Tf = i × K f × m
n2
(c) P = P o
n1 + n 2
n1
Po − P
=
o
P
n1 + n 2
P
n1
1− o =
P
n1 + n 2
P
n1
= 1−
Po
n1 + n 2
P
n2
=
P o n1 + n 2
observed value
theoretical value
n2
P= P o
n1 + n 2
3+
2−
For
Al 2 (SO 4 )3 = 2Al + 3SO 4 (i = 5)
497. 31 g of ethylene glycol (C2H6O2) is mixed with
+2
−
500 g of solvent (Kf of the solvent is 2 K kg molBaCl 2 = Ba + 2Cl
(i = 3)
1
). What is the freezing point of the solution (in
K)? (Freezing point of solvent =273 K)
AlCl3 = Al3+ + 3Cl−
(i = 4)
(a) 272
(b) 271
NH 4Cl = NH +4 + Cl−
(i = 2)
(c) 270
(d) 274
(d) 275
Hence, NH4Cl, has i = 2 which show highest freezing
Kerala-CEE-2016
point.
Ans.
(b)
:
From
formula,
495. What will be the boiling point of 30 g benzene
containing 0.75 g of benzoic acid which
W = mass of solute = 31g
undergoes 75% dimerization? Boiling point of ∆Tf= K WB ×1000 WB = mass of solvent = 500g
f
A
o
-1
pure benzene = 80.1 C and Kb= 2.53 K mol kg
M B × WA M = Ethylene glycol ( C H O )
2 6 2
B
(a) 90.52o C
(b) 104.35o C
o
o
2
×
31
×
1000
(c) 76.12 C
(d) 80.42 C
∆Tf =
= 2K
62 × 500
J & K CET-(2017)
Now, ∆Tf = Tfo − Tf
Ans. (d) : Given data,
Weigth of benzene = 30 g
∴Tf = 273 – 2
Weight of benzoic acid = 0.75 g
Tf = 271K
From formula
498. At 100°C the vapour pressure of a solution of
6.5 g of a solute in 100 g water is 732 mm. if Kb
∆Tb = iKb m …..(i)
= 0.52, the boiling point of this solution will be
α
i = 1– α+
(a) 102°C
(b) 103°C
n
(c) 101°C
(d) 100°C
0.75 ∴α = 0.75
(NEET-I-2016)
i = 1 − 0.75 +
n=2
2
Ans. (c) :
i = 0.625
Po − P
n
∵ A o S = B ( nA > nB )
Now putting the values in equation. (i)
PA
nA
0.75 /122
∆Tb = 0.625×2.53×
× 1000
760 − 732 WB × M A
30
⇒
=
760
M B × WA
∆Tb = 0.324
Where, i =
(
Objective Chemistry Volume-II
)
102
YCT
∴M B = 31.75
We know that,
∆Tb =Kb×m
6.5 × 1000
∆Tb = 0.52 ×
= 1.06
19.2 × 100
∴ Boiling point = 100+1.06 =101.06oC ≈ 101o C
499. Which of the following statements about the
composition of the vapour over an ideal 1:1
molar mixture of benzene and toluene is
correct?
Assume that the temperature is constant at
25°C.(Given, vapour pressure data at 25°C,
benzene = 12.8kPa, toluene = 3.85kPa)
(a) The vapour will contain amounts of benzene
and toluene.
(b) Not enough information is given to make a
prediction.
(c) The vapour will contain a higher percentage
of benzene.
(d) The vapour will contain a higher percentage
of toluene.
(NEET-I-2016)
°
Ans. (c) : PBenzene = X Benzene P Benzene
°
PToluene = X Toluene PToluene
For an ideal 1:1 molar mixture of benzene and toluene,
1
1
X Benzene = and X Toluene =
2
2
1 °
1
PBenzene = PBenzene = ×12.8kPa = 6.4kPa
2
2
1 °
1
PToluene = PToluene = ×3.85kPa = 1.925kPa
2
2
Thus, the vapour will contain a high percentage of
benezene as the partial pressure of benzene is higher as
compared to that of toluene.
500. For a non–volatile solute,
(a) Vapour pressure of solution is more than
vapour pressure of solvent
(b) Vapour pressure of solvent is zero
(c) Vapour pressure of solute is zero
(d) All of the above
UPTU/UPSEE-2016
Ans. (c): Non-Volatile solute always has zero vapour
pressure till no vapour is form from solute.
501. What is the freezing point of a 10% (by
weight) solution of CH3OH in water? [ Kf of
CH3OH = 1.86oC/m]
(a) 90oC
(b) 10oC
o
(c) 6.45 C
(d) – 6.45oC
[BITSAT – 2016], Panjab PMET-2011
10
Ans. (d) : 10% Weight CH3OH = l0g =
mol
32
So, Water will be present = 100 – 10
= 90g
moles of solute
Molality =
mass of solvent
Objective Chemistry Volume-II
10
Now, Molality = 32 ×1000
90
= 3.47m
∆Tf = K f .m
= 1.86o × 3.47
= 6.45o C
Then freezing point of the solution = 0oC –6.45oC
= – 6.45oC
502. The vapour pressure of pure benzene and
toluene are 160 and 60 mm Hg respectively.
The mole fraction of benzene in vapour phase
in contact with equimolar solution of benzene
and toluene is
(a) 0.073
(b) 0.027
(c) 0.27
(d) 0.73
AP-EAMCET – 2016
Ans. (d) : Vapour pressure of pure benzene = 160 of Hg
Vapour pressure of pure toluene = 60 mm Hg
For equimolar solution,
XB = XT = 0.5
PB = XB×PB0 = 0.5 ×160 = 80 mm
PT = XT × PT0 = 0.5 × 60 = 30 mm
Ptotal = 80 + 30 = 110 mm
∴ Mole fraction of benzene in vapour phase
P
80
= B =
= 0.73
Ptotal 110
Mole fraction of toluene in vapour phase
P
30
= T =
= 0.27
PTotal 110
503. Partial vapour pressure of a solution
component is directly proportional to its mole
fraction. This statement is known as
(a) Raoult’s law
(b) distribution law
(c) Henry’s law
(d) Ostwald’s dilution law
COMEDK-2016
Ans. (a) :Partial vapour pressure of a solution
component is directly proportional to its mole fraction.
This statement is know as Raoult's law.
i.e. PA = PAo X A
Where, PAo = Partial pressure of pure component.
504. Assertion : If a liquid solute more volatile than
the solvent is added to the solvent, the vapour
pressure of the solution may increase i.e., ps>p°.
Reason : In the presence of a more volatile
liquid solute, only the solute will form the
vapours and solvent will not.
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
103
YCT
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.
AIIMS-2016
Ans. (c): In the presence of volatile solute, solvent will
also form vapours. So, assertion is correct but its reason
incorrect.
505. Lowering of vapour pressure is highest for
(a) 0.1 M BaCl2
(b) 0.1 M glucose
(c) 0.1 M MgSO4
(d) 0.1 M Urea
UPTU/UPSEE-2015
Ans. (a) : Lowering of vapour pressure ∝ Vant Hoff
factor,
For BaCl2, i = 3(maximum)
So, BaCl2 has highest value for relative lowering of
vapour pressure.
506. 12 g of a non-volatile solute dissolved in 108 g
of water produces the relative lowering of
vapour pressure of 0.1. The molecular mass of
the solute is
(a) 80
(b) 60
(c) 20
(d) 40
JCECE - 2015
Ans. (c) : Given,
w = 12 g; W = 108 g, m = ?
∆P
M = 18 g, o = 0.1
P
From Raoult's law, relative lowering in vapour pressure
∆P Pº −P n w M
=
= = ×
Pº
Pº
N m W
12 18
0.1 = ×
m 108
12 × 18
m=
= 20
10.8
507. 18 g of glucose is dissolved in 178.2 g of water.
The vapour pressure of the solution at 100o C is
(vapour pressure of pure water at 100o C is 760
mm Hg.)
(a) 767.6 mm Hg
(b) 760 mm Hg
(c) 752.4 mm Hg
(d) 725.4 mm Hg
(e) 745.2 mm Hg
Kerala-CEE-2015
Ans. (c) : Total number of moles
= x glucose + xwater
18 178.2
=
+
18
180
= 10 moles
Now,
∆P 0.1
=
∆Pº 10
∆ P = 0.01 Pº = 0.01 × 760
= 7.6 mm Hg
Vapour pressure of solution = (760 – 7.6) mm Hg
= 752.4 mm Hg
508. The vapour pressure of pure benzene at certain
temperature is a 1 bar. A non-volatile, non
electrolyte solid weighing 2 g when added to 39
g of benzene (molar mass 78 g mol–1) yields
solution of vapour pressure of 0.8 bar, The
molar mass of solid substance is
Objective Chemistry Volume-II
(a) 32
(c) 64
(b) 16
(d) 48
J & K CET-(2015)
Ans. (b) : Given,
Vapour pressure of pure benzene(P o ) = 1bar
Vapour pressure of solution (Ps ) = 0.8bar
Molar mass of benzene = 78 g mole –1
Weight of non-volatile, non-electrolyte solid = 2 g
From formula,
P o − Ps n 2
=
Ps
n1
1 − 0.8 W2 M1
=
0.8
M 2 W1
0.8 × 2 × 78
M2 =
0.2 × 39
M 2 = 16g mol−1
509. 0.5 molal solution of a solute in benzene shows
a depression in freezing point equal to 2K.
Molal depression constant for benzene is 5 K
kg mol–1. If the solute forms dimer in benzene,
what is the % association?
(a) 40
(b) 50
(c) 60
(d) 80
J & K CET-(2015)
Ans. (a) : Given,
Molality =0.5m,
Molal depression constant of benzene (Kf) = 5K kg
mol–1,
Depression in freezing point (∆Tf ) = 2K and α = ?
By Vant Hoff factor, ∆Tf = K f × m
∆Tf
2
=
K f × m 5 × 0.5
2
i=
= 0.8
2.5
% Association of acid (α ) , while solute from dimer in
benzene, i.e, 2AA 2
∴ value of n = 2
i − 1 0.8 − 1
α association =
=
1
1
−1
−1
n
2
% of α association = 0.4 × 100 = 40%
i=
510. Molar enthalpy change for melting of ice is 6
kJ/mol. Then the internal energy change (in
kJ/mol) when 1 mole of water is converted into
ice at 1 atm at 0°C is
(a) RT/1000
(b) 6
(c) 6 – (RT/1000)
(d) 6 + (RT / 1000)
J & K CET-(2015)
Ans. (b) : Given,
Molar enthalpy change (∆H) = 6kJ / mol
In this dissociation H2O(l) ↽ ⇀ H2O(s)
104
YCT
From formula,
∆H = ∆U + ∆ngRT
∆ng = 0
So,
∆H = ∆U = 6 kJ/mol
511. 18 g of glucose is dissolved in 90 g of water. The
relative lowering of vapour pressure of the
solution is equal to :
(a) 6
(b) 0.2
(c) 5.1
(d) 0.02
AP-EAMCET (Engg.) 2015
Ans. (d) : Given that,
Weight of glucose (w1) = 18 gm
Weight of water = w2 = 90 gm
Molar mass of glucose (M2) = 180
Molar mass of water (M1) = 18
From Raoult's law for non volatile solute–
Pº –Ps
n2
=
Pº
n1 + n 2
For dilute solution–
Pº –Ps n 2 w 2 × M1
=
=
Ps
n1 m 2 × W1
Putting these value, we get–
Pº –Ps
18 × 18
=
= 0.02
Ps
180 × 90
513. The vapour pressure of pure liquid solvent A is
0.80 bar. When a non-volatile substance B is
added to the solvent, its vapor pressure drops
to 0.60 bar. What is the mole fraction of
component B in the solution?
(a) 0.75
(b) 0.50
(c) 1.5
(d) 0.25
SCRA-2015, J & K CET-2010
Ans. (d): Given that,
Partial pressure of pure liquid A (P) = 0.60 bar
Vapor pressure of component (Po) = 0.80 bar
According to Raoult's law
P = Po × X
Putting these value we get,
0.60 = 0.80 × X
or
X = 0.75
∴ The mole fraction of component (B) = 1 – 0.75
= 0.25
514. To observe an elevation of boiling point of
0.05oC, the amount of a solute (mol. wt = 100)
to be added to 100 g of water (Kb = 0.5) is
(a) 2 g
(b) 0.5 g
(c) 1 g
(d) 0.75 g
WB-JEE-2014
Ans. (c) : We know that,
W × K f × 1000
∆Tb =
M×W
W × 0.5 × 1000
∆Tb =
100 × 100
Putting the value:–
W × 0.5 × 1000
0.05 =
100 × 100
So,
W=1g
512. The vapour pressure of acetone at 20°C is 185
torr. When 1.2g of a non-volatile substance was
dissolved in 100 g of acetone at 20°C, its vapour
pressure was 183 torr. The molar mass (g mol–
1
) of the substance is
515. A solution of 1.25 of 'P' in 50g of water lowers
(a) 32
(b) 64
freezing point by 0.3oC. Molar mass of 'P' is 94.
(c) 128
(d) 488
Kf (water) = 1.86 K kg mol–1. The degree of
[JEE Main 2015]
association of 'P' in water is
Ans. (b) : Given,
(a) 80%
(b) 60%
Vapour pressure of acetone = 185 torr
(c) 65%
(d) 75%
Karnataka-CET-2014
Weight of non-volatile substance = 1.2 g
Ans. (a) : From formula
Weight of acetone = 100 g
∆T f = i × Kf × m
Vapour pressure of non-volatile substance = 183 torr
w ×1000
From formula,
= i × Kf × 1
W1 × M1
Psolvent − Psolute n Where, n = moles of solute
=
1.25 ×1000
N = moles of solvent
Psolute
N
0.3 = i × 1.86 ×
50 × 94
1.2
i
=
0.6064
185 − 183 M
=
i −1
100
183
now, degree of association of 'P' in water α =
1
58
−1
n
185 −183 1.2 58
=
×
0.6064 − 1
183
M 100
=
1
183 ×1.2 × 58
−1
M=
2
2 × 100
= 0.7872
M = 63.684
= 78.72%
≃ 64g / mol.
≃ 80%
Objective Chemistry Volume-II
105
YCT
516. An aqueous solution containing 3 g of a solute
∆Tf = 1.80, m=0.001
of molar mass 111.6 g mol-1 in a certain mass of
0.0054
i=
=3
water freezes at -0.125 oC. The mass of water in
0.001× 1.80
grams present in the solution is (Kf = 1.86 K kg
water
[Pt(HN 3 ) 4 Cl 2 ]Cl2 ↽ ⇀[Pt(NH 3 ) 4 Cl2 ]2+ + 2Cl−
mol-1)
(a) 300
(b) 600
519. A pressure cooker reduces cooking time for
(c) 500
(d) 400
food because
(e) 250
(a) heat is more evenly distributed in the cooking
Kerala-CEE-2014
space
(b) boiling point of water involved in cooking is
Ans. (d) : From formula–
increased
K f × w × 1000 w and M = weight and molar
(c)
the
higher pressure inside the cooker crushes
∆T f =
mass of solute
the
food
material
M×W
W
=
weight
of
solvent
or
water
(d) cooking involves chemical changes helped by
Putting the value,
a rise in temperature.
∆Tf = 0 − (−0.125)
SRMJEEE-2014, AIEEE-2003
= 0.125
1.86 × 3 × 1000
0.125 =
111.6 × W
1.86 × 3 × 1000
W=
0.125 ×111.6
= 400 g
517. If M is molecular weight of solvent, kb is molal
elevation constant, Tb is its boiling point, po is
its vapour pressure at temperature T and Ps is
vapour pressure of its solution having a non–
volatile solute at T K, then
(a)
P o − Ps ∆Tb
=
×M
Po
Kb
(b)
P o − Ps K b
=
×M
Po
∆Tb
(c)
P o − Ps K b
M
=
×
∆Tb 1000
Po
(d)
P o − Ps ∆Tb
M
=
×
Po
K b 1000
P° − Ps
w M
.
m W
P
24 − PS
18
So,
= 0.1×
24
180
24 – Ps = 0.01×24
Ps = 24 – 0.24 = 23.76 mm Hg
521. Vapour pressure of water at 293 K is 17.535
mm Hg. The vapour pressure of water at 293 K
containing 25 g of glucose dissolved in 450 g of
water is
(a) 17.439 mm Hg
(b) 17.535 mm Hg
(c) 0.097 mm Hg
(d) 34.973 mm Hg
COMEDK-2014
Ans. (a) :
W1
P° – Ps
M1
W1 → solute
=
W → solvent
W
W
P°
1
+ 2 2
M1 M 2
25
17.535 – Ps
180
=
25 450
17.535
+
180 18
25
17.535 – Ps
180
=
25 + 4500
17.535
180
°
UPTU/UPSEE-2014
Ans. (d) :
P o − Ps n Molality × M
= =
Po
N
1000
∆Tb
Molality =
(∵∆Tb = K b × m )
Kb
So,
P o − Ps ∆Tb
M
=
×
o
K b 1000
P
518. A 0.001 molal solution of [Pt(NH3)4 Cl4] in
water had a freezing point depression of
0.00540C. If Kf for water is 1.80 the correct
formulation of the above molecule is
(a) [Pt(NH3)4Cl3]Cl
(b) [Pt(NH3)4Cl2]Cl2
(c) [Pt(NH3)4Cl]Cl2
(d) [Pt(NH3)4Cl4]
UPTU/UPSEE-2014
Ans. (b) : According to the depression in freezing point
∆ T f= i × K f × m
Given that,
Objective Chemistry Volume-II
Ans. (b) : A pressure cooker reduces cooking time for
food because at altitudes by raising the pressure and
boiling point the pressure cooker cooks the food at a
faster rate.
520. Vapour pressure in mm Hg of 0.1 mole of urea
in 180 g of water at 25° C is (The vapour
pressure of water at 25°C is 24 mm Hg)
(a) 2.376
(b) 20.76
(c) 23.76
(d) 24.76
AP-EAMCET (Engg.)-2014
Ans. (c) : Given that, P° = 24, M = 18, W = 180
w
Number of moles urea = = 0.1
m
According to the Raoult's law of dilute solution –
106
=
YCT
17.535 – Ps
25
=
17.535
4525
17.535–Ps = 0.00 552 × 17.535
– Ps = 0.0968 – 17.353
Ps = 17.4381
522. Which one of the following is the ratio of the
lowering of vapour pressure of 0.1M aqueous
solutions of BaCl2, NaCl and Al2(SO4)3
respectively?
(a) 3 : 2 : 5
(b) 5 : 2 : 3
(c) 5 : 3 : 2
(d) 2 : 3 : 5
BCECE-2014
AP EAMCET (Engg.)-2011
Ans. (a) : Lowering of vapour pressure depends on the
number of particles of the solute (i).
∆P
= i Kf m
P°
∆P
⇒
∝i
P°
BaCl2 → Ba2+ + 2Cl–
3 moles of ions
NaCl → Na+ + Cl–
2 moles of ions
Al2 (SO4)3 →2Al3+ + 3SO 24−
5 moles of ions
So, the ratio will be
3: 2: 5
523. 45 g of ethylene glycol (C2H6O2) is mixed with
600 g of water, what is the depression of
freezing point?
(a) 7.9 K
(b) 2.5 K
(c) 6.6 K
(d) 2.2 K
J & K CET-(2014)
Ans. (d) : Given,
K f = 1.86 g mol −1 , Weight of ethylene glycol(w) = 45g
Weight of water (W) = 600g ,
Molecular mass of C6 H 6 O 2 ( m ) = 62
From formula–
K × w × 1000
∆Tf = f
m×W
1.86 × 45 × 1000
∆Tf =
62 × 600
∆Tf = 2.25k.
(a) 50
(c) 37.5
(b) 25
(d) 53.5
SRMJEEE-2014
Ans. (a) : Given that,
P° = 75 torr (for benzene)
P° = 22 torr (for toluene)
Pbenzene = ?
Amount of benzene = 78 g
78
∴ Moles of benzene =
= 1mole
78
Amount of toluene = 46 g
46 1
∴ Moles of toluene =
= mole
92 2
According to the Raoult’s law –
P = P° X
Where, P = Partial vapour pressure
P° = Vapour pressure.
X = Mole fraction
1
∴ P = 75 ×
1
1+
2
2
or
P = 75 ×
3
or
P = 50 torr
526. The measured freezing point depression for a
0.1 m aqueous CH3COOH solution is 0.19oC.
The acid dissociation constant Ka at this
concentration will be (Given, Kf the molal
cryoscopic constant = 1.89 K kg mol-1)
(a) 4.76 × 10-5
(b) 4 × 10-5
-5
(c) 8 × 10
(d) 2 × 10-5
WB-JEE-2013
Ans. (b) : According to the depression in freezing point
∆Tf= i×Kf×m
∆Tf
So ,
i=
Kf × m
0.19
=
1.86 × 0.1
= 1.02
i − a 1.02 − 1
Since, α =
=
= 0.02 = 2× 10-2
n −1
2 −1
Now, CH3COOH ↽ ⇀ CH3COO– + H+
Ka = Cα2
524. Henry’s law constant value for O2 in water is
= 0.1 ×(2×10–2)2
(a) 46.82
(b) 43.86
= 4×10–5
(c) 88.84
(d) 76.48
527. The vapour pressure in mm of Hg, of an
J & K CET-(2014)
aqueous solution obtained by adding 18g of
glucose (C6H12O6) to 180 g of water at 1000C is
Ans. (a) : Henry’s law constant value for O2 in water is
(a) 7.60
(b) 76.0
46.82K bar at 303K and 34.86K bar at 293K.
(c)
759
(d)
752.4
525. Benzene and toluene form nearly ideal solution.
AP-EAMCET (Engg.)-2013
At 20°C, the vapour pressure of benzene is 75
torr and that of toluene is 22 torr. The partial Ans. (d) :
vapour pressure of benzene at 20°C for a
W 180
solution containing 78 g of benzene and 46 g of Moles of solvent ( n1 ) = 1 =
= 10 mol
M
18
toluene (in torr) is
1
Objective Chemistry Volume-II
107
YCT
W2 18
=
= 0.1 mol
M 2 180
According to Raoult's law.
P 0 − PS
n2
=
0
n1 + n 2
P
Moles of solute ( n 2 ) =
Moles of solvent (n1) =
P 0 − Ps
0.1
=
0
10 + 0.1
P
10.1 P 0 − Ps = 0.1 P 0
(
)
10P = 10.1 Ps
0
10 × 760
= 752.4 mm Hg.
10.1
528. The boiling point of a solution containing 68.4 g
of sucrose (molar mass = 342 g mol–1) in 100 g
of water is
(Kb for water = 0.512 K kg mol–1)
(a) 98.98°C
(b) 101.02°C
(c) 100.512°C
(d) 100.02°C
AP-EAMCET (Medical)-2013
Ans. (b) : Given that,
Amount of solvent (w1) = 100 g
Amount of solute (w2) = 68.4 g
Molar mass of solute (M2) = 342 g mol–1
Kb = 0.512 K kg mol–1
Tb° = 100°C
or Ps =
100
mole
18
Now, From Rault's laww2
1.8
60
=
100 w 2 100
+
60 18
1.8
w2
or
=
3w
100
2 + 1000
60
180
1.8 w 2
180
or
=
×
100 60 ( 3w 2 + 1000 )
or
3w 2
1.8
=
100 3w 2 + 1000
or
100
1000
= 1+
1.8
3w 2
or
w 2 = 6.10
or
w 2 ≈ 6g
530. At a particular temperature, the vapour
pressures of two liquids A and B are
respectively 120 and 180 mm of mercury. If 2
moles of A and 3 moles of B are mixed to form
an ideal solution, the vapour pressure of the
solution at the same temperature will be (in
1000 × K b × w 2
mm of mercury)
Now, ∆Tb =
(a) 156
(b) 145
w1 × M 2
(c) 150
(d) 108
1000 × 0.512 × 68.4
or
∆Tb =
AIIMS-2013
100 × 342
Ans. (a): Given data,
or
∆Tb = 1.024°C
Vapour pressure of liquid A = 120 mm Hg
Thus, boiling point of solutionVapour pressure of liquid B = 180 mm Hg
Tb = Tb° + ∆Tb
Number of moles of liquid A = 2
Tb = 100 + 1.024 = 101.024°C
Number of moles of liquid B = 3
529. The weight in gram, of the non-volatile solute
Vapour pressure of solution
urea (NH2CONH2) to be dissolved in 100 g of
PA + PB = X A PAo + X B PBo
water in order to decrease its vapour pressure
2
3
by 1.8% is
= ×120 + × 180
(a) 6.0
(b) 0.3
5
5
(c) 3.0
(d) 0.6
= 48 + 108 = 156 mm Hg
AP-EAMCET (Medical)-2013 531. What will be the freezing point of a 1%
Ans. (a) : According to the Roult's lawsolution of glucose in water, given that molal
depression constant for water is 1.84 K kg
P o − Ps
n2
=
mol–1?
n1 + n 2
Po
(a) 272.898 K
(b) 0.102oC
Where, Po = Vapour pressure of pure water
(c) 273 K
(d) 0.108oC
n2 = Moles of solute
J & K CET-(2013)
n1 = Moles of solvent
Ans. (a) : Mass of glucose = 1 g
Let vapour pressure of pure water = 100 mm
Mass of solvent = 100 –1 = 99 g
lowering in vapour pressure = 1.8 mm (Given)
Mass of solution = 100 g
w
Moles of solute (n2) = 2 mole
So,
60
Objective Chemistry Volume-II
108
YCT
= 1.84×
1
1
×
180 99 /100
w
1
∵ ∆Tf × ×
m
W
/1000
= 0.103
Now,
Freezing point of solution = 273–0.103
= 272.897 K
532. The aqueous solution that has the lowest
vapour pressure at a given temperature is
(a) 0.1 molal sodium phosphate
(b) 0.1 molal barium chloride
(c) 0.1 molal sodium chloride
(d) 0.1 molal glucose
UP CPMT-2013
Ans. (a) : The same concentrations of relative lowering
of vapour pressure depends upon the number of ions or
particles present in different solutions.
Na 3PO 4 → 3Na + + PO34− = 4 ions
BaCl2 → Ba 2 + + 2Cl− = 3 ions
NaCl → Na + + Cl− = 2 ions
Glucose → no dissociation = 1 ions
As Na3PO4 gives maximum ions (4) on dissociation,
hence relative lowering of vapour pressure is maximum
for sodium phosphate (Na3PO4) and it has lowest
vapour pressure.
533. 0.4 g of H2, 22 g of CO2 and 6.4 g of SO2 are
taken in a container. Total pressure is found to
be 1.6 atm. What is the partial pressure of CO2
gas in the mixture?
(a) 1.0 atm
(b) 0.8 atm
(c) 0.6 atm
(d) 1.6 atm
SRMJEEE-2013
Ans. (a) : Given that, PTotal = 1.6 atm
0.4
Amount of H2 = 0.4 g, moles of H2 =
= 0.2 mole
2
22
Amount of CO2 = 0.22 g, moles of CO2=
= 0.5mole
44
6.4
Amount of SO2 = 6.4g, moles of SO2 =
= 0.1mole
64
Now, P = Ptotal × X CO2
Where, P = Partial pressure
PTotal = Total pressure
X CO2 = Moles fraction of CO2
0.5
0.5 + 0.2 + 0.1
0.5
PCO2 = 1.6 ×
0.8
PCO2 = 1atm
PCO2 = 1.6 ×
Ans. (d) : On mixing acetone and chloroform shows
negative deviation from Raoult's law due to the
formation of hydrogen bond with acetone molecule by
chloroform molecule.
535. 58.4 g of NaCl and 180 g of glucose were
separately dissolved in 1000ml of water.
Identify the correct statement regarding the
elevation of boiling point (b.p) of the resulting
solutions.
(a) NaCl solution will show higher elevation of
boiling point
(b) Glucose solution will show higher elevation
of boiling point
(c) Both the solutions will show equal elevation
of boiling point
(d) The boiling point elevation will be shown by
neither of the solutions
WB-JEE-2012
Ans. (a) :
58.4g
Molality of NaCl =
× 1000g / kg
58.4g ×1000 × 1
=1m
For NaCl Solution, using the formula
∆Tb ∝ i×m
= i×m
= 2×1 = 2
Now, Molality of glucose
180g
=
×1000
180g / mol ×1000 ml × 1g / ml
=1m
For glucose ∆T∝ i×m
= 1×1= 1
Now, we can say that NaCl solution will show higher
elevation of boiling point.
536. PA and PB are the vapour pressures of pure
liquid components, A and B, respectively of an
ideal binary solution. If xA represents the mole
fraction of component A, the total pressure of
the solution will be
(a) PA +xA(PB–PA)
(b) PA +xA(PA–PB)
(c) PB +xA(PB–PA)
(d) PB +xA(PA–PB)
(AIPMT -2012)
Ans. (d) : According to Roult's law, if volatile liquid
added in pure solvent, then total pressure equal to sum
of the partial pressure of volatile liquid and solvent.
Total Pressure–
P T = XA P A + XB P B
For fraction (XA + XB) = 1(binary Solution)
So XB = 1–XA
Now putting the value.
PT = XA PA + (1–XA)PB
=XA PA+ PB–XA PB
PT = PB +X A (PA -PB )
534. Which of following will show a negative
deviation from Raoult’s law?
(a) Acetone-benzene
537. Vapour pressure of chloroform (CHCl3) and
(b) Acetone -ethanol
dichloromethane (CH2Cl2) at 25°C are 200 mm
(c) Benzene-methanol
Hg and 41.5 mm Hg respectively. Vapour
pressure of the solution obtained by mixing
(d) Acetone -chloroform
25.5 g of CHCl3 and 40 g of CH2Cl2 at the same
WB-JEE-2012
temperature will be (Molecular mass of
Objective Chemistry Volume-II
109
YCT
(a) 11.4 g
(b) 9.8 g
CHCl3 = 119.5 u and molecular mass of CH2Cl2
= 85 u)
(c) 12.8 g
(d) 10 g
(a) 173.9 mm Hg
(b) 615.0 mm Hg
Karnataka-CET-2012
(c) 347.9 mm Hg
(d) 285.5 mm Hg
Ans. (d) : From Raoult’s law(AIPMT -Mains-2012)
Pº −P w M
= ×
40
P
m W
Now,
putting
the
value
85
Ans. (c) : Mole fraction of CH2Cl2 =
40 25.5
100 − 80 w 114
+
= ×
85 119.5
80
40 114
0.47
20
×
40
=
w=
0.47 + 0.213
80
=0.688
w = 10g
Now, mole fraction of CHCl3=1 – 0.688
541. Freezing point of an aqueous solution is
= 0.312
–0.186oC. If the values of Kb and Kf of water
From formula,
are respectively 0.52K kg mol–1 and 1.86 K kg
mol–1 then the elevation of boiling point of the
PT = P1o + (P2o − P1o )X 2
solution in K is
= 200+ (415-200)×0.688
(a) 0.52
(b) 1.04
= 200+147.9
(c)
1.34
(d)
0.134
=347.9mm Hg
(e)
0.052
–1
538. Kf for water is 1.86 K kg mol . If your
Kerala-CEE-2012
automobile radiator holds 1.0 kg of water, then
how many grams of ethylene glycol (C2H6O2) Ans. (e) : Elevation of boiling point = Kbm
T 2 – T 1 = K bm
must you add to get the freezing point of the
solution lowered to –2.8°C?
Depression in freezing point = Kfm = T3 – T4
(a) 72 g
(b) 93 g
Depression in freezing point = 0.186ºC
(c) 39 g
(d) 27 g
So,
[AIEEE-2012] 0.186 = 1.86 × m
Ans. (b) Glycol is a non-electrolyte and it is coolant
m = 0.1
Now,
∆Tf = 0 − (−2.8°C)
T 2 – T 1 = K bm
∆Tf = 2.8°C, K f = 1.86
= 0.52 × 0.1
1000K f W1
= 0.052
∆Tf =
M1W2
542. Assertion: Lowering of vapour pressure is
directly proportional to osmotic pressure of the
W1 = Weight of ethyleneglycol
solution
W = Weight of water
Reason:
Osmotic pressure is a colligative
2
property.
M1 = Molecular weight of ethylene glycol
(a) If both Assertion and Reason are correct and
1000 × 1.86 × W1
the Reason is the correct explanation of
2.8 =
62 × 1000
Assertion.
2.8 × 62
(b) If both Assertion and Reason are correct, but
W1 =
Reason is not the correct explanation of
1.86
Assertion.
W1 = 93.33g ≈ 93g
(c) If Assertion is correct but Reason is incorrect.
539. The dissolution of a gas in a liquid is governed
(d) If both the Assertion and Reason are
by
incorrect.
(a) Raoult's law
AIIMS-2012
(b) Henry's law
Ans. (b): According to Van't Hoff equation for dilute
(c) Dalton's law of partial pressure
solutions is(d) Van't Hoff factor.
J & K CET-(2012) Π = n RT
V
Ans. (b) : The dissolution of a gas in a liquid is
N → number of moles of solvent
governed by Henry’s law. Henry’s law states that the
NM
M → molecular weight
weight of the gas dissolved in liquid is proportional to V =
ρ → density
ρ
the pressure of the gas upon the liquid.
V → volume
540. The mass of a non-volatile solute of molar mass
40g mol–1 that should be dissolved in 114g of n = ΠM
N ρRT
octane to lower its vapour pressure by 20% is
Objective Chemistry Volume-II
110
YCT
From Raoult 's law
Po − P n
=
Po
N
Po − P ΠM
=
Po
ρRT
Π
M
Po − P =
× Po
ρRT
MP o
factor is constant at constant temperature
ρRT
(Po–P) ∝ Π
lowering of vapour pressure ∝ osmotic pressure.
The reason osmotic pressure is a colligative property is
also true but not the correct explanation of the assertion.
543. 12 g of urea is dissolved in 1 litre of water and
68.4 g of sucrose is dissolved in 1 litre of water.
The lowering of vapour pressure of first case is
(a) equal to second
(b) greater than second
(c) less than second
(d) double of that of second
AIIMS-2012
Wt.of substance
Ans. (a): Number of moles =
Molecular mass
WA
nA
MA
XA =
=
n A + n B WA + WB
MA MB
WA
60
0.02=
WA 90
+
60 18
WA
WA
+5 =
60
0.02 × 60
WA
WA
+5=
60
1.2
WA WA
5=
−
1.2 60
WA = 6.122
WA ≈ 6
545. The mass of a non-volatile solute of molar mass
60 g mol–1 that should be dissolved in 126 g of
water to reduce its vapour pressure to 99% will
be
(a) 4 g
(b) 8 g
(c) 5.6 g
(d) 3 g
COMEDK-2012
Ans. (a) : Mole fraction of the non-volatile solute
= lowering in vapour pressure.
P° – Ps
W2 / M 2
X2 =
=
P°
W1 / M1 + W2 / M 2
100 – 99
W2 / 60
=
100
126 /18 + W2 / 60
W2 = 4.2gm ≈ 4gm
Molecular mass of urea ( NH 2 CONH 2 )
= 14 + 2 + 12 + 16 + 14 + 2
= 16 + 12 + 16 + 16
= 60
Number of moles of urea in 1 liter of water
12 1
=
= = 0.2
60 5
546. When 1.20 g of sulphur is melted with 15.00 g
Molecular mass of sucrose [C12 H 22 O11 ]
of naphthalene, the solution freezes at 77.2oC.
= 12×12 + 1× 22 + 16×11
What is the molar mass of this form of
= 342
sulphur?
[Data for Naphthalene:
Number of moles of sucrose in 1 liter of water
Melting point = 80°C
68.4 1
=
= = 0.2
Freezing point depression constant,
342 5
Kf = 6.80oC m-1]
Urea and sucrose are non-electrolytic solutes and have
(a) 160 g mol–1
(b) 190 g mol–1
same concentration.
–1
(c) 260 g mol
(d) 450 g mol–1
So, both witll have equal lowering of vapour pressure.
SCRA-2012
544. The weight in grams of a non-volatile solute
–1
(mol. wt. 60) to be dissolved in 90g of water to Ans. (b) : Given that, Kf = 6.80°Cm , W2 = 1.2 gm
produce a relative lowering of vapour pressure
W1 = 15 gm, ∆Tf = Tf° − Tf = 80 − 77.2 = 2.8o C
of 0.02 is
M=?
(a) 4
(b) 8
We know that,
(c) 6
(d) 10
AP-EAMCET (Engg.)-2012
K × W2 × 1000
∆Tf = f
Ans. (c) : Given that,
M × W1
Relative lowering of vapour pressure
Where ∆Tf = depression in freezing point
Po − P
0.02
=
K f = Molaldepression constant.
Po
Molecular weight (MA) = 60
W2 = Weight of solute.
Weight of water (WA) = 90 gram
Wl = Weight of solvent
Po − P
=
X
M = Molar mass
A
Po
Objective Chemistry Volume-II
111
YCT
∴
M=
Tf × W2 × 1000
∆Tf × W1
Ans. (d) : From formula,
∆Tf = iKfm
6.80 × 1.2 × 1000
2.8 × 15
M =194.28g mol−1
M=
or
M ≈ 190 g mol–1
547. The vapour pressure lowering caused by the
addition of 100g of sucrose (molecular mass =
342) to 1000 g of water the vapour pressure of
pure water at 25oC 23.8 mm Hg is
(a) 0.12 mm Hg
(b) 0.125 mm Hg
(c) 1.15 mm Hg
(d) 1.25 mm Hg
UP CPMT-2011
Ans. (b) : Moles of sucrose, (n) =
100
= 0.292 mol
342
1000
= 55.5 mol
18
Vapour pressure of pure water, (Po) = 23.8 mmHg
According to Raoult’s law,
∆P
n
=
o
P
n+N
∆P
0.292
=
23.8 0.292 + 55.5
23.8 × 0.292
∆P =
= 0.125mm Hg
55.792
548. The freezing point depression constant for
water is –1.86 °C m–1 . If 5.00 g Na2SO4 is
dissolved in 45.0 g H2O, the freezing point is
changed by –3.82 °C. Calculate the Van't Hoff
factor for Na2SO4.
(a) 2.05
(b) 2.63
(c) 3.11
(d) 0.381
(AIPMT -2011)
Ans. (b) : Given that, ∆Tf = −3.82°C
Moles of water, (N) =
K f = −1.86°C m−1
Weight of solvent (H 2 O) WA = 45g
From formula,
i=
∆Tf × WA
K f × n B ×1000
(nB = no. of moles of solute (Na2SO4))
−3.82 × 45
=
5
−1.86 ×
× 1000
142
= 2.63
549. A 0.1 molal aqueous solution of a weak acid is
30% ionized. If Kf for water is 1.86°C/m, the
freezing point of the solution will be
(a) –0.18 °C
(b) –0.54°C
(c) –0.36°C
(d) –0.24°C
(AIPMT-2011)
Objective Chemistry Volume-II
= 1.3 × 1.86 × 0.1 (∴ i = 1 − α + nα )
i = 1.3
∆Tf = +0.24ºC
Tf = 0 – 0.24°C = – 0.24°C
550. 200mL of an aqueous solution of a protein
contains its 1.26 g. The osmotic pressure of this
solution at 300K is found to be 2.57×10–3bar .
The molar mass of protein will be (R = 0.083 L
bar mol–1K–1)
(a) 5122 g mol–1
(b) 122044 g mol–1
(c) 31011 g mol–1
(d) 61038 g mol–1
(AIPMT -Mains-2011)
Ans. (d) : Given that,
R=0.083 L bar mol–1 K–1 , W = 1.26g
V = 200 mL, T = 300K
We know that,
Π = CRT
w
Π V = RT
M
200 1.26
2.57 × 10−3 ×
=
× 0.083 × 300
1000
M
–1
M = 61038 g mol
551. A solution containing 1.8 g of a compound
(empirical formula CH2O) in 40 g of water is
observed to freeze at – 0.465oC. The molecular
formula of the compound is (Kf of water = 1.86
kg K mol-1)
(a) C2H4O2
(b) C3H6O3
(c) C4H8O4
(d) C5H10O5
(e) C6H12O6
Kerala-CEE-2011
Ans. (e) : From formula:
∆T f = Kf × M
1.8 1000
0.465 = 1.86 ×
×
M
40
M = 180
Now,
molar mass
180
n=
=
=6
empiricalformula weight 30
So, molecular formula of compound will be C6H12O6
552. 34.2 g of cane sugar is dissolved in 180 g of
water. The relative lowering of vapour
pressure will be
(a) 0.0099
(b) 1.1597
(c) 0.840
(d) 0.9901
MHT CET-2011
Ans. (a) : From formula,
w2
Pº −Ps
m2
=
w1 w 2
Pº
+
m1 m 2
112
YCT
34.2
342
=
34.2 180
+
342 18
0.1
=
0.1 + 10
0.1
=
= 0.0099
10.1
553. The freezing point of a solution composed of
10.0g of KCl in 100 g of water is 4.5°C.
Calculate the Van’t Hoff factor, i for this
solution.
(a) 2.50
(b) 1.8
(c) 1.2
(d) 1.3
VITEEE- 2011
Ans. (b) : Given that, W1 = 10 gm of KCl, ∆ Tf =
4.5°C
W2 = 100 gm of water ,
i=?
W × 1000
molality = 1
∴
M × W2
10 × 1000
molality =
= 1.34 m
74.55 ×100
Now, we use the following equation –
∆Tf = i. Kf.m
Where , ∆Tf = depression in freezing point
Kf = Cryoscopy constant
∆Tf
i=
or
K f .m
4.5°C
i=
(1.86°C / m)(1.34m)
i = 1.80
554. When a solution containing non-volatile solute
is diluted with water
(a) its osmotic pressure increases
(b) its boiling point increases
(c) its freezing point decreases
(d) its vapour pressure increases
J & K CET-(2011)
Ans. (d) : When a non-volatile solute is added to water,
the vapour pressure of the solution decreases. As the
solution is diluted, the vapour pressure of the solution
increases since more number of molecules of solution
can be vaporized.
555. Which one of the following liquid pairs will
exhibit a positive deviation from Raoult’s law?
(a) n-hexane and n-heptane
(b) Ethanol and chloroform
(c) Phenol and aniline
(d) chloroform and acetone
J & K CET-(2011)
Ans. (b) : From Raoult’s law, in case of positive
deviation the intramolecular force A–B are weaker than
those between A–A or B–B.
Example:– Acetone and ethanol, water and ethanol,
ethanol and chloroform etc.
Hence, ethanol and chloroform liquid pairs shows
positive deviation from Raoult’s law.
Objective Chemistry Volume-II
556. A solution of sucrose (molar mass = 342 g mol–
1
) has been prepared by dissolving 68.5 g of
sucrose in 1000 g of water. The freezing point
of the solution obtained will be (Kf for water =
1.86K kg mol–1)
(a) –0.372°C
(b) –0.520°C
(c) +0.372°C
(d) –0.570°C
(AIPMT -2010)
Ans. (a) : Given that, Kf= 1.86 K kg mol–1
Molar mass of sucrose (M1) = 342 g mol–1
Weight of solute (W1) = 68.5g
Weight of solvent (W2) = 1000g
We know that,
∆Tf = Kf × molality
W × K f ×1000
∆Tf = 1
M1 × W2
68.5 × 1000
∆Tf = 1.86 ×
342 ×1000
o
= 0.372 C
Tf= TFo − ∆TF
= 0–0.372oC
= –0.372oC
557. The amount of solute (molar mass 60 g mol-1)
that must be added to 180 g of water so that the
vapour pressure of water is lowered by 10% is
(a) 30 g
(b) 60 g
(c) 120 g
(d) 12 g
(e) 24 g
Kerala-CEE-2010
Ans. (b) : From Raoult’s lawP ° − PS w A M B
=
×
P°
MA w B
10 w A 18
So,
=
×
100 60 180
wA = 60g
558. Dissolution of 1.5 g of a non-volatile solute
(molecular weight = 60) in 250 g of a solvent
reduces its freezing point by 0.01°C. Find the
molal depression constant of the solvent.
(a) 0.01
(b) 0.001
(c) 0.0001
(d) 0.1
MHT CET-2010
Ans. (d) : From formula ∆Tf=Kf×m,
Weight of solute × 1000
Molality(m) =
Molecular weight of solute ×
Weight of solvent
Now,
1.5 × 1000
m=
= 0.1
60 × 250
∆T f = Kf × m
0.01 = Kf × 0.1
0.01
Kf =
0.1
= 0.1ºC molal –1
113
YCT
559. Ratio of loss in solvent to gain in CaCl2 tube is
p°
p
(a)
(b)
p
p°
p° − p
p − p°
(c)
(d)
p°
p
MHT CET-2010
p° − p
Ans. (c) :
p°
∵ Loss in the weight of solvent = p°– p
Gain in the weight of CaCl2 tube = p°
Ans. (d) : When a non-volatile solute is added to a
volatile solvent, the solute cover up some of the surface
of solvent.
Thus, the less surface area is available for vapourisation
of solvent and hence vapour pressure decreases. As the
amount of non-volatile solute increases, vapour pressure
decreases.
∴ The order of vapour pressure is P3 > P1 > P2
564. What is the freezing point of a solution
containing 8.1 g HBr in 100 g water assuming
the acid to be 90% ionised? (Kf for water = 1.86
K mol–1)
Loss in the weight of solvent
p° – p
Required ratio =
=
(a) 0.85ºC
(b) –3.53ºC
Gain in the weight of CaCl2 tube
p°
(c) 0ºC
(d) –0.35ºC
560. What is the activity in microcurie?
JCECE - 2010
(a) 1.6 × 104
(b) 1.6 × 107
Ans. (b) : Given, Kf = 1.86 K mol–1
4
7
(c) 3.2 × 10
(d) 3.2 × 10
mass of HBr = 8.1 g
SCRA-2010
mass of H2O = 100 g
Ans. (c) : 1µCi = 3.2 × 104 disintegration per second.
degree of ionization (α) = 90% m (molality)
561. In winter season specially in cold countries,
mass of solute/ mol. wt. of solute
m=
ethylene glycol is added to water in the
mass of solvent in kg
radiators of car. It results in
(a) reducing the viscosity of water
8.1/ 81
=
(b) lowering the freezing point of water
100 /1000
(c) lowering the boiling point of water
HBr
→ H+ + Br−
(d) reducing the specific heat of water
Ions at equilibrium 1− α
α
α
SCRA-2010 ∴
Total ions = 1 – α + α + α
Ans. (b) : Ethylene glycol is added to water in the
=1+α
radiator of cars during winter because it has much lower ∴
i = 1 + α = 1 + 0.9 = 1.9
freezing point than water. It’s role in an automobile is to
∆Tf = i × Kf × m
absorb heat from the engine ethylene glycol as an
8.1/ 81
antifreeze.
i = 1.9 × 1.86 ×
562. An ideal solution is formed when its
100 /1000
components
i = 3.534º C
(1) can be converted into gases
Depression in freezing point
(2) obey Raoult’s law
(∆Tf) = Freezing point of water – Freezing point of
(3) have no change of volume
solution
(4) have zero heat of mixing
3.534 = 0 – freezing point of solution
Which of the above statements is/are correct?
∴ Freezing point of solution = –3.534ºC
(a) 1 only
(b) 2, 3 and 4
565.
Which of the following will increase with the
(c) 2 only
(d) 3 and 4 only
increase
in temperature?
SCRA-2010
(a)
Surface
tension
(b) Viscosity
Ans. (b) : Conditions for formation of an ideal solution(c)
Molality
(d) Vapour pressure
(i) An ideal solution is a homogeneous mixture
UPTU/UPSEE-2009
(ii) The formation of ideal solution must have no
volumetric or thermal effects. i.e, ∆Hmix=0 and ∆Vmix=0 Ans. (d) : Vapour pressure increases with increases in
(iii) This should obey Raoult’s law.
temperature.
563. X is a non-volatile solute and Y is a volatile 566. The difference between the boiling point and
solvent. The following vapour pressures are
freezing point of an aqueous solution
observed by dissolving X in Y.
containing sucrose (molecular weight = 342 g
X/mol L-1
Y/mm of Hg
mol-1) in 100 g of water is 105.0oC. If Kf and Kb
of water are 1.86 and 0.51 K kg mol-1
0.10
p1
respectively, the weight of sucrose in the
0.25
p2
solution is about
0.01
p3
(a) 34.2 g
(b) 342 g
The correct order of vapour pressures is
(c)
7.2
g
(d)
72 g
(a) p1 < p2 < p3
(b) p3 < p2 < p1
(e)
68.4
g
(c) p3 < p1 < p2
(d) p2 < p1 < p3
Kerala-CEE-2009
AP-EAMCET- (Engg.) - 2010
Objective Chemistry Volume-II
114
YCT
Ans. (d) : It is clear that
∆Tb = ∆Tf
or
Kb × m = Kf × m
Then,
∆Tb = ∆Tf = (Kb + Kf)m
Now,
o
o
Tb – Tf = (Tb + ∆Tb ) + (Tf − ∆Tf )
So,
105 = (∆Tb + ∆Tf) + 100
∴
∆Tb + ∆Tf = 5
now,
∆T + ∆Tf
m= b
kb + kf
5
=
1.86 + 0.51
= 2.11
Now, moles of solute
= 2.11 × 0.1
= 0.211
∴ Mass of solute = 0.211 × 342
= 72.16g
567. A 0.0020 m aqueous solution of an ionic
compound [Co(NH3)5(NO2)]Cl freezes at –
0.00732 °C. Number of moles of ions which 1
mol of ionic compound produces on being
dissolved in water will be (Kf = –1.86 °C/m)
(a) 3
(b) 4
(c) 1
(d) 2
(AIPMT -2009)
Ans. (d) : Given that,
Molality of solution (m) = 0.0020 m
Freezing of solution = 0.0073°C
From formula,
∆T f = i × Kf × m
or,
∆Tf
i=
Kf × m
0.00732
i=
1.86 × 0.0020
i =1.967 ≈ 2
568. Correct order of freezing point of 1 M solution
of sucrose, KCl, BaCl2 and AlCl3 is
(a) Sucrose > KCl > BaCl2 > AlCl3
(b) AlCl3 > BaCl2 > KCl > Sucrose
(c) BaCl2 > KCl > AlCl3 > Sucrose
(d) KCl > BaCl2 > AlCl3 > Sucrose
UP CPMT-2009
Ans. (a): Depression in freezing point is a colligative
property, it means that as the number of particles (ions)
of solute increases, freezing point of the solution
decreases.
Sucrose
→ not ionised
KCl ↽ ⇀ K+ + Cl–
2 ions
Objective Chemistry Volume-II
BaCl2 ↽ ⇀ Ba2++2Cl–
3 ions
AlCl3 ↽ ⇀ Al3++3Cl–
4 ions
Hence, the correct order of freezing point is –
Sucrose > KCl > BaCl2 > AlCl3
569. At 300K, the vapour pressure of an ideal
solution containing 3 mole of A and 2 mole of B
is 600 torr. At the same temperature, if 1.5
mole of A and 0.5 mole of C (non-volatile) are
added to this solution the vapour pressure of
solution increases by 30 torr. What is value of
PBo ?
(a) 940
(b) 405
(c) 90
(d) None of these
AIIMS-2009
Ans. (c): From Raoult’s law
P = PAº X A + PBº X B
2
3
º
600=PAº
+ PB
3+ 2
2+3
1
600 = 3PA° + 2PB°
5
º
3PA + 2PBº = 3000...........(1)
Now, after addition of 1.5 mole of A and 0.5 mole of C
in the given solution
4.5
2
º
630 = PAº
+ PB
4.5 + 2 + 0.5
4.5 + 2 + 0.5
4.5PAº + 2PBº = 4410 ..........(2)
Subtracting equation (1) from equation (2), we get,
1.5 PA° =1410
PA° =940 torr
Putting the value of PA° in equation (1) then we get,
3 × 940 + 2 PB° =3000
PB° = 90 torr
570. A solution with negative deviation among the
following is
(a) Ethanol-Acetone
(b) Chlorobenzene-Bromobenzene
(c) Chloroform-Acetone
(d) Benzene-Toluene
J & K CET-(2009)
Ans. (c) : In case of negative deviation from Raoult’s
law, the attraction forces between the two components
of a solution are greater than that between the
components themselves, the obtained solution exhibit
negative deviation from Raoult’s law.
So, solution of chloroform and acetone is greater than
the exists between the molecules of chloroform or
between the molecules of acetone, hence solution shows
negative deviation.
115
YCT
571. At 300K, two pure liquids A and B have vapour Ans. (b) : We know that, ∆Tf = m × K f
pressures 150 mm Hg and 100 mm Hg Where, m = morality
respectively. In an equimolar liquid mixture of
W × K f ×1000
∆Tf = B
A and B, the mole fraction of B in the vapour
M B × WA (kg.)
mixture at this temperature is
0.6 × 1000 ×1.86
(a) 0.6
(b) 0.5
273 − 272.187 =
M B × 21.7
(c) 0.8
(d) 0.4
0.6
×
1000
×
1.86
J & K CET-(2009)
MB =
21.7 × 0.813
Ans. (d) :
WB = mass of solutein gram
Consider, moles of A = x
(∵ equimolar mixture)
M B = molar mass of solute
moles of B = x
1,116
Mole fraction of A and BMB =
17.6421
x
MB = 63.25 gram
xA =
= +0.5
x+x
574. Consider the following statements in respect of
xA = xB
an ideal solution :
From formula–
(1) Raoult's law valid for an ideal solution over
the whole concentration range.
PT = PAo x A + PBo x B
(2) Enthalpy of mixing is zero i.e. ∆Hmix= 0.
PT = 150 × 0.5 + 100 × 0.5 = 125
(3) Volume of mixing is zero i.e. ∆Vmix ≠ 0.
PBo x B 100 × 0.5
(4) The components of ideal solution cannot be
Mole fraction of B =
=
= 0.4
separated by fractional distillation.
PT
125
Which of the statements given above is/are
572. Consider lowering of vapour pressurs (∆p),
correct ?
(a) 3 and 4
(b) 1 and 4
elevation in boiling point (∆Tb) and depression
(c)
1
and
2
(d)
2 and 3
in freezing point (∆Tf) of a solvent for the same
SCRA-2009
molar concentration of each of the following
Ans.
(c)
:
1.
Raoult's
law
is
valid
for
an
ideal
solution
three solutes :
over
the
whole
concentration
range
and
pressure
range.
1. BaCl2
2. For an ideal solution, ∆H and ∆V for mixing should
2. NaCl
be zero i.e, ∆H mix = 0
3. MgCl2
∆Vmix = 0
Which of the following is/are the correct
Here, ∆S = +ve and ∆G ≠ 0
sequence(s) ?
3. Component of non-ideal binary solution can’t be
(a) ∆p : 3 < 2 < 1
(b) ∆Tb : 1 < 2 < 3
completely separated by fractional distillation.
(c) ∆Tb : 3 < 2 < 1
(d) None of these
575. The freezing point of water is depressed by
SCRA-2009
0.37oC in a 0.01 molal NaCI soln. The freezing
2+
–
point of 0.02 molal soln. of urea is depressed by
Ans. (d) : BaCl2 = Ba + 2Cl , i = 3
+
–
(a) 0.37oC
(b) 0.74oC
NaCl = Na +Cl
, i=2
o
(c) 0.185 C
(d) 0oC
MgCl2 = Mg2+ + 2Cl– , i = 3
WB-JEE-2008
Since, ∆Tb ∝ i × C
Ans. (a) : According to the depression in freezing point
1
∆Tf ∝ m and ∆Tf = Kf m×i
∆Tf ∝
i×C
∆Tf
So, K f =
Molar concentrations are same. So, ∆Tf and ∆Tb depend
i×m
upon Van’t Hoff factor.
∆Tf NaCl
∆Tf urea
Now,
=
So, none of these given sequence are correct.
I NaCl X M m urea X urea
573. An organic compound of 0.6 g when it dissolves
∆Tf urea
0.37
in water of 21.7g freezes at 272.187 K. The
=
2 × 0.01 0.02 × 1
molar mass of the organic compound is close to
∆
Tf urea = 0.37oC
: (Kf of water is 1.86 deg/molality; freezing
point is 273 K)
576. Which of the following can be measured by the
(a) 61 g mol–1
(b) 63 g mol–1
Ostwald-Walker dynamic method?
–1
–1
(a)
Relative lowering of vapour pressure
(c) 65 g mol
(d) 67 g mol
(b) Lowering of vapour pressure
SCRA-2009
Objective Chemistry Volume-II
116
YCT
(c) Vapour pressure of the solvent
(d) All of the above
Karnataka-CET, 2008
Ans. (a) : Ostwald- walkar dynamic method is used for
measurement of relative lowering of vapour pressure.
577. The chemical decomposition of XY2 occurs as
XY2 ( g ) ↽ ⇀ XY ( g ) + Y ( g )
The initial vapour pressure of XY2 is 600 mm
of mercury and at equilibrium it is 800 mm of
mercury. Find out the value of K for this
reaction when the volume of the system
remains constant.
(a) 50
(b) 100
(c) 166.6
(d) 150
CG PET -2008
Ans. (b) : From mechanism,
XY2 ↽ ⇀ XY +Y
Initial 600
0 0
Final 600-P
P
P
Now,
PT = 600–P+P+P.
800 = 600– P+P+P
P = 200 mm
∴ value of KP when the volume of system remain
constant
P×P
KP =
600 − P
200 × 200
=
400
K P = 100
579. Maximum lowering in vapour pressure is
observed in the case of
(a) 0.1 M Glucose
(b) 0.1 M BaCl2
(c) 0.1 M MgSO4
(d) 0.1 M NaCl
J & K CET-(2008)
Ans. (b) : For maximum, lowering vapour pressure
(P° − P) is directly proportional to the Van’t Hoff’s
factor i . So for BaCl2 ,i = 3 maximum.
Thus, BaCl2 has highest value relative lowering of
vapour pressure .
0.1M BaCl2 → Ba 2+ + 2Cl− ⇒ 3ions
0.1M MgSO 4 → Mg 2+ + SO 24− ⇒ 2ions
0.1M Glucose →1ions
( P − P = lowering in vapour pressure ).
°
580. The relative lowering of vapour pressure is
equal to
(a) Ratio of the number of solute molecules to
the total number of molecules in solution
(b) Ratio of the number of solvent molecules to
the number of solute molecules
(c) Ratio of the number of solute molecules to
the number of solvent molecules
(d) Ratio of the number of molecules in solution
AP EAMCET-2008
Ans. (a) : According to Raoult's law the relative
lowering of vapour pressure is equal to ratio of the
number of solute molecules to the total number of
molecules in solution.
581. A solution that obeys Raoult's law is called.
(a) Normal solution
(b) Molar solution
(c) Ideal solution
(d) Saturated solution
AP EAMCET-2008
Ans. (c) : Raoult's law states that the partial vapour
pressure of a solvent in a solution is equal to the vapour
pressure of the pure solvent multiplied by its mole
fraction in the solution.
P1 = X P1o
Where, P1 = Vapour pressure of solute
X = Mole fraction of solute
P1o = Vapour pressure of pure solvent
578. The vapour pressure of water at 20°C is 17.5
mmHg. If 18 g of glucose (C6H12O6) is added to
178.2 g of water at 20°C, the vapour pressure
of the resulting solution will be
(a) 17.675 mmHg
(b) 15.750 mmHg
(c) 16.500 mmHg
(d) 17.325 mmHg
[AIEEE 2008]
Ans. (d) : Given,
Vapour pressure of water = 17.5 mm Hg
Weight of glucose (solute) = 18 g
582. When 25 g of a non- volatile solute is dissolved
Weight of water (solvent) = 178.2 g
in 100 g of water, the vapour pressure is
18
lowered by 2.25 ×10-1 mm. If the vapour
180
pressure of water at 200C is 17.5 mm, what is
Mole fraction of glucose =
178.2 18
the molecular weight of the solute?
+
18
180
(a) 206
(b) 302
Mole fraction of glucose = 0.01
(c) 350
(d) 276
According to Raoult's law
AP EAMCET-2008
Xsolute =
P o − PS
Po
17.5 − PS
17.5
PS = 17.5 − 0.175
= 17.325mmHg
0.01 =
Objective Chemistry Volume-II
Ans. (c) : Given that,
Non volatile solute (w) = 25 gm.
Weight of solvent (W) = 100 gm.
Lowering vapour pressure (P°–Ps) = 0.225 mm
Vapour pressure of pure solvent (Po) = 17.5 mm
Molecular weight of H2O (M) = 18 gm
Molecular weight of solute (m) = ?
117
YCT
(a) 1.0 K kg mole–1
(c) 10 K gram mol–1
According to Raoult's law –
P o − Ps
w M
=
×
m W
Po
0.225
25 × 18
=
m ×100
17.5
25 × 18 ×17.5
m=
0.225 × 100
m = 350 gm
583. The vapour pressure of water at 230°C is
19.8mm. 0.1 mole of glucose is dissolved in
178.2g of water. What is the vapour pressure
(in mm) of the resultant solution.
(a) 19.0
(b) 19.602
(c) 19.402
(d) 19.202
AP-EAMCET-2008
Ans. (b) : Given that,
Vapour pressure of H2O (Po) =19.8 mm
nA = 0.1 moles
178.2 gm
nB =
= 9.9
18
According to Raoult's law–
P o − Ps
nA
=
nA + nB
Po
(b) 10 K kg mol
(d) 10 K kg mole –1
GUJCET-2008
Ans. (d) : Given data: w = 0.6g
Molar mass of Urea (M) = 60
W = 200g
∆TB = 0.50º C
Molal elevation constant (Kb) = ?
Now, from the given equation–
∆Tb = K b × m
or
Kb =
∆Tb
m
given mass
1000
×
Molality(m)=
molar mass weight of solvent(gm)
0.50
or
Kb =
0.6 ×1000
60 × 200
or
K b = 10K kg mol –1
586. 0.5 molal aqueous solution of a weak acid (HX)
is 20% ionised. If Kf for water is 1.86 K kg
mol–1, The Lowering in freezing point of the
solution is
(a) 0.56 K
(b) 1.12 K
(c) –0.56 K
(d) –1.12 K
(AIPMT -2007)
Ans. (b) : From reaction,
HX ↽ ⇀ H + +X−
initial 1
0 0
19.8 − Ps
0.1
=
19.8
0.1 + 9.9
198–10Ps = 0.1×19.8
10Ps = 198 – 1.98
196.02
Ps =
= 19.602
10
final 1 − α
α α
584. A non-volatile solute (A) is dissolved in a
volatile solvent (B). The vapour pressure of the So,
resultant solution is Ps. The vapour pressure of
1+ α
i=
= 1 + 0.2
pure solvent is PBo , If XB is the mole fraction of
1
the solvent, which of the following is correct?
Now,
(a) Ps = XA⋅ PBo
(b) PBo = Ps⋅XA
∆Tf = i × Kfm
o
o
= 1.2 × (1.86) × (0.5)
(c) Ps = XB ⋅ PB
(d) PB = Ps⋅XB
= 1.12K
AP-EAMCET-2008
587. Fractional distillation is a process by which the
Ans. (c) : Given, non-volatile solute = A
separation of different fractions from mixture
Volatile solvent = B
of solution is carried by making us of the
Vapour pressure of resultant solution = Ps
following property of the fractions
Vapour pressure of pure solvent = PBo
(a) freezing point
(b) boiling point
(c) melting point
(d) solubility
Mole fraction of the solvent = XB
J & K CET-(2007)
Now, according to Raoult's law–
°
°
Ans.
(b)
:
Fractional
distillation
is
a
process by boiling
PS = PA X A + PB X B
point of the separation of different temperature
For non-volatile solution
vapourise from a liquid mixture at least 30°C difference
PA° = 0
in boiling point.
P s ∝ XB
588. Freezing point of urea solution is –0.6º C. How
much urea ( M.W. = 60 gm/mole ) required to
or
Ps = XB PBo
dissolved
in
300gm
water
?
The proportionality constant is equal to the vapor
0
–1
o
( K f = 1.5 C kg mol )
pressure of pure solvent, PB .
585. The increase in boiling point of a solution
containing 0.6 gram Urea in 200 gram water is
0.50ºC. Find the molal elevation constant.
Objective Chemistry Volume-II
118
(a) 3.6 gm
(c) 7.2 gm
(b) 2.4 gm
(d) 6.0 gm
GUJCET-2007
YCT
Ans. (c) : Given that,
∆Tf = −0.6º C
WA = ?
WB = 3kg, K f = 1.5º C
Molecular Weight (M) = 60 gm/mole
Now, ∆Tf = K f × m
Ans. (d) :
Objective Chemistry Volume-II
∆H(Changein enthalpy)
T
6 ×103
=
= 0.0219 ×103 JK −1 mol−1
273
= 21.98JK–1mole–1
591. The molar heat capacity (C) of water at
constant pressure, is 75 JK−1 mol−1. When 1.0
kJ of heat is supplied to 100 g of water which is
free to expand, the increase in temperature of
water is
(a) 1.2 K
(b) 2.4 K
(c) 4.8 K
(d) 6.6 K
CG PET -2007
Ans. (b) : Given,
Q = 1kJ = 1000J
Molar heat capacity = 75 JK–1 mol–1
Weight of water = 100 g.
Molar mass of H2O = 18g
So,
100
Number of moles of water =
18
= 5.5
Heat, Q = nC∆T
(Change in entropy) ∆S =
WA
M × WB
Where, WA = Weight of solute
WB = Weight of solvent
M = Molecular weight of solute
∆T × M × WB
or
WA = f
Kf
0.6 × 60 × 3
WA =
1.5
WA = 72gm
If we consider the mass of water given to 300 gm–
1000gkg −1 ×1.5°C kgmol−1 × Weight of urea
So, 0.6°C =
60gmol−1 × 300g
or Weight of Urea = 7.2 g
589. The vapour pressure of pure benzene at a
certain temperature is 0.850 bar. A nonvolatile, non-electrolyte solid weighing 0.5g is
added to 39.0 g of benzene (molar mass 78
g/mol). The vapour pressure of the solution
then is 0.845 bar. What is the molecular mass
of the solid substance?
(a) 58
(b) 180
(c) 170
(d) 145
AIIMS-2007
Ans. (c): Given,
Vapour pressureof pure benzene(P° ) = 0.850 bar ,
Vapour pressure of solution (P) = 0.845bar ,
Weight of benzene (W1 ) = 39g ,
Weight of solid substance(W2 ) = 0.5g ,
Molar mass of benzene (M l ) = 78g / mol
From formula –
P o − P W2 M1
=
Po
W1M 2
0.850 − 0.845 0.5g × 78g / mol
=
0.850
39g × M 2
0.005 0.5g × 78g / mol
=
39g × M 2
0.850
0.5g × 78g / mol × 0.850
M2 =
39g × 0.005
M2= 170g/mol
590. What is the entropy change (in JK−1 mol−1)
when one mole of ice is converted into water at
0ºC? (The enthalpy change for the conversion
of ice to liquid water is 6.0 kJ mol−1 at 0ºC).
(a) 20.13
(b) 2.013
(c) 2.198
(d) 21.98
CG PET -2007
∆Tf = K f ×
∆H = 6kJ mol–1 = 6 × 103 J mol–1
1000 = 5.5 × 75 × ∆T
1000
∆T =
5.5 × 75
∆T = 2.42K ≈ 2.4K
592. 1.00 g of a non-electrolyte solute (molar mass
250g mol–1) was dissolved in 51.2 g of benzene.
If the freezing point depression constant, Kf of
benzene is 5.12 K kg mol–1 , the freezing point
of benzene will be lowered by
(a) 0.2 K
(b) 0.4 K
(c) 0.3 K
(d) 0.5 K
(AIPMT -2006)
Ans. (b) : Given,
Weight of the solute (W1) = 1 g
Weight of solvent (W2) = 51.2 g
Molar mass of solute (M1) = 250 g mol–1
Freezing point depression constant (Kf) = 5.12 K kg
mol–1
From the formula,
W ×1000
∆Tf = K f × 1
W2 × M1
Then,
1× 1000
∆Tf = 5.12 ×
51.2 × 250
∆Tf = 0.4K
593. The volume strength of 1.5N H2O2 solution is:
(a) 4.8
(b) 8.4
(c) 3.0
(d) 8.0
AP-EAMCET (Medical), 2006
119
YCT
Ans. (b) : The volume strength of 1.5N H2O2 Solution
will be
Volume strength = 5.6 × normality
= 1.5×5.6 = 8.4 L Solution.
o
594. At 25 C, the total pressure of an ideal solution
obtained by mixing 3 moles of A and 2 moles of
B, is 184 torr. What is the vapour pressure (in
torr) of pure B at the same temperature?
(vapour pressure of pure A, at 250C, is 200 torr).
(a) 180
(b) 160
(c) 16
(d) 100
J & K CET-(2006)
Ans. (b) : Given data,
XA = 3 and XB = 2, P = 184 Torr
PAo = 200Torr and PBo = ?
P = X A PA° + X B PB°
nA
3
2 o
So, 184 =
× 200 +
× PB ∵ X A =
nA + nB
3+ 2
3+ 2
2
184 = 120 + PB°
5
PB° = 160 Torr
595. Relative lowering of vapour pressure of a dilute
solution is 0.2. What is the mole fraction of non
volatile solute?
(a) 0.8
(b) 0.5
(c) 0.3
(d) 0.2
J & K CET-(2006)
Ans. (d) : According to Raoult’s law, the relative
lowering in vapour pressure of an ideal solution
containing the non-volatile solute is equal to the mole
fraction of the solute.
Given–
Relative lowering of vapour pressure = 0.2
So,
Mole fraction will be also 0.2
596. Which of the following does not show negative
deviation from Raoult's law?
(a) Acetone—Chloroform
(b) Acetone—Benzene
(c) Chloroform—Ether
(d) Chloroform—Benzene
CG PET-2005
Ans. (b) : Negative deviation from the law of Raoult's
means that as opposed to the total vapour pressure of
the solution, the total vapour pressure of the orginal
liquid would be greater when both the liquid are mixed.
Acetone-Benzene does not show negative deviation
from Raoult’s law.
597. Which of the following is incorrect?
(a) Relative lowering of vapour pressure is
independent of the nature of the solute and
the solvent.
(b) Vapour pressure is a colligative property.
(c) Vapour pressure of a solution is lower than
the vapour pressure of the solvent.
(d) Relative lowering of vapour pressure is
directly proportional to the original pressure.
J & K CET-(2005)
Objective Chemistry Volume-II
Ans. (d) : By Raoult’s law the relative lowering in
vapour presuure of a dilute solution is equal to mole
fraction of the solute present in the solution.
P° − P
= XB
P°
598. A solution of urea (mol. mass 56g mol–1) boils
at 100.18°C at the atmospheric pressure. If Kf
and Kb for water are 1.86 and 0.512 K kg mol–1
respectively the above solution will freeze at
(a) 0.654°C
(b) –0.654°C
(c) 6.54°C
(d) –6.54°C
(AIPMT -2005)
Ans. (b) : Given,
∆Tb = T2 − T1 = 100.18 −100 = 0.18°C
Kb for water = 0.512 K kg mol–1
K f for water = 1.86 K kg mol–1
From formula
∆Tb K b m K b
=
=
∆Tf K f m K f
0.18 0.512
=
1.86
∆Tf
∆Tf = 0.654
Tf = ( 0 − 0.654 ) °C
= −0.654°C
599. If a solution containing 0.072g atom of sulpur
in 100 g of a solvent (Kf=7.0) gave a freezing
point depression of 0.84°C, the molecular
formula of sulphur in the solutions is:
(a) S6
(b) S7
(d) S9
(c) S8
UPTU/UPSEE-2005
Ans. (a) : Given that,
Mass of solvent (w) = 100g
Depression in freezing point (∆Tf) = 0.84oC, Kf = 7.0
1000 × K f w
So, ∆Tf =
n×w M
1000 × 7.0 × 0.072
0.84 =
n × 100
n =6
∴ Sn isinS6 formin solution.
600. Which has the minimum freezing point?
(a) One molal NaCl aq. solution
(b) One molal CaCl2aq. solution
(c) One molal KCl aq. solution
(d) One molal urea aq. solution
UPTU/UPSEE-2005
Ans. (b) : It is clear that, having the more ions, will be
responsible for ∆Tf maximum and minimum freezing
point.
1 mole CaCl2 solution has the minimum freezing point .
120
YCT
4.
Osmosis and Osmotic Pressure
of the Solution
601. Which one is not correct mathematical
equation for Dalton’s law of partial pressure?
Here p=total pressure of gaseous mixture.
(a) p i = x i p io , where xi = mole fraction of ith gas
(c) From solution having higher concentration
only
(d) None of the above
CG PET -2007, (AIPMT -2006)
Ans. (b) : Osmosis is a process by which molecules of a
solvent tend to pass through a semi-permeable
membrane from a less concentrated solution into more
concentrated one.
605. The osmotic pressure of a 5% (wt./vol) solution
of cane sugar at 150º C is
(a) 3.078 atm
(b) 4.078 atm
(c) 5.078 atm
(d) 2.45 atm
JCECE - 2015, UP CPMT-2002
Ans. (c) : Given,
Weight of solution = 5% of sugarcane
Molecular weight of sugar cane = 342
5
Weight of solution w =
= 0.0146g
342
Temperature, T = 150ºC = 150 + 273
= 423K
R = 0.0821 L bar mol–1 K–1
We know that,
w
Π = ×R×T
W
0.0146 × 0.082 × 423
Π=
0.1
Π = 5.064 atm
606. A 3 mL of solution was made by dissolving 20
mg of protein at 00C. The osmotic pressure of
the resulting solution is 3.8 torr. The molecular
weight of the protein is approximately (in
g/mol)
(a) 300
(b) 3 × 105
4
(d) 3 × 103
(c) 3 × 10
TS-EAMCET (Engg.), 05.08.2021 Shift-II
Ans. (c) : Mass of protein = m
20 mg
0.02
3mL →
⇒
mol
m
m
0.02 1000
6.6
1000 ml →
×
⇒
mol
m
3
m
3.8 torr = 3.8 × 0.00132 atm
= 5.0 × 10–3 atm
∏ = CRT
in gaseous mixture p io = pressure of ith gas in
pure state
(b) p = p1 + p2 + p3
RT
RT
RT
+ n2
+ n3
(c) p = n1
V
V
V
(d) pi = xip, where pi = partial pressure of ith gas
xi = mole fraction of ith gas in gaseous
mixture
NEET-17.06.2022
Ans. (a) : Dalton’s law of partial pressures according to
this law, the total pressure by a mixture of gases is equal
to the sum of the partial pressures of each of the
constituent gases.
i.e. partial pressure of gas = mole fraction of gas in
gaseous mixture × Total pressure of gaseous mixture.
P 1 = X1 P
P 2 = X2 P
P 3 = X3 P
∴ Total pressure P = P1 + P2 + P3
602. What is the effect of external pressure on the
osmotic pressure (OP) of a solution?
(a) OP decreases with increases of pressure
(b) OP decreases initially, then increases
(c) OP remained nearly same with increase/
decrease of external pressure
(d) OP increases with increase of pressure
TS-EAMCET-19.07.2022, Shift-I
Ans. (c) : The effect of external pressure on the osmotic
pressure (OP) of a solution then OP remained nearly
same with increase/decrease of external pressure
because osmotic pressure is directly related to the
concentration of the solutes. If increase solute
concentration, then osmotic pressure osmotic pressure
decrease.
603. Which of the following condition is correct for
reverse osmosis?
[C = Concentration of solution in mol L-1; T =
Temperature in Kelvin scale ; n = no of moles
(Where, Π = Osmotic pressure, C = Concentration, R =
of solute]
Gas constant, T = Temperature)
(a) Pext = CRT
(b) Pext > CRT
6.6
(c) Pext < CRT
(d) Pext > nRT
5.0 × 10−3 =
× 0.0821× 273
m
AP-EAMCET-05.07.2022, Shift-II
6.6 × 0.0821× 273
Ans. (b) : Reverse osmosis occurs a pressure higher
m=
⇒ 29.8 × 103 ≈ 3 × 104
than osmotic pressure is applied on the solution.
5.0 × 10−3
Pexternal > CRT
m = 3 × 104 g/mol
604. During osmosis, flow of water through a 607. pH of a 0.1M monobasic acid is 2. Its osmotic
semipermeable membrane is
pressure at a given temperature T(K) is (Given
(a) From both sides of semipermeable membrane
that the effective concentration for osmotic
(b) From solution having lower concentration
pressure is (1+α).x concentration of acid: α is
only
the dissociation factor)
Objective Chemistry Volume-II
121
YCT
(a) RT
(b) 0.11 RT
Ans. (b) : Given that- Π = 7.65
(c) 0.01 RT
(d) 0.001RT
R = 0.08206
TS-EAMCET (Engg.), 07.08.2021 Shift-II
T = 273 +37 = 300K
We know thatAns. (b) : Given, pH = 2
Π = CRT
pH = –log [H+]
+
7.65
=
C
× 0.08206 × (273 + 37)
or
[H ] = 0.01
C = 0.3M
For monobasic acid
Here, 1L of solution will contain 0.3 moles of glucose
= 0.3 × 180 = 54g
610. The osmotic pressure of a solution of NaCl is
0.10 atm and that of a glucose solution is 0.20
atm. The osmotic pressure of a solution formed
by mixing 1L of the sodium chloride solution
with 2L of the glucose solution is X × 10–3 atm.
x is .......(nearest integer).
[JEE Main 2020, 4 Sep Shift-II]
Ans. (167) : From formula
By Vant Hoff’s principle–
Π = iCRT
i=1+α
= i[n/V]RT
i = 1 + 0.1
Π final = ( Π 1 V1) + ( Π 2 V2)/ (V1 + V2)
i = 1.1
Π final = (0.1 × 1) + (0.2 × 2)/ 3
Osmotic pressure–
= (0.1 + 0.4)/ 3
Π = iCRT
= 0.5/3
Π = 1.1× 0.1× RT
= 500/3 × 10–3 atm
x = 167 × 10–3 atm
Π = 0.11RT
So,
x = 167
608. 1.46 g of a biopolymer dissolved in a 100 mL
611. The size of a raw mango shrinks to a much
water at 300 K exerted an osmotic pressure of
smaller size when kept in a concentrated salt
2.42×10–3 bar. The molar mass of the
solution. Which one of the following processes
4
–1
biopolymer is.........×10 g mol . (Round off to
can explain this?
the nearest integer)
(a) Osmosis
(b) Dialysis
[Use : R = 0.083 L bar mol–1 K–1]
(c) Diffusion
(d) Reverse osmosis
[JEE Main 2021, 27 July Shift-I]
[JEE Main 2020, 2 Sep Shift-II]
Ans. (15) : Given,
Ans. (a) : Under the osmosis rule, Raw mango shrink in
m = 1.46g
salt solution due to net transfer of water molecules from
V = 100 ml water = 0.1L water
mango to salt solution.
T = 300K
612.
The osmotic pressure of a dilute solution of an
R = 0.083 L bar mol–1 K–1
ionic compound XY in water is four times that
Π = 2.42 ×10–3 bar
of a solution of 0.01 M BaCl2 in water.
Assuming complete dissociation of the given
Π = iCRT
ionic compounds in water, the concentration of
Where, Π = osmotic pressure
XY (in mol L–1) in solution is
i = Van't Hoff index
(a) 4×10–2
(b) 16×10–4
C = molar concentration of solute/ molarity
–4
(c) 4×10
(d) 6×10–2
R = Ideal gas constant
[JEE Main 2019, 9 April Shift-I]
T = Temperature in Kelvin
Ans.
(d)
:
For
xy
1
×
1.46
×
1000
×
0.083
×
300
2.42 ×10−3 =
i = 2 as complete dissociation takes place.
M × 100
4
∏ xy = 2 × CRT
M = 15.02 ×10 g/mol
For BaCl2
So,
i = 3, ∏ BaCl2 = 3 × 0.01 × RT = 0.03RT
x = 15
As
∏
xy
= 4 × ∏ BaCl2
609. At 37 °C osmotic pressure of human blood is
7.65 atm. Tell how much glucose can be used in
∏ xy
2 × CRT
=
1 lit of water for intravenous injection so that
∏ BaCl2 0.03RT
osmotic pressure of this glucose solution
4 ∏ BaCl 2 2 × C
becomes equal to osmotic pressure of human
=
blood.
∏ BaCl2 0.03
(a) 22.2 gm
(b) 54.2 gm
4 × 0.03
(c) 15 gm
(d) 59.8 gm
= 6 × 10−2 mol L−1
C=
2
Tripura JEE-2021
Objective Chemistry Volume-II
122
YCT
613. Molal depression constant for a solvent is 4.0 K
kg mol–1. The depression in the freezing point
of the solvent for 0.03 mol kg–1 solution of
K2SO4 is (Assume complete dissociation of the
electrolyte)
(a) 0.18 K
(b) 0.36 K
(c) 0.12 K
(d) 0.24 K
[JEE Main 2019, 9 April Shift-II]
Ans. (b) : Ionic Solution–
K2SO4 →2K+ + SO42–
Here i = 3, m = 0.03 mol kg–1
∆ Tf
= i × kf × m
= 3 × 4 × 0.03
= 0.36K
614. A solution is prepared by dissolving 0.6 g of
urea (molar mass = 60 g mol–1) and 1.8 g of
glucose (molar mass = 180g mol–1) in 100 mL of
water at 27°C. The osmotic pressure of the
solution is (R = 0.08206L atm K–1 mol–1)
(a) 8.2 atm
(b) 2.46 atm
(c) 4.92 atm
(d) 1.64 atm
[JEE Main 2019, 12 April Shift-II]
Ans. (c) : From formula,
Π 1 = C1RT
Π 2 = C2RT
Hence,
Π = Π1+ Π2
0.6 / 60
1.8 /180
RT +
RT
=
0.1
0.1
Π = 2 (0.1) RT
= 0.2 × 24.6
= 4.92 atm.
615. 18 gram glucose (Molar mass = 180) is
dissolved in 100 mL of water at 300 K. If R =
0.0821 L-atm mol-1K–1 what is the osmotic
pressure of solution?
(a) 2.463 atm
(b) 24.63 atm
(c) 8.21 atm
(d) 0.821 atm
MHT CET-03.05.2019, SHIFT-I
Ans. (b) : Given,
m = 18g
V = 100 ml water = 0.1L water
T = 300K
R = 0.083 L bar mol–1 K–1
Π =?
Where Π = osmotic pressure
We know that,
w RT
Π =
MV
18 × 0.0821× 300
=
180 × 0.1
= 24.63 atm
616. Assertion: Reverse osmosis is used for
desalination of sea water.
Reason: Reverse osmosis occurs when a pressure
larger than the osmotic pressure is applied to the
solution.
Objective Chemistry Volume-II
(a) If both Assertion and Reason are correct and
Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.
AIIMS 25 May 2019 (Evening)
Ans. (a) : The salt solution is subjected to pressure and
pressure against the semi-permeable membrane.
The applied pressure is greater than the osmotic
pressure.
When exerted pressure is greater than osmotic pressure
H2O molecules go from lower concentration of H2O to
higher concentration of water through semi-permeable
membrane and this process is known as reverse
osmosis.
Hence, question is correct reason is correct explanation
of assertion.
617. pH of a 0.1 M monobasic acid is found to be 2.
Hence, its osmotic pressure at a given
temperature T is
(a) 0.1RT
(b) 0.11RT
(c) 1.1RT
(d) 0.01RT
[BITSAT – 2018]
Ans. (b) : Given–
HA ↽ ⇀ H + + A −
pH=2
∴ [H+] = 10–2
cα = 10–2
10−2
α = −1 = 0.1
10
i = 1+ α
i = 1+ 0.1
i = 1.1
Π v = inRT
n
Π = i
v
Π = 1.1 × 0.1 × RT
Π = 0.11RT
618. If osmotic pressure of 4% (w/v) solution of
sucrose is same as 2% (w/v) solution of 'X',
then the molecular mass of X (g/mol) is
(a) 171
(b) 205.2
(c) 570
(d) None of these
JIPMER-2018
Ans. (a) : We know that,
i1 C1 = i2 C2
4 ×1000
2 1000
= ×
342 ×100 M 100
M = 171
619. Sea water is converted into fresh water based
on the phenomenon of
(a) diffusion
(b) osmosis
(c) plasmolysis
(d) reverse osmosis
COMEDK 2018
123
YCT
Ans. (d) : If external pressure of light on solution is
more than osmotic pressure then solvent molecules
move from solution to solvent and this phenomenon is
known as reverse osmosis.
Sea water is converted into fresh water based on the
phenomenon of reverse osmosis.
620. 1 g of polymer having molar mass 1,60,000 g
dissolves in 800 mL water, So calculate osmotic
pressure in Pascal at 27ºC?
(a) 0.78
(b) 0.90
(c) 0.50
(d) 19.4
[AIIMS-26 May, 2018 (E)]
Ans. (d): From formula;
Π =CRT
1× 1000
Π=
× 0.082 × 300
160000 × 800
Π = 1.92 × 10 −4 atm
= 1.92×10–4×101325 Pascal
= 19.47 Pascal
621. The osmotic pressure of 0.2 molar solution of
urea at 300 K is :
(R = 0.082 L atm mol–1K–1)
(a) 4.92 atm
(b) 1 atm
(c) 0.25 atm
(d) 27 atm
Manipal-2017
Ans. (a) : Given,
Temperature, T = 300K
C = molar concentration of solute = 0.2
R = ideal gas constant = 0.082
From formula,
Π = CRT
= 0.2 × 0.082 × 300
= 4.92 atm
622. The osmotic pressure of solution containing
34.2 g of cane sugar (molar mass = 342 g mol–1)
in 1L of solution at 20°C is (Given R = 0.082 L
atm K–1 mol–1)
(a) 2.40 atm
(b) 3.6 atm
(c) 24 atm
(d) 0.0024 atm
MHT CET-2017
Ans. (a) : From formula,
W
Π = CRT =
RT
M×V
Given thatW = 34.2g
M = 342g/mol
V = 1L
34.2g
So, Π =
×0.0821×293K
342g / mol
Π = 2.40 atm
623. Osmotic pressure of the solution can be
increased by
(a) increasing the temperature of the solution
(b) decreasing the temperature of the solution
(c) increasing the volume of the vessel
(d) diluting the solution
Karnataka-CET-2016
Objective Chemistry Volume-II
Ans. (a) : On increase temperature, the osmotic
pressure is also increase.
Π = kCT where,
Π = osmotic pressure
k = constant
C = Molar concentration
T = Temperature
624. The osmotic pressure of equimolar solutions of
Urea, BaCl2 and AlCl3 will be in order of –
(a) AlCl3 > Urea > BaCl2
(b) Urea > BaCl2 > AlCl3
(c) AlCl3 > BaCl2 > Urea
(d) BaCl2 >AlCl3 > Urea
BCECE-2016
Ans. (c) : Osmotic pressure ∝ No. of solute particles in
the solution.
∴ Dissociation of AlCl3 is maximum i.e. (4)
& that of urea is minimum i.e. (1)
∴ AlCl3 > BaCl2 > Urea
625. 1% (w/v) solutions of KCl is dissociated to the
extent of 82%. The osmotic pressure at 300K
will be
(a) 3.2 atm
(b) 5.824 atm
(c) 4.0 atm
(d) 6.0 atm
CG PET- 2016
1× 1000
Ans. (d) : Molarity=
74.5 ×100
= 0.134 M
From formula,
Π = CRT
= 0.134×0.0821×300
= 3.30 atm
Now
observed
1 + 0.8 observe
i=
⇒
=
calculated
1
3.30
= 1.8×3.30
= 5.94 atm
≈ 6.00 atm
626. At a certain temperature, the value of the slope
of the plot of osmotic pressure ( Π ) against
concentration (C in mol L-1) of a certain
polymer solution is 291R. The temperature at
which osmotic pressure is measured, is (R is
gas constant)
(a) 271oC
(b) 18oC
(c) 564 K
(d) 18 K
WB-JEE-2015
Ans. (b) :
124
YCT
A plot of osmotic pressure vs concentration will be
linear with slope = RT
As
Π = CRT
So,
RT = 291R
T = 291 K
= (291 – 273)ºC
= 18ºC
627. If M, W and V represent molar mass of solute,
then mass of solute and volume of solution in
litres respectively, which among following
equations is true?
MWR
TMR
(a) Π =
(b) Π =
TV
WV
TWR
TRV
(c) Π =
(d) Π =
VM
WM
MHT CET-2015
Ans. (c) : Let C = concentration
Now, Π = osmotic pressure
Relation between osmotic pressure and molecular mass
Π = CRT
M
× RT
Π=
V
TWR
=
MV
628. Calculate the osmotic pressure of 0.01 M
solution of cane sugar at 300 K
(R = 0.08212 atm degree–1 mol–1)
(a) 0.3568 atm
(b) 0.2463 atm
(c) 0.1562 atm
(d) 0.5623 atm
UPTU/UPSEE-2015
Ans. (b) : Given,
solution weight (w) = 0.01 M
Temperature = 300K
ideal gas constant R = 0.08212
We know that,
Π = CRT
= 0.01 × 0.0821 × 300
= 0.2463 atm
629. Beta-carotene is the most important of the A
vitamins. Its molar mass can be determined by
measuring the osmotic pressure generated by a
given mass of beta-carotene. If 10 mL of a
solution containing 7.68 mg of beta-carotene,
has an osmotic pressure of 3.54 x 10-2 bar at
250C (R = 0.083 dm3 bar mol-1k-1), the molar
mass of beta-carotene is
(a) 109 g/mol
(b) 54 g/mol
(c) 537 g/mol
(d) 768 g/mol
S. C. R. A - 2014
Ans. (c) : Given that,
V=10ml
W = 7.68mg = 7.68 × 10–3g
Π =3.54 × 10-2 bar
T = 25 + 273 = 298 K
R = 0.083 dm3 bar mol–1 K–1
M=?
Objective Chemistry Volume-II
Now, Π = C. RT
Where, Π = Osmotic pressure
C = Concentration
R = Gas constant
T = Absolute temperature
W × 1000 RT
∴
Π=
M× V
or
M=
W × R × T ×1000
Π×V
7.68 × 0.083 × 298 ×1000 × 10-3
3.54 × 10−2 × 10
M = 536.6 g/mol
or
M ≈ 537 g/mol
630. The osmotic pressure of blood is 8.21 atm at
37oC. How much glucose would be used for an
injection that is at the same osmotic pressure as
blood?
(a) 22.17 gL-1
(b) 58.14 gL-1
-1
(c) 61.26 gL
(d) 75.43 gL-1
JIPMER-2013
Ans. (b) : Given,
The osmotic pressure ( Π ) = 8.21 atm
Temperature, (T) = 37ºC
= 37 + 273
= 310 K
Molecular weight of glucose (M) = 180gm
We know that,
Π = CRT
Π
and C =
RT
m
8.21 = × 0.082×310
M
M=
8.21 =
m
× 0.082×310
180
8.21×180
0.082×310
m = 58.13gm
631. Arrange the following solutions in the
increasing order of their osmotic pressures.
i. 34.2 g/L sucrose (M = 342)
ii. 60 g/L urea (M = 60)
iii. 90 g/L glucose (M = 180)
iv. 58.5 g/L NaCl (M = 58.5)
(a) Sucrose < Urea < Glucose < NaCl
(b) Sucrose < Glucose < NaCl < Urea
(c) Sucrose < Glucose < Urea < NaCl
(d) NaCl < Urea < Glucose < Sucorse
AMU-2013
Ans. (c) : Osmotic pressure property of the solution
which depends only on the number of solute particles.
So,
Sucrose (0.1 mole) <Glucose (0.5 mole)<
Urea (1.0 mole) < NaCl (1 mol)
125
m=
YCT
632. Osmotic pressure of insulin solution at 298K is
found to be 0.0072 atm. Hence, height of water
column due to this pressure is (given d (Hg) =
13.6 g/mL)
(a) 0.76 cm
(b) 0.70 cm
(c) 7.4 cm
(d) 76 cm
BCECE-2013
Ans. (c) : From formula,
P = 0.0072 atm
P = 0.0072 × 76cm hdg
P = 0.0072×76 × 13.6 cm
Also, P = h × 1cm of water column
∴
h × 1 = 0.0072 × 76 × 13.6cm
h = 7.4 cm
633. The most suitable property for molecular
weight determination of polymers is?
(a) Osmotic pressure
(b) Lowering of vapour pressure
(c) Elevation in boiling point
(d) Depression in freezing point
MPPET - 2012
Ans. (a) : Most suitable property for molecular weight
determination of polymer is Osmotic pressure.
Osmotic pressure is the colligative property is used to
determined the molecular mass of polymer.
634. The highest osmotic pressure corresponds to
the following solution
(a) M/10 urea
(b) M/10 glucose
(c) M/10 HCl
(d) M/10 BaCl2
AMU-2012
Ans. (d) : Osmotic pressure, Π = iCRT
i for Urea & glucose is 1 where as isonisation BaCl2 is a
higher than HCl
i BaCl2 > iHCl
637. The order of osmotic pressure of three
equimolar aqueous solutions of CaCl2, NaCl
and C6H12O6 (glucose) is
(a) CaCl2 > NaCl > C6H12O6
(b) NaCl > CaCl2 > C6H12O6
(c) C6H12O6 > CaCl2 > NaCl
(d) C6H12O6 > NaCl > CaCl2
UP CPMT-2012
Ans. (a) : Colligative properties depend only on the
number of solute particles in the solution. For different
solutes of same molar concentration, the colligative
properties (osmotic pressure) have greater value for the
solution which gives more number of particles on
ionisation.
CaCl 2 (aq) ↽ ⇀ Ca 2 + (aq) + 2Cl− (aq) = 3ions
NaCl(aq) ↽ ⇀ Na + + Cl − = 2ions
C6 H12 O6 (aq) → No ions
Hence, the order of osmotic pressure of equimolar
solutions of CaCl2, NaCl and glucose will be
CaCl2 > NaCl > glucose (C6H12O6)
638. If 20 g of a solute was dissolve in 500 mL of
water and osmotic pressure of the solution was
found to be 600 mm of Hg at 15o C , then
molecular weight of the solute is
(a) 1000
(b) 1200
(c) 1400
(d) 1800
UPTU/UPSEE-2012
Ans. (b) : Mass of solution (m) = 20g
Volume of the solution (V) = 500 mL = 0.5L
Osmotic pressure of solute (P) = 600mm of Hg
600
P=
= 0.7895atm
760
From formula,
Π V = nRT
n
mRT
Π = RT =
V
MV
Putting the value
600
20 × 0.0821× (273 + 15)
atm =
760
M × 0.5L
So,
M = 1200
639. 0.1 M NaCl and 0.1 M CH3COOH are kept in
separate containers. If their osmotic pressures
are p1 and p2 respectively then what is the
correct statement?
(a) p1 > p2
(b) p1 = p2
(c) p1 < p2
(d) p1 = p2 = 0 atm
JCECE - 2011
Ans. (a) : Π = iCRT
Π NaCl = 2 × 0.1 × RT = p1
(i = 2 for NaCl)
ΠCH3COOH = 1.3 × 0.1× RT = p2
∴ Osmotic pressure is in following order
BaCl > HCl > Urea = glucose
635. Desalination of sea water can be done by
(a) osmosis
(b) reverse osmosis
(c) filtration
(d) diffusion
J & K CET-(2012)
Ans. (b) : In reverse osmosis desalination, water is
taken from the sea and receives a first treatment to
eliminate impurities, oil, seaweed, rubbish and so on.
The reverse osmosis process can be used for
desalination of sea water for getting drinking water.
636. Which of the following would exert maximum
osmotic pressure?
(a) Decinormal aluminium sulphate
(b) Decinormal barium chloride
(c) Decinormal sodium chloride
(d) A solution obtained by mixing equal volumes
of (b) and (c) and filtering
JIPMER-2012
Ans. (a) : Osmotic pressure depends upon the number
of particles of solute.
in solution, which solute have maximum number of ions
exerts maximum osmotic pressure.
Decinormal aluminium sulphate exerts maximum
osmotic pressure.
⇒
Objective Chemistry Volume-II
126
(i < 2 for CH3COOH as it is a weak acid.)
p1 > p2
YCT
640. Assuming the salts to unionised solution which
of the following has highest osmotic pressure?
(a) 1% CsCl
(b) 1% RbCl
(c) 1% KCl
(d) 1% NaCl
AMU-2011
Ans. (d) : ∵ Salts are unionised
∴ i is same for all
Now, Π = CRT
w
Π =
RT
m×v
1
Π ∝
m
1% of salt means 1gram solute dissolved in 100 gram of
solvent.
∵ NaCl have lowest molecules mass.
Hence, has highest osmotic pressure.
641. The relationship between the values of osmotic
pressure of 0.1M solutions of KNO3(p1) and
CH3COOH(p2) is
p1
p2
=
(b) p1 > p 2
(a)
p1 + p 2 p1 + p 2
(c) p 2 > p1
(d) p1 = p 2
BCECE-2010
Ans. (b) : KNO3 is a strong electrolyte but CH3 COOH
is weak electrolyte and it cannot be totally ionised.
So, option (b) p1 > p2 is correct answer.
642. For getting accurate value of molar mass of a
solute by osmotic pressure measurement
(a) the solute must be volatile
(b) the solution concentration must be high
(c) the solute should undergo dissociation
(d) the solute must be non-volatile.
J & K CET-(2010)
Ans. (d) : For getting accurate value of molar mass of a
solute by osmotic pressure measurement, the solute
should be non-volatile.
643. Vapour pressure of pure ‘A’ is 70 mm of Hg
at 20oC. It forms an ideal solution with ‘B’
in which mole fraction of A is 0.8. If the vapour
pressure of the solution is 84 mm of Hg at 25oC,
the vapour pressure of pure ‘B’ at 25oC is
(a) 28 mm
(b) 56 mm
(c) 70 mm
(d) 140 mm
KARNATAKA-CET, 2009
Ans. (d) :
Vapour pressure of pure 'A', PAº = 70mm of Hg
Mole fraction of A = 0.8
Mole fraction of B = 0.2
Then,
Vapour pressure of solution, (P) = 84mm
p = p oA x A + p ob x B
⇒
84 = 70 × 0.8 + p oB × 0.2
84 = 56 + p oB × 0.2
p oB =
28
= 140mm
0.2
Objective Chemistry Volume-II
644. An 1% solution of KCl (I), NaCl (II), BaCl2
(III) and urea (IV) have their osmotic pressure
at the same temperature in the ascending order
(molar masses of NaCl, KCl BaCl2 and urea
are respectively 58.5, 74.5, 208.4 and 60 g. mol1
) Assume 100% ionization of the electrolytes at
this temperature
(a) I < III < II < IV
(b) III < I < II < IV
(c) I < II < III < IV
(d) I < III < IV< II
(e) III < IV < I <II
Kerala-CEE-2009
Ans. (e) : 1% solution means 1g solute is present in 100
mL of water.
iw × RT
Osmotic pressure ( Π ) =
M×V
2 × 1× 1000 × RT
(I) Π KCl =
74.5 × 100
Π KCl = 2 × 0.134RT
2 × 1× 1000 × RT
58.5 ×100
= 2 × 0.171RT
(II) Π NaCl =
Π NaCl
3 × 1× 1000 × RT
208.5 ×100
= 3 × 0.048RT
(III) Π BaCl 2 =
1 ×1 ×1000 × RT
60 × 100
= 1 × 0.167RT
(IV) Π urea =
Since, temperature is same in all cases, then the
ascending order of osmotic pressure is III < IV < I <II.
645. After removing the hard shell of an egg by
dissolving in dil. HCl, a semipermeable
membrane is visible. If such an egg is kept in a
saturated solution of common salt, the size of
the egg will
(a) shrink
(b) grow
(c) remain the same
(d) first shrink and then grow larger
AP - EAMCET(MEDICAL) - 2009
Ans. (a) : During the process of osmosis, solvent flows
from the solution of lower concentration to of higher
concentration of solution. Since, saturated solution of
common salt has higher concentration than solution of
NaCl. Solvent flows outside from the egg and the size
of the egg will shrink.
646. What is the osmotic pressure of the solution
obtained by mixing 300 cm3 of 2% (massvolume) solution of urea with 300 cm3 of 3.42%
solution of sucrose at 20C°? (R = 0.082 L atm
K–1 mol–1)
(a) 5.0 atm
(b) 5.2 atm
(c) 2.6 atm
(d) 4.5 atm
SCRA - 2009
127
YCT
Ans. (b) : We know that, Π = iCRT
According to question, Osmotic Pressure ( Π ) = I (C1 +
C2) RT ….(i)
n
C1 i.e. concentration of urea =
V ( in lit )
2
× 300
= 100
60 × 600 × 10−3
1
=
6
3.42
× 300
C2 i.e. concentration of sucrose = 100
342 × 600 ×10−3
1
=
20
Put, value of C1 and C2 in equation (i), We get1 1
Π = + × 293 × 0.082
6 20
Π = 5.2 atm
647. A solution containing 10 g of urea (M = 60) per
litre is isotonic with a solution containing 5% of
solute X. What is the molar of the solute X ?
(a) 60
(b) 100
(c) 300
(d) 600
SCRA - 2009
Ans. (c) : We know that, Osmotic Pressure π = iCRT
i×n×R ×T
Π=
V ( inL )
i × WB × R × T × 1000
M B × V ( in ml )
According to question, Π 1(urea) = Π 2 (unknown
solute)
∵Π ∝ C
So, C1 (urea) = C2 (unknown solute)
Π=
Now,
WB × 1000
WB × 1000
=
×
M
V
B
urea M B × V unknown solute
5 × 1000
10 × 1000
60 × V = 100 × M × V
urea
unknown solute
B
5 × 60
⇒
MB =
10 × 100
300
⇒
MB =
1000
⇒
M B = 0.3
M B = 300 gram
So, option (c) is correct.
648. An aqueous solution of hydrochloric acid
(a) obeys Raoult’s law
(b) shows negative deviation from Raoult’s law
(c) show positive deviation from Raoult’s law
(d) obey Henry’s law at all compositions
[AIIMS-2009]
⇒
Objective Chemistry Volume-II
Ans. (b): An aqueous solution of hydrochloric acid
shows negative deviation from Raoult's law.
Raoult's Law state that the vapour pressure of a solvent
above a solution is equal to the vapour pressure of the
pure solvent at the same temperature scaled by the mole
fraction of the solvent present.
649. The molar mass of the solute sodium hydroxide
obtained from the measurement of the osmotic
pressure of its aqueous solution at 270C is 25 g
mol–1. Therefore, its ionization percentage in
this solution is
(a) 75
(b) 60
(c) 80
(d) 70
J & K CET-(2009)
Ans. (b) : Given,
NaOH → Na+ + OH −
Normal molecular mass of NaOH = 40
Observation molecular mass = 25
Temperature = 27ºC
According to,
Van't Hoff factor
observed
i=
calculated
40
=
= 1.6
25
∴
i=1+α
Now,
i = [1 + (2 – 1) α], where α-degree of ionisation
1.6 = (1 + α) α
= 1.6 – 1
= 0.6
0.6
The % ionisation of NaOH =
× 100 = 60%
1
650. The osmotic pressure (At 270C) of an aqueous
solution (200 ML) containing 6 g of a protein is
2×10-3 atm. If R=0.080 L atm mol-1K-1 the
molecular weight of protein is
(a) 7.2×105
(b) 3.6×105
5
(c) 1.8×10
(d) 1.0×105
UPTU/UPSEE-2008
Ans. (b) : From formula,
Π = iCRT
6 × 1000
× 0.08 × 300
2 × 10–3 =
M × 200
30 × 0.08 × 300
M=
2 × 10−3
M = 360 × 103
M = 3.6 × 105
651. Osmotic pressure of a solution at a given
temperature
(a) increases with concentration
(b) decreases with concentration
(c) remains same
(d) initially increases and then decreases
UPTU/UPSEE-2008
128
YCT
Ans. (a) : Boyle-Van't Hoff law, at constant
temperature the osmotic pressure of a solution is
directly proportional to its concentration and inversely
proportional to its dilution.
Thus,
Π ∝ C, where C = concentration
652. A solution containing 4 g of polyvinyl chloride
polymer in one litre of dioxane was found to
have an osmotic pressure of 4.1×10–4 atm. at
270C. The approximate molecular weight of the
polymer is
(a) 1500
(b) 10,000
(d) 2 × 1012
(c) 2.4× 105
J & K CET-(2008)
Ans. (c) : Given,
Weight of polyvinyl chloride polymer, w = 4g
Volume of solution, (v) = 1L
Osmotic pressure, Π = 4.1 × 10–4
Temperature, T = 27ºC = 27 + 273 = 300K
R = 0.0821 L bar mol –1 K–1
w
We know that, Π V = nRT =
RT
M
wRT 4 × 0.0821× 300
M=
=
ΠV
4 ×10−4 ×1
5
M = 2.4 × 10
653. Two solutions of KNO3 and CH3COOH are
prepared separately. Molarity of both is 0.1 M
and osmotic pressures are P1 and P2
respectively. The correct relationship between
the osmotic pressures is
(a) P2 > P1
(b) P1 = P2
P1
P2
(c) P1 > P2
(d)
=
P1 + P2 P1 + P2
SRMJEEE – 2007
Ans. (c) : KNO3 is a strong electrolyte and gets 100%
ionized while CH3COOH is weak electrolyte and does
not get's completely ionized due to which KNO3 have
the more osmotic pressure than the CH3COOH i.e. P1 >
P 2.
654. The osmotic pressure is expressed in units of
(a) MeV
(b) calories
(c) cm/sec
(d) atmosphere
J & K CET-(2007)
Ans. (d) : Osmotic pressure is the minimum pressure
which needs to be applied to a solution to prevent the
inward flow of water across a semi permeable
membrane. It is denoted by ' Π ' and the unit of osmotic
pressure is atmosphere.
Π = CRT
655. Osmotic pressure of 0.4% urea solution is 1.64
atm, and that of 3.42% cane sugar is 2.46 atm.
When the above two solutions are mixed, the
osmotic pressure of the resulting solution is:
(a) 0.82 atm
(b) 2.46 atm
(c) 1.64 atm
(d) 4.10 atm
UPTU/UPSEE-2006
Ans. (b) : Given,
Π 1 = 1.64 atm, Π 2 = 2.46atm
Objective Chemistry Volume-II
We know that,
Π = CRT
n
Π = RT
v
Π = nRT
(here V1 = V2)
VT = V1 + V2
VF = 2V
Now,
Π F VF = Π 1 V1 + Π 2 V2
2 Π F V = 1.64V + 2.46V
2 Π F V = 4.10V
4.10
ΠF=
= 2.05 atm
2
656. Two solutions with equal osmotic pressure are
(a) normal solutions
(b) isotonic solutions
(c) hypotonic solutions (d) hypertonic solutions
AMU–2006
Ans. (b) : Solutions having equal molar concentration
and hence equal osmotic pressure are called isotonic or
iso-osmotic solutions.
657. Four solution A, B, C, D has glucose 0.5 M,
NaCl 0.1M,BaCl2 0.5M and MgCl2 0.1M, then
which of the following will have highest osmotic
pressure?
(a) Glucose
(b) BaCl2
(c) MgCl2
(d) NaCl
CG PET -2006
Ans. (b) : ∵ NaCl, BaCl2 & MgCl2 is an electrolyte
∴ 1 mole of NaCl 2 Mole ions
molar concentration of particle in solution = 0.2M
1 mole of BaCl2 give 3 mole ions
molar concentration of particle in solution = 1.5M
1 mole of MgCl2 give 3 mole ions
molar concentration of particle in solution = 0.3M
∵ Osmotic α number of particle in solution
So, increasing order of osmotic pressure is
Glucose < NaCl < MgCl2 < BaCl2
5.
Solubility
CH4(g) respectively are 40.39,1.67, 1.83 × 10–5
and 0.413. then identify the correct increasing
order of their solubility’s.
(a) HCHO < CH4 CO2 < Ar
(b) HCHO < CO2 < CH4 < Ar
(c) Ar < CO2 < HCHO < CH4
(d) Ar < CO2 < CH4 < HCHO
AP EAPCET 20.08.2021 Shift-I
Kerala-CEE-29.08.2021
Ans. (d) : Given that KH value for this compound.
Ar(g) = 40.39
CO2(g) = 1.67
HCHO(g) = 1.83 × 10–5
CH4(g) = 0.413
Decrease the value of KH then increasing order of their
solubilites.
658.
129
YCT
659. The pair of components which cannot exist
together in solution is
(a) NaHCO3 and NaOH
(b) NaHCO3 and H2O
(c) NaHCO3 and Na2CO3
(d) Na2CO3 and NaOH
Assam CEE-2014
UPTU/UPSEE-2013
Ans. (a) : From the given options, NaHCO3 and NaOH
is the only pair cannot exist together in the solution.
NaHCO3 is acidic and is decomposed by NaOH forming
Na2CO3. The reaction is a follows:
NaHCO3 + NaOH → Na2CO3 + H2O
660. Identify the correct order of solubility of Na2S.
CuS and ZnS in aqueous medium
(a) CuS > ZnS > Na2S (b) ZnS > Na2S > CuS
(c) Na2S > CuS > ZnS (d) Na2S > ZnS > CuS
[BITSAT – 2014]
NEET-2013
Ans. (d) : The more the ionic soluble in water than
covalent compounds. Increasing covalent character
leads to less solubility. Ionic character is directly
proportional to size of the cation.
The order of size of cation Na+ > Zn+2 > Cu+2
Therefore the correct order of solubility is
Na2S > ZnS > CuS
Note- Sodium salts are highly soluble in water thus of
Na2S shows highest solubility.
661. In which solution/solvent the solubility of AgCl
is minimum?
(a) 0.01 M NaCl
(b) 0.01 M CaCl2
(c) Pure water
(d) 0.01 M AgNO3
CG PET -2007
AIPMT -1995
Ans. (b) :
(a) 0.01 M NaCl solutionNaCl ↽ ⇀ Na + + Cl −
(b) 0.01 M CaCl2 solutionCaCl 2 ↽ ⇀ Ca 2+ + 2Cl −
(c) Pure waterH 2 O ↽ ⇀ H + + OH −
(d) 0.01 M AgNO3 solutionAgNO3 ↽ ⇀ Ag + + NO3−
∵ CaCl2 in solution is dissociated into three ions
(Maximum).
∴ Due to common ion effect solubility of AgCl will be
minimum in 0.01 M CaCl2 solution.
662. The water having more dissolved O2 is
(a) Boiling water
(b) Water at 80°C
(c) Polluted water
(d) Water at 4°C
[JEE Main 2021, 22 July Shift-II]
Ans. (d) : Dissolved oxygen means amount of Oxygen
that is present in water. Cold water has the possibility to
hold more dissolved oxygen in comparison of warm
water. So, the water having more dissolved O2 is water
at 4oC.
Objective Chemistry Volume-II
663. Henry's constant (in kbar) for four gases α, β, γ
and δ in water at 298 K is given below :
α
β
γ
δ
–5
0.5
KH
50
2
2×10
(density of water = 103 kg m–3 at 298 K). This
table implies that
(a) α has the highest solubility in water at a given
pressure
(b) Solubility of γ at 308 K is lower than at 298 K
(c) The pressure of a 55.5 molal solution of γ is 1
bar
(d) The pressure of a 55.5 molal solution of δ is
250 bar
[JEE Main 2020, 3 Sep Shift-I]
Ans. (d) : Pγ = KH Xγ
55.5
Pγ = 2×10–15×
1000
55.5 +
18
= 2×10–2 bar
Now for,
P δ = KH Xδ
55.5
Pδ= 0.5×
1000
55.5 +
18
= 0.249 K bar
= 249 bar
So, it is clear that P = KH
Where X = mole fraction of gas in liquid.
On increasing temperature solubility of gases decreases.
664. At 35°C, the vapour pressure of CS2 is 512 mm
Hg and that of acetone is 344 mm Hg. A
solution of CS2 in acetone has a total vapour
pressure of 600 mm Hg. The false statement
amongst the following is
(a) Raoult's law is not obeyed by this system
(b) CS2 and acetone are less attracted to each
other than to themselves
(c) A mixture of 100 mL CS2 and 100 mL
acetone has a volume < 200 mL
(d) Heat must be absorbed in order to produce the
solution at 35°C
[JEE Main-07.01.2020, Shift-I]
Ans. (c) : We know that,
PT = PA0 X A + PB0 X B
Now,
XA = 1
PT = 512 mHg < PT
It show's positive deviation from Raoult's law is
observed. So, it will be non ideal solution.
665. Which one of the following statements
regarding Henry's law is not correct?
(a) Different gases have different KH (Henry's
law constant) values at the same temperature
(b) Higher the value of KH at a given pressure,
higher is the solubility of the gas in the
liquids
130
YCT
(c) The value of KH increases with increase of
temperature and KH is function of the nature
of the gas
(d) The partial pressure of the gas in vapour
phase is proportional to the mole fraction of
the gas in the solution
[JEE Main-09.01.2019, Shift-I]
Ans. (b) : For liquid solution,
PGas = KH × XGas
So, higher the value of KH at 0 given pressure, higher is
the solubility of the gas in the liquids.
666. For the solution of the gases w, x, y and z in
water at 298 K, the Henry's law constants (KH)
are 0.5, 2.35 and 40 K bar, respectively. The
correct plot for the given data is
(a)
(b)
(c)
(d)
667. AgCl is soluble in NH4OH. The solubility is due
to the formation of
(a) AgOH
(b) Ag2O
(c) [Ag(NH3)2]+
(d) NH4Cl
JCECE - 2019
Ans. (c) : AgCl is soluble in NH4OH. The solubility is
due to formation of [Ag(NH3)2]+. Silver chloride (AgCl)
is a white crystalline solid. AgCl reacts with ammonium
hydroxide and yields diamine.
AgCl(s) + 2NH 4 OH(aq) → [Ag(NH 3 ) 2 ]Cl
White ppt
Solublecomplete
668. Out of BeF2, MgF2, CaF2, SrF2 which has
maximum solubility:
(a) BeF2
(b) MgF2
(c) CaF2
(d) SrF2
AIIMS 25 May 2019 (Evening)
Ans. (a): Among the all options BeF2 → Be2+ have high
hydration energy so BeF2 has maximum solubility.
669. Which of the following compound is least
soluble?
(a) Na2S
(b) MgS
(c) MgCl2
(d) NaCl
AIIMS 26 May 2019 (Morning)
Ans. (b): It is clear that higher the lattice energy lower
the solubility.
MgS is a bi-valent ionic solid and It has higher lattice
energy.
670. Which of the following is most soluble in
water?
(a) CsClO4
(b) NaClO4
(c) KClO4
(d) LiClO4
AMU-2018
Ans. (d) : The high solubility of LiClO4 is mainly due
to high heat of hydration of Li+ ion.
671. Which of the following conditions are correct
for real solutions showing negative deviation
from Raoult's law?
(a) ∆H Mix < 0; ∆VMix > 0
(b) ∆H Mix > 0; ∆VMix > 0
(c) ∆H Mix > 0; ∆VMix < 0
(d) ∆H Mix < 0; ∆VMix < 0
TS EAMCET-2017
Ans. (d) : Conditions that are real solution showing
negative deviation from Raoult's law are –
PA < PA° X A
PB < PB° X B
So, ∆ Mix H = − ve
and dissolution is exothermic heating decreases
[JEE Main 2019, 8 April Shift-II] solubility ∆
Mixing V = − ve .
Ans. (a) : From fig.(a)
672. You are supplied with 500 mL each of 2N HCl
Pgas = KH.Xgas
and 5 N HCl. What is the maximum volume of
Pgas = K H .(1 − X H2O ) (∵ Xgas+ X H2O = 1 )
3M HCl that you can prepare using only these
two solutions?
Pgas = KH–KH. X H2O
(a) 250 mL
(b) 500 mL
Pgas = – KH. X H2O + KH (This is type of y = mx + c)
(c) 750 mL
(d) 1000 mL
where slope is –KH and intercept is KH.
WB-JEE-2017
Objective Chemistry Volume-II
131
YCT
Ans. (c) : From Normality equation
N1V1 + N2V2 = N (V1+V2)
2×500 + 5x = 3(500 + x)
1000 + 5x = 1500 + 3x
x = 250 mL
∵ maximum volume = 500 + x = 500 + 250 = 750 mL
673. When BaCI2 is added to an aqueous salt
solution, a white precipitate is obtained. The
anion among CO 23− ,SO 23− and SO42− that was
present in the solution can be
(a) CO32− but not any of the other two
(b) SO32− but not any of other two
(c) SO 24− but not any of other two
(d) Any of them
676. At 800C, the vapour pressure of pure liquid A
is 520mm Hg and that of pure liquid B is
1000mm Hg. If a mixture solution of A and B
boils at 800C and 1 atm pressure. The amount
of A in the mixture is–
(a) 50 mol percent
(b) 54 mol percent
(c) 32 mol percent
(d) 44 mol percent
BCECE-2015
Ans. (a) : Given:
PAo = 520 mmHg, PBo = 1000 mmHg
Let mole fractions of A = xA
Let mole fractions of B = xB
From formula,
PAo XA + PBo XB = 760 mmHg
PAo XA + PBo (1– XA) = 760
= 520XA + 1000 – 1000 XA
= 760 mm
WB-JEE-2017
Ans. (d) : Reaction,
BaCl2 + Na2SO3 →BaSO3 + 2NaCl
(White PPt)
So,
From above reaction, use can say if CO32− , SO32− and
SO 24− are present along with BaCl2, then they will show
white precipitate and any of them will be present in
reaction.
674. PbCI2 is insoluble in cold water. Addition of
HCI increases its solubility due to
(a) formation of soluble complex anions like
[PbCI3]–
(b) oxidation of Pb (II) to PB (IV)
(c) formation of [Pb(H2O)6]2+
(d) formation of polymeric lead complex
WB-JEE-2017
Ans. (a) : Required Chemical Reaction,
Cold
PbCl2+Cl–
→ [PbCl3]–
1
or 50 mol%
2
677. Which of the following alcohol has highest
solubility in water?
(a) Secondary butyl alcohol
(b) Tertiary butyl alcohol
(c) Ethelene glycol
(d) Glycerol
GUJCET-2014
Ans. (d) : Among the all options only glycerol has more
number of hydroxyl groups so, glycerol has tendency to
form hydrogen bond with water. So we can say glycerol
has highest solubility in water.
678. 50 g of saturated aqueous solution of potassium
chloride at 30oC is evaporated to dryness, when
13.2 g of dry KCl was obtained. The solubility
of KCl in water at 30oC is
(a) 35.87 g
(b) 25.62 g
(c) 28.97 g
(d) 27.81 g
UPTU/UPSEE-2013
Ans. (a) : The Solubility of KCl in water at 30oC
Mass of KCl
=
× 100
Mass of water
XA =
increase
PbCl2+2Cl– →
[PbCl4]2–
From above reaction it is clear that addition of HCl
increases it's solubility due to form of soluble complex
anion like [PbCl3]–.
675. At 20oC the solubility of N2 gas in water is
0.015 g/L when the partial pressure of N2 is 580
13.2
torr. What is the solubility of N2 in H2O at 20oC
=
×100
50
− 13.2 )
(
when its partial pressure is 800 torr?
(a) 0.207 g/L
(b) 0.0207 g/L
= 35.87g
(c) 0.414 g/L
(d) 0.0414 g/L
679. 100 mL of 0.1 M acetic acid is completely
AMU-2016
neutralized using a standard solution of NaOH.
The volume of ethane obtained at STP after the
Ans. (b) : Applying Henry's law
complete electrolysis of the resulting solution is
m∝P
(a) 56 mL
(b) 224 mL
m = K HP
(c) 560 mL
(d) 112 mL
0.015 = KH × 580
Karnataka-CET-2012
At solubility of N2 (20ºC)
Ans. (d) : CH3COOH + NaOH → CH3COONa + H2O
800 800
No. of moles of 1 sodium acetate CH3COONa =
=
=
× 0.015
K H 580
0.1×100
= 0.01mol
= 0.0207g/L
1000
Objective Chemistry Volume-II
132
YCT
hydrolysis
2CH 3COONa
→ CH 3 − CH 3 + 2CO 2 + 2NaOH + H 2 ↑
electrolysis
Ans. (c) : Normal solution has one gram equivalent of
solute dissolved in 1L of solution.
ethane
No. of moles of ethane =
1
no. of moles of CH3COONa
2
1
= × 0.001 = 0.005 moles
2
∵ At STP, 1 mole of ethane → 22400 mL volume
∴ 0.005 mole of ethane → 0.005 × 22400
= 112 mL
680. Assuming fully decomposed, the volume of CO2
released at STP on heating 9.85 g of BaCO3
(atomic mass, Ba = 137) will be–
(a) 1.12 L
(b) 0.84 L
(c) 2.24 L
(d) 4.06 L
BCECE-2012
Ans. (a) : Reaction
BaCO3 → BaO + CO2↑
∵ It is clear that 197g produces 22.4L at S.T.P
Then,
22.4
9.85 will produce =
×9.85
197
= 1.12L at STP
681. On dissolving a non- volatile solute in a solvent,
the vapor pressure of the solvent is decreased by
10 mm of mercury. The mole fraction of the
solute in this solution is 0.2. If the vapour
pressure of the solvent is decreased by 20 mm of
mercury by dissolving more solute, what is the
mole fraction of solvent in this solution now?
(a) 0.2
(b) 0.4
(c) 0.6
(d) 0.8
CG PET -2008
Ans. (c) : We know that,
P o – Ps = P o × mole fraction of solute
10 = P o × 0.2
………. (1)
20 = P o × X2
………. (2)
From (1) and (2)
X2 = 0.4
So,
X1 = 1 – X2
= 1 – 0.4
= 0.6
682. Which one of the following salts give an acidic
solution in water?
(a) CH3COONa
(b) NH4Cl
(c) NaCl
(d) CH3COONH4
CG PET -2007
−
⇀
Ans. (b) : NH 4 Cl ↽ NH 4 + Cl
6.
Isotonic Solution
684. A solution containing 6.0 g of urea is isotonic
with a solution containing 10g of a nonelectrolytic solute X. The molar mass of X (in g
mol-1) is
(a) 50.0
(b) 100
(c) 75.0
(d) 68.0
AP-EAMCET-04.07.2022, Shift-II
Ans. (b) : Weight of area = 6.0 gm
Molecular weight of area = 60 g mol–1
Weight of solute × = 10 g
For,
isotonic solution λ1 = λ2
Or C1 = C2 (conc. In mol/lit )
weight
Where, C =
molecular weight
So,
6.0 10
=
60 m w
60×10
6.0
m w = 100g mol−1
mw =
685. x% (w/v) solution of urea is isotonic with 4%
(w/v) solution of a non-volatile solute of molar
mass 120 g mol-1. The value of x is
(a) 2
(b) 4
(c) 3
(d) 5
AP-EAMCET-04.07.2022, Shift-II
Ans. (a) : For isotonic solution λ1 = λ2
C1 = C2 (∵ λ = CRT)
x ×1000
2×1000
=
60×100 120×100
x 2
=
6 6
x=2
686. Which of the following pair of solutions is
isotonic?
(a) 18g/L of glucose of solution and 6g/L of Urea
solution
(b) 10g/L of glucose of solution and 10g/L of
Urea solution
(c) 0.01 M NaOH solution and 0.02 M glucose
solution
NH 4 Cl + H 2 O ↽ ⇀ NH 4 OH + H + Cl−
(d)
0.01
M NaCl solution and 0.01 M glucose
So, NH4Cl salt gives an acidic solution in water.
solution
683. Normal solution is
(Assume that NaCl undergoes complete
(a) inert solution
dissociation)
(b) acidic solution
(a)
A and B
(b) A and C
(c) one litre containing one equivalent
(c) B and D
(d) B and C
(d) basic solution
AP-EAMCET-05.07.2022, Shift-II
UP CPMT-2005
Objective Chemistry Volume-II
133
YCT
Ans. (b) : Isotonic solution are those which have the
same osmotic pressure (π = iCRT). But were we have
different concentration of the solution and also they
have different vant's huff factors (i).
So 10g/L liter of glucose of solution and 10g/L of urea
solution is Isotonic solution.
687. Solutions A, B, C and D are 0.1 M glucose, 0.05
M NaCl, 0.05 M BaCl2 and 0.1 M AlCl3
respectively. Which one of the following pairs is
isotonic?
(a) B and C
(b) A and B
(c) A and D
(d) A and C
JIPMER-2017
SRMJEEE – 2016
CG PET- 2011
MHT CET–2008
Ans. (b) : The solutions which have same osmotic
pressure is known as isotonic solution.
∏ = i C.R.T
Where - ∏ = Osmotic pressure
C = Concentration of solution
R = Gas constant
T = Temperature
i = Van’t Hoff factor.
For solution A –
∏A = 1 × 0.1 RT = 0.1RT
For solution B –
∏B = 2 × 0.05 RT = 0.1RT
For solution C –
∏C = 3 × 0.05 RT = 0.15RT
For solution D –
∏D = 4 × 0.1 RT = 0.4RT
Thus, solution A and B have the same osmotic pressure
therefore isotonic solution.
688. Isotonic solutions have same
(a) temperature
(b) vapour pressure
(c) freezing point
(d) osmotic pressure
COMEDK 2017
JCECE - 2007
Ans. (d) : When two solutions have same osmotic
pressure and salt concentration are said to be isotonic
solution. Iso (same) and tonic (concentration) .
689. A 0.5 g/L solution of glucose is found to be
isotonic with a 2.5 g/L solution of an organic
compound. What will be that molecular weight
of the organic compound?
(a) 300
(b) 600
(c) 900
(d) 1200
JCECE - 2004
UPCPMT–2008
0.5g
Ans. (c) : Molarity of glucose solution =
180 × 1L
= 0.00277 M
Isotonic solution have equal molar concentrations,
hence, molarity of organic compound solution =
0.00277 M
2.5
∴ Molacular weight of compound =
= 900
0.00277 ×1
Objective Chemistry Volume-II
690. Red blood cells are placed in a solution and
neither haemolysis nor crenation occurs.
Therefore, the solution is
(a) hypertonic
(b) hypotonic
(c) isotonic
(d) isotopic
JIPMER-2018
J & K CET-(2000)
Ans. (c) : When red blood cells are placed in a solution
and neither haemolysis nor erenation occurs. This
means that solution is isotonic in nature, since in
isotonic solution, there is no process of osmosis takes
place.
691. Blood cells will remain as such in
(a) hypertonic solution (b) hypotonic solution
(c) isotonic solution
(d) None of the above
UP CPMT-2004
(AIPMT -1991)
Ans. (c) : (i) In hypertonic solution, blood cells (living
cells) shrink due to plasmolysis (because water comes
out of them).
(ii) In hypotonic solution, blood cells burst due to
endosmosis.
(iii) In isotonic solution, blood cells remain as it is
because water neither enters in them nor comes out.
692. A 25% solution of cane-sugar (mol mass = 342
g mol–1) is isotonic with 5% solution of a
substance A. Then find the molecular weight of
A?
(a) 6.84 g mol–1
(b) 68.4 g mol–1
–1
(c) 25 g mol
(d) 684 g mol–1
AP-EAMCET 25-08-2021 Shift - I
Ans. (b) : Given,
25% solution of cane sugar means 25g of solute present
in 100mL of solution.
∴ For isotonic solutions,
∏1 = ∏2
or
C1 = C2
25
5
=
342 × 0.1 M 2 × 0.1
342
= 68.4g / mol
5
693. Acetaldehyde upon treated with dil. NaOH
forms.
or
134
M2 =
(a) CH3
CH2
OH
OH
(b) CH3
CH
CH2
CHO
(c) CH3COOH
(d) CH3–CH3
Assam CEE-2021
YCT
Ans. (d) : Here–
∏ urea = ∏ sucrose
C1 = C2
So,
Ans. (b)
CH3
C
H + CH3
O
C
H
O
dil. NaOH
CH2
C
H + CH3
+
C
H
O-
O
↓
H
CH3
C
CH2
OH
C
H
O
694. Which of the following pairs of solutions is
expected to be isotonic at same temperature?
(a) 0.1 M urea and 0.1 M NaCl
(b) 0.1 M glucose and 0.2 M NaCl
(c) 0.1 M NaCl and 0.1 M CH3COOH
(d) 0.1 M NaCl and 0.1 M KNO3
AP EAMCET (Engg.) 17.09.2020, Shift-II
Ans. (d) : Two solutions of same osmotic pressure (∏)
are isotonic solutions.
∏1 (0.1 M NaCl) = 0.1 × RT × 2
∏2 (0.1 M KNO3) = 0.1 × RT × 2
So, ∏1 (0.1 M NaCl) = ∏2 (0.1 M KNO3) at same
temperature.
695. Which of the following pair of solutions is
isotonic?
(a) 0.01 M BaCl2 and 0.015 M NaCl
(b) 0.001 M Al2(SO4)3 and 0.001 M BaCl2
(c) 0.001 M CaCl2 and 0.001 M Al2 (SO4)3
(d) 0.01 M BaCl2 and 0.001 M CaCl2
Karnataka-CET-2020
Ans. (a) : Isotonic solution are those which have same
osmotic pressure at the given temperature
For 0.01 M BaCl2, i=3
∏= iCRT
= 3×0.01RT
= 0 .03RT
For 0.015 M NaCl, i=2
∴ ∏ = 2×0.015RT
= 0.03RT
Therefore, 0.01 M BaCl2 and 0.015 M NaCl are
isotonic.
696. Which of the following sets of solutions of urea
(mol. mass 60 g mol–1) and sucrose (mol. mass
342 g mol–1) is isotonic?
(a) 9.1 gL–1 urea and 6.0 gL–1 sucrose
(b) 3.0 gL–1 urea and 3.0 gL–1 sucrose
(c) 6.0 gL–1 urea and 9.0 gL–1 sucrose
(d) 3.0 gL–1 urea and 17.1 gL–1 sucrose
MHT CET-02.05.2019, SHIFT-II
Objective Chemistry Volume-II
3 ( M 2 ) w.t m1 m 2
=
=
∴
60
342 M1 M1
M2 = 17.1gL–1
697. Which solution is isotonic with 6% w/v aqueous
solution of urea? [Mole mass of Urea = 60 gm.
mol-1]
(a) 0.25 M NaCl
(b) 0.5 M NaCl
(c) 0.1 M NaCl
(d) 1 M NaCl
GUJCET-2018
Ans. (b) : From formula of osmotic pressure–
∏ = iCRT
Where i = 2(NaCl) and
1 for urea
Given, 60g urea in 100 mL solution.
Now,
60
C=
= 1M
60 × 1L
then
(i×C)urea = (i×C)NaCl
1×1M = (2×C)NaCl
(C)NaCl = 0.5 M.
698. Isotonic solutions are solutions having the same
(a) surface tension
(b) vapour pressure
(c) osmotic pressure
(d) viscosity
Karnataka-CET-2018
Ans. (c) : Isotonic solution having the same osmotic
pressure (∏ = iCRT) at the same temperature.
699. At certain temperature 1.6% solution of an
unknown substance is isotonic with 2.4%
solution of Urea. If both the solutions have the
same solvent and both the solutions have same
density 1 gm/cm3, what will be the molecular
mass of unknown substance in gm./mol.
[Molecular mass of urea = 60 gm/mol]
(a) 30
(b) 40
(c) 80
(d) 90
GUJCET-2017
Ans. (b) From formula,
∏ = iCRT
∏1 = ∏2 , C1= C2 (isotonic Solution have same
osmotic pressure and concentration)
Now,
For same volume,
n1 = n2
1.6 2.4
=
M 60
M = 40 g/mol.
700. Assuming the compounds to be completely
dissociated in aqueous solution, identify the
pair of the solution that can be expected to be
isotonic at the same temperature.
135
YCT
(a)
(b)
(c)
(d)
0.01M urea and 0.01 M NaCl
0.02 M NaCl and 0.01 M Na2SO4
0.03 M NaCl and 0.02 M MgCl2
0.01 M sucrose and 0.02 M glucose
WB-JEE-2017
Ans. (c) : In the isotonic solution at same temperature∏1 = ∏ 2
i1C1RT = i2C2RT
i1C1 = i2C2
For 0.03 M NaCl
i1 = 2
i1C1 = 2 × 0.03
= 0.06
For 0.02 M MgCl2 i2 = 3
i2C2 = 3 × 0.02
= 0.06
Therefore, 0.03M NaCl and 0.02M MgCl2 are isotonic.
701. 5% solution of cane sugar is isotonic with
0.877% of X.
The molecular weight of substance X is
(a) 126.98
(b) 119.96
(c) 95.5
(d) 59.98
JCECE - 2016
Ans. (d) : ∵ Both solutions are isotonic.
∴
or,
Given,
W1
W2
= 1,
=5
V1
V2
M2 = 342g/mol, M1 = ?
For isotonic solution–
W1
W2
=
V1M1 V2 M 2
Ans. (a) : Given
1
5
=
M1 342
342
5
M1 = 68.4g mol–1
Hence, the molar mass of solute x = 68.4 g mol–1
704. 0.6% solution of urea will be isotonic with
(a) 0.1 M glucose
(b) 0.1 M KCl
(c) 0.6% glucose solution
(d) 0.6% KCl solution
Assam CEE-2014
0.6 1000
×
Ans. (a) : 0.6 urea has molarity =
60 100
= 0.1M glucose
705.
M1 =
ncane sugar = n(x)
W
W
=
M cane sugar( = w1 ) M X( = w2 )
W1 W2
=
M1 M2
5
0.877
=
342
X
(Mcane sugar = 342, X = ?)
0.877 × 342
∴
X=
= 59.98 g/mol
5
702. 0.06% (w/v) aqueous solution of urea is isotonic
with
(a) 0.06% glucose solution
(b) 0.6% glucose solution
(c) 0.01 M glucose solution
(d) 0.1 M glucose solution
Karnataka-CET-2015
Ans. (c) : 0.06 % w/v solution means 100 ml 06
solution contains 0.06 gm of urea–
0.06
Moles of urea =
= 0.001mol
60
0.001
Molarity of urea solution =
× 1000
100
= 0.01 M
703. The molar mass of a solute X in g mol–1, if its
1% solution is isotonic with a 5% solution of
cane sugar (molar mass = 342 g mol–1), is
(a) 68.4
(b) 34.2
(c) 136.2
(d) 171.2
AP-EAMCET (Engg.) - 2014
Objective Chemistry Volume-II
Compartments A and B have the following
combinations solution
List-I
List-II
A 0.1 M KCl
0.2 M KCl
B 0.1% (m/V) NaCl
10% (m/V) NaCl
C 18 gL–1 glucose
34.2 gL–1 sucrose
D 20% (m/V)
10% (m/V)
glucose
glucose
Indicate the number of solutions which is/are
isotonic
(a) I only
(b) III only
(c) IV only
(d) II only
BCECE-2014
Ans. (b) : In the isotonic solution having the same
molarity.
18
Molarity of 18 g glucose =
180 × 1
18
=
= 0.1 g/L
180
34.2
Molarity of 34.2 g sucrose=
342 × 1
34.2
=
= 0.1 g/L
342
The concentration of 18 g L–1 glucose is same to 34.2 g
L–1 sucrose.
136
YCT
(c) 100 mL of 0.2M AgNO3 + 100 mL of 0.1M
KI
(d) 100 mL of 0.15M AgNO3 + 100 mL of 0.15M
KI
BCECE-2013
Ans. (b) : Negatively charged colloidal solution is
formed when AgNO3 is completely precipitated as AgI
and extra KI is absorbed on AgI.
Ag+ + I– → AgI
Thus, [Ag+] < [I–]
710. At 25oC, at 5% aqueous solution of glucose
(molecular weight = 180 g mol-1) is isotonic with
a 2% aqueous solution containing and
unknown solute, What is the molecular weight
of the unknown solute?
(a) 60
(b) 80
(c) 72
(d) 63
(e) 98
Kerala-CEE-2011
Ans. (c) : Here osmotic pressure is equal
So, ∏1 = ∏ 2
C1 = C2
Then molecular weight of unknown
MPPET-2013 Solute, m1 = m 2
M1 M 2
Ans. (b) : The hydrolysis of sodium hydride occurs in
the reaction where it gets broken into to Na+ cation and
5
2
=
⇒ M 2 = 72
H– anion. The Na+ forms sodium hydroxide and H–
180 M 2
forms hydrogen gas. The formation of sodium
hydroxide reduces the risk of fires. Sodium Hydroxide 711. Pure water can be obtained from sea water by
(a) centrifugation
(b) plasmolysis
is corrosive in nature. The reaction takes place as
(c) reverse osmosis
(d) sedimentation
follows:
(AIPMT -2011)
NaH(s) + H2O(l) → NaOH + H2(g)
The hydrolysis of NaH is violent. The reaction is an Ans. (c) : Pure water can be obtained from the sea water
by the process of reverse osmosis. In this method,
exothermic reaction in which the energy is released.
708. Which one of the following is an isotonic pair of pressure is applied and the solute is retained at the
pressurised side by allowing the solvent to pass through
solution?
the membrane.
(a) 0.15 M NaCl and 0.1 M Na2SO4
712. A 6% solution of urea is isotonic with
(b) 0.2 M Urea and 0.1 M Sugar
(a) 1 M solution of glucose
(c) 0.1 M BaCl2 and 0.2 M Urea
(b) 0.05 M solution of glucose
(d) 0.4 M MgSO4 and 0.1 M NH4Cl
(c) 6% solution of glucose
AP-EAMCET (Engg.) 2013
(d) 25% solution of glucose
Ans. (a) : Two solution having the same osmotic
Karnataka-CET, 2009
pressure across a semi permeable membrane is referred
6
to as isotonic solution.
For 0.15M NaCl and 0.1M Na2SO4, NaCl is an Ans. (a) : Molarity of urea = 60 = 1M
100
electrolyte which dissociates to give 2 ions, thus
concentration of ions in the solution 0.30 M. Similarly
1000
Na2SO4 (3 ions) the concentration of ions in the solution Hence, 1 M solution of glucose is isotonic with 6% urea
= 0.30 M
solution.
Hence, both are isotonic.
713. A 5.25% solution of a substance is isotonic with
a 1.5% solution of urea (molar mass=60 g mol–
709. On adding AgNO3 solution into KI solution, a
1
) in the same solvent. If the densities of both
negatively charged colloidal sol is obtained
the solutions are assumed to be equal to 1.0 g
when they are in–
cm–3, molar mass of the substance will be
(a) 100 mL of 0.1M AgNO3 + 100 mL of 0.1M
(a) 90.0 g mol–1
(b) 115.0 g mol–1
KI
–1
(c) 105.0 g mol
(d) 210.0 g mol–1
(b) 100 mL of 0.1M AgNO3 + 100 mL of 0.2M
[AIEEE 2007]
KI
706. Concentrated hydrochloric acid when kept in
open air sometime produces a cloud of white
fumes. The explanation for it is that
(a) concentrated hydrochloric acid emits atrongly
smelling HCl gas all the time
(b) oxygen in air reacts with the emitted HCl gas
to form a cloud of chlorine gas
(c) strong affinity of HCl gas for moisture in air
results in forming of droplets of liquid
solution which appears like a cloudy smoke
(d) due to strong affinity for water, concentrated
hydrochloric acid pulls moisture of air
towards itself. This moisture forms droplets
of water and hence, the cloud.
BCECE-2014
Ans. (b) : Oxygen in air reacts with the emitted HCl gas
to form a cloud of chlorine gas.
1
2HCl + O 2
→ H 2 O + Cl 2
2
707. Sodium hydride when dissolved in water
produces :
(a) Acidic solution
(b) Basic solution
(c) Neutral solution
(d) Cannot say
Objective Chemistry Volume-II
137
YCT
Ans. (d) : For same osmotic pressure
∏1 = ∏2
So,
C1RT = C2RT
5.25
1000 1.5 × 1000
⇒
×
=
molar mass 100
60 × 100
Molar mass =
5.25 × 60
1.5
W1
10
=
60 × 1 342 × 0.1
or
W1 = 17.54 g/L
717. At certain temperature a 5.12% solution of
cane sugar is isotonic with a 0.9% solution of
an unknown solute. The molar mass of solute
is;
(a) 60
(b) 46.17
(c) 120
(d) 90
(e) 92.34
Kerala-CEE-2006
∴
= 210g / mol
714. Which of the following make up an isotonic
342
X
Ans. (a) : Molar mass of solute =
=
⇒ 60.117
triad?
5.12
0.9
78
77
74
40
40
40
(a) 32 Ge, 33 As, 31 Ga
(b) 18 Ar, 19 K, 20 Ca
718. A solution containing 10 g per dm3 of urea
233
232
239
13
12
14
(molecular mass = 60 g mol–1) is isotonic with a
(c) 92 U, 90 Th, 94 Pu
(d) 6 C, 7 C, 7 N
5% solution of a non-volatile solute. The
16
15
(e) 14
6 C, 8 O, 7 N
molecular mass of this non-volatile solute is
Kerala-CEE-2007
(a) 200 g mol–1
(b) 250 g mol–1
–1
(c) 300 g mol
(d) 350 g mol–1
Ans. (e) : Isotonic triad indicates the number of
neutrons same for the three species.
(AIPMT -2006)
As we know,
Ans. (c) : For Isotonic solution at same temperature so
14
concentration are equal.
C6 ⇒ As e– = p = 6
Concentration of urea = Concentration of non volatile
n = 14 – 6
solute
=8
16
10
5 1000
O8 ⇒ e– = p = 8
=
×
n = 16 – 8
60 100 M
=8
1 50
=
15
N7 ⇒ e– = p = 7
6 M
n = 15 – 7
M = 300g / mol
=8
Hence, option (a) is correct.
7. Coagulation
715. Blood cells do not shrink in blood because
blood is
719. Given below are two statements:
(a) hypotonic
(b) isotonic
Statement I:
(c) equimolar
(d) hypertonic
In the coagulation of a negative sol, the
UP CPMT-2007
flocculating power of the three given ions is in
Ans. (b) : Two solutions of different substances having
the ordersame osmotic pressure at same temperature are known
Al3+ > Ba2+ > Na+
as isotonic solutions.
Statement II:
Blood cells do not shrink in blood because blood is
In the coagulation of a positive sol, the
isotonic. When a cell is placed in a solution having an
flocculating power of the three given salts is in
isotonic pressure equal to its own, the cell maintains its
the ordernormal volume. As isotonic solution has the same
NaCl > Na2SO4 > Na3PO4
osmotic pressure as the cells contained in it, there is no
In the light of the above statements, choose the
concentration gradient.
most appropriate answer from the options
given below:
716. What is the amount of urea dissolved per litre
(a) Statement I is incorrect but statement II is
if aqueous solution is isotonic with 10% cane
correct.
sugar solution? (mol. wt. of urea = 60)
(b)
Both
Statement I and Statement II are correct.
(a) 200 g/L
(b) 19.2 g/L
(c)
Both
Statement I and Statement II are
(c) 17.54 g/L
(d) 16.7 g/L
incorrect
AP-EAMCET (Medical), 2006
(d) Statement I is correct but Statement II is
Ans. (c) : Given- W2= 10g, W1 = ?
incorrect.
NEET-17.06.2022
V2 = 100mL = 0.1L, V1 = 1L
Isotonic solutions have same osmotic pressure.
Ans. (d) : The order of coagulating power of PO34− ,
W1
W2
SO 24− and Cl− in the coagulation of a positive solution is
=
( Molecular weight )urea × V1 ( Molecular weight )canesugar × V2
– coagulation
Objective Chemistry Volume-II
138
YCT
↓
Negatively, charged ions
PO34− , SO 24− , Cl −
So coagulating power α magnitude of charge.
Hardy schulze Rule opposite charged ions coagulate
solution particles and greater the magnitude of charge
higher is the coagulating power. So, correct order is
PO34− , SO 24− , Cl −
i.e.
Na3PO4 > Na2SO4 > NaCl
720. The correct order of coagulating power of the
following ions to coagulate the positive sol is
[Fe (CN)3]4-, Cl(-), SO 24
I
II
III
(a) I > II > III
(b) III > II > I
(c) I > II > III
(d) I > III > II
AP-EAMCET-05.07.2022, Shift-I
Ans. (d) :
(I) [Fe(CN)3]4–– Coagulation power is 4
(II) Cl–1 – Coagulation power is –1
(III) SO−4 2 – coagulating power is 2
Charge
Coagulating power ∝ polarizing power ∝
Size
Hence, the correct order of coagulating power –
I > III > II .
721. The coagulation of 200 ml of a positive colloid
took place when 0.73 g of HCl was added to it
without changing the volume much. The
flocculation value of HCl for the colloid is
(a) 1000
(b) 0.365
(c) 200
(d) 100
TS-EAMCET-19.07.2022, Shift-II
Ans. (d) : Given that –
Mass of HCl = 0.73 g
0.73
Number of moles in 0.73 g =
36.5
= 20 m mol
200 mL of solution required = 20 m mol
20
1 L solution required =
×1000
200
= 100 m mol.
Flocculation value of HCl = 100
722. Which of the following substances show the
highest colligative properties?
(a) 0.1 M BaCl2
(b) 0.1 M AgNO3
(c) 0.1 M urea
(d) 0.1 M (NH4)3 PO4
TS-EAMCET-19.07.2022, Shift-II
Ans. (d) : Colligative properties depend upon number
of solute particles. Here 0.1M (NH4)3 PO4 contains the
maximum particles. So, it has the highest colligative
properties.
So, option (d) is correct.
723. For As2S3 sol, the most effective coagulating
agent is
(a) CaCO3
(b) NaCl
(c) FeCl3
(d) Clay
TS-EAMCET-18.07.2022, Shift-I
Objective Chemistry Volume-II
Ans. (c) : AS 2 S3 is negative sol. It is obvious that
captions are effective in coagulating negative sols.
According to hardy Schulze rule
greater the valency of the coagulating ion greater is its
coagulating power the given FeCl3 Fe3+ is most
(
)
effective for causing coagulation of As 2S3 sol.
724. The coagulating power of an electrolyte for
arsenious sulphide solution decreases in the
order,
(a) Al3+ > Ba 2+ > Na + (b) Cl − > SO 24− > PO34−
(c) PO34− > Cl − > SO 24−
(d) Na + > Al3+ > Ba 2+
JEE Main -2013
GUJCET-2007
AMU-2017
CG PET-2010
JCECE - 2005
Ans. (a) : It has been observed that, generally, the
greater the valence of the flocculating ion added, the
greater is its power to cause precipitation. This is known
as Hardy-Schulze rule. In the coagulation of a negative
solution, the flocculating power is in the order.
Al3+ > Ba 2+ > Na + .
725. Which of the following ions will have maximum
flocculating power for coagulation of As2S3
solution?
(a) Na+
(b) Al3+
2+
(c) Mg
(d) Ba2+
GUJCET-2021
JEE Main 2019, 9 Jan Shift-II
Kerala-CEE-2009
Ans. (b) : According to Hardy Schulze rule, greater
the valency of the coagulating ion, greater is its
coagulating power. Al3+ has maximum flocculating
power for coagulation of As2S3 solution.
• The flocculation power in the coagulation of negative
solution decreases in the order Al3+ > Ba2+ > Na+.
726. Match the entries in column-I with the terms
used to show the effect of process in ColumnII?
Column-I
Column-II
(i) Ferric hydroxide is (a) Double
decomposition
mixed with arsenic
sulphide solution
(ii) FeCl3 is mixed with (b) Coagulation
freshly prepared ppt
of Fe (OH)3
(iii) H2S gas is passed (c) Tyndall effect
through arsenic oxide
solution
(iv) A beam of light is (d) Peptization
passed through milk
(a) (i→ d), (ii → a), (iii → b), (iv → c)
(b) (i→ b), (ii → d), (iii → a), (iv → c)
(c) (i→ d), (ii → b), (iii→ a), (iv → c)
(d) (i →b), (ii → a), (iii → d), (iv → c)
AP- EAPCET- 07-09-2021, Shift-I
139
YCT
Ans. (b) : (i) Ferric hydroxide is mixed with arsenic
sulphide solution is coagulation
(ii) FeCl3 is mixed with freshly prepared ppt of
Fe(OH)3 is peptization.
(iii) H2S gas is passed through arsenic oxide solution is
double decomposition.
(iv) A beam of light is passed through milk is Tyndall
effect.
727. Which of the following solution has the lowest
osmotic pressure?
(a) 200 ml of 2 M NaCl solution
(b) 200 ml of 1 M glucose solution
(c) 200 ml of 2 M urea solution
(d) 200 ml of 1 M KCl solution
TS EAMCET 10.08.2021, Shift-II
Ans. (b) : Osmotic pressure is a colligative property it
depends upon the number of constituents of solution.
And temperature
∏ = C R T or
∏ =
n
Now,
0.0507
1
= 0.507
the coagulation value of Cl– =
So, Nearest integer will be 1.
731. Zeta potential is
(a) Potential required to bring about coagulation
of a colloidal solution.
(b) Potential required to give the particle a speed
of 1 cm s–1
(c) Potential difference between fixed charged
layer and the diffused layer having opposite
charges
(d) Potential energy of the colloidal particles.
Kerala-CEE-29.08.2021
Ans. (c) : The charges of opposite signs on the fixed
and diffused part of the double layer results in a
difference in potential between these layers in the same
manner as potential difference is developed in a
capacitor. This potential difference between the fixed
layer and the diffused layer of opposite charges is called
the zeta potential.
732. A colloidal solution is subjected to an electric
field than colloidal particles more towards
anode. The amount of electrolytes of BaCl2,
and NaCl required to coagulate the given
colloid is in the order
(a) NaCl > BaCl2 > AlCl3
(b) BaCl2 > AlCl3 > NaCl
(c) AlCl3 = NaCl = BaCl2
(d) AlCl3 > BaCl2 > NaCl
Kerala-CEE-29.08.2021
Ans. (d) : According to the Hardy Schulze rule, the
effective ions of the electrolytes in bringing about
coagulation are those which carry a charge opposite to
that of colloidal particles. Greater the valency of
coagulation, greater is its power to bring about
coagulation.
Hence, the electric field subjected to the coagulation
power should be in the order of AlCl3 > BaCl2 > NaCl
733.
If 0.2 moles of sulphuric acid is poured in to
250 ml of water. Calculate the concentration of
the solution?
(a) 0.8 N
(b) 0.8 M
(c) 8 M
(d) 0.2 N
AP EAPCET 19-08-2021, Shift-II
Ans. (b) : Given, n = 0.2 mol
V = 250 ml
M=?
n × 1000
Concentration of the solution =
V ( in liter )
RT
v
200 ml of l M glucose solution have lower molar
concentration and unionised.
728. The most effective coagulating agent among the
options for Sb2S3 solution is
(a) Na2SO4
(b) Al2(SO4)3
(c) CaCl4
(d) NH4Cl
TS EAMCET 05.08.2021, Shift-I
Ans. (b) : The most effective coagulating agent among
the options for Sb2S3 solution is Al2(SO4)3 because
Sb2S3 is a negative solution. Al2(SO4)3 will be the most
effective coagulant due to higher charge density on Al3+
by the Hardy-Schulze rule.
729. For the coagulation of a negative solution, the
species below, that has the highest flocculation
power is:
(a) Ba 2 +
(b) Na +
3–
(c) PO 4
(d) SO 2–
4
JEE Main 17.03.2021, Shift-II
Ans. (a) : The coagulation of negative solution, the
higher charge containing ions have the highest
flocculating power. So, Ba2+ has highest flocculating
power.
730. 100 mL of 0.0018% (w/v) solution of Cl– ion
was the minimum concentration of Cl–
required to precipitate a negative sol in one h.
0.2 × 1000
The coagulating value of Cl– ion is ........
=
(Nearest integer)
250
[JEE Main 2021, 20 July Shift-II]
= 0.8 M
Ans. (1) : Given:734. The flocculation value of HCl for arsenic
sulphide solution is 30 m mol L–1. If H2SO4 is
weight of Cl– ion = 0.0018 g
used for the flocculation of arsenic sulphide,
So,
the amount in grams of H2SO4 is 250 mL
0.0018
required for the above purpose is
Millimoles of Cl– ion e =
× 1000
35.5
.......(molecular mass of H2SO4 = 98 g/mol)
= 0.0507
[JEE Main 2020, 7 Jan Shift-II]
Objective Chemistry Volume-II
140
YCT
Ans. (0.37) :
Given:1L solution →30 m.mol HCl
30
1mL solution →
m.mol HCl
1000
Now, for 250 ml solution15
=
× 98 × 10−3
4
= 0.3675g
735. In the coagulation of a positive sol, the
flocculating power of the ions PO43–, SO42– and
Cl– decreases in the order
(a) PO43– > Cl– > SO42– (b) PO43–> SO42– > Cl–
(c) Cl–>SO42–> PO43–
(d) Cl–> PO43– > SO42–
2–
3–
–
(e) SO4 > PO4 > Cl
Kerala-CEE-2020
Ans. (b) : The minimum amount of an electrolyte
required to cause precipitation of one litre of a colloidal
solution is called coagulation value of flocculation
value of the electrolite for the solution. The reciprocal
of coagulation value is regarded as the coagulating
power.
For coagulation of positive sol, the flocculating power
depends on two factors:–
(1) Large size of anion
(2) High (–ve) charge
So, PO43–> SO42– > Cl–
736. Among the following the ion which will be
more effective for flocculation of Fe(OH)3
solution is
(a) PO34−
(b) SO 24 −
(c) SO32 −
(d) NO3−
WB-JEE-2020
Ans. (a) : Fe(OH)3 solution is positively charged
solution. It will be coagulated or flocculation by
negatively charged ions.
PO34− have maximum negative charge among the given
electrolytes.
Hence, PO34− is most effective for coagulation of
Fe(OH)3
737. The formation of association of colloidal
particles by addition of electrolyte to form as
insoluble precipitate is called______.
(a) Flocculation
(b) Emulsification
(c) Coagulation
(d) Micelle
GUJCET-2019
Ans. (c) : The formation of association of colloidal
particles by addition of electrolyte to form as insoluble
precipitate is called coagulation.
Coagulation involves coming together of colloidal
particle.
738. Bleeding due to a cut can be stopped by
applying ferric chloride solution in the
laboratory. This is due to
Objective Chemistry Volume-II
(a) coagulation of negatively charged blood
particles by Fe3+ ions.
(b) coagulation of positively charged blood
particles by Cl − ions.
(c) reaction taking place between ferric ions and
the hemoglobin forming a complex.
(d) common element, iron, in both FeCl3 and
hemoglobin.
JIPMER-2018
Ans. (a) : Bleeding to cut can be stopped by applying
ferric chloride solution. This is due to coagulation of
negatively charged colloidal blood particles by
positively charged Fe3+ ions.
739. Which of the following electrolytes will have
maximum coagulating value for AgI/Ag+ sol?
(a) Na2S
(b) Na3PO4
(c) Na2SO4
(d) NaCl
Karnataka-CET-2018
Ans. (b) : Na3PO4 is the electrolytes that has maximum
coagulating value of AgI/Ag+ solution.
740. On which of the following properties does the
coagulating power of an ion depend?
(a) The magnitude of the charge on the ion alone
(b) Size of the ion alone
(c) Both magnitude and sign of the charge on the
ion
(d) The sign of charge on the ion alone
(NEET 2018)
Ans. (c) : The coagulating power of an electrolyte
depends on the magnitude and sign of the charge
present on the ion.
741. Ferric chloride is used to stop bleeding in cuts
because
(a) Fe3+ coagulated blood which is positively
charged solution
(b) Fe3+ coagulated blood which is negatively
charged solution
(c) Cl– coagulated blood which is positively
charged solution
(d) Cl– coagulated blood which is negatively
charged solution
UPTU/UPSEE-2018
Ans. (b) : Ferric chloride is used to stop bleeding in
cuts because Fe3+ coagulated blood which is negatively
charged sol.
742. The process which is responsible for the
formation of delta at a place where rivers meets
the sea is
(a) coagulation
(b) colloid formation
(c) peptisation
(d) emulsification
Karnataka-CET-2017
Ans. (a) : The coagulation is responsible for the
formation of delta at a place where rivers meets the sea.
Sea water contains electrolytes which have properly to
coagulate the colloidal particles.
743. The coagulation values in millimoles per litre of
the electrolyes used for the coagulation of As2S3
are given below:
141
YCT
I. ( NaCl ) = 52,
II. ( BaCl 2 ) = 0.69,
Ans. (e) : Gold solution can be prepared by reduction
of gold (III) chloride with formalin solution,
So, Reaction will be:
III. ( MgSO 4 ) = 0.22
Reduction
The correct order of their coagulating power is 2AuCl + 3HCHO + 3H O
→ 2Au + 3HCOOH + 6HCl
3
2
(a) I > II > III
(b) II > I > III
748. Which ion has least flocculation value for a
(c) III > II > I
(d) III > I > II
positive sol?
(NEET-II 2016)
(a) [Fe(CN)6]4–
(b) Cl–
1
(c) SO 24−
(d) PO 24−
Ans. (c) : Coagulation power ∝
Coagulation values
UP CPMT-2013
Higher the coagulation power, lower is the coagulation Ans. (a) : Greater the valency of the coagulating or the
values in millimoles per liter.
flocculating ion, greater is its power to bring about
MgSO4 > BaCl2 > NaCl
coagulation and smaller is its coagulation or
744. Which of the following is the most effective in flocculation value. Thus for coagulation of positively
causing coagulation of ferric hydroxide charged sol, tetravalent [Fe(CN)6]4– anions are more
solution?
effective than trivalent anions ( PO34− ) which are more
(a) KCl
(b) KNO3
(d) K3[Fe(CN)6]
(c) K2SO4
effective than divalent ( SO 24 − ) anions which in turn are
AP-EAMCET (Engg.) 2015 more effective than monovalent (Cl–) anions.
Ans. (d) : More the number of ions of a compound after 749. The gold numbers of A, B, C and D are 0.04,
dissociation causing most effective coagulation of ferric
0.0002, 10 and 25 respectively. The protective
hydroxide solution.
powers of A, B, C, and D are in the order
(i) KCl ↽ ⇀ K + + Cℓ –
(a) A > B > C > D
(b) B > A > C > D
+
–
(c)
D
>
C
>
B
>
A
(d)
C>A>B>D
⇀
(ii) KNO3 ↽ K + NO3
AMU-2011
(iii) K 2SO 4 ↽ ⇀ 2K + + SO 4 2 –
Ans. (b) : The gold number is inversely proportional to
the protective power of the colloid. According to the
(iv) K 3 [Fe(CN)6 ] ↽ ⇀ 3K + + [Fe(CN)6 ]3–
given data in question we can say that option (b) are
Thus, K3[Fe(CN)6] has more coagulation of ferric
correct then B > A > C > D.
hydroxide solution.
750.
Which one of the following does not involve
745. The pair of compound which cannot exist
coagulation?
together solution is
(a) Formation of delta regions
(a) NaHCO3 and NaOH
(b)
Peptization
(b) NaHCO3 and H2O
(c) Treatment of drinking water by potash alum
(c) NaHCO3 and Na2CO3
(d) Clotting of bloos by the use of ferric chloride
(d) Na2CO3 and NaOH
KARNATAKA-CET, 2010
Karnataka-CET-2015
Ans. (a) : NaHCO3 and NaOH compound cannot exist Ans. (b) : Coagulation is the phenomenon of
conversion of colloidal solution into precipitate while in
together in solution aspeptization,
a fresh precipitate is converted into solution
NaHCO3 + NaOH → Na2CO3 + H2O
by
adding
electrolyte.
Hence, it is clear that peptization
Neutral salt
does not involve coagulation.
746. Which of the following colloids cannot be easily
751. Coagulation is not done by
coagulated?
(a) persistant dialysis
(b) boiling
(a) Lyophobic colloids
(c) electrophoresis
(d) peptisation
(b) Multimolecular colloids
UPTU/UPSEE-2010
(c) Macromolecular colloids
Ans. (d) : Coagulation is the settling of colloidal
(d) Irreversible colloids
Karnataka-CET-2015 particles or precipitation of the solution.
Ans. (c) : In macromolecular colloids dispersed phase Thus, boiling, persistant dialysis and electrophoresis all
have size in the colloidal range that is why, they can not result in coagulation.
Hence, option (d) is not done by coagulation.
be coagulated easily.
752. Select wrong statement.
747.
Gold sol can be prepared by
(a) If a very small amount of AlCl3 is added to
(a) hydrolysis of gold (III) chloride
gold solution, coagulation occurs, but if a
(b) oxidation of gold by aqua-regia
large quantity of AlCl3 is added, there is no
(c) peptization
coagulation
(d) treating gold (III) chloride with metallic zinc
(b)
Organic
ions are more strongly absorbed on
(e) reduction of gold (III) chloride with formalin
charged surface in comparison to inorganic
solution
ions
Kerala-CEE-2014
Objective Chemistry Volume-II
142
YCT
(c) Both emulsifier and peptising agents stabilize
colloidal but their actions are different
(d) Colloidal solution are thermodynamically
stable.
AMU – 2009
Ans. (a) : When an excess of an electrolyte is added,
the colloidal particles are precipitated. This is because
colloidal particles take ions carrying charge opposite to
that present on themselves. This causes neutralization
leading to their coagulation.
753. Which is more powerful to coagulate the
negative colloid?
(a) ZnSO4
(b) Na3PO4
(c) AlCl3
(d) K4[Fe(CN)6]
UPTU/UPSEE-2009
Ans. (c) :
• Negative colloid is coagulation by positive ion or
vice-versa.
• Greater the valency of coagulating ion, greater will be
the coagulating power.
(i) ZnSO4 → Zn2+ + SO 24−
(ii) Na3PO4 → 3Na+ + PO34−
(iii) AlCl3 → Al3+ + 3Cl–
(iv) K4[Fe(CN)6] → 4K+ + [Fe(CN)6]4–
Hence, In AlCl3, the valency of positive ion
(coagulating ion) is highest, it is the most powerful
coagulating agent among the given to coagulate the
negative colloid.
754. Which of the following will be most effective in
the coagulation of Fe(OH)3 Sol?
(a) KCN
(b) BaCl2
(c) NaCl
(d) Mg3(PO4)2
UPTU/UPSEE-2005
Ans. (d) : Mg3(PO4)2 will be the most effective in the
coagulation of Fe(OH)3m solution.
8.
Types of Solution
755. 20 mL of 0.02 M hypo solution is used for the
titration of 10 mL of copper sulphate solution,
in the presence of excess of KI using starch as
an indicator. The molarity of Cu2+ is found to
be ____ × 10−2 M [nearest integer]
Given : 2Cu2+ + 4I− → Cu2I2 + I2
−
2I2 + 2S2 O 23 → 2I + S4 O 6
JEE Main-26.07.2022, Shift-II
Ans. (4) :
n
eq of I 2 = n eq of Na 2S2O3 = 20 × 0.002 ×1
2 × n mol of I 2 = 0.4
n mol of I 2 = 0.2m mol
n mol of Cu +2 = 0.2 × 2 × 10 −3
0.4 × 10−3
Cu +2 =
= 0.04 = 4 × 10−2 M.
10 × 10−3
756. The no. of positively and negatively charged
sols respectively in the following are
Objective Chemistry Volume-II
TiO 2 , blood, CdS, Cu, Ag, Clay, SiO 2 , Fe 2O 3 .
(b) 3, 5
(d) 4, 4
AP-EAMCET-07.07.2022, Shift-II
Ans. (d) : Colloidal particles always carry an electric
charge, it may be either positive or negative e.g.
Positively Charge
Negatively Charge
• Blood, Cu, Ag, Clay
• TiO2, Cds, SiO2, Fe2O3
757. Which of the following form an ideal solution?
(I) Chloroethane and bromoethane
(II) Benzene and toluene
(III) n - Hexane and n – heptanes
(IV) Phenol and aniline
(a) I & II only
(b) I, II & III only
(c) II III & IV only
(d) I & IV only
AP-EAMCET-05.07.2022, Shift-I
Ans. (b) :
(I)
Chloroethane and bromoethane – It is an ideal
solution
(II)
Benzene and Toluene – these are non-polar,
operating intermolecular forces are almost
similar hence, they form an ideal solution.
(III)
n- Hexane and n-Heptanes – ideal solution.
(IV)
Phenol and Aniline – A liquid mixture of
aniline and phenol exhibit negative deviation
from ideal behavior. Hence, these are not ideal
solution.
The correct option is (b) I,II and III only.
758. Which of the following statement is not correct
for azeotropes?
(a) At the point, the composition of a binary
mixture is same as the composition in vapour
phase.
(b) Minimum boiling azeotrope shows positive
deviation from Raoult's Law.
(c) Maximum boiling azeotrope shows positive
deviation from Raoult's Law.
(d) Nitric acid and H2O can form maximum
boiling azeotrope
AP-EAPCET-12.07.2022, Shift-I
Ans. (c) : Here, statement is not correct about
azeotropes. The correct statement is –
Minimum boiling azetrope show positive durations
from Raoult's law.
and other statements are correct about azeotropes.
759. Which of the following mixtures show negative
deviation from Raoult's law?
(I) CHCL3+ (CH3)2CO
(II) C6H5OH+C6H5NH2
(III) C2H5OH+(CH3)2CO
(IV) C6H6+C6H5CH3
(a) II& III only
(b) III & IV only
(c) I & II only
(d) I & III only
AP-EAPCET-11.07.2022, Shift-I
Ans. (c) : Negative deviation from Raoult's law is
exhibited by a mixture of chloroform (CHCl3) and
acetone (CH3)2(O) and phenol (C6H5OH) and aniline
(C6H5NH2) because of the formation of hydrogen
bonding between the two molecular species.
143
(a) 5, 3
(c) 6, 2
YCT
02.
ELECTROCHEMISTRY
3.
1.
1.
Electrode Potential
For the reaction taking place in the cell:
Pt(s) H 2 (g) H + (aq) Ag + (aq) Ag(s)
Eocell = +0.5332 V.
Cu(s) + Sn2+(0.001 M)→Cu2+(0.01M)+Sn(s)
The Gibbs free energy change for the above
reaction at 298 K is X × 10–1kJ mol–1. The value
of X is ––––––. [Nearest integer]
[Given
o
o
E Cu
= 0.34V;ESn
= −0.14V;F
2+
2+
/ Cu
/ sn
= 96500C mol−1 ]
The value of ∆rG ○– is______ kJ mol–1 [in
nearest integer]
JEE Main-26.06.2022, Shift-II
JEE Main-27.06.2022, Shift-II Ans. (983) : Given that,
Ans. (51.4538) : Given the cell reaction–
E oCu 2+ / Cu = 0.34V
+
+
Pt(s) H 2 (g) H (aq) Ag (aq) Ag(s)
o
ESn
= − 0.14V
2+
/ Sn
E ocell = + 0.5332V.
F =96500 C mol−1
1
+
−
→ H (aq) + e
At anode, H 2 (g)
o
o
∵
E ocell = E Sn
− E Cu
2+
2+
2
/ Sn
/ Cu
+
−
At
cathode,Ag
+
e
→
Ag(s)
=− 0.14 − 0.34
____________________________
= − 0.48
1
n =1
H 2 (g) + Ag + (aq) →
H + (aq) + Ag(s)
2
0.059
[Cu 2+ ]
E cell = E ocell −
log
∆G o = −nFE ocell = −1× 96500 × 0.5332
2
[Sn 2+ ]
= 5145.38J mol−1
−1
= 51.4538kJ mol
The cell potential for the given cell at 298 K Pt|
H2(g)1 bar)| H+(aq) || Cu2+(aq)|Cu(s) is 0.31V.
The pH of the acidic solution is found to be 3.
whereas the concentration of Cu2+ is 10-xM.
The value of x is_____.
2.303RT
⊕
= 0.06 V
Given: E Cu 2+ / Cu = 0.34 V and
F
JEE Main-29.06.2022, Shift-II 4.
Ans. (7) : Anode : H2(g)→2H+ + 2e−
Cathode : Cu2+ + 2e− → Cu
Overall : H2 + Cu2+ →2H+ + Cu
2.
+ 2
Ecell = E
o
cell
0.059
[H ]
−
log
2
[Cu 2+ ]
2
H+
0.06
0.31 = 0.34 −
log 2+
2
Cu
or,
0.31 = 0.34 + 0.03 [−log[H+]2 + log [Cu2+]]
0.31 = 0.34 + 0.03 [2pH + log[Cu2+]]
∵ pH = − log[H + ]
2+
−0.03 = 0.03[2pH + log [Cu ]]
−1 = 6 + log [Cu2+]
–7 = log[Cu2+]
[Cu2+] = 10−7
10–x = 10–7
x = 7.
Objective Chemistry Volume-II
0.059
10−2
log −3
2
10
= − 0.48 − 0.029log10
= − 0.48 −
= − 0.5095
∆G = − nFE cell = 2 × 96500 × 0.509
= 98333.5J
∆G = 983.3 × 10−1 kJ mol−1
In a cell, the following reactions take place
E°Fe3+ /Fe2+ = 0.77V
Fe2+ Fe3++e–
2I– I2+2e–
E°I /I- = 0.54V
2
The standard electrode potential for the
spontaneous reaction in the cell is X × 10–2 V at
298 K. The value of X is_____. (Nearest
Integer)
JEE Main-25.06.2022, Shift-I
Ans. (23) :
Fe3+ + 2I − → I 2 + Fe+2
Cathode
E
o
cell
Anode
= E ocathode − E oanode
= 0.77 − 0.54
= 0.23
= 23 × 10 −2 V
The value of x is 23.
144
YCT
A diluted solution of sulphuric acid is
Overall:H 2 (g) + Cu 2+ (aq) → 2H + (aq) + Cu(s)
electrolyzed using a current of 0.10 A for 2
o
E ocell = E Cu
− E oH+ / H = 0.34 V
2+
hours to produce, hydrogen and oxygen gas.
/ Cu
2
The total volume of gases produced at STP is –
+ 2
3
0.06
[H
]
o
––– cm . (Nearest integer)
E
=
E
−
log
cell
cell
2+
–1
2
[Cu ]
[Given: Faraday constant F = 96500 C mol
at STP, molar volume of an ideal gas is 22.7 l
0.576 = 0.34 + 0.03 − log[H + ]2 + log[Cu 2+ ]
mol–1]
JEE Main-29.06.2022, Shift-I
0.576 = 0.34 + 0.03 2pH + log[Cu 2+ ]
Ans. (127) : Reaction at anode
0.236 = 0.03[ 2pH − 2]
2H2O → O2(g) + 4H+ + 4e−
7.866 = 2pH − 2
Reaction at cathode
pH = 4.93 ≈ 5
2H+ + 2e− → H2(g)
i.t
8.
The quantity of electricity of Faraday needed
Number of equivalent =
96500
to reduce 1 mole of Cr2 O72− to Cr 3+ is
0.1× 2 × 60 × 60
JEE Main-28.06.2022, Shift-I
=
96500
Ans. (6) :
= 0.00746
→ 2Cr 3+ + 7H 2O + 6Fe3+
14H + + Cr2 O72− + 6Fe 2+
0.00746
VO2 =
× 22.7
→ 2Cr 3+ + 7H 2 O
14H + + Cr2 O72−
4
= 0.0423
∴ Quantity of electricity in faraday needed to reduce 1
mole of Cr2 O72− to Cr3+ is 6F.
0.00746
VH2 =
× 22.7
9.
For the given reactions
2
Sn2+ + 2e- → Sn
= 0.0846
Sn4+ + 4e- → Sn
3
VTotal =127 cm
The
electrode
potentials
are;
o
o
ESn2+ / Sn = −0.140V and ESn 4+ / Sn = 0.010V. The
6.
The limiting molar conductivities of Nal,
NaNO3 and AgNO3 are 12.7, 12.0 and 13.3 mS
magnitude
of standard electrode potential for
m2 mol–1, respectively (all at 25oC). the limiting
4+
2+
o
Sn / Sn i.e. ESn
is _______ x 10-2 V.
4+
/ Sn 2+
molar conductively of Agl at this temperature
is _____ mS m2 mol-1.
(Nearest integer)
JEE Main-27.06.2022, Shift-I
JEE Main-28.06.2022, Shift-II
Ans. (14) : Given,
Ans. (16) : Sn2+ + 2e− → Sn, ∆G1o = 2 × 0.140 × F
The limiting molar conductivities of
Sn+4 + 4e− → Sn, ∆G o2 = − 4 × 0.01 × F
NaI = 12.7
o
NaNO3 = 12.0
×F
Sn+4 + 2e− → Sn+2, ∆G 3o = − 2 × ESn
4+
/Sn 2+
AgNO3 = 13.3
∆G 3o = ∆G o2 + ∆G1o
λ om (NaI) =12.7m Sm 2 mol−1
−2 × E o × F = − ( 0.04 + 0.28 ) × F
λ o (AgNO ) =13.3m Sm 2 mol −1
5.
m
3
E o = 0.16 Volt =16 × 10 −2 V
λ om (NaNO3 ) =12m Sm 2 mol−1
λ om (AgI) =λ om (AgNO3 ) + λ om (NaNO3 )
= 13.3+12.7 − 12
10.
=14mSm 2 mol −1
7.
The cell potential for the following cell
1×10-4 2 ×10-4 0.1×10-4
0.2 ×10-4
(Where E is the electromotive force)
Which of the above half cells would be
preferred to be used as reference electrode?
(a) A
(b) B
(c) C
(d) D
JEE Main-26.06.2022, Shift-I
Pt H 2 (g) H + (aq) Cu 2+ (0.01M) Cu(s)
is 0.576 V at 298 K. The pH of the solution is
________. (Nearest integer)
2.303 RT
0
= 0.06V)
(Given: ECu2+ /Cu = 0.34V and
F
JEE Main-24.06.2022, Shift-I
Ans. (5) : Anode:H 2 (g)
→ 2H + (aq) + 2e −
Cathode:Cu 2+ (aq) + 2e −
→ Cu(s)
Objective Chemistry Volume-II
∂E
The
of different types of half cells are as
∂T P
follows:
A
B
C
D
∂E
Ans. (c) : Lowest value of
will be preferred to
∂T P
be used as reference electrode.
145
YCT
11.
The resistance of a conductivity cell containing
0.01 M KCl solution at 298 K is 1750 Ω. If the
conductivity of 0.01 M KCl solution at 298 K is
0.152 ×10–3 S cm–1, then the cell constant of the
conductivity cell is_____×10–3 cm–1.
JEE Main-24.06.2022, Shift-II
Ans. (266) : For KCl Solution ⇒ R =1750Ω
K = 0.152 × 10−3 S cm−1
ℓ 1
K =
AR
l
Cell constant = = K × R
A
= [0.152×10−3×1750]
= 266 × 10−3 cm−1
Hence, the cell constant of the conductivity cell is 266 ×
10−3 cell cm−1 .
12. In Fuel cells ____ are used as catalysts.
(a) Zinc - Mercury
(b) Lead - Manganese
(c) Platinum - Palladium (d) Nickel - Cadmium
Karnataka CET-17.06.2022, Shift-II
Ans. (c) : In fuel cells typically used of platinumpalladium catalyst to speed up the reaction at the
oxygen electrode, but platinum is expensive. The start
up and operation of the fuel cell also introduces side
reactions that damage the efficiency of the catalyst.
13. The molar conductivity is maximum for the
solution of concentration
(a) 0.005M
(b) 0.001M
(c) 0.004M
(d) 0.002M
Karnataka CET-17.06.2022, Shift-II
1
Ans. (b) : Molar conductance ∝
Molarity
The molar conductivity of electrolytes increases with
decreasing concentration, so the molar conductivity of
the solution will be maximum at a concentration of
0.001 M.
14. Half-life of a reaction is found to be inversely
proportional to the fifth power is initial
concentration, the order of reaction is
(a) 5
(b) 6
(c) 3
(d) 4
Karnataka CET-17.06.2022, Shift-II
1
Ans. (b) : t1/ 2 ∝ n −1 , n = 6
a
1
t1/ 2 = 6−1
a
1
t1/ 2 = 5
a
Hence, the order of reaction is 6.
15. The correct order of reduction potentials of the
following pairs is
–
–
(A) Cl2/Cl
(B) I2/I
(C) Ag+/Ag
(D) Na+/Na
+
(E) Li /Li
Objective Chemistry Volume-II
Choose the correct answer from the options
given below:
(a) A > C > B > D > E (b) A > B > C > D > E
(c) A > C > B > E > D (d) A > B > C > E > D
JEE Main-25.06.2022, Shift-II
Ans. (a) : (A)E
o
Cl2 / Cl −
=1.36 V
= 0.54 V
(B)E
o
I2 / I−
(C)E
o
Ag + / Ag
= 0.80V
(D)E oNa + / Na = − 2.71V
(E) E oLi+ / Li = − 3.05V
The correct order of reduction potential of the following
pairs is
A>C>B>D>E
16. 96.5 amperes current is passed through the
molten AlCl3 for 100 seconds. The mass of
aluminum deposited a the cathode is (Atomic
weight of Al - 27 u)
(a) 0.90 g
(b) 0.45 g
(c) 1.35 g
(d) 1.8 g
AP-EAMCET-05.07.2022, Shift-I
Ans. (a) : Given, current (i) = 96.5 amperes
Time (t) = 100 second
m=2
E = 27
Faraday second law,
E
m = ×it
F
27
m=
×96.5×100
3×96500
m = 0.9 g
17. The Eo of Ce4+/Ce+3 = 1.6 V
Fe3+/Fe2+ = 0.76 V
o
the E of Fe3+ oxidising Ce3+ is
(a) +0.84 V
(b) − 0.84 V
(c) −2.32 V
(d) +1.5 V
TS-EAMCET-20.07.2022, Shift-II
Ans. (b) : Given that,
Eo of Ce4+ / Ce3+ = 1.6 V
Fe3+ / Fe2+ = 0.76 V
To find Eo of Fe3+ oxidising Ce3+ is –
Eo = Reduction potential of cathode – Reduction
Potential of anode
Eo = (Fe3+)Red – (Ce3+)Red
= 0.76 – 1.6
Eo = – 0.84 V
18.
146
Given
Eo Mn7+ /Mn2+ = 1.51V, Eo Mn4+ /Mn2+ = 1.23 V
Calculate the Eo Mn7+ /Mn4+
(a) 0.28 V
(c) 1.7 V
(b) −0.28 V
(d) 0.48 V
TS-EAMCET-19.07.2022, Shift-I
YCT
21. The standard reduction potentials for Zn2+/Zn,
Ans. (c) : Mn +7 + 5e −
→ Mn 2+
Ni2+ / Ni and Fe2+/Fe are –0.76, –0.23 and –0.44
Mn +4 + 2e −
→ Mn 2 +
V, respectively. The reaction
Required equation
X + Y2+ → X2+ + Y will be spontaneous when
+7
−
4+
Mn + 3e → Mn
(a) X = Ni, Y = Fe
(b) X = Ni, Y= Zn
(c) X = Fe, Y = Zn
(d) X = Zn, Y = Ni
∴ ∆G° = –5F × 1.51
JIPMER-2017, Kerala CEE-2016
∆G° = – 2F × 1.23
[AIEEE 2012]
∆G° = – 3FE°
Ans. (d) : For a spontaneous reaction ∆G must be –ve
∴ – 3FE° = –5F × 1.51 + 2F × 1.23
Since, ∆G = –nFE°
E° = 1.69V ≈ 1.7V
Hence for ∆G to be –ve ∆E° has to be positive, which is
19. The cell potential, Ecell the following cell possible when X = Zn, Y = Ni
notation (in Volts) is
Zn + Ni++
→ Zn ++ + Ni
A(s) | A + (aq, 0.1M || B 2+ (aq, 0.01M) | B(s)
E ° Zn / Zn 2+ + E ° Ni2+ / Ni
E0A+ /A = 1V and E0B2+ /B = 3 V
⇒ 0.76 + (–0.23)
(a) 1.0
(b) 3.0
⇒ +0.53 (positive)
(c) 2.0
(d) 2.5
22. During the electrolysis of copper sulphate
AP-EAPCET-12.07.2022, Shift-II
aqueous solution using copper electrode, the
reaction taking place at the cathode is
Ans. (c) :
+
2+
A(s) |
A (aq), (0.1M) || B (aq), (0.01M) | B(s)
→ Cu 2+ ( aq ) + 2e −
(a) Cu
Given that,
(b) Cu 2+ ( aq ) + 2e −
→ Cu ( s )
o
E A+ / A = 1V
1
(c) H + ( aq ) + e−
→ H2 ( g )
E oB2+ / B = 3V
2
1
o
o
2−
E = E cathode − E Anode
→ SO3 ( g ) + O 2 ( g ) + 2e −
(d) SO 4 ( aq )
∵ cell
2
AP-EAMCET (Engg.)-2013
= 3–1
AP-EAMCET (Medical)-2001
Ecell = 2 volt
20. The standard reduction potentials at 298 K for Ans. (b) : When an aqueous solution of CuSO4 is
the following half cell reactions are given subjected to electrolysis using copper electrodes the
following reaction take place at cathode is
below:
2+
−
CuSO 4
→ Cu 2+ + SO 4 2−
Zn ( aq ) + 2e ↽ ⇀ Zn ( s ) , −0.762V
Reduction at cathode
Cr 3+ ( aq ) + 3e − ↽ ⇀ Cr ( s ) , −0.74V
Cu 2+ ( aq ) + 2e−
→ Cu ( s )
+
−
2H ( aq ) + 2e ↽ ⇀ H 2 ( g ) , + 0.00V
• When copper sulphate solution is electrolysed with
3+
−
2+
copper electrodes copper metal is deposited on the
⇀
Fe ( aq ) + e ↽ Fe ( aq ) ,+0.77V
cathode in a reddish pink color.
Which one of the following is the strongest 23. For the following cell reaction,
reducing agent?
Ag | Ag+ / AgCl | Cl– |Cl2, Pt
(a) Zn (s)
(b) Cr (s)
o
∆G f (AgCl) = –109 kJ / mol
(c) H2(g)
(d) Fe 2+ ( aq )
∆G of (Cl–) = –129 kJ / mol
Karnataka-CET-2017
∆G of (Ag+) = 78 kJ / mol
AIIMS-2013
E° of the cell is
MHT CET-2008
(a) –0.60 V
(b) 0.60 V
VITEEE-2007
(c) 6.0 V
(d) none of these
J & K CET-(2000)
AP EAMCET (Engg.)-2009
AP EAMCET (Medical) -1998
VITEEE–2009
AP-EAMCET-1996
Ans.
(a):
From
the
given
cell
reaction–
Ans. (a): Zinc has higher negative value of standard
AgCl
→ Ag + + Cl −
reduction potential. Therefore, it is the stronger
reducing agent because the metals having higher We know that,
negative value of standard reduction potential are
o
o
o
∆G (reaction
) = G (product ) − G (reactant)
placed above hydrogen in electrochemical series. The
=(78–129) – (–109)
metals placed above hydrogen has a great tendency to
= 78 – 129 + 109
donate electrons or oxidizing power. The metals having
greater oxidizing power are strongest reducing agent.
= + 58kJ / mol
Objective Chemistry Volume-II
147
YCT
∴
o
∆G o = − nFE cell
58000 = − 1 × 96,500 × E
E cell = 1.10 −
o
cell
58000
−96500
= − 0.60 V
=1.10 −
E ocell =
24.
Corrosion of iron is essentially an
electrochemical phenomenon where the cell
reactions are
(a) Fe is oxidised to Fe2+ and dissolved oxygen in
water is reduced to OH–
(b) Fe is oxidised to Fe3+ and H2O is reduced to
O −2
(c) Fe is oxidised to Fe2+ and H2O is reduced to
O −2 .
(d) Fe is oxidised to Fe2+ and H2O is reduced to
O 2.
COMEDK 2018
JIPMER 2013
Ans. (a):
Fe → Fe 2+ + 2e− (anode)
0.0591
1
log
2
0.1
0.0591
= 1.10 − 0.0295
2
=1.07 V
The standard electrode potential of three
metals X, Y and Z are – 1.2 V, + 0.5 and – 3.0 V
respectively. The reducing power of these
metals will be
(a) X > Y > Z
(b) Y > Z > X
(c) Y > X > Z
(d) Z > X > Y
JIPMER-2014
JCECE - 2012
Ans. (d) :
27.
E ox = −1.2V
E oy = + 0.5V
E oz = −3.0V
∵ Higher the reduction potential lesser the reducing
power.
∴ Order of reducing power is:
−
Z>X>Y
O 2 + 2H 2O + 4e− → 3OH (cathode)
28.
Consider
the
following electrodes
The overall reaction is:
2+
P
=
Zn
(0.0001
M)/Zn Q = Zn2+ (0.1 M)/Zn
2Fe + O2 + 2H2O → 2Fe(OH)2
2+
R = Zn (0.01 M)/Zn
S = Zn2+ (0.001 M)/Zn
Hence, Fe is oxidized to 2Fe2+ and dissolved oxygen in
º
EZn|Zn2+ = -0.76 V Electrode potentials of the
water is reduced to OH − .
above electrodes in volts are in the order
(a) P > S > R > Q
(b) S > R > Q > P
(c) Q > R > S > P
(d) P > Q > R > S
Kerala-CEE-29.08.2021
Karnataka-CET-2021
2+
2+
2+
2+
Ans. (c) : Electrode potential is directly proportional to
(c) Cu Zn Zn Cu
(d) Cu Zn Zn Cu
concentration of Zn2+ ions.
BCECE-2008, 2005 29. In the electrochemical cell:
Ans. (a) : For cell reaction
Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu, the emf
of this Daniell cell is E1. When the
Zn + Cu 2+
→ Zn 2+ + Cu
concentration of ZnSO4 is changed to 1.0 M
At cathode: Cu 2+ + 2e −
→ Cu
and that of CuSO4 changed to 0.01 M, the emf
2+
–
At anode : Zn →Zn+2e
changes to E2. From the following which one is
∴Cell representation
the relationship between E1 and E2? (Given,
RT|F = 0.059)
Zn| Zn2+ || Cu2+ |Cu
(a) E1< E2
(b) E1 > E2
26. For the redox reaction
2+
2+
(c)
E
=
0
≠
E
(d) E1 = E2
2
1
Zn(s) + Cu (0.1M) → Zn (1M) + Cu(s)
(NEET- 2017, 2003)
o
Taking place in a cell Ecell is 1.10 V. Ecell for
0.059
[Zn 2+ ]
Ans. (b) : E Cell = E °Cell −
log
RT
the cell will be 2.303
= 0.0591
n
[Cu 2+ ]
F
0.059
0.01
(a) 2.14 V
(b) 1.80 V
E1 = E ° −
log
2
1
(c) 1.07 V
(d) 0.82 V
0.059
JCECE - 2014, 2008
E1 = E ° −
log10−2
2
Ans. (c) : From Nernst equation,
0.059
E1 = E° −
(−2)
2.303RT
[Zn 2+ ]
Ecell = E ocell −
log
2
2+
nF
[Cu ]
E1 = (E ° + 0.059) V
For the given cell reaction, n = 2
25.
For cell reaction
Zn + Cu2+ → Zn2+ + Cu
cell representation is :
(a) Zn Zn 2 + Cu 2 + Cu (b) Cu Cu 2 + Zn 2 + Zn
Objective Chemistry Volume-II
148
YCT
Ans. (a) : Mg(s) + Cl2(g) → Mg(2aq+ ) + 2Cl(–aq )
and
0.059
1
E 2 = E° −
log
2
0.01
0.059
°
E2 = E −
log102
2
2× 0.059
E 2 = E° −
2
E 2 = ( E° − 0.059) V
The symbolic representation of cell reaction is-
Mg Mg (2aq+ ) (1M ) Cl(–aq ) (1M ) Cl2( g ) (1bar ) Pt
33.
Using the data given below find out the
strongest reducing agent?
EoCr O 2- /Cr 3+ = 1.33V
EoCl /Cl – = 1.36V
2
Hence, E1 > E2
30. On the basis of the following E0 values, the
strongest oxidizing agent is
[Fe(CN)6]4– → [Fe(CN)6]3– + e–; E0 = –0.35V
Fe2+ → Fe3+ + e–; E0 = –0.77 V
(a) Fe3+
(b) [Fe(CN)6]3–
4–
(c) [Fe(CN)6]
(d) Fe2+
[AIIMS-2017]
(AIPMT -2008)
Ans. (a) : The substance which has lower reduction
potential are stronger reducing agent while the
substances which have higher reduction potential are a
stronger oxidizing agent.
[Fe(CN)6 ]3+ + e− → [Fe(CN)6 ]4 ; E o = 0.35V
E
2
7
o
MnO 4– /Mn2+
–
(a) Cl
(c) Cr
= 1.51V
E
o
Cr 3+ /Cr
= – 0.74V
3+
(b) Cr
(d) Mn2+
GUJCET-2021
Ans. (c) : A reducing agent is typically one of its lower
possible oxidation states. The reduction potential
increases, it tends to undergo reduction easily and acts
as oxidizing agent. Lower the value of reduction
potential, stronger is the reducing agent i.e., undergoes
oxidation most easily.
Cr → Cr3+ + 3e–
Cr/Cr3+ has the highest oxidation potential i.e. (+0.74)
so, it is oxidized most easily and, therefore it is the
strongest reducing agent.
34. Emf of the following cell at 298 K in V is x ×
10–2 .
Fe3+ + e − → Fe 2+ ; E o = 0.77V
The reduction potential of Fe3+/Fe2+ is higher.
Hence, Fe3+ is a strongest oxidsing agent.
31. Zinc is used to protect iron from corrosion
Zn Zn 2+ (0.1M) Ag + (0.01 M) Ag
because
(a) Ered of Zn < Ered of iron
The value of x is _____ . (Rounded off to the
(b) Eoxidation of Zn < Eoxidation of iron
nearest integer)
(c) Eoxidation of Zn = Eoxidation of iron
o
Given : EZn
= –0.76V;E Ago +/Ag = +0.80V;
2+
/Zn
(d) Zinc is cheaper than iron
UPTU/UPSEE- 2013, 2010
2.303RT
= 0.059
Ans. (a) : Zinc metal is used to protect iron from
F
corrosion because zinc is more electropositive than iron.
JEE Main 26.02.2021,Shift-II
The process of coating the iron surface by zinc is called
galvanisation. Zinc metal present on the surface of iron Ans. : Given the cell reaction
forms a thin protective layer of basic zinc carbonate due Zn / Zn 2+ (0.1M) Ag + (0.01M) Ag(s)
to the reaction between zinc, oxygen, CO2 and moisture
in air.
E° = 0.80 − (−0.76) = 1.56V
Since, standard reduction potential of zinc is less than
Zn 2+
iron, hence iron will not undergo corrosion.
0.059
∴ E = E° –
log + 2
°
E Zn 2+ / Zn = −0.76V
n
(Ag )
°
E Fe2+ / Fe = −0.44V
0.059
0.1
E = 1.56 –
log
2
(0.01) 2
32. Which is symbolic representation for following
cell reaction, Mg(s) + Cl2(g) → Mg (2+aq ) + 2Cl (–aq ) .
E=1.56–0.0295×3
E=1.4715
(a) Mg Mg (2+aq ) (1M ) Cl(–aq ) (1M ) Cl2( g ) (1bar ) Pt
E=147.15×10–2 Volt
So, x=147
(b) Pt Cl(–aq ) (1M ) Cl2( g ) (1bar ) Mg (2aq+ ) (1M ) Pt
35. The element with highest standard reduction
potential (in Volt) [M2+ → M] among the Ist
2+
–
(c) Mg Mg ( aq ) (1M ) Cl 2( g ) (1bar ) Cl( aq ) (1M ) Pt
row of transition elements is
(a) Ti
(b) Ni
–
2+
(d) Pt Cl 2(g ) (1bar ) Cl( aq ) (1M ) Mg ( aq ) (1M ) Mg
(c) Cr
(d) Cu
GUJCET-2021
TS-EAMCET (Engg.), 06.08.2021
Objective Chemistry Volume-II
149
YCT
Ans. Cell reaction is
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Given, E1 = 0.3095 V
We know that0.0591
[Cu 2+ ]
–0.25
Ni
Ni +2 + 2e − → Ni
E = E ocell −
log
2
[Ag + ]2
Cu
Cu +2 + 2e − → Cu
+0.337
0.591
0.1
–0.163
Ti
Ti +2 + 2e− → Ti
E1 = E ocell −
log
2
2
(0.01)
36. The electrode potential of M2+/M for 3d series
0.0591
0.1
elements shows positive value for :
0.3095 = E ocell −
log
(a) Fe
(b) Co
2
(0.01) 2
(c) Zn
(d) Cu
E ocell = 0.398
JEE Main 24.02.2021, Shift-I
0.0591
0.01
E 2 = E ocell −
log
Ans. (d) :
2
2
(0.001)
–
(1) Zn 2+ + 2e
→ Zn (s) E ° = –0.76V
0.0591
0.01
–
= 0.398 −
log
→ Co(s) E ° = –0.26V
(2) Co 2+ + 2e
2
(.001) 2
–
E 2 = 28 ×10 −2 V
(3) Fe 2+ + 2e
→ Fe(s) E ° = –0.44V
Ans. (d):
Element Electrode Reaction
(Reduction)
Cr
Cr +2 + 2e − → Cr
Standard Electrode
Potential (E0 volt)
–0.91
(4) Cu 2+ + 2e
→ Cu (s) E ° = +0.34V
39.
–
Hence, only copper shows positive value for electrode
potential of M2+/M of 3d-series elements. Hence, the
correct option is (d).
37. Given below are two statements.
Statement I : The Eº value of Ce4+/Ce3+ is
1.74V.
Statement II : Ce is more stable in Ce4+ state
than Ce3+ state.
In the light of the above statements, choose the
most appropriate answer from the options
given below.
(a) Both statement I and statement II are correct.
(b) Statement I is incorrect but statement II is
correct.
(c) Both statement I and statement II are
incorrect.
(d) Statement I is correct but statement II is
incorrect.
(JEE Main 2021, 16 March Shift-I)
Ans. (d) : The E0 value for Ce+4/Ce+3 is +1.74 V
because the most stable. Oxidation state of lanthanide
series elements is +3.
It means Ce3+ is more stable than Ce4+.
38. For the cell
Cu(s)|Cu2+ (aq)(0.1)M|| Ag+ (aq)
(0.01M)| Ag(s)
the cell potential, E1 = 0.3095 V
For the cell,
Cu(s)|Cu2+ (aq) (0.01M)|| Ag+ (aq)
(0.001 M)| Ag(s)
the cell potential
= ................×10–2 V.
(Round off the nearest integer)
2.303RT
= 0.059
Use :
F
[JEE Main 2021, 27 July Shift-II]
Objective Chemistry Volume-II
For the galvanic cell,
Zn(s) + Cu2+ (0.02 M)→ Zn2+(0.04M)+ Cu(s),
Ecell = ..........×10–2 V.
(Nearest integer)
[Use EoCu/Cu2+ = – 0.34V,EoZn/Zn2+ = + 0.76V ]
[JEE Main 2021, 26 Aug Shift-II]
Ans. (109) : Cell reaction
Cu 2+ (aq)+ Zn(s) ↽ ⇀ Cu(s) + Zn 2+ (aq)
0.02M
E
°
cell
=E
0.04M
°
Cu 2+ / Cu
− E
°
Zn 2+ / Zn
= 0.34 − (− 0.76) = 1.10 V
0.059
[Zn 2+ ]
log
2
[Cu 2+ ]
0.059
[0.04]
= 1.10 −
log
2
[0.02]
= 1.10 − 0.03 log 2 = 1.10 − 0.03 × 0.30
E cell = E °cell −
E cell
E cell
E cell = 1.09 V = 109 × 10 −2 V
40.
Potassium chlorate is prepared by electrolysis
of KCl in basic solution as shown by following
equation
6OH – + Cl – → ClO 3– + 3H 2O + 6e –
A current of xA has to be passed for 10h to
produce 10.0 g of potassium chlorate. The
value of x is.......(Nearest integer)
(Molar Mass of KClO3 = 122.6 g mol–1, F =
96500 C)
[JEE Main 2021, 20 July Shift-II]
→ ClO3− + 3H 2 O + 6e −
Ans. (1) : 6OH − + Cl−
Given that, KClO3 =122.6 g mol−1 , F = 96500C
E
W = Q × I × t ∵ Q =
F
E
W = × I× t
F
150
YCT
Ans. (a) : Cr 2+ → Cr 3+ + e−
Eº = – 0.72 V (Oxidation)
Fe 2+ + 2e − → Fe
Eº = −0.42V ( Reduction )
Where, W = weight of potassium chlorate
E = molar mass of potassium chlorate
F = Faraday's constant
Q = current
t = time for which current is pass
122.6
10 =
× x × 10 × 3600
96500 × 6
10 × 6 × 96500
x=
10 × 3600 × 122.6
x =1.311 ≈ 1
41.
Eº cell = E right − E left = − 0.42 + 0.72 = 0.30V
44.
EoCr 3+ / Cr = 0.91V . Find the standard reduction
potential of Cr3+/Cr
(a) –1.31V
(b) –1.71V
(c) –0.74V
(d) –0.51 V
A hydrogen electrode is made by dipping
AP- EAPCET- 07-09-2021, Shift-I
platinum wire in a solution of nitric acid of pH
o
= 9 and passing hydrogen gas around the Ans. (c) : E Cr3+ / Cr 2+ = − 0.4V
platinum wire at 1.2 atm pressure. The
o
oxidation potential of such an eletrode equals E Cr3+ / Cr = − 0.91V
____V.
E oCr3+ / Cr = ?
(a) +0.59
(b) –0.531
Cr 3+ + Cr 2+ + 3e −
→ Cr 2+ + Cr
(c) –0.59
(d) +0.531
AP-EAMCET 25-08-2021, Shift - I
Ans. (d) : pH of HNO3 solution = 9
Then, [H+] = 10–pH = 10–9M
According to Nernst equation for hydrogen electrode–
( )
+
E cell = E° −
H
0.059
log
n
PH 2
( )
n = 1
For hydrogen
electrode i.e. E° = 0
0.059
10−9
log
1
1
= – 0.059 × (–9 )
= 0.531
For
the
cell
reaction
Cu|Cu2+(0.1M)
2+
Cu (1.0M) | Cu; the emf of the cell at 25 °C is
Ecell = 0 −
Ecell
Ecell
42.
Given half-cell potentials EoCr 3+ / Cr 2+ = 0.4V and
E°Cu2+ /Cu = 0.34V
(a) 0.059V
(b) 0.311V
(c) 0.369V
(d) 0.029 V
TS-EAMCET (Engg.), 07.08.2021 Shift-II
Ans. (d) : Given, E °Cu 2+ / Cu = 0.34V
We can write the Nernst equation for the given cell–
0.059
[Product]
E = E° –
log
n
[Reactant]
0.059
[0.1]
E = 0–
log
2
[1.0]
0.059
E=0–
log10−1
2
0.059
E= 0+
2
E = 0.029 V
43. Find the emf of the following cell reaction,
given
EoCr 3+ /Cr 2+ = − 0.72V and EoFe2+ /Fe = − 0.42V
at
3+
2+
25º C is Cr|Cr (0.1M)||Fe (0.1 M)|Fe
(a) 0.30 V
(b) 0.25 V
(c) 1.14 V
(d) 1.56 V
AP EAPCET 19-08-2021, Shift-II
Objective Chemistry Volume-II
Cr 3+ + 3e −
→ Cr
o
∆G = ∆G1 + ∆G o2
= −1× −0.4 ( F ) + −2 × ( −0.91) F
− nFE ocell = +2.22F
− 3E ocell = 2.22
2.22
E ocell = −
3
E ocell = − 0.74V
45. In the electrolysis of a CuSO4 solution, how
many grams of Cu are plated out on the
cathode, in the time that is required to liberate
5.6 L of O2(g), measured at 1 atm and 273. K,
at the anode?
(a) 31.75 g
(b) 14.2 g
(c) 4.32 g
(d) 3.175 g
AP EAPCET 20.08.2021 Shift-II
Ans. (a): Given that, P = 1 atm, VO2 = 5.6 lit.,T = 273 K
At anode
2H2O → 4H++ O2+4e–
V(STP) 5.6
=
= 0.25 moles
Since, Molesof O2 =
22.4 22.4
+2
–
Cathode:- Cu +2e → Cu
Anode:- 4OH– → 2H2O + O2 + 4e–
Equivalent of Cu = Equivalent of O2
Mass of Cu
= Moles × n Factor
Equivalent mass
Mass of copper
= 0.25 × 4
63.5 / 2
Mass of copper = 31.75 gm
46. The correct order of Eº values of the given
2+
–
elements is M + 2e → M ( s )
151
(
(a)
(b)
(c)
(d)
)
Cu2+ > Ni2+ > Pb2+ > Fe2+
Ni2+ > Fe2+ > Cu2+ > Pb2+
Cu2+ > Pb2+ > Ni2+ > Fe2+
Pb2+ > Cu2+ > Ni2+ > Fe2+
TS EAMCET 10.08.2021, Shift-II
YCT
Ans. (c)
Cathode Reduction
Standard Potential
Half reaction
Eº (Volts)
M 2+ + 2e – → M(s)
Cu 2+ + 2e – → Cu(s)
+ 0.337
Pb + 2e → Pb(s)
– 0.13
Ni + 2e → Ni(s)
– 0.25
2+
–
2+
–
Fe + 2e → Fe(s)
– 0.44
The correct order is
Cu2+ > Pb2+ > Ni2+ > Fe2+
47. For
the
reaction
Sn(s)|Sn2+(0.10M)||
Pb2+(0.5M)|Pb(s).
What
will
be
the
approximate Ecell when [Pb2+] is 0.1M
o
(given ESn
= – 0.136 V : EoPb2+ /Pb = – 0.126 V)
2+
/Sn
2+
–
(a) – 0.011V
(c) – 0.03 V
(b) – 0.022 V
(d) + 0.022 V
TS EAMCET 10.08.2021, Shift-I
Ans. (c) : Cell reaction:Sn → Sn2+ + 2e–
.....(i)
Pb2+ + 2e– → Pb
.....(ii)
Adding equation (i) and (ii) we get–
Sn + Pb2+ → Sn2+ + Pb
Nernst equation:
0.0591 [oxidation]
o
E cell = E cell
−
log
n
[reduction]
o
E ocell = ESn
− E oPb2+ / Pb
2+
/ Sn
(a) Zn to Cu
(b) Cu to Zn
(c) no current flows
(d) data insufficient
AP EAMCET (Engg.) 21.09.2020, Shift-I
Ans. (a) : On applying an external voltage greater than
1.1 V in a Daniell cell, the current flows in the reverse
direction i.e., from Zn to Cu (cathode to anode) and
electrons flow from Cu to Zn.
Zn is deposited at Zn electrode and Cu dissolves at Cu
electrode. The reaction is
Zn 2+ + Cu → Zn + Cu 2+
50. The equation that is incorrect is
− Λ om
= Λ om
− Λ om
(a) Λ om
(b)
(c)
(d)
0.0591
× 0.6989
2
= −0.0306
E cell
48.
or
∴
∴
1
log = 0.6989
5
49.
or
∴
and
∴
∴
51.
∆Gº = – nFE ºcell
–3
–1
= – 6 × 96500 × 0.89 × 10 kJ mol
= – 515.31 kJ mol–1
In the Daniell cell, Zn |Zn+||Cu2+| Cu, when an
external voltage is applied such that Eexternal >
Ecell , current flows from .......... .
Objective Chemistry Volume-II
o
m KCl
o
m NaBr
o
m H O
2
( )
− (Λ )
− (Λ )
= (Λ )
( )
= (Λ )
= (Λ )
+ (Λ )
NaCl
o
m NaCl
o
m NaI
o
m HCl
( )
− (Λ )
− (Λ )
− (Λ )
KBr
o
m KBr
o
m KBr
o
m NaOH
KCl
o
m NaBr
o
m NaBr
o
m NaCl
λ ° m (Br − ) + λ ° m (Cl − ) = λ ° m (Br − ) − λ ° m (Cl− )
LHS = RHS
The LHS of option (a)
λ ° m (Br − ) − λ° m (Cl− )
and also RHS of that
= λ ° m (Br − ) − λ ° m (Cl − )
LHS = RHS
For option (b)
λ ° m (K + ) + λ ° m (Cl− ) − λ ° m (Na + ) − λ ° m(Cl − )
= λ ° m (K + ) + λ° m (Br − ) − λ ° m (Na + ) − λ ° m(Br − )
For the cell reaction,
3Sn 4+ + 2Cr → 3Sn 2+ + 2Cr 3+ , Eocell is 0.89 V.
Then ∆Gº for the reaction is ....... .
(a) –515.31 kJ/mol
(b) –125.41 kJ/mol
(c) –457.41 kJ/mol
(d) –347.40 kJ/mol
AP EAMCET (Engg.) 17.09.2020, Shift-II
Ans. (a) :
3Sn4+ + 6e– → 3Sn2+
2Cr → 2Cr3+ + 6e–
So, n = 6, E ocell = 0.89 V
Now,
NaBr
= λ ° m (K + ) + λ° m (Br − ) − λ ° m (K + ) − λ ° m (Cl− )
0.0591
0.10
log
2
0.5
E cell = −0.01 −
)
)
)
)
[JEE Main 2020, 7 Jan Shift-II]
Ans. (c) : According to Kohlrausch's law,
Limiting molar conductivity of an electrolyte in the sum
of the individual contributions of the cation and the
anion of the electrolyte. Therefore,
For option (a)
λ ° m (Na + ) + λ ° m (Br − ) − λ° m (Na + ) − λ ° m (Cl− )
= −0.136 + 0.126
= −0.01
E cell = −0.01 −
(
(Λ
(Λ
(Λ
152
λ ° m (K + ) − λ ° m (Na + ) = λ ° m (K + ) − λ ° m(Na + )
LHS = RHS
Similarly for option (c)
LHS = λ ° m (Br − ) − λ ° m (I − )
RHS = λ° m (K + ) − λ° m (Na + )
LHS ≠ RHS
And for option (d) also
λ ° m (H + ) − λ ° m (OH − ) = λ ° m (H + ) + λ ° m (OH − )
LHS = RHS
Thus the correct answer is (c)
108 g of silver (molar mass 108 g mol–1) is
deposited at cathode from AgNO3(aq) solution
by a certain quantity of electricity. The volume
(in L) of oxygen gas produced at 273K and 1
bar pressure from water by the same quantity
of electricity is..............
[JEE Main 2020, 9 Jan Shift-II]
YCT
Ans.(5.765) (n Ag ) deposited =
108
= 1mole
108
Ag + + e−
→ Ag
1Faraday charge is required to deposit 1 mole of Ag
1
H 2 O
→ O 2 + 2H + + 2e−
2
1
2Faraday charge deposit → moleof O 2
2
1
1Faraday charge deposite
→ mole of O2
4
Given that,
P = 1 bar, T = 273 K, R = 0.0823 L bar mol–1 K–1
Using the ideal gas equation–
PV = nRT
nRT
VO2 =
P
1 0.082× 273
VO2 = ×
4
1
VO2 = 5.675 litre
Ans. (a) : Given data,
E oFe+3 / Fe+2 = +0.76V
E oI / I− = + 0.55V
2
∴
E ocell = E oC − E oA = 0.76 − 0.55 = 0.21
Total reaction,
2Fe3+ + 2I −
→ 2Fe 2+ + I 2
∴
54.
52.
0.059
0.059
log K c ⇒ 0.21 =
log K c
2
2
log K c = 7
E ocell =
K c = 1× 107
The correct statement about Cr2+ and Mn3+
among the following is (Given, atomic numbers
of Cr = 24 and Mn = 25)
(a) Cr2+ is a reducing agent
(b) Mn3+ is a reducing agent
(c) Both Cr2+ and Mn3+ exhibit d4 outer electronic
configuration
(d) When Cr2+ is used as reducing agent, it attains
d5 electronic configuration
AP EAMCET (Engg.) 17.09.2020 Shift-I
250 mL of a waste solution obtained from the
workshop of a goldsmith contains 0.1 M AgNO3
2+
4
º
and 0.1 M AuCl. The solution was electrolysed at Ans. (a) : Cr ⇒ [ Ar ] 3d ,E Cr3+ / Cr 2+ = −0.41V
2 V by passing a current of 1 A for 15 minutes.
º
Mn 3+ ⇒ [ Ar ] 3d 3 ,E Mn
= 1.51V
3+
/ Mn 2 +
The metal/metals electrodeposited will be
2+
So, reducing power Cr > Mn2+
EoAg+ /Ag = 0.80V,EoAu+ /Au = 1.69
55. Which of the following statements is correct for
(a) only silver
the cell Zn|Zn+2||Cu+2|Cu ?
(b) silver and gold in equal mass proportion
(a) Zn is reducing agent.
(c) only gold
(b) Cu is anode.
(d) silver and gold in proportion to their atomic
(c) Cu is oxidising agent.
weights
(d) The cell reaction is Zn + Cu+2 → Zn+2 + Cu
[JEE Main 2020, 4 Sep Shift-II]
AP EAMCET (Engg.) 21.09.2020, Shift-II
it
Ans.
(d)
:
In the Daniell cell, copper and zinc
Ans. (c) : Charge (q) =
F
electrodes
are
immersed in a solution of copper (II)
96500
sulphate
and
zinc
sulphate, respectively.
1× 15 × 60
900
=
=
At the anode (negative electrode), zinc is oxidised per
96500
96500
the following half reaction.
9
=
F = 0.0093F
Zn ( s ) → Zn 2+ ( aq ) + 2e −
965
At the cathode (positive electrode), copper is reduced
Number of moles of Au+ = 0.1
+
per
the following reaction.
Number of moles of Ag = 0.1
o
Cu 2+ ( aq ) + 2e − → Cu ( s )
The value of standard reduction potential (E Au + / Au =
1.69 V) will be higher, which deposited first at cathode. Overall cell reaction is
Zn ( s ) + Cu 2+ ( aq ) → Zn 2+ ( aq ) + Cu ( s )
Au + (aq)+ e− → Au(s)
0.01
0.0093mol
These processes result in the accumulation of solid
So, only Au will get deposited.
copper at the cathode and the corrosion of the zinc
53. Given EoFe3+ / Fe2+ = + 0.76 V and EoI / I− = + 0.55 V. electrode into the solution as zinc cations.
2
56. Which one of the following has a potential
The equilibrium constant for the reaction
more than zero?
taking place in galvanic cell consisting of above
1
two electrodes is
(a) Pt, H 2 (1atm) HCl(1 M)
2
2.303 RT
= 0.06
1
F
(b) Pt, H 2 (1atm) HCl(2 M)
2
(a) 1 × 107
(b) 1 × 109
8
12
1
(c) 3 × 10
(d) 5 × 10
(c) Pt, H 2 (1atm) HCl(0.1 M)
2
Karnataka-CET-2020
(
Objective Chemistry Volume-II
)
153
YCT
1
(d) Pt, H 2 (1atm) HCl(0.5 M)
2
Then
−2.303 RT log K c = −nFE ocell
COMEDK 2020
Ans. (b) :We know that–
Pt H
0.0591
log +2
Ecell = E ocell –
n
[H ]
For
0.0591 1
(i) Ecell = 0 –
log
1
1
(ii) E cell = – 0.0591× ( −0.3010 )
= 0.177
(iii) E cell = – 0.0591
(∵ n=1)
2.303 RT
= 0.059V
F
2.303RT
log Kc
F
0.590 = 0.059 log Kc
10 = log Kc
Kc = 1x1010
Eocell =
The standard electrode potential ( Eo ) values of
Al3+/Al, Ag+/Ag, K+/K and Cr3+/Cr are –1.66 V,
0.88 V, –2.93 V and –0.74V, respectively. The
correct decreasing order of reducing power of
the metal is
(a) Ag >Cr >Al > K
(b) K > Al > Cr >Ag
(iv) E cell = –0.0591 × 0.3010
(c) K > Al > Ag > Cr
(d) Al > K > Ag > Cr
(Odisha NEET 2019)
= –0.177
So, option (b) has more potential
Ans. (b) : More negative the value of standard
57.For which of the following electrolytes the graph of reduction potential higher is the reduction power.
1
Λm against C gives a negative slope.
Reducing power ∝
Standard reduction power
(a) Acetic acid
(b) Sodium acetate
Thus, the correct decreasing order of reducing power of
(c) Ammonium hydroxide
the metals is
(d) Water
K (E °K+ / K = −2.93V ) > Al ( E°Al3+ / Al = −1.66V )
GUJCET-2020
°
°
Ans. (b) : λm v/s c is a straight line with negative > Cr (E Cr3+ / Cr = −0.74V ) > Ag E Ag+ / Ag = +0.80V
slope for strong electrolyte sodium acetate.
61. If electrolysis of aqueous CuSO4 solution is
58. What would be the electrode potential for the
carried out using Cu-electrodes, the reaction
given half-cell reaction at pH = 5?......
taking place at the anode is
2H2O → O2 + 4H⊕ + 4e–;
(a) H+ + e–→H
o
Ered = 1.23V
(b) Cu2+(aq) + 2e–→Cu(s)
–1
–1
(R = 8.314 J mol K ; Temp = 298
(c) SO 42− (aq) − 2e− → SO 4
K; oxygen under std. atm.
(d) Cu(s) – 2e–→Cu2+(aq)
pressure of 1 bar)
WB-JEE-2019
[JEE Main 2020, 8 Jan Shift-I]
2+
−
Ans. (d) : Cu → Cu + 2e At anode (oxidation)
0.0591
Ans. E = −1.23 −
log[H + ]4
Cu 2+ + 2e− → Cu At cathode
4
0.059
Copper dissolves into the solution from the anode and it
= −1.23 −
× log(10−5 )4
deposited at the cathode.
4
º
0.059
62. Calculate emf of cell at 25ºC if value of Ecell
is 4
= −1.23 −
× (−20)
4
RT
volt Given
in10 = 0.06
= −1.23 + 0.059 × 5
F
= – 0.935V
2
2+
+
Cell notation : M M
M M
59. For a cell involving one electron, E0cell =0.59 V
0.01
0.0001
at 298 K, the equilibrium constant for cell
(a) 3.94V
(b) 4.06V
2.303RT
(c)
2.03V
(d) 8.18V
reaction is [Given that
= 0.059 V at
AIIMS 25 May 2019 (Morning)
F
T = 298 K]
RT
[Product]
Ans. (a): E cell = E ocell −
log
(a) 1.0 ×1030
(b) 1.0 × 102
nF
[Reactant]
(c) 1.0 × 105
(d) 1.0 × 1010
0.059
(10−4 )
(NEET- 2019)
= 4−
log −2
(∵ n = 2)
2
(10 )
Ans. (d) : Given, E o = 0.59V
60.
(
cell
= 4 − 0.03 × log10−2
We know
△G = −2.303 RT log K c and △G = −nFE ocell
Objective Chemistry Volume-II
)
= 4 − 0.03 × (−2) log10 = 4 − 0.06 = 3.94 V
154
YCT
At 298 K temperature, a hydrogen gas
Zn + Cu2+ → Zn2+ + Cu
electrode is made by dipping platinum wire in a
2+
solution of HCl of pH = 10 and by passing E cell = E ocell − 0.0591 log [Zn 2+ ]
n
[Cu ]
hydrogen gas around the platinum wire at one
atm pressure. The potential of electrode would
0.0591
0.01
=2−
log
be?
2
1
(a) 0.59V
(b) 0.118V
= 2.0592 V
(c) 1.18V
(d) 0.059V
AIIMS 26 May 2019 (Morning) 66. Standard hydrogen electrode (SHE) is a
(a) primary reference electrode
Ans. (a): For hydrogen electrode, oxidation half
(b) secondary reference electrode
reaction is
(c) metal - sparingly soluble salt electrode
H2→ 2H+ + 2e–
+
–pH
–10
(d)
metal - metal ion electrode
If pH = 10,[H ] = 1 × 10 = 1×10 M
MHT CET-02.05.2019, SHIFT-II
From Nernst equation,
+ 2
Ans.
(a)
:
The
electrode
potential of a single electrode
0.0591
[H ]
E cell = E ocell −
log
can only be measured by using some reference electrode
2
[H 2 ]
the reference electrode used is the standard or normal
o
hydrogen electrode (S.H.E or N.H.E).
For hydrogen electrode, E cell = 0
67.
Given,
0.0591
(10−10 )2
E cell = −
log
Co3+ + e– → Co2+; Eº = + 1.81V
2
1
Pb4+ + 2e– → Pb2+ ; Eº = + 1.67 V
= 0.0591× log1010 = 0.0591 ×10 = 0.591V
Ce4+ + e– → Ce3+ ; Eº = + 1.61 V
64. Ecell of the following cell is
Bi3+ + 3e– → Bi; Eº = + 0.20 V
+
+
Pt(s) H2(g), 1 bar H (1M) H (0.1 M)
Oxidising power of the species will increase in
the order
H2(g), 1 bar Pt(s)
(a) Ce4+ < Pb4+ < Bi3+ < Co3+
−2.303RT
2.303RT
(a)
(b)
(b) Bi3+ < Ce4+ < Pb4+ < Co3+
F
F
(c) Co3+ < Ce4+ < Bi3+ < Pb4+
−2.303RT
2.303RT
(d) Co3+ < Pb4+ < Ce4+ < Bi3+
(c)
(d)
2F
2F
[JEE Main 2019, 12 April Shift]
RT
Ans. (b) : Oxidizing power ∝ tendency to undergo
(e)
reduction potential values. Higher the emf value,
2F
power of
Kerala-CEE-2019 stronger the oxidising power. So, oxidising
species will increase in the order of Bi3+ < Ce4+ < Pb4+ <
Ans. (a) : Given Cell
3+
Pt(s) H2(g), 1 bar H+ (1M) H+ (0.1 M) H2(g),1bar Co .
68. If the standard electrode potential for a cell is
Pt(s)
2V at 300 K, the equilibrium constant (K) for
2.303RT
1
the reaction
For Nernst equation, Ecell = E ocell −
log +
nF
H
Zn ( s ) + Cu 2+ ( aq ) ⇌ Zn 2+ ( aq ) +Cu(s)
2.303RT
1
at 300 K is approximately
E ocell = 0 −
log
F
0.1
(R = 8 JK–1 mol–1, F = 96000 C mol–1)
(a) e–160
(b) e160
2.303RT
=
log10
–80
(c) e
(d) e320
F
[JEE Main 2019, ( Jan Shift-II]
2.303RT
E cell = −
Ans.
(b)
:
We
know
that,
F
o
o
65. Calculate E.M.F of following cell at 298 K ∆G = –RT lnK = – nFE cell
Zn(s)ZnSO4(0.01M)CuSO4(1.0M) Cu(s) if
n × F× E °cell
2×96000× 2
lnK =
⇒
Eocell = 2.0 V
R ×T
8×300
(a) 2.0 V
(b) 2.0592 V
lnK = 160
(c) 2.0296 V
(d) 1.0508 V
K = e160
MHT CET-03.05.2019, SHIFT-I
69. Consider the following reduction processes:
Ans. (b): Given, E ocell = 2.0V, Ecell = ?
Zn2+ + 2e– → Zn(s); Eº = – 0.76 V
At Anode :
Ca2+ + 2e– → Ca(s); Eº = – 2.87 V
Zn → Zn2+ + 2e–
…..(i)
Mg2+ + 2e– → Mg(s); Eº = – 2.36 V
At Cathode :
Ni2+ + 2e– → Ni(s); Eº = – 0.25 V
Cu2+ + 2e– → Cu
…..(ii)
The reducing power of the metals increases in
Adding equation (i) and equation (ii)
the order
63.
Objective Chemistry Volume-II
155
YCT
(a) Zn < Mg < Ni < Ca (b) Ni < Zn < Mg < Ca
(c) Ca < Zn < Mg < Ni (d) Ca < Mg < Zn < Ni
[JEE Main 2019, 10 Jan Shift-I]
Ans. (b) : Reducing power of an element
1
∝
Standard reduction potential
Hence, E °M2+ / M values of the given metals are as
Metals
Ni
Zn
Mg
Ca
Reducing power
→
Thus, the correct order of increasing reducing power of
the given metal is
Ni < Zn < Mg < Ca.
70. In the cell,
Pt(s)| H2 (g, 1 bar)|HCl|(aq)AgCl|(s)
|Ag(s)|Pt(s) the cell potential is 0.92 V when a
10–6 modal HCl solution is used. The standard
electrode potential of (AgCl/Ag, Cl–) electrode
is
2.303RT
= 0.06V at 298K
Given,
F
(a) 0.40 V
(b) 0.20 V
(c) 0.94 V
(d) 0.76 V
[JEE Main 2019, 10 Jan Shift-II]
Ans. (b) : At anode:
1
H 2 (g )
→ H + (aq) + e− E ° = 0.0V
2
At Cathode : –
AgCl(s) + e−
→ Ag(s) + Cl− E ° = 0.92V
0.92 = E °cell + 0.72
E °cell = 0.92 − 0.72
E °cell = 0.20V
71.
(ii) For Ag + / Ag : E °cell = 0.80 − (−0.76) = 1.56V
(iii) For Fe3+ / Fe 2+ : E°cell = 0.77 − (−0.76) = 1.53V
(iv) For Fe 2+ / Fe : E °cell = −0.44 − (−0.76) = 0.32V
Note: The transfer of electron in the cell would not
affect the E ocell value.
E (V ) − 0.25 − 0.76 − 236 − 287
0.06
log[H + ][Cl− ]
1
0.92 = E °cell − (12× 0.06)
(i) For Au 3+ / Au : E °cell = 1.40 − (−0.76) = 2.16V;
E °cell is maximum for E °Au3+ (aq)/ Au(s) .
°
E cell = E °cell −
°
− E °anode
Ans. (c) : E °cell = E cathode
72.
Given, that EOo 2 /H 2O = + 1.23V;
EΘS O 2- /SO2- = 2.05V;
2
8
−
E○
Br
2
/ Br
4
−
= +1.09V,
EΘAu3+ /Au = +1.4V
The strongest oxidising agent is
(a) Au3+
(b) O2
2−
(c) S2 O8
(d) Br2
[JEE Main 2019, 8 April Shift-I]
Ans. (c) : The more positive the reduction potential,
higher is the oxidizing power.
The decreasing order of oxidising agent –
S2 O82− > Au 3+ > O 2 > Br2
73.
o
o
Given, E Cr3+ /Cr = – 0.72 V, E Fe2+ / Fe = – 0.42 V.
The potential for the cell
Cr | Cr3+ (0.1 M) || Fe2+ (0.01 M) Fe is
(a) – 0.339 V
(b) – 0.26 V
(c) 0.26 V
(d) 0.3 V
JIPMER-2019
o
Ans. (c) : E cell = E cathode − E anode = − 0.42 − (−0.72) = 0.3
At anode[Cr
→ Cr 3+ + 3e − ] × 2
At cathode[Fe 2+ + 2e −
→ Fe] × 3
2Cr + 3Fe 2+
→ Cr 3+ + 3Fe
From Nernst equation,
0.0591 [Cr 3+ ]2
E cell = E ocell −
log
For the cell,
n
[Fe 2+ ]3
Zn(s)|Zn2+ (aq)||Mx+ (aq)|M(s), different half
0.0591
(0.1) 2
cells and their standard electrode potentials are
E
=
0.3
−
log
= 0.3 – 0.00985 log(104)
cell
3
given below.
6
(0.01)
Mx+
Au3+
Ag+
Fe3+(aq)/ Fe2+
= 0.3 – 0.00985× 4 = 0.3 – 0.0394 = 0.2606 V
(aq)/
(aq)/
(Ag)/
Fe2+(aq) (aq)/ 74. One litre solution of MgCl is electrolysed
2
M(s)
Au(s) Ag(s)
Fe(s)
completely by passing a current of 1A for 16
x+
min 5 sec. The original concentration of MgCl2
EºM /M/V 1.40
0.80
0.77
–0.44
solution was
°
If EZn2+ /Zn = – 0.76 V, which cathode will give a
(Atomic mass of Mg = 24)
maximum value of Eº cell per electron
(a) 5 × 10–3 M
(b) 5 × 10–2 M
–3
transferred?
(c) 0.5 × 10 M
(d) 1.0 × 10–2 M
+
2+
Karnataka-CET-2019
Ag
Fe
(b)
(a)
Ans.
(a)
:
Given
that,
Ag
Fe
Current, I = 1A, time (t) = 16 min 5 sec = 965 sec.
3+
Au
Fe3+
According to Faraday first law,
(c)
(d)
Quantity of electricity passed (Q) = I.t
Au
Fe 2 +
= 1 × 965 = 965 C
[JEE Main 2019, 11 Jan Shift-I]
Objective Chemistry Volume-II
156
YCT
W = Z.i.t (where W = amount of substance deposited)
Mg +2 + 2e−
→ Mg
12
12
So ratio will be
W=
× 965 =
g
1
1
96500
100
mol. Al : 1 mol Ag : mol Mg = 2 : 6 : 3
E
At.wt / 2 24 / 2
3
2
=
=
∵ Z =
78.
Given
the
equilibrium
constant (Kc) of the
96500
96500
96500
reaction:
Conc. of MgCl2 i.e. molarity
Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s)
Weight
12
=
M=
is 10 × 1015, calculate the Eocell of this reaction at
Mol.Mass × V(L) 100 × 24 × 1
298 K.
1
=
= 5 × 10−3 M
RT
200
2.303 F at 298K = 0.059V
75. The Eº values of the following half cells are
given: (25ºC)
Eº
(a) 0.4736 V
(b) 0.04736 mV
Fe 3+ (aq) + e – → Fe 2+ (aq) + 0.771V
(c) 0.4736 mV
(d) 0.04736 V
Fe 2+ (aq) + 2e – → Fe(s) – 0.447V
[JEE Main 2019, 11 Jan Shift-II]
Ans.
(a)
:
Given,
K
=
10 × 1015
c
The∆Gº of Fe (aq) + 3e → Fe(s)
We know that,
(a) 160.65kJ mol−1
(b) −74.39kJ mol−1
2.303RT
(c) 86.26kJ mol−1
(d) 11.87kJ mol−1
E °cell =
log K c
nF
CG PET -2019
0.059
log (10×1015 )
⇒
Ans. (d) : ∆Go=–nFEocell
2
Fe3+ (aq) + e − → Fe(aq), ∆G1o = − F × 0.771
0.059×16
⇒ 0.472V
⇒
Fe 2+ (aq) + 2e − → Fe(s), ∆G o2 = − 2 × F × (− 0.447)
2
79. For the following transformation, the reduction
= 0.894F
3+
−
o
half reaction is
Fe + (aq) + 3e → Fe(s), ∆G 3 = ?
Sn2+ +2Hg2+ → Sn4+ + Hg 2+
2
o
o
o
∆G 3 = ∆G1 + ∆G 2
(a) Sn2+ → Sn4+
(b) Sn2+ → Hg2+
∆G 3o = 0.771F + 0.894F
(c) Hg2+ → Hg 2+
(d) Hg2+ → Sn4+
2
o
∆G 3 = 0.123F = 0.123 × 96485
Assam CEE-2019
o
Ans. (c) : Oxidation of Half-reaction
∆G 3 = 11867.6 J / mol
o
Sn 2+ (aq)
→ Sn 4+ (aq) + 2e − (loss of 2 electrons)
∆G = 11.87 kJ / mol
3+
3
76.
–
2+
+
Zn (s) | Zn 2( aq
) ( 1M ) || Ni ( aq ) ( 1M ) | Ni ( s )
(Reductant )
Reduction of Half- reactions
→ Hg 22+ (aq) (gain of 2 electrons)
2Hg 2+ (aq) + 2e −
Which is incorrect for the above given cell?
(a) Electrochemical cell (b) Voltaic Cell
80.
(c) Galvanic Cell
(d) Daniel Cell
GUJCET-2019
Ans. (d) : Cell reaction in Daniel cell is
Zn(s) + Cu 2+(aq)
→ Zn 2+(aq) + Cu(s)
(Oxidant )
Net cell reaction of Pt | H2 (640 mm) | HCl | H2
(510mm) pt.
(a) 0.89V
(b) 0.93V.
(c) 2.91 × 10–3V
(d) 2.5 × 10–2V
[BITSAT – 2019]
77. If one mole electrons is passed through the Ans. (c) : Pt |H2 (640mm) | HCl | H2 (510 mm)| Pt
solutions of AlCl3, AgNO3 and MgSO4, in what
°
ratio Al, Ag and Mg will be deposited all the E cell = 0
electrodes?
(a) 3 : 6 : 2
(b) 2 : 6 : 3
(c) 1 : 2 : 3
(d) 3 : 2 : 1
GUJCET-2019
Ans. (b) : From the second law of electrolysis
n eq ( Al) = n eq (Ag) = n eq ( Mg )
…….(i)
n eq = moles × n − factor
Now,
Al+3 + 3e−
→ Al
Ag + + e−
→ Ag
Objective Chemistry Volume-II
⇒ 2.91 × 10–3 V
157
YCT
The standard reduction potential E0 for half 84. For the electrochemical cell,
reactions are
Ag+|AgCl|KCl||AgNO3|Ag+, the overall cell
Zn→Zn2+ + 2e– ;Eo = + 0.76V
reaction is
Fe →Fe2+ +2e–
;Eo = + 0.41V
(a) Ag+ + KCl → AgCl(s) + K+
The EMF of the cell reaction
1
(b) Ag + AgCl → 2Ag + Cl2
Fe2+ + Zn→ Zn2+ + Fe is
2
(a) – 0.35V
(b) + 0.35V
(c) AgCl(s) → Ag+ + Cl −
(c) +1.17V
(d) –1.17V
(d) Ag+ + Cl − → AgCl(s)
WB-JEE-2018, 2011
JIPMER-2018
Ans. (b) : Zn → Zn 2+ + 2e− ; E ° = +0.76V
Ans.
(c)
:
For
the
electrochemical
cell,
Fe → Fe 2+ + 2e− ; E ° = +0.41V
Ag–|AgCl|KCl||AgNO3|Ag+
We know,
The cell reaction is,
E °cell = E °(cathode) − E °(anode)
AgCl(s) + e– → Ag + Cl– (aq)
Ag → Ag+(aq) + e–
= (0.76 − 0.41)V
Overall reaction
= 0.35 V
AgCl(s) → Ag+ + Cl–
82. Assertion: When l M CusO4 (aq) solution is
2+
2+
electrolysed using copper electrodes, copper is 85. If Eº (Zn , Zn) = – 0.763 V and Eº (Fe ,2+Fe) =
–0.44 V, then the emf of the cell Zn Zn (a =
dissolved at anode and copper gets deposites at
0.001)||Fe2+ (a = 0.005) Fe is
cathode.
(a) equal to 0.323 V
(b) less than 0.323 V
Reason: The standard oxidation potential of
(c)
greater
than
0.323
V
(d) equal to 1.103 V
copper is less than the standard oxidation
JIPMER-2018
potential of water and standard reduction
potential of copper is
greater than the Ans. (c) : The cell reaction isstandard reduction potential of water.
Zn + Fe2+
→ Zn2+ + Fe
(a) If both Assertion and Reason are correct and
From Nernst equation
the Reason is the correct explanation of
a 2+
0.0591
Assertion.
Ecell = E ocell −
log Zn
(b) If both Assertion and Reason are correct, but
n
a Fe2+
Reason is not the correct explanation of
0.0591
0.001
Assertion.
= (0.763 − 0.44) −
log
(c) If Assertion is correct but Reason is in
1
0.005
correct.
= 0.364 V
(d) If both the Assertion and Reason are 86. For a cell involving two electron changes,
incorrect.
E ocell = 0.3V at 25oC. The equilibrium constant
AIIMS-26 May, 2018 (M)
of the reaction is
Ans. (c) : Copper is dissolved at anode and deposited at
(a) 10–10
(b) 3 × 10–2
cathode when Cu electrodes are used and electrolyte is
(c) 10
(d) 1010
1M CuSO4(aq) solution as standard oxidation potential
Karnataka-CET-2018
of Cu is higher than standard oxidation potential of
Ans.
(d)
:
water and standard reduction potential of Cu is greater
than standard reduction potential of water.
For ∆G o is related to Ksp by the equation,
83. In following cell reaction
∆G o = – 2.303 RT log Ksp
+
2+
Mg(s)+2Ag (0.001M) → Mg (0.20M)+2Ag(s)
nFE ocell
−∆G o
Calcuate Ecell for the reason [E0 =3.17]
log K sp =
=
(at 25o C)
2.303RT
2.303RT
(a) 2.63 V
(b) 3.04 V
2E ocell
2 × 0.3
(c) 3.33V
(d) 3.51 V
=
=
=10
[AIIMS-27, May, 2018 (M)]
0.0591 0.0591
Ans. (b) : We know thatlog Ksp = 10 ⇒ K sp = 1010
RT
87. The voltage of the cell consisting of Li (s) and
E cell = E °cell −
lnQ
nF
F2 (g) electrodes is 5.92 V at standard condition
at 298 K. What is the voltage if the electrolyte
0.059
[Mg 2+ ]
= 3.17 −
log
consists of 2M LiF.
+ 2
2
[Ag ]
(ln 2= 0.693, R = 8.314 JK-1 mol-1 and F = 96500
C mol-1)
0.059
[0.2]
= 3.17 −
log
(a) 5.90 V
(b) 5.937 V
2
[0.001]2
(c) 5.88 V
(d) 4.9 V
= 3.17 − 0.1563
(e) 4.8 V
81.
≃ 3.04 V
Objective Chemistry Volume-II
Kerala-CEE-2018
158
YCT
Ans. (c) : Given that, E °cell = 5.92V,T = 298K
R = 8.314 JK −1mol −1 , F = 96500 Cmol −1
Now, the cell reaction is
1
Li(s) + F2 (g)
→ Li + + F−
2
We know that,
RT
[Product]
o
E cell = E cell
−
log
nF
[Reactant]
= E ocell −
2.303RT
[Li + ][F− ]
log
nF
PF0.5
2
[PF2 = 1atm]
2.303 × 8.314 × 298
log(2 × 2)
1× 96500
0.059
= 5.92 −
× 2log 2
1
= 5.92 − 0.035 = 5.88 V
= 5.92 −
Ans. (a) : In both galvanic and electrolytic cells,
oxidation takes place at the anode and electrons flow
from the anode to the cathode and reduction takes place
at cathode.
91. What will be the Ecell for the given cell?
Zn|Zn2+(0.1 M)|| Cu2+(0.01M)|Cu
Given : EoZn2+ /Zn = 0.76 V and EoCu2+ /cu = 0.34 V
Also predict whether the reaction is
spontaneous or non-spontaneous.
(a) 1.07 V and spontaneous
(b) –1.13 V and non-spontaneous
(c) –1.07 V and non-spontaneous
(d) 1.13 V and spontaneous
J & K CET-(2018)
Ans. (a) : Zn | Zn 2+ (0.1M) || Cu 2+ (0.01M) | Cu
0.0591
[Zn 2+ ]
log
n
[Cu 2+ ]
°
E °cell = E (Cu
− E °(Zn 2+ / Zn )
2+
/ Cu )
E cell = E °cell −
88.
During galvanisation of iron, which metal is
used for coating iron surface?
⇒ 0.34 − (−0.76) = 1.10V
(a) Copper
(b) Zinc
(c) Nickel
(d) Tin
0.0591
0.1
E cell = 1.10 −
log
MHT CET-2018
2
0.01
Ans. (b) : Galvanization is the process of applying a
E cell = 1.10 − 0.03 = 1.07V
protective zinc coating to iron or steel, to prevent
rusting. The most common method is hot dip As E cell is positive the reaction is spontaneous.
galvanizing, in which iron or steel sections are 92. What will be the half-cell potential of a
submerged in a bath of molten zinc.
hydrogen electrode acting as an anode and
°
dipped in a solution of pH = 2?
89. The Ered
of Ag, Cu, Co and Zn are 0.799, 0.337,
(a) 0 V
(b) 0.0196 V
– 0.277, –0.762 V respectively, which of the
(c)
0.276
V
(d) 0.118 V
following cells will have maximum cell emf ?
2+
2+
J & K CET-(2018)
(a) Zn | Zn (1M) || Cu (1M) | Cu
2+
+
Ans.
(d)
:
(b) Zn | Zn (1M) || Ag (1M) | Ag
pH = 2
(c) Cu | Cu2+ (1M) || Ag+ (1M) | Ag
(d) Zn | Zn2+ (1M) || Co2+ (1M) | Co
H2(g) ↽ ⇀ 2H+ + 2e–
UPTU/UPSEE-2018
K = [H+]2 = [10–2]2 = 10–4M
°
Ans. (b) : A cell with high E Red at cathode and low
0.0591
E = Eo −
log10−4
E °Red at anode, will have high E °Cell among given half
2
cells. Ag half cell has highest E °Red and Zn half cell has
E = 0.1182 V
lowest E °Red and this cell will have highest E °Cell value.
93. For the following, cell Zn Zn 2 + Cd 2 + Cd
According to option,
o
Ecell = 0.30V,Ecell
= 0.36V, then the value of
E o = E ocathode − E oanode
2+
[Cd ]
(a)
E o = 0.337 – (–0.762) = 1.099 V
is–
o
[Zn 2 + ]
(b)
E = 0.799 – (–0.762) = 1.561 V
(a) 10
(b) 0.01
(c)
E o = 0.799 – 0.337 = 0.462 V
(c) 0.1
(d) 100
o
(d)
E = – 0.277 – (–0.762) = 0.485 V
BCECE-2018
90. Which of the following statements is/are true
2+
Ans. (b) : Cell reaction is Zn + Cd
→ Cd + Zn 2+
for an electrochemical cell?
(a) Oxidation occurs at the anode only.
Zn 2+
K
=
(b) Reduction occurs at the anode only
eq
Cd 2+
(c) Oxidation occurs at both the anode and
°
cathode.
E cell = 0.30V, E cell
= 0.36 V
(d) Reduction occurs at both the anode and
0.0591
°
cathode.
Nernst equation, E cell = E cell
−
log K eq
n
J & K CET-(2018)
Objective Chemistry Volume-II
159
YCT
97.
0.0591
log K eq = E °cell − E cell
2
0.0591
log K eq = 0.06
2
log K eq =
0.06× 2
⇒
0.0591
Zn 2+
= 100,
Cd 2+
EoFe/Fe2+ = 0.44 V,EoAl/Al 3+ = 1.66 V
Zn 2+
log 2+ = 2
Cd
EoNi/Ni 2+ = 0.25 V,EoAg – /Ag = 0.80 V
(a) Ag
(c) Fe
Cd 2+
= 0.01
Zn 2+
(b) Ni
(d) Al
GUJCET-2018
2+
94.
Calculate the standard free energy change for
the reaction
2Ag + 2H+ → H2 + 2Ag+
E° for Ag+ + e– → Ag is 0.80 V.
(a) + 308.8 kJ
(b) + 154.4 kJ
(c) – 308.8 kJ
(d) – 154.5 kJ
CG PET -2018
°
°
Ans. (b) : E cell = −0.80V, ∆G ° = −nFE cell
95.
In which metal container, the aqueous solution
of CuSO4 can be stored?
EoCu2- /Cu = 0.34 V
→ Cu + 2M
Ans. (a) : Cu + M
For occurrence of the reaction,
E °cell > 0
+
For M = Fe, E °cell = E °Fe / Fe2+ + E °Cu 2+ / Cu
⇒ 0.44 + 0.34 = 0.78
For M = Ag E °cell = E °Ag / Ag + E °cell2+ / Cu
⇒ 0.80 + 0.34 = 0.46
So when CuSO4 is Kept in Ag container, no reaction
occurs.
∆G° = −2×96500×(−0.80) J = +154.4kJ
98. How long (approximate) should water be
The standard reduction potential of Pb and Zn
electrolysed by passing through 100 amperes
electrodes are – 0.126V and – 0.763V
current so that oxygen released can completely
respectively. The cell equation will be:
burn 27.66 g of diborane?
(a) Pb2+ + Zn → Pb + Zn2+
(Atomic weight of B = 10.8µ)
(b) Pb4+ + 2Zn→ Pb + 2Zn2+
(a) 6.4 hours
(b) 0.8 hours
(c) Zn2+ + Pb → Zn + Pb2+
(c) 3.2 hours
(d) 1.6 hours
(d) None
[JEE Main JEE Main 2018]
HP CET-2018 Ans. (c) : B H + 3O
→ B O + 3H O
Ans. (a) : The E ored of Zn is lower than that of Pb, hence,
Zn is a better reducing agent than Pb.
Pb2++Zn → Pb+Zn2+
96. What will be the oxidation potential for the
following hydrogen half cell at 1 bar pressure
and 25o C temperature?
2
6
2
2
3
2
27.66g of B2H6 (1mole) requires 3 moles of oxygen
(O2) for complete burning.
27.66
No. of mole B2H6 =
gm = 1mole
27.66
2H2O → O2 + 4H+ + 4e–
No. of factor = 4
So,
3O
mole
required
2
Pt H 2( g ) HCl (aq )pH = 3
According
to
Faraday's
first law1bar
Meq
charge
=
Meq
substance
(a) 0.059 V
(b) 0.188 V
100 × t
(c) 0.177 V
(d) 0.000 V
× 1000 = 3 × 4 × 1000
GUJCET-2018
96500
t =11580secor 3.2 hr
Ans. (c) : The reduction reaction is
+
−
2H + 2e
→ H2
99. The electrode potential, Eo, for the reduction of
Pressure P = 1 bar = 0.9869 atm
Mn O4− to Mn2+ in acidic medium is + 1.51 V.
pH = 3
Which of the following metal (s) will be
[H+] = 10–pH = 10–3 M
oxidised? The reduction reaction and standard
o
electrode potentials for Zn2+, Ag+, and Au+ are
E SHE = 0.0 V
given as
0.0591
PH 2
°
o
E HE = ESHE
−
log10
2
Zn 2+( aq ) + 2e- → Zn ( s ) ,E = -0.762 V
n
H+
o
Ag + ( aq ) + e- ↽ ⇀ Ag( s ), E = +0.80 V
0.0591
0.9869
E HE = 0.0V −
log10
2
o
2
10−3
Au + ( aq ) + e- ↽ ⇀ Au ( s ) ,E = +1.69 V
E HE = −0.177V
(a) Zn and Au
(b) Ag and Au
This is the reduction potential the oxidation potential is
(c) Au
(d) Zn and Ag
+ 0.177 V.
AMU-2018
Objective Chemistry Volume-II
160
YCT
Ans. (d) : More the value of reduction potential more
will be the tendency to undergo reduction.
the E °red (MnO4– / Mn2+) = + 1.51 V value is greater
then E °red (zn2+/zn) = –0. 762 V and
E °red (Ag +/Ag) = + 0.80V. Thus only Zn and Ag will be
oxidized.
100. The standard emf of the cell ( Eocell ) and
equilibrium constant (Keq) of the following
reaction,
Cd2+ + 4NH3 ⇌ Cd ( NH 3 )4 at 298 K is
2+
(a) E cell = 1.0V; K eq = 1.26 × 107
(b) E cell = 0.21V; K eq = 1.26 ×107
(c) E cell = 1.0V; K eq = 6.60 × 1033
(d) E cell = 0.21 V; K eq = 6.60 × 1033
Ans. (b) : For the given reaction–
Cd 2+ + 4NH 3 ↽ ⇀ Cd(NH3)42+
At equilibrium–
E °cell = 0
2.303RT
log K c
nF
Where Kc is unknown
2.303 × 8.314 × 298
E ocell =
2 × 96500
E °cell = 0.21V
Hence, E ocell =
0.059
log K c
2
0.21× 2
log K c =
0.0591
0.42
log K c =
= 7.1065
0.059
7
Kc= 1.27×10
0.21 =
E °MnO4 + 5e−
→ Mn 2+ ; E ° = 1.51V
As we know that,
lower the reduction potential stronger the reducing
agent.
Order of reducing power, Cr > Cr3+ > Cl– > Mn2+'
So, Cr is the strongest reducing agent.
102. The standard emf of galvanic cell involving 3
moles of electrons in its redox reaction is 0.59
V. The equilibrium constant for the reaction of
the cell is
(a) 1025
(b) 1020
15
(c) 10
(d) 1030
JCECE - 2017
Ans. (d) :
Galvanic cell redox reaction is
Zn(s) + Cu2+(aq) ↽ ⇀ Zn2+(aq) + Cu(s)
E cell = 0, because Cu2+ and Zn2+ ions are equilibrium.
Then Kc (equilibrium constant)
AMU-2018 E o = 0.059 log K
cell
c
n
0.059
0.59 =
log K c
3
0.59 × 3
log K c =
= 30
0.059
K c = antilog 30
K c = 1030
103. The value of reaction quotient (Q), for the
following cell
Zn(s)|Zn2+(0.01 M) | | Ag+ (1.25 M) | Ag(s) is
(a) 156
(b) 125
(d) 6.4 × 10–3
(c) 1.25 × 10–2
JIPMER-2017
Ans. (d) :
Zn(s) → Ζn2+(0.01Μ) + 2e–
[Ag+(1.25M) + e– → Ag(s)] × 2
Zn(s) + 2Ag+ (1.25M)→Zn+(0.01M) + 2Ag(s)
[Zn 2+ ]
[Ag + ]2
2
0.01
0.01
EoCr 3+ /Cr = -0.74V
Q=
=
2
(1.25)
1.5625
EoCr O2 /Cr 3+ = 1.33V,
0 7
1
Q=
= 0.0064
EoMnO- /Mn2+ = 1.51V
156.25
4
Among the following, the strongest reducing
Q = 6.4 ×10 –3
agent is
104. The reaction is spontaneous if the cell potential
(a) Cr
(b) Mr2+
is :
(d) Cl–
(c) Cr3+
(a) positive
(b) negative
[JEE Main-2017]
(c) zero
(d) infinite
Manipal-2017
Ans. (a) : O.A. + ne−
→ R.A
o
−
−
°
Ans.
(a):
If
E
=
0,
Reaction
isin
equilibrium
Cl 2 + 2e
→ 2Cl ;
E = 1.36V
cell
Q=
101. Given EoCl /Cl - = 1.36V,
Cr +3 + 3e−
→ Cr;
If E ocell > 0 , Reaction is spontaneous
E ° = −0.74V
If E ocell < 0 , Reaction is non spontaneous
Cr2 O72− + 6e−
→ 2Cr +3 ; E ° = 1.33V
Objective Chemistry Volume-II
161
YCT
105. During electrolysis of molten NaCl, some water
was added. What will happen?
(a) Electrolysis will stop
(b) Hydrogen will be evolved
(c) Some amount of caustic soda will be formed
(d) A fire is likely
WB-JEE-2017
Ans. (b, c, d) :
During electrolysis of molten NaCl:
Reaction at cathode :
Reaction at anode:
1
Na+ + e− → Na
Cl− → Cl2 + e −
2
1
By adding some water, Na + H2O→NaOH + H 2
2
A fire is likely to take place due to vigorous reaction of
sodium with water.
106. Consider the single electrode process
4H + + 4e – ⇌ 2H 2 catalysed by platinum black
electrode in HCl electrolyte. The potential of
the electrode is –0.059 V. SHE. What is the
concentration of the acid in the hydrogen half
cell if the H2 pressure is 1 bar ?
4H + + 4e – ⇌ 2H 2
Ans. (a) :
Given, MCl + e – → M + Cl – (reduction) …..(i)
1
Cl – → Cl2 + e – (oxidation)…..(ii)
2
From equation (i) and (ii) we get–
1
MCl → M + Cl2
2
From Nernst Equation,
0.059
E cell = E°cell −
log K c
n
0.059
−1.140 = −0.55 −
log K c
1
–0.59 = –0.059 log Kc
0.59
log K c =
= 10
0.059
10
Kc = 10
1
1
∴ K sp =
=
K c 1010
= 10–10
108. What is the standard potential of the Tl3/Tl
electrode?
Tl3+ + 2e– → Tl+, Eo = 1.26V
Tl3+ + e– → Tl, Eo = –0.336V
(a) 0.924V
(b) −0.924V
(a) 1 M
(b) 10 M
(c) 0.728V
(d) −0.728V
(c) 0.1 M
(d) 0.01 M
TS EAMCET-2017
CG PET -2017
Ans. (c) : Given the single electrode process–
Ans. (c) : Given
4H + + 4e – ⇌ 2H 2
Tl3+ + 2e − → Tl+ , Eº = 1.26V
....(i)
Tl+ + e − → Tl,Eº = −0.336V ....(ii)
Required relation
Tl3+ + 3e− → Tl, Eº = ?
...(iii)
0.059
1
log + 4
−0.059 = 0 −
AS,
∆
Gº
=
−
nFEº
4
[H ]
∴
∆G 3º = ∆G1º + ∆G 2º
1
1
1 = log
–nFEo = [(–nEo)1 + (–nEo)2]F
4
4
H +
–3Eo = [–(2 × 1.26) – (1 × –0.336)]
–3Eo = [–2.52 + 0.336]
4 = log[H + ]−4
–3Eo = –2.184
4 = −4log[H + ]
2.184
Eo = −
= 0.728V
−1 = log[H + ]
−3
109. Which option is incorrect for the working cell?
⇒ [H + ] = 10−1 = 0.1M
Pt | Cl 2(g) | Cl – ( c1 ) || Cl – ( c2 ) | Cl 2(g) | Pt
107. Consider the following electrode processes of a
1 bar
1bar
cell,
(a) ∆G = − v
(b) C2 > C1
1
–
–
–
–
Cl → Cl 2 + e
[MCl + e → M + Cl ]
(d) C1 > C2
(c) Eocell = 0
2
GUJCET-2017
If EMF of this cell is –1.140 V and E° value of
∴ E = E° −
PH
0.059
log +2 4
n
[H ]
the cell is –0.55 V at 298 K, the value of the
equilibrium constant of the sparingly soluble
salt MCl is in the order of
(a) 10–10
(b) 10–8
–7
(c) 10
(d) 10–11
TS EAMCET-2017
Objective Chemistry Volume-II
162
−
Ans. (b) : Cl → Cl
Cl
−
C2
0.0591
C
log 1
1
C2
They, cell reaction is spontaneous
When C1 > C2
E cell =
YCT
110. The standard reduction potentials of Cu2+ /Cu
and Cu2+ /Cu+ are 0.337 and 0.153 volts
respectively. The standard electrode potential
for Cu+ /Cu half cell will be
(a) 0.490 V
(b) 0.980 V
(c) 0.827 V
(d) 0.521 V
JCECE - 2016
Ans. (d) :
→ Cu ; E Cu 2+ / Cu = 0.337 V
Cu 2+ + 2e−
....(i)
log
[Pb 2+ ]
0.01
=
= − 0.3
2+
[Sn ] −0.0291
Pb 2 +
= antilog (– 0.3)
Sn 2 +
Pb 2 +
1
= 0.5 =
2
Sn 2 +
113. Electrolytic reduction of alumina to aluminium
by Hall-Heroult process is carried out :
(a) in the presence of NaCl
Overall equation (i) and (ii),
+
−
(b)
in the presence of Hnorite
→ Cu ;E Cu + / Cu = ?
Cu + e
....(iii)
(c) in the presence of cryolite which forms a melt
From the above equation–
with lower melting temperature
nE Cu + / Cu = nE Cu 2+ / Cu − nE Cu 2+ / Cu +
(d) in the presence of cryolite which forms a melt
with higher melting temperature
E Cu + / Cu = 2 × 0.337 − 1× 0.153 = 0.521
Manipal-2016
E Cu + / Cu = 0.521V
Ans. (c): Electrolytic reduction of alumina to
111. The Eº values for Mn and Zn are more aluminium by Hall-Heroult process is carried out in the
presence of cryolite which forms a melt with lower
negative than expected because
(a) they have either half-filled and fully-filled melting temperature.
114. In the cell represented by
configurations
(b) they can easily donate electrons
Pb ( s ) Pb 2+ (1M ) Ag + (1M ) Ag ( s ) ,
the
(c) it is quite easy to remove electrons from their
reducing agent is
orbitals
(a) Pb
(b) Pb2+
(d) None of the above
(c) Ag
(d) Ag+
JIPMER-2016
MHT CET-2016
Ans. (a) : The value of standard electrode potentials for
(a) : Pb is the reducing agent. It itself oxidized to
Mn and Zn are more negative than expected. The reason Ans.
2+
behind this is that they have exactly half-filled and fully Pb .
filled configuration ions.
Pb → Pb 2+ + 2e−
5 2
25Mn : [Ar]18 3d 4s
The loss of electrons is oxidation. The species that is
10 2
oxidized reduces other species.
30Zn : [Ar] 3d 4s
112. For the cell reaction
115. The pressure of H2 required to make the
potential of H2 electrode zero in pure water at
Pb + Sn2+
→ Pb2+ + Sn
298 K is
2+
–
°
Given that, Pb
→ Pb + 2e , E = 0.13 V
(a) 10–10 atm
(b) 10–4 atm
–14
Sn2+ + 2e–
→ Sn; E° = – 0.14 V
(c) 10 atm
(d) 10–12 atm
What would be the ratio of cation
(NEET-I 2016)
concentration for which E = 0?
Ans. (c) : We know, for hydrogen electrode ( E ocell )=0
(a) 1/4
(b) 1/2
(c) 1/3
(d) 1/1
− log[H + ] = 7 ⇒ [H + ] = 10−7
JIPMER-2016
2H + (aq) + 2e− → H 2 (g)
Ans. (b) : For the given cell reaction,
PH
0.0591
E 0cell = – 0.14V + 0.13V
E Cell = E °Cell −
log +2 2
2
[H ]
= – 0.01V
PH
0.0591
According to Nernst equation,
0 = 0−
log −72 2
2+
2
(10
)
Pb
0.0591
Ecell = E 0cell −
log 2 +
PH
2
Sn
log −72 2 = 0
(10 )
0.0591 [Pb 2+ ]
0 = − 0.01 −
log
PH2
2
[Sn 2+ ]
1=
[∵ log1 = 0]
(10−7 ) 2
2+
0.0591 [Pb ]
−0.01 =
log
PH2 = 10−14 atm
2
[Sn 2+ ]
Cu 2+ + e−
→ Cu + ; E Cu 2+ / Cu + = 0.153V
Objective Chemistry Volume-II
....(ii)
163
YCT
116. Given that the standard reduction potentials C
–0.46V
E °red
so
Moderated
for M+/M and N+/N electrodes at 298 K are 0.52
medium
reducing
agent
V and 0.25 V respectively.
Which of the following is correct in respect of So, order reducing agent
B > C > A.
the following electrochemical cell?
+
+
120.
The EMF of a galvanic cell by coupling two
M/M N /N
electrodes M1 | M 12+ (0.1M) | | M 2+
2 (0.01M) | M 2
(a) The overall cell reaction is a spontaneous
is + 1.47 V. if the E° value (reduction potential)
reaction
of M2 electrode is 0.9V, E° (reduction potential)
(b) The standard EMF of the cell is –0.27V.
value of M1 electrode in volts would be
(c) The standard EMF of the cell is 077V.
2.303RT(T = 298K)
(d) The standard EMF of the cell is –0.77 V.
[Assume
= 0.06]
F
[AIIMS-2016]
(a) – 0.57
(b) – 0.60
Ans. (b) : The EMF
(c)
+
0.57
(d)
+ 0.60
E oCell = E°cathode − E°anode = E oright − E° left
J & K CET-(2016)
Ans. (b) : The overall cell reaction can be represented
E oCell = 0.25 − 0.52
as
= – 0.27 V
M1 + M 22+ → M12+ + M 2
117. During electrolysis of H2O, molar ratio of H2
and O2 produced is –
M 2+
2.303RT
E cell = E ° −
log 12+
(a) 2 : 1
(b) 1 : 2
nF
M2
(c) 1 : 1
(d) 1 : 4
0.06
0.1
BCECE-2016
1.47 = E° −
log
2
0.01
1
Electrolysis
°
°
Ans. (a) : H 2 O
→ H 2 + O2
⇒
1.47
=
E
−
0.03
⇒
E
=
1.47
+ 0.03 = 1.5V
2
1
2
=2:1
118. The EMF of the cell Tl/Tl+ (0.001M)||
Cu2+(0.01M)/ Cu is 0.83. The cell EMF can be
increased by
(a) Increasing the concentration of Tl+ ions.
(b) Increasing the concentration of Cu2+ ions
(c) Increasing the concentration of Tl+ and
Cu2+ ions
(d) None of these
[BITSAT – 2016]
1
Ans. (b) : oxidation potential ∝
Concentration of ions
reduction potential ∝ concentration of ions.
The cell voltage can be increased by decreasing the
concentration of ions around anode or by increasing the
concentration of ions around cathode.
∴ Ratio = 1 :
119. The value Eorad for metal, A,B and C are 0.34
volt -0.80 and -0.46 volt respectively state the
correct order for their ability to act a reducing
agent.
(a) C>A>B
(b) A>B>C
(c) B>C>A
(d) C>B>A
GUJCET-2016, 2015
Ans. (c) : B>C>A
Metal
Note
E °cell
A
0.34V
B
–0.80V
Highest E °red so highest
reducing agent.
°
red
Least E
no strong,
reducing agent
Objective Chemistry Volume-II
Now
°
E ° = E cathode
− E °anode
1.5 = 0.9 − E °anode
E oanode = 0.9 − 1.5
= −0.6 V
121. In an electrochemical cell, anode and cathode
are respectively
(a) positively and negatively charged ions
(b) positively and negatively charged electrodes
(c) negatively and positively charged electrodes
(d) negatively and positively charged ions.
SRMJEEE – 2016
Ans. (c) : In both kinds of electrochemical cells, the
anode is the electrode at which the oxidation halfreaction occurs, and the cathode is the electrode at
which the reduction half reaction occurs. The anode is
considered positive and the cathode is considered
negative.
122. The products obtained at the cathode and
anode respectively during the electrolysis of
aqueous K2SO4 solution using platinum
electrodes are
(a) O2, H2
(b) H2, O2
(c) H2, SO2
(d) K, SO2
AP-EAMCET – 2016
Ans. (b) : During the electrolysis of aqueous K2SO4
solution using platinum electrodes are K 2SO 4 → 2K + + SO 24−
Cathode : 2H + + 2e − → H 2
Anode : 4OH − → 2H 2 O + O 2 + 4e −
Hence, H2 gas will generate at cathode and O2 gas will
generate at anode.
164
YCT
(a) –1.532 V
(b) –1.503 V
123. For the cell Ag(s) | Ag + (aq) || Cu 2+ (aq) | Cu(s),
(c)
1.532
V
(d)
–3.06 V
the reduction potentials of the left and right
WB-JEE-2015
hand electrodes are 0.337 and 0.799 volts, the
Ans. (a) : The Nernst equation is
cell e.m.f. is
(a) –1.136 volt
(b) 1.136 volt
0.0591 [Ag + ]2
o
E
=
E
−
log
(c) –0.462 volt
(d) 0.462 volt
cell
cell
nF
[Zn 2+ ]
AMU-2016
Therefore we put the value of E °cell
Ans. (d) : Ag (s) |Ag+(aq) ||Cu2+(aq)| Cu (s)
(0.0591)
(0.1) 2
2Ag
→ 2Ag + + 2e − (Anode)
….(i)
E cell = (−1.562) −
log
2
(0.1)
→ Cu (Cathode)
….(ii)
Cu 2+ + 2e −
0.0591
The sum of equation (i) and (ii)
E cell = (−1.562) −
log10−1
2+
+
2
→ 2Ag + Cu
2Ag + Cu
0.0591
°
°
Ecell = −1.562 +
E cell = E cathode – E °anode = E °right – E °left
2
= (0.799 – 0.337) V
Ecell = −1.562 + 0.02955
= 0.462 V
Ecell = −1.532V
124. Given
Eo Cr 3+ / Cr = −0.72V, Eo Fe2+ / Fe = –0.42V . 127. A hydrogen electrode is immersed in a solution
with pH = 0 (HCl). By how much will the
The potential for the cell
potential (reduction) change if an equivalent
Cr|Cr3+ (0.1M) || Fe2+ (0.01 M) | Fe is
amount of NaOH is added to the solution.
(a) 0.26 V
(b) 0.399 V
(Take PH 2 = 1 atm, T = 298 K)
(c) –0.399 V
(d) – 0.26 V
UPTU/UPSEE-2015
(a) increase by 0.41 V (b) increase by 59 mV
Ans. (a) : Half cell reactions are:
(c) decrease by 0.41 V (d) decrease by 59m V
[AIIMS-2015]
Anode : [Cr→Cr3+ + 3e−] × 2
Cathode : [Fe2+ + 2e– → Fe] × 3
Ans. (c) : The pH changes from 0 to 7
Over all reaction : 2Cr + 3F2+ → 2Cr3+ + 3Fe
∴ [H+] changes from 1 to 10–7 M.
o
Applying Nernst equationE cell = Oxidation potential + Reduction potential
0.059
= 0.75 + (−0.45) = 0.30
E cell = E ocell +
log[H + ]
nF
0.0591 [Cr 3+ ]2
E cell = E ocell −
log
0.059
n
[Fe 2+ ]3
E cell = E ocell +
log(10−7 )
1
2
0.0591
(0.1)
= 0.30 −
log
E cell = E ocell − 0.41
6
(0.01)3
So, the potential decreases by 0.41V.
0.24
128. The standard reduction potentials for the
= 0.30 −
= 0.26 V
6
following reactions are
o
Fe3+ + 3e–→ Fe with Eo = – 0.036 V
125. The E M 3+ / M 2+ values for Cr, Mn, Fe and Co are
Fe2+ + 2e– → Fe with Eo = – 0.44 V
–0.41, +1.57, + 0.77, and +1.97 V, respectively.
What would be one standard electrode
For which one of these metals the change in
potential for the reaction
oxidation state from +2 to + 3 is easiest?
Fe3++ e– →Fe2+ ?
(a) Cr
(b) Mn
(a) 0.772V
(b) 0.077 V
(c) Fe
(d) Co
(c) –0.404 V
(d) –0.772V
UPTU/UPSEE-2015
°
SCRA-2015
Ans. (a) : E Cr3+ / Cr 2+ = −0.41V
Ans. (a): Given that,
E °Fe3+ / Fe2+ = +0.77V
Fe+3 + 3e–→ Fe, Eo = – 0.036V
………(i)
o
o
°
∵
∆G
=
–
nFE
E Mn3+ / Mn 2+ = +1.57V
∴
∆G1 = –3×F×(–0.036)
E °Co3+ / Co2+ = +1.97V
∆G1 = 0.108 F
Negative value of E °Red indicates oxidation of Cr2+ to
2+
Fe + 2e– → Fe, Eo = – 0.44V
………(ii)
Cr3+ is easiest.
Reversing equation (ii)
126. At temperature of 298K, the emf of the
Fe → Fe2+ + 2e–, Eo = – 0.44V ………(iii)
following electrochemical cell,
∴
∆G 2 = –2 × F × – 0.44
Ag(s)| Ag+(0.1 M) || Zn2+ Zn(s)(0.1 M)
∆G 2 = 0.88 F
Will be (Given, Eocell = –1.562V)
Objective Chemistry Volume-II
165
YCT
Subtracting equation (iii) and (i), we get
Fe3++ e– → Fe2+ ;
∴
∆Go = ∆G1 + ∆G 2
– n FEo = 0.108 F – 0.88 F
– 1× FEo = – 0.772 F
or
Eo = 0.772 volt.
129. What pressure of H2 would be required to
make emf of the hydrogen electrode zero in
pure water at 25oC?
(a) 10−7 atm
(b) 10–14 atm
(c) 1 atm
(d) 0.5 atm
AMU-2015
Ans. (b) : For water at 298K, [H+] = 10–7 M
1
H++ e–
→ H2 ,
2
0.0591
P(H )
E = Eo −
log + 22
n
[H ]
Relation between standard free energy and electrode
potential.
–∆Go = –nFEo
∴
∆Go ∝ Eo
∆G1o + ∆G o2 = ∆G
–5 × F×1.51–2 × F × (–1.23) = –3 × F × Eo
2 × 1.23 − 5 × 1.51
Eo =
−3
−5.09
Eo =
−3
E o = 1.70 V
132. Standard electrode potential data are useful for
understanding the suitability of an oxidant in a
redox titration. Some half cell reactions and
their
standard
potentials
are
given
below.
MnO 4– (aq) + 8H + (aq) + 5e – → Mn 2+ (aq) + 4H 2O(l);Eo
P(H 2 )
=0 is
[H + ]2
∵ For hydrogen electrode Eo=0
For E° to be equal to zero, log
= 1.51V
Cr2O 72– (aq) + 14H + + 6e – → 2Cr 3+ (aq) + 7H 2O(l);
P ( H 2 )
= 1 Or
[H + ]2
2
P ( H 2 ) = H +
Eo = 1.38V
Fe3+ (aq) + e– → Fe2+ (aq); Eo = 0.77 V
Cl2(g)+2e– → 2Cl– (aq); Eo = 1.40 V
Identify the incorrect statement regarding the
quantitative estimation of gaseous Fe (NO3)2
(a) MnO −4 can be used in aqueous HCl
P ( H 2 ) =10–14 atm
130. The standard electrode potentials of Zn and Ni
respectively are – 0.76 V and –0.25 V. The
(b)
standard emf of the spontaneous cell by
coupling these under standard conditions is
(c)
(a) +1.01 V
(b) – 0.51 V
(d)
(c) + 0.82 V
(d) + 0.25 V
(e) +0.51 V
Kerala-CEE-2015
Ans. (a) :
2+
2+
Ans. (e) : Zn |Zn || Ni | Ni
−
MnO −4 can be used in aqueous H2SO4
Cr2 O72− can be used in aqueous H2SO4
UPTU/UPSEE-2014
−
4
−
MnO can also oxidise Cl to Cl2
2MnO 4 (aq) + 16H + + 10Cl− → 2Mn 2+ (aq) + 8H 2 O(l) + 5Cl 2 (g)
E °cell = E °cathode − E °anode
= − 0.25 − (−0.76)
= − 0.25 + 0.76
= 0.51 V
131.
Cr2 O72− can be used in aqueous HCl
The cell corresponding to this reaction is as follows,
Pt,Cl2 (1atm) Cl− MnO 4− , Mn 2+ , H + Pt
E °cell = E °R − E °O
= E °Mn − E °Cl
=1.51 −1.40 = 0.11V
MnO 4– + 8H + + 5e - → Mn 2+ + 4H 2 O,E o = 1.51V
MnO 2 + 4H + + 2e - → Mn 2+ + 2H 2 O, Eo = 1.23V
As E ocell is positive, the above reaction is feasible. Thus,
Eo MnO- /MnO is
MnO −4 will not oxidize Fe2+ ion but also Cl– ion
(b) 0.91 V
simultaneously.
(d) 0.548V
UP CPMT-2014 133. The
two
half-cell
reactions
of
an
electrochemical
cell
is
given
as
Ans. (a) :
Ag+ + e− → Ag; EoAg+/Ag = – 0.3995V
MnO −4 + 8H + + 5e −
→ Mn 2+ + 4H 2 O; E o = 1.51V ....(i)
4
2
(a) 1.70V
(c) 1.37 V
MnO 2− + 4H + + 2e −
→ Mn 2+ + 2H 2 O; E o = 1.23V ....(ii)
nd
Inverting 2 reaction,
Mn 2+ + 2H 2 O
→ MnO −2 + 4H + + 2e− ; E o = −1.23V ...(iii)
Adding equation (i) and (iii)
MnO +4 + 3e − + 4H +
→ MnO 2 + 2H 2O; E o = ?
Objective Chemistry Volume-II
166
Fe2+→Fe3+ + e– ; E Fe3+/Fe2+ = – 0.7120V
The value of cell EMF will be
(a) –0.3125V
(b) 0.3125 V
(c) 1.114V
(d) –1.114V
WB-JEE-2014
YCT
Ans. (b) : Ag + + e− → Ag : E °Ag+ / Ag = −0.3995V
Ans. (a) : ∆G ° = −nFE ° for both half cell reactions.
Mn 2+ + 2e−
→ Mn, E ° = −1.18
Fe 2+ → Fe3+ + e− : E °Fe3+ / Fe2+ = −0.7120V
So, ∆G = −2F(−1.18)
We know that,
E °cell = E °Ag+ / Ag − E °Fe3+ / Fe2+
Hence, answer for (i) is 2.36 F
= −0.3995 − (−0.7120)
= 0.3125 V
134. Assertion: During electrolysis of CuSO4(aq)
using copper electrodes, copper is dissolved at
anode and deposited at cathode.
Reason: Oxidation takes place at anode and
reduction at cathode.
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is in
correct.
(d) If both the Assertion and Reason are
incorrect.
[AIIMS-2014]
Ans. (a) :
At Cathode
Cu2+ (aq) + 2e– → Cu (s) (Reduction)
At anode
Cu (s) → Cu2+ (aq) + 2e– (Oxidation)
So, it is clear that mass of Cu deposited on cathode =
mass of Cu decreased at anode.
135. A 1.0 M with respect to each of the metal
halides AX3 BX2, CX3 and DX2 is electrolysed
using platinum electrodes. If
E 0A3+ / A = 1.50V, E 0B2+ / B = 0.3V,
E 0C3+ / C = −0.74V, E 0D2+ / D = −2.37V
The correct sequence in which the various
metals are deposited at the cathode is
(a) A,B,C,D
(b) A,B,C
(c) D,C,B,A
(d) C,B,A
[AIIMS-2014]
Ans. (b) : The more the reduction potential the more is
the deposition of metals at cathode. Cation having E°
value less than –0.83V (reduction potential of H2O).
The cations will be liberated in the sequence of
decreasing reduction potentials. Cations having E°
value less than – 0.83V will not be liberated from
aqueous solution.
136. Given below are the half-cell
reactions
Mn2+ + 2e– → Mn, Eº = – 1.18V
2(Mn3+ + e– → Mn2+), Eº = + 1.51 V
The Eº for 3Mn2+ → Mn + 2Mn+3 will be
(a) –2.69 V, the reaction will not occur
(b) –2.69 V the reaction will occur
(c) –0.33V, the reaction will not occur
(d) –0.33V, the reaction will occur
[JEE Main 2014]
Objective Chemistry Volume-II
…..(i)
°
1
→ Mn 2+ ) E ° = +1.51V
2( Mn 3+ + e−
…..(ii)
°
2
So, ∆G = −F(−1.51)
Answer for (ii) is – 1.51 F
Now, equation (ii) – equation (i)
So, that would mean
3Mn 2+
→ Mn + 2Mn 3+
∆G 3o = ∆G1o − ∆G 2o
……(iii)
o
–nFE = (2.36+3.02) F
–2 FEo= 5.38F
E = –2.69
As Eo < 0
Reaction less not take place.
137. Zn | Zn2+ (a = 0.1 M) || Fe2+ (a = 0.01 M) | Fe.
The emf of the above cell is 0.2905 V.
Equilibrium constant for the cell reaction is
(a) 100.32/0.0591
(b) 100.32/0.0295
0.26/0.0295
(c) 10
(d) 100.32/0.295
JCECE - 2014
2+
Ans. (b) : For cell Zn|Zn (a = 0.1M)|| Fe2+ (a =
0.01M)|Fe
The half-cell reactions are
(i) Zn(s)
→ Zn2+(aq) + 2e–
(ii) Fe2+ (aq) + 2e–
→ Fe(s)
Zn(s) + Fe2+ (aq)
→ Zn 2+ ( aq ) + Fe ( s )
On applying Nernst equation, n = 2
º
Ecell = Ecell
−
0.0591
log10
n
Zn2+
Fe2+
0.0591
0.1
log10
2
0.01
0.2905 = Eºcell − 0.0295 × log10 10
º
0.2905 = Ecell
−
º
0.2905 = Ecell
− 0.0295 × 1
∴
Eºcell = 0.2905 + 0.0295 = 0.32 V
At equilibrium, (Ecell = 0)
0.0591
º
Ecell = Ecell
−
log10 Kc
n
0.0591
∴
0 = E0cell −
log10 Kc
n
0.0591
º
or
Ecell
=
log10 Kc
2
0.0591
0.32 =
log10 Kc
2
or
Kc = 100.32/0.02955
167
YCT
138. For hydrogen-oxygen fuel cell at 1 atm and 298
K
1
H 2 (g) + O 2 (g) → H 2 O(l ); ∆G o = −240 kJ
2
Eo for the cell is approximately, (Given F =
96500 C)
(a) 2.48 V
(b) 1.24 V
(c) 2.5 V
(d) 1.26 V
Karnataka-CET-2014
Ans. (b) : From, ∆G° = –nFE°
Where, n = number of moles of electrons transferred = 2
F = 96500 C
∆G° = –240 kJ = –240 × 1000J
−240000
E° =
= 1.243 V
−2 × 96500
139. The change in potential of the half-cell
Cu 2+ Cu, when aqueous Cu2+ solution is diluted
141. In the cell reaction
→ Cu2+ (aq) + 2Ag(s),
Cu(s) + 2Ag+ (aq)
Eocell = 0.46 V. By doubling the concentration of
Cu2+, Eocell is
(a) Doubled
(b) Halved
(c) Increased but less than double
(d) Decreases by a small fraction
[BITSAT – 2014]
Ans. (d) : Applying Nernst equation,
2+
RT Cu
°
E cell = E cell −
ln
nF Ag + 2
Doubling (Cu2+) decreases the emf by a Small fraction.
142. Cu + (aq) is unstable in solution and undergoes
simultaneous
oxidation
and
reduction
according to the reaction:
2Cu+(aq) ⇌ Cu2+ (aq) + Cu(s)
Choose correct Eo for above reaction if
ΕoCu2+ /Cu = 0.34 V and ΕoCu2+ /Cu+ = 0.15 V
2.303RT
100 times at 298 K?
= 0.06
F
(a) increases by 120 mV
(b) decreases by 120 mV
(a) – 0.38 V
(b) + 0.49 V
(c) increases by 60 mV
(c) + 0.38 V
(d) –0.19 V
(d) decreases by 60 mV
[BITSAT – 2014]
(e) no change
Ans. (c) : Given data,
Kerala-CEE-2014
E °cu 2+ / cu = 0.34V, E °cu 2+ / cu + = 0.15V
Ans. (d) :
We know that, ∆Go = – nEoF
∆E = EInitial – EFinal
2Cu + (aq) ↽ ⇀ Cu 2+ (aq) + Cu(s)
0.06
1 °
0.06
100
∆E = E °cell −
log
−
E
−
log
cell
2
[Cu 2+ ]
2
[Cu 2+ ]
Cu(s)
→ Cu 2+ (aq)+2e– ,∆G1° = − 2×( − 0.34)×F........(i)
0.06
0.06
∆E =
log[100] =
×2
Cu 2+ (aq)+e–
→ Cu + (aq), ∆G °2 = − 1×(0.15)×F......(ii)
2
2
On addition eqn (i) and eqn (ii)
⇒ 0.06 V = 60 mV
Cu(s)
→ Cu + (aq)+e– , ∆G 3° = –1× E °3 × F
140. Match the following–
∆G °3 = ∆G1° + ∆G °2
Column I
Column II
A
Potential of hydrogen
1 0.76 V
–1× E °3 × F = (–2 × (–0.34) × F) + (–1 × 0.15 × F)
electrode at pH = 10
–E °3 × F = 0.68F – 0.15F
B
2 0.059 V
Cu 2+ Cu
E °3 = –0.53V
C
3 –0.591 V
Zn Zn 2 +
Reaction 2Cu + (aq) ↽ ⇀ Cu 2+ (aq) + Cu(s) , E° = ?
D
4
0.337
V
2.303RT
So, Cu+(aq) + e– ↽ ⇀ Cu(s), E° = 0.53V
F
5 –0.76 V
Cu+(aq) ↽ ⇀ Cu2+(aq) + e–, E° = –0.15V
A
B
C
D
2Cu+(aq) ↽ ⇀ Cu2+(aq) + Cu(s), E° = +0.38V
(a)
3
1
2
5
143. Which of the following will give H2(g) at cathode
(b)
2
5
1
4
and O2(g) at anode on electrolysis using
(c)
3
4
1
2
platinum electrodes?
(d)
5
1
4
2
(a) Molten NaCl
BCECE-2014
(b)
Concentrated aq. Solution of NaCl
Ans. (c) : Potential of Hydrogen electrode at pH = 10 is
(c)
Dilute aq. Solution of NaCl
–0.591V
2+
(d)
Solid
NaCl
Reduction potential of Cu |Cu is 0.36V
2+
GUJCET-2014
Oxidation potential of Zn|Zn is 0.76 V
+
–
Ans.
(c)
:
NaCl
→
Na
+
Cl
2.303RT 2.303 × 8.314 × 298
∴ Valueof
=
= 0.059
H 2 O → H + + OH−
F
96500
Objective Chemistry Volume-II
168
YCT
147. The main function of the salt bridge is
(a) to allow ions to go from one half-cell to
another
(b) to provide link between two half-cells
(c) to keep the solution electrically neutral in two
half-cells
(d) None of the above
Assam CEE-2014
Ans. (c) : The main function of the salt bridge is to keep
the solution electrically neutral in two half-cells.
148. A hydrogen gas electrode is made by dipping
platinum wire in a solution of HCl of pH = 10
and by passing hydrogen gas around the
platinum wire at one atm pressure. The
oxidation potential of electrode would be
(a) 0.118 V
(b) 1.18 V
(c) 0.059 V
(d) 0.59 V
(NEET- 2013)
Ans. (d) : Given, pH=10
∵
pH = –log[H+]
At Anode.
2Cl − → Cl 2 + 2e − ;E ooxd = 1.36V
4OH − → O 2 + 2H 2 O + 4e − ;E ooxd = 1.23V
At Cathode
Na + + e − → Na;E ored = −2.71V
o
2H + + 2e − → H 2 ;E red
= −0.83V
144. In electrolytic refining of copper, the electrodes
used are
(a) Anode is impure copper plate and cathode is
pure nickel plate
(b) Anode is pure copper plate and cathode is
impure copper plate
(c) Anode is impure copper plate and cathode is
pure graphite plate
(d) Anode is impure copper plate and cathode is
pure copper plate
SCRA-2014
Ans. (d) : In electrolytic refining of copper, the
∴
electrodes used are–
For
(i) Anode is impure copper plate.
(ii) Cathode is pure copper plate.
145. The
standard
emf
of
the
cell
[H+] = 10– pH = 10–10M
2H + + 2e−
→ H2
E Reduction = E o −
Zn Zn 2+ Ag + Ag is 1.56 V. If the standard
reduction potential of Ag is 0.8 V, the standard
oxidation potential of Zn is
(a) –0.76 V
(b) +0.76 V
(c) –2.36 V
(d) +2.36 V
COMEDK 2014
Ans. (b) : Given the cell reaction-
= 0−
PH2
0.0591
log
2
2
H+
o
= 0)
(∵ ESHE
0.0591
1
log
−10 2
2
(10 )
0.0591
log1020
2
0.0591
=
× 20log10
2
EReduction = −0.591V
Zn Zn 2+ Ag + Ag
EOxidation = + 0.59V
E °cell = 1.56 V
149. Consider the helf-cell reduction reaction
Mn2+ + 2e–→ Mn, E0 = - 1.18 V
E °anode = ?
Mn2+→ Mn3+ + e–, Eo = – 1.51 V
°
E cathode = 0.8V
The Eo for the reaction,
o
o
3Mn2+ → Mn0 + 2Mn3+,
E ocell = E cathode
– E anode
and possibility of the forward reaction are
1.56 = 0.8 − ( –E°anode ) ∴ E°anode = –0.8 + 1.56 = + 0.76 V
respectively
(a) – 4.18 Vand yes
(b) + 0.33 V and yes
146. Given the reduction potentials of Na+, Mg2+,
3+
+
(c)
+
2.69
V
and
no
(d) – 2.69 V and no
Al and Ag as
(Karnatak-NEET- 2013)
o
o
E Na+/Na = −2.71V
E Mg2+/Mg = − 2.37V
Ans. (d) : Mn 2+ + 2e− → Mn E ° = −1.18V ____(i)
o
o
E Al3+/Al = − 1.66V
E Ag+/Ag = 0.08V
2Mn 2+ → 2Mn 3+ + 2e− E ° = −1.51V ____(ii)
The least stable oxide is
For the cell, adding the equation (i) and (ii)
(a) Ag2O
(b) Al2O3
Then,
(c) MgO
(d) Na2O
3Mn 2+ → Mn + 2Mn 3+ E ° = −2.69V
AMU-2014 E° value is negative, since the process is nonAns. (a) : In electrochemical series, as the spontaneous.
electropositivity decreases from top to bottom, the 150. H2 gas is liberated at cathode and anode both by
thermal stability of the oxide also decreases from top to
electrolysis of the following solution except inbottom. The oxides of metals having positive reduction
(a) NaCl
(b) NaH
potentials are not stable towards heat. Hence, Ag2O is
(c) LiH
(d) HCOONa
the least stable oxide.
BCECE-2013
Objective Chemistry Volume-II
169
= 0−
YCT
Ans. (d) : Higher the SRP, better is oxdising agent
Among the given E °MnO / Mn 2+ is highest. Hence, MnO−4 is
4
the strongest oxidising agent.
153. An electrochemical cell has two half cell
reactions as
EoA2+ /A = 0.34V
A2+ + 2e– →Α ;
Ans. (a) : (a) NaCl ↽ ⇀ Na + + Cl−
At Cathode same as above
At anode
1
Cl−
→ Cl 2 + e−
2
+
⇀
(b) NaH ↽
Na + H−
X → X2+ + 2e– ;
Eox2+ /x = –2.37 V
The cell voltage will be
(a) 2.71 V
(b) 2.03 V
(c) –2.71 V
(d) –2.03 V
J & K CET-(2013)
Ans. (a) : E cell = E A2+ / A − E X2+ / X
At Cathode
Na + + e−
→ Na
1
Na + H 2 O
→ NaOH + H 2
2
At Anode
1
H + + e−
→ H 2 + e−
2
(c) LiH ↽ ⇀ Li+ + H−
E °cell = [0.34 − (−2.37)]V
At Cathode
Li+ + e−
→ Li
1
Li + H 2O
→ LiOH + H 2
2
At Anode
1
H−
→ H 2 + e−
2
(d) HCOONa ↽ ⇀ HCOO− + Na +
At Cathode same as above
At anode
1
HCOO−
→ H 2 + CO 2 + e−
2
151. A quantity of electrical charge brings about the
deposition of 4.5 g Al from Al3+ at the cathode
will also produce the following volume at (STP)
of H2(g) from H+ at the cathode–
(a) 44.8 L
(b) 22.4 L
(c) 11.2 L
(d) 5.6 L
BCECE-2013
+
−
→ H2
Ans. (d) : 2H + 2e
Al3+ + 3e−
→ Al
W(H2 ) Eq(H 2 ) 1
=
=
W(Al)
Eq(Al) 9
1
1
W(H2 ) = 4.5× = 0.5g = mol H2
9
4
= 5.6L H 2 at STP
152. Given, EoCr 3+ /Cr = −0.74V;
EoMnO- /Mn2+ = 1.51V
E °cell = 2.71V
154. Which of the following statement is not correct
about an inert electrode in a cell?
(a) It does not pariticipate in the cell reaction
(b) It provides surface either for oxidation or for
reduction reaction,
(c) It provides surface for conduction of electrons
(d) It provides surface for redox reaction
JIPMER-2013
Ans. (d) : Inert electrodes do not take part in cell
reaction. Inert electrode provides surface for oxidation
or reduction reaction but not for redox reaction.
155. The emf of a galvanic cell constituted with the
electrodes Zn2+|Zn (–0.76 V) and Fe2+| Fe (–0.41
V) is
(a) –0.35 V
(b) + 1.17 V
(c) + 0.35 V
(d) – 1.17 V
Karnataka-CET-2013
Ans. (c) : As Zn2+ ion has lower reduction potential
than Fe2+ ion.
∴ Oxidation takes place on zinc electrode,
Ecell = Eright – Eleft
Ecell = – 0.41 – (– 0.76) = + 0.35V
156. Consider the half-cell reduction reaction,
Mn 2+ + 2e – → Mn,E° = –1.18 V
Mn 2+ → Mn 3+ + e – ,E° = –1.51 V
The
E°
for
the
reaction
3Mn 2+ → Mn 0 + 2Mn 3+ , and possibility of the
forward reaction are respectively
(a) –4.18 V and yes
(b) +0.33 V and yes
(c) +2.69 V and no
(d) –2.69 V and no
Karnataka NEET 2013
Ans. (d) : Given that,
Mn 2+ + 2e – → Mn, E1o = –1.18 Volt
Mn 2+ → Mn 3+ + e – ,E o2 = –1.51 Volt
The E° for the reaction is3Mn 2+ → Mn + 2Mn +3
EoCl/Cl - = 1.36V
o
o
o
Based on the data given above, strongest E = E1 + E 2
oxidising agent will be
E o = −1.18 + (−1.51)
3+
(a) Cl
(b) Cr
E o = −2.69
(c) Mn2+
(d) MnO -4
Since, E°= –2.69 V is negative, so the process is
[JEE Main 2013] spontaneous.
4
E
o
Cr2 O72- /Cr 3+
= 1.33V;
Objective Chemistry Volume-II
170
YCT
157. Which one of the following has a potential
more than zero?
1
(a) Pt, H 2 (1atm) | HCl (2M)
2
1
(b) Pt, H 2 (1atm) | HCl (0.1 M)
2
1
(c) Pt, H 2 (1atm) | HCl (0.5 M)
2
1
(d) Pt, H 2 (1atm) | HCl (1M)
2
Karnataka-CET-2012
Ans. (a) : According to Nernst equation, electrode
potential is given by–
0.0591
[Product]
E cell = E ocell −
log
(n = 2 and
n
[Reactant]
E ocell = 0 )
1
(a) For Pt, H2 (1 atm)| HCl (2M)
2
0.0591
Ecell =
log 2
2
0.0591
=
× 0.3010 = 0.0089
2
1
(b) From Pt, H2 (1 atm)| HCl (0.1M)
2
0.0591
Ecell =
log(0.1)
2
0.0591
=
× (–1) = – 0.0295
2
1
(c) For Pt, H2 (1 atm)| HCl (0.5M)
2
0.0591
Ecell =
log(0.5)
2
0.0591
=
×(– 0.3010) = – 0.0089
2
1
(d) From Pt, H2 (1 atm)| HCl (1M)
2
0.0591
Ecell =
log1 = 0
2
From the given option (a) has potential more than zero.
158. Given that the standard reduction potentials
for M+/M and N+/N electrodes at 298 K are 0.52
V and 0.25 V respectively. Which of the
following is correct in respect of the following
electrochemical cell?
M|M+ || N+|N
(a) The overall cell reaction is a spontaneous
reaction
(b) The standard EMF of the cell is –0.27 V
(c) The standard EMF of the cell is 0.77 V
(d) The standard EMF of the cell is –0.77 V
(e) The standard EMF of the cell is 0.27 V
Kerala-CEE-2012
Objective Chemistry Volume-II
Ans. (b) :
E °cell = E °cathode − E °anode = E °right − E °left
E °cell = (0.25 − 0.52) V
= –0.27 V
159. For the disproportionation of copper
2Cu+ → Cu2+ + Cu, E0 is ( Given : E0 for Cu2+ /
Cu is 0.34 and E0 V for Cu2+ / Cu+ is 0.15 V)
(a) 0.49 V
(b) –0.19 V
(c) 0.38 V
(d) –0.38 V
(AIPMT -Mains- 2012)
Ans. (c) : For the reaction 2Cu +
→ Cu 2+ + Cu
+
The cathode is Cu /Cu and anode is. Cu+/Cu2+
Given, Cu 2+ + 2e
→ Cu; E1° = 0.34V _______(i)
Cu + + e
→ Cu; E °2 = 0.15V ________(ii)
Cu + + e
→ Cu; E °3 = ?_____________ (iii )
Now
∆G1° = −nFE1° = −2× 0.34× F
∆G °2 = −1× 0.15× F, ∆G 3° = −1× E 3° × F
Again ∆G1o = ∆G 2o + ∆G 3o
⇒
−0.68F = −0.15F − E °3 × F
E 3° = 0.68 − 0.15
= 0.53V
°
E °Cell = E Cathode
(Cu + / Cu) − E °anode (Cu 2+ / Cu)
= 0.53 − 0.15
= 0.38V
160. Standard reduction potentials of the half
reactions are given below:
F2(g)+2e- → 2F-(aq);E0 = +2.85V
Cl2(g)+2e- → 2Cl-(aq);E0 = +1.36V
Br2(l)+2e- → 2Br-(aq);E0 = +1.06V
I2(s)+2e- → 2I-(aq);E0 = +0.53V
The strongest oxidising and reducing agents
respectively are
(a) F2 and I–
(b) Br2 and Cl–
–
(c) Cl2 and Br
(d) Cl2 and I2
(AIPMT -Mains 2012)
Ans. (a) : Given,
As,
E oF / F− = +2.85V,
2
o
E Cl
2
/ Cl −
= +1.36V
E oBr / Br− = +1.06V, E oI / I− = + 0.53V
2
2
Electrochemical series:• Negative to positive → oxidizing power increase
• Positive to negative → reducing power increase
According to question, strongest oxidizing F2 and
strongest reducing I2.
161. In a cell that utilizes the reaction
Zn(s) + 2H+(aq)
→ Zn 2+ ( aq ) + H 2 ( g )
171
addition of H2SO4 to cathode compartment
will.
(a) lower the E and shift equilibrium to the left
(b) lower the E and shift the equilibrium to the
right
YCT
(c) increase the E and shift the equilibrium to the 164. The reduction potential of an electrode can be
increased by
right
(a) Increasing the area of electrode
(d) increase the E and shift the equilibrium to the
(b) Decreasing the temperature
left.
(c) Increasing the temperature
UPTU/UPSEE-2012
(d) Decreasing the concentration of metal ions.
+
2+
→ Zn (aq) + H 2 (g)
Ans. (c) : Zn(s) + 2H (aq)
COMEDK-2012
2+
The
reduction
potential
of
an
electrode can be
Ans.
(b)
:
0.0591
[Zn ]
E o = E ocell −
log + 2
increased by decreasing the temperature.
n
[H ]
For the reduction reaction,
Increasing H+ will shift the equilibrium towards right
M n + + ne –
→M
o
they also increase E ,
RT
[M]
E 2H+ / H , E o will also increase.
E= E° –
log n +
2
nF
[M ]
162. The emf (in V) of a Daniell cell containing So, the value of E will increase if [Mn+] is increased or
0.1 M ZnSO4 and 0.01 M CuSO 4 solutions
at decrease the temperature.
165. From the following data at 25oC
their respective electrodes is
Cr 3+ (aq ) + e − → Cr 2+ (aq) , E o = −0.424 V
Eo
= +0.34 V; Eo
= -0.76 V
(
Cu2+ / Cu
Zn2+ /Zn
(a) 1.10
(c) 1.13
(b) 1.16
(d) 1.07
AP EAMCET (Engg.) 2012
Ans. (d) : For reaction
CuSO 4 + Zn
→ ZnSO 4 + Cu
or
→ Zn 2+ + Cu
Cu 2+ + Zn
Given that,
Eo
= 0.34V
E°
= − 0.76V
Cu 2+ / Cu
Zn 2+ / Zn
∴ E o = Eo
Cu 2+ / Cu
)
− Eo
Cr 2+ (aq) + 2e − → Cr(s) , E o = −0.900 V
Find Eo at 25o C for the reaction,
Cr 3+ (aq) + 3e − → Cr(s)
(a) -0.741 V
(b) -1.324 V
(c) -0.476 V
(d) +0.741 V
AMU-2012
3+
−
2+
Ans. (a) : Cr( aq ) + e
→ Cr( aq ) ; − 1× F × ( −0.424 )
= 0.424 F ––––––– (i)
Cr(2aq+ ) + 2e −
→ Cr( s) ; −2 × F ( − 0.900 )
=1.8 F ––––––––(ii)
Adding eq (i) and (ii),
Cr 3+ + 3e −
→ Cr(s);
Zn +2 / Zn
= +0.34 − ( −0.76 )
= 1.1 Volt
0.059
[Product]
log
n
[Reactant]
0.059
[0.1]
E cell = 1.1 −
log
2
[0.01]
E cell = 1.1 − 0.0295 × log[10]
Ecell= 1.1 – 0.0295
E cell = 1.0705 V ≃ 1.07V
Thus, E cell = E o cell −
0.424F + 1.8F
E° = −
= −0.741V
3F
166. 1 M solution each of Cu(NO3)2,AgNO3,
Hg2(NO3)2 and Mg (NO3)2 is electrolysed using
Pt-electrodes. The values of standard electrode
potentials in volts are
Ag+/Ag=+0.80V,Cu2+/Cu=0.34 V
Hg 22+ /2Hg=+0.79V,Mg2+/Mg= −2.37 V
The sequence of deposition of metals on the
cathode will be
(a) Mg,Ag, Cu
(b) Mg,Cu,Ag
(c) Ag,Hg, Cu
(d) Cu,Hg,Ag
AMU-2012
Ans. (c) : According to electrochemical series, metals
having more standard electrode potential will deposit
rapidly. Therefore, sequence of deposition of metals on
the cathode will be
Ag > Hg > Cu
163. Zn(s) + Cl2 (1 atm) → Zn2+ + 2Cl–; E° = 2.0V.
To increase the e.m.f. of the above cell
(a) [Zn2+] should be increased
(b) [Zn2+] should be decreased
(c) [Cl–] should be increased
(d) partial pressure of Cl2 should be decreased.
( + 0.80V)
( +7.09V ) ( + 0.34V )
COMEDK 2012 167. The equilibrium constant of the reaction;
Ans. (b) :Given data,
Cu(s) + 2Ag + (aq)
→ Cu 2 + (aq) + 2Ag(s)
2+
−
o
Zn(s) + Cl2 (1 atm) → Zn + 2Cl , E = 2.0V
E = 0.46V at 298K
0.0592
°
2+
− 2
(a) 2.4 × 1010
(b) 2.0 × 1010
Ecell= E cell −
log[Zn ][Cl ]
10
2
(c) 4.0 × 10
(d) 4.0 × 1015
Therefore, [Zn2+] is decrease and increase the Ecell.
BCECE-2012
Objective Chemistry Volume-II
172
YCT
→ Cu 2+ ( aq ) + 2Ag ( s )
Ans. (d) : Cu ( s ) + 2Ag + ( aq )
(a)
(b)
(c)
(d)
E° = 0.46 V at 298k
0.059
log K c
2
0.059
0.46 =
log K c
2
log Kc = 15.59
Kc = antilog 15.59
Kc = 3.92×1015
= 4× 1015
168. In electrolysis of NaCl when Pt electrode is
taken then H2 is liberated at cathod while with
Hg cathode it forms sodium amalgam because
(a) Hg is more inert than Pt
(b) more voltage is required to reduce H+ at Hg
than at Pt
(c) Na is dissolved in Hg while it does not
dissolved in Pt
(d) concentration of H+ ions is larger when Pt
electrode is taken
BCECE-2012
Ans. (b) : In electrolysis of NaCl when Pt electrode is
taken then H2 liberated at cathode while with Hg
cathode it forms sodium amalgam because more voltage
is required to reduce H+ at Hg than Pt.
169. The standard reduction potential Eº for halfreactions are
Zn → Zn2+ + 2e–; Eo = +0.76 V
Fe → Fe2+ + 2e–; Eo = +0. 41 V
The EMF of the cell reaction
Fe 2+ + Zn → Zn 2+ + Fe is
(a) + 1.28V
(b) − 1.28 V
(c) + 0.35V
(d) − 0.35 V
CG PET- 2012
Ans. (d) : These are oxidation potentials. Reduction
potentials are equal and opposite. Fe forms cathode and
Zn forms anode.
o
E ocell = E o(red)c + E (oxid)
a
E° =
X2 is a stronger oxidizing agent than Y2
Y2 is a stronger oxidizing agent than X2
X Θ is a stronger reducing agent than Y Θ ion.
The reducing strength of X Θ and Y Θ ions is
the same.
SCRA 2012
Ans. (a) : E oX2 = 2.87V
E oY2 = 1.36V
• Y2 is a stronger reducing agent than X2
• X2 is a stronger oxidizing agent than Y2
• Oxidation = electron loss, Reduction = electron gain.
172. In a galvanic cell the following reaction takes
place at 298K
Cr2O 72− + 14H + + 6Fe 2 + → 2Cr 3 + + 6Fe 3+ + 7H 2O
(
)
Given that : Eº Cr2O 72- ,H + ,Cr 3+ /Pt = 1.33V
(
3+
)
Eº Fe ,Fe /Pt = 0.77V
The standard e.m.f. of the cell is
(a) (1.33 + 0.77) V
(b) (1.33 – 0.77) V
(c) –(1.33 + 0.77) V
(d) (–1.33 + 0.77) V
MPPET - 2012
Ans. (b) : For the reaction–
Cr2 O72− + 14H + + 6Fe2+
→ 2Cr 3+ + 6Fe3+ + 7H 2 O
We can write half cell reaction–
2+
2x + 7(–2) = –2
x = +6
And
∴
x = +3
E °cell = E °cathode − E °anode
E °cell = (1.33 – 0.77)V
173. The standard emf of a galvanic cell involving 2
moles of electrons in its redox reaction is 0.59
V. The equilibrium constant for the redox
E ocell = 0.41 – 0.76 = – 0.35V
reaction of the cell is
(a) 1020
(b) 105
170. Standard electrode potential of half cell
(c) 10
(d) 1010
reactions are given below:
Karnataka-CET-2011
Cu 2+ + 2e – → Cu;Eo = 0.34 V
Ans. (a) : Given, E = 0.59, n = 2
Zn 2+ + 2e – → Zn;Eo = – 0.76 V
In a galvanic cell, ∆Go = – nFE ocell
… (i)
What is the EMF of the cell?
Relationship between ∆Go and equilibrium constant
(a) + 1.10 V
(b) – 1.10 V
∆Go = – 2.303RT logKp
… (ii)
(c) – 0.42 V
(d) + 0.42 V
From equation (i) and (ii), We get
J & K CET-(2012)
– 2.303 RT log Kp = – nFEo
Ans. (a): For Half cell reaction
nFE o
nE o
=
log K p =
E cell = E cathode − E anode
2.303 RT 2.303RT
E cell = 0.34 − (−0.76)V
F
2 × 0.59
E cell = + 1.10 V
=
= 20
0.059
171. The standard electrode potential for half-cell
Kp = Antilog (20)
reduction of X2(g) and Y2 (g) are 2.87 and 1.36
Kp = 1020
Eo/V. Which one of the following is correct?
Objective Chemistry Volume-II
173
YCT
174. The standard redox potentials for the reactions Ans. (c) : The reaction will take place for which E ° is
Cell
Mn2+ + 2e–→ Mn and Mn3+ + e–→ Mn2+ are –
positive.
1.18 V and 1.51 V respectively. What is the
I 2 + 2e−
→ 2I− ,E °Red = 0.536V
redox potential for the reaction Mn3+ + 3e–→
Mn?
Fe 2+
→ Fe3+ + e− , E°ox = −0.770V
(a) 0.33 V
(b) 1.69 V
Redox reaction will occur
(c) – 0.28 V
(d) – 0.85 V
2I−
→ I2 + 2e− , E °ox = −0.536V
(e) 0.85
Fe3+ + e−
→ Fe 2+ , E °red = + 0770V
Kerala-CEE-2011
Ans. (c) :
E °Cell is + Ve, Hence there will be no redox reaction.
Mn 2+ + 2e− → Mn; ∆G° = − nFE° = 2.36F......(i)
178. Standard electrode potentials of three metals
X,Y and Z are –1.2V,+0.5 V and – 3.0 V
Mn 3+ + e− → Mn 2+ , ∆G o = −nFE° = −1.51F.........(ii)
respectively. The reducing power of these
3+
–
Add equation (i) and (ii), Mn + 3e →Mn
metals will be
∆G° = 2.36F + (−1.51F) = 0.85F = −nFE°
(a) Y > Z > X
(b) Y > X > Z
0.85F = – 3× F×E° ⇒ E° = −0.28 V
(c) Z > X > Y
(d) X >Y > Z
(AIPMT -2011)
175. The standard reduction potential for Fe2+/Fe
2+
and Sn /Sn electrodes are –0.44 and –0.14V Ans. (c) : As the value of standard reduction potential
respectively. For the cell reaction, Fe2+ + Sn → decreases, the reducing power increases.
Fe+Sn2+ the standard emf is
1
Reducing power ∝
(a) + 0.30 V
(b) – 0.58 V
Electrode potential
(c) + 0.58 V
(d) – 0.30 V
Therefore the order of reducing power is :MHT CET-2011
Z>X>Y
Ans. (d) : Given, E °Cathode = −0.44V E °anode = −0.14V
179. A Saturated solution of KNO3 is used to make
For the cell reaction Fe acts as cathode and Sn as
salt bridge because
anode
(a) the velocity of K+ is greater than that of NO3−
°
E °Cell = E °Cathode − E anode
(b) the velocity of is NO3− greater than that of K+
= − 0.44 − (−0.14)
(c) the velocity of both K+ and NO3− are nearly
= − 0.30V
same
176. The electrode potentials for,
(d)
KNO3 is soluble in water
Cu2+ (aq) + e– → Cu+(aq) + e– →Cu(s) are +0.15 V
UP CPMT-2011
and + 0.50 V respectively. The value of
O 2+
Ans.
(c)
:
Saturated
solution
of
KNO
is used to make
3
E Cu /Cu will be
salt bridge because velocity of both K+ and NO3− are
(a) 0.500 V
(b) 0.325 V
nearly same.
(c) 0.650 V
(d) 0.150 V
(AIPMT -2011) 180. Given the following in E.q. (i) and (ii), claculate
the EMF of the cell given in Eq. (iii)
Ans. (b) :
2+
−
+
°
°
°
CuI(s) + e–→Cu(s) + I–
Cu + e → Cu ; E1 = 0.15V, ∆G 2 = −n1FE1
E0 = –0.16
.......(i)
Cu + + e− → Cu; E °2 = 0.50V, ∆G °2 = −n 2 FE °2
2+
Zn (aq) +2e–→ Zn(s)
Cu 2+ + 2e− → Cu; ∆G ° = ∆G1° + ∆G °2
E0 = –0.76
.........(ii)
−nFE ° = −n1FE1° +−n 2 FE °2
Zn Zn 2+ (1.0M) I - (1.0M) CuI Cu
..........(iii)
−
O
⊕
n1E1° + n 2 E °2 0.15×1 + 0.50×1
°
E =
=
(a) 1.08V
(b) 0.44V
n
2
(c)
0.92V
(d) 0.60V
E o = 0.325V
UPTU/UPSEE-2011
177. A solution contains Fe2+, Fe3+ and I– ions. This Ans. (d) : We know that,
solution was treated with iodine at 350C.E0 for
−
−
°
Fe3+/Fe2+ is +0.77V and E0 for I2/2I– =0.536 V. CuI(s) + e → Cu(s) + I ; E = 0.16 ____(i)
The favourable redox reaction is
Zn 2+ (aq) + 2e− → Zn(s) ; E ° = 0.76 _____(ii)
(a) I2 will be reduced to I–
Substracted equation (ii) from (i)
(b) there will be no redox reaction
Then EMF of the cell = E °(Cathode) − E°(Anode)
(c) I– will be oxidised to I2
= 0.16 − (−0.76)
(d) Fe2+ will be oxidised to Fe3+
= 0.60V
(AIPMT -Mains - 2011)
Objective Chemistry Volume-II
174
YCT
(a)
(b)
(c)
(d)
181. Given E k + / k = –2.93V;
o
EoFe2+ /Fe = −0.44V;E Zn
2+/Zn = −0.76V;
EoCu2+ /Cu = 0.34V
For lead +4, for tin + 2
For lead +2, for tin + 2
For lead +4, for tin + 4
For lead +2, for tin + 4
[AIEEE 2011]
Based on this data, which of the following is the Ans. (d) :
PbO 2 + Pb
→ 2PbO
strongest reducing agent?
Oxidation state +4
0
+2
+
(a) Cu(s)
(b) K (aq
Since,
∆
G°
<
0,
hence
+2
state
of
lead
is favourable.
)
r
2+
SnO2 + Sn
→ 2SnO
(d) Fe(s)
(c) Zn (aq )
Oxidation state +4
0
+2
[AIIMS-2011] Since, ∆rG° > 0, it means forward reaction is not
spontaneous.
Ans. (b) :
2SnO
→ SnO2 + Sn
+2
+4
0
Since, ∆rGo > 0, it means forward reaction is not
spontaneous.
2SnO → Sn2 + Sn
So, K+ is the strongest reducing agent.
+2
+4
0
182. Which of the following reactions is correct for a
For this, ∆r G° < 0, thus + 4 state of tin is favourable
given electro chemical cell at 25ºC?
185. The reduction potential of hydrogen half-cell
Pt Br2 (g) Br – (aq) Cl – (aq) Cl 2 (g) Pt.
will be negative if
(a) PH2 = 1 atm and [H+] = 2.0 M
−
−
(a) 2Br (aq) + Cl2 (g) → 2Cl (aq) + Br2 (g)
(b) PH2 = 1 atm and [H+] = 1.0 M
(b) Br2 (g) + 2Cl− (aq) → 2Br − (aq) + Cl 2 (g)
(c) PH2 = 2 atm and [H+] = 1.0 M
(c) Br2 (g) + Cl2 (g) → 2Br − (aq) + 2Cl− (aq)
(d) PH2 = 2 atm and [H+] = 2.0 M
(d) 2Br − (aq) + 2Cl− (aq) → Br2 (g) + Cl 2 (g)
[AIEEE 2011]
GUJCET-2011
Ans. (c) : The reduction potential of hydrogen half cells
Ans. (a): The correct reaction of the electrochemical will be negative if : (c) P = 2 atm and [H+] = 1.0M
H2
cell at 25ºC.
+
−
2H + 2e
→ H2
2Br − (aq) + Cl2 (g) → 2Cl− (aq) + Br2 (g) .
PH2
0.0591
183. The emf of the cell involving the following
E red = E°red −
log
2
+
+
n
H+
reaction 2Ag +H2
→ 2Ag+2H is 0.80 volt.
The standard oxidation potential of silver
0.0591
2
E red = 0 −
log 2
electrode is
2
1
()
(a) –0.80 volt
(b) 0.80 volt
Ered will only be negative when PH 2 > [H+]
(d) –0.40 volt
AMU-2011 186. Which one the following will increase the
voltage of the cell? (T = 298 K)
Ans. (a) : The cell reaction for the given cell is
Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s)
(a) Increase in the size of silver rod
→ 2H + + 2e − ,
H 2
(b) Increasing the size of plate
Oxidation
(c) Increase in the concentration of Ag+ ions
Ag + + e−
→ Ag
(d) Increase in the concentration of Sn2+ ions
Reduction
JCECE - 2011
Given E.m.f of cell is .8 volt
2+
0.0591
[Sn ]
º
E °cell = E °cathode – E °anode
−
log
Ans. (c) : Ecell = Ecell
+ 2
2
[Ag
]
0.8 = 0 + E °cathode
+
Increase in the [Ag ] will decrease the logarithmic
Since, Ag + + e −
→ Ag is reduced at cathode the factor and hence, increase the Ecell.
187. The emf of a cell with 1 M solutions of
standard oxidation potential is – 0. 80 volt
reactants and products in solution measured at
184. In view of the signs of ∆rGº for the following
25°C is called
reactions
(a) half cell potential
PbO2 + Pb → 2PbO, ∆rGº < 0
(b) standard emf
(c) single electrode potential
SnO2 + Sn → 2SnO, ∆r Gº > 0
(d) redox potential.
Which oxidation states are more characteristic
SRMJEEE – 2011
for lead and tin?
(c) 0.40 volt
Objective Chemistry Volume-II
175
YCT
190. Consider the following relations for emf of an
electrochemical cell
(i) EMF of cell = (Oxidation potential of anode)
– (Reduction potential of cathode)
(ii) EMF of cell = ( Oxidation potential of anode)
+ (Reduction potential of cathode)
(iii) EMF of cell = (Reduction potential of anode)
+ (Reduction potential of cathode)
(iv) EMF of cell = (Oxidation potential of anode)
– (Oxidation potential of cathode)
Which of the above relations are correct?
(a) (iii) and (i)
(b) (i) and (ii)
(c) (iii) and (iv)
(d) (ii) and (iv)
(AIPMT -Mains - 2010)
Ans. (d) : The EMF of a cell is given by the following
expression.
EMF of a cell = Reduction potential of cathode –
reduction potential of anode = Reduction potential of
cathode + oxidation potential of anode.
Ans. (b) : Given – Concentration of reactant = 1M
Concentration of product = 1M
T = 25°C + 273 = 298K
E=?
Now, For general reaction –
Reactant
→ Product
This equation can be written as –
RT
[ Product ]
E = E° –
log
nF
[ Reactant ]
E = E° –
R × 298
1
log
nF
1
(∵ log1= 0)
E = E°
188. Match the following lists.
List I
(A) Potential of
hydrogen
electrode at pH =10
List II
(I) 0.76 V
(B) Cu 2+ Cu
(II) 0.059
(C) Zn Zn 2+
(III) –0.591 V
o
o
E ocell = E cathode
− E anode
(Reduction )
(D)
or E
o
cell
(IV) 0.337 V
2.303 RT
F
(V) – 0.76 V
A
B
C
D
(a) (III)
(I)
(II)
(V)
(b) (II)
(V)
(I)
(IV)
(c) (III)
(IV)
(I)
(II)
(d) (V)
(I)
(IV)
(II)
AP-EAMCET- (Engg.)-2011
Ans. (c) : List-I
List-II
(i) Potential of hydrogen
–0.591V
electrode at pH =10
(ii) Reduction potential
0.337 V
of Cu2+/Cu
(iii) Oxidation potential
0.76 V
of Zn/Zn2+
2.303 RT
(iv)
0.059
F
189. Given, for Sn4+ / Sn2+, standard reduction
potential is 0.15 V and for Au3+ /Au, standard
reduction potential is 1.5 V.
For the reaction,
3Sn 2+ + 2Au 3+
→ 3Sn 4+ + 2Au,
°
the value of Ecell
is,
(a) + 1.35
(c) – 1.35
or E
o
cell
(b) + 2.55
(d) – 2.55
MHT CET-2010
o
= 0.15V
Ans. (a) : Given, E oAu3+ / Au = 1.5V, ESn
4+
/ Sn 2+
Standard Potential ( E ocell ) ,
E ocell = E ocathode − E oanode
=1.5 − 0.15
= 1.35V
Objective Chemistry Volume-II
(Reduction )
=E
o
Oxidation
(anode)
+ E oReduction
=E
o
Oxidation
(anode)
− E oReduction
(cathode)
(cathode)
o
red
191. The standard E values of A, B, C are 0.68 V,
–2.54 V, – 0.50 V respectively, The order of
their reducing power is
(a) A > B > C
(b) B > C > A
(c) A > C > B
(d) C > B > A
UP CPMT-2010
Ans. (b): Higher the negative value of standard
reduction potential (E red), higher will be the reducing
power.
1
E oreduction ∝
Reducing power of element
The order of E red of A, B, C is B > C > A.
∴ The order of reduction power is B > C > A.
O
O
o
–
192. The standard electrode potential (E ) for OCl /
–
–
1
Cl 2 respectively are 0.94 V and
Cl and Cl /
2
o
–
1
–1.36 V. The E value for OCl / Cl 2 will be
2
(a) – 0.42 V
(b) – 2.20 V
(c) – 0.52 V
(d) 1.04 V
UPTU/UPSEE-2010
°
Ans. (a) : Electrode potential (E ) = 0.94V
−
−
OCl → Cl ,E ° = 0.94V
1
−
Cl → Cl 2 , E ° = −1.36V
2
Adding equation (i) & (ii) we get.
1
−
OCl → Cl 2
2
°
E = 0.94 −1.36
= −0.42V
176
....(i)
....(ii)
YCT
193. Consider the following Eo values
EFe3+ / Fe2+ = +0.77V
(a) 0.653 V
(c) 0.683 V
ESn2+ / Sn = −0.14V
(b) 0.889 V
(d) 2.771 V
J & K CET-(2010)
RT
[Fe 2+ ]
°
× 2.303log
Under standard conditions the potential for the Ans. (a) : E cell = E cell −
nF
[Fe3+ ]
reaction,
0.771− 8.314× 2.303× 298
2
Sn(s) + 2Fe 3 + (aq) → 2Fe2 + (aq) + Sn 2 + (aq) isE cell =
log
96500
0.02
(a) 1.68V
(b) 1.40V
= 0.771− (0.0591× 2)
(c) 0.91V
(d) 0.63V
= 0.771− 0.1182
BCECE-2010
°
°
°
E cell = 0.6528V
Ans. (c) : E = E
2+ + E
3+
2+
cell
Oxd (Sn / Sn
)
Red( Fe
/ Fe
)
197. The emf of the cell
= 0.14 + 0.77 = 0.91V
Ni | Ni2+ (1.0 M) || Au3+ (1.0 M) | Au
194. Copper sulphate solution is electrolysed using
o
o
is [ E(Ni
= – 0.25V and E(Au
= +1.5 V]
2+
3+
copper electrode. The reaction taking place at
/ Ni)
/Au)
anode is
(a) 2.00 V
(b) 1.25 V
(a) H + + e − → H
(c) –1.25 V
(d) 1.75 V
JCECE - 2010
(b) SO 24− ( aq ) → SO 4 + 2e −
2+
3+
Ans.
(d)
:
For
the
cell,
Ni
|
Ni
||
Au
|
Au
2+
−
(c) Cu + 2e → Cu
Given,
EºNi2+ / Ni = − 0.25 V
(d) Cu ( s ) → Cu 2+ ( aq ) + 2e −
EºAu3+ / Au = +1.5 V
CG PET- 2010
Ans. (d) : Given,
Copper sulfide solution is electrolysed using
copper electrodes.
2H2O → 4H+ + O2 + 4e–
(Oxidation)
Cu+2e– → Cu2+
(Reduction)
195. The Gibbs energy for the decomposition of
Al2O3 at 500ºC is as follows
2
4
Al 2 O 3 → Al + O 2 '
3
3
G
=
+
966
kJ mol–1
∆r
The potential difference needed for electrolytic
reduction of Al2O3 at 500ºC is at least
(a) 4.5V
(b) 3.0V
(c) 2.5V
(d) 5.0V
[AIEEE-2010]
2
4
Ans. (c) : Al 2 O3
→ Al + O 2
3
3
4
2
Al + O 2
→ Al 2 O3
3
3
∵ ∆r G = −nFE cell
Here, Ni is anode and Au is cathode.
∴
Ecell = Ec – Ea
= 1.5 – (–0.25)
= 1.5 + 0.25
= 1.75 V
198. How long it will take to deposit 1.0 g of
chromium when a current of 1.25 A flows
through a solution of chromium (III) sulphate?
(Molar mass of Cr = 52)
(a) 1.24 min
(b) 1.24 h
(c) 1.24 s
(d) None of these
JIPMER-2010
→ Cr(s)
Ans. (b): Cr 3+ +3e –
3 mol or 3 × 96500C of electricity are needed to deposit
1 mol or 52g of Cr.
52g of Cr require current = 3 × 96500C
3 × 96500
1g of Cr will require current =
C = 5567.3C
52
Now, number of coulombs = Current (ampere) × t
(seconds)
no.of coulombs
Time (s) required =
current (ampere)
5567.3
−966×1000 = −(4)(96500) E cell
=
1.25
966 (10)
E cell =
4453.8
= 4453.8s =
hr =1.24 hr
965( 4)
3600
E cell = 2.5V
199. The standard emf of galvanic cell involving 3
moles of electrons in its redox reaction is 0.59
196. Calculate the reduction potential of a half-cell
V. The equilibrium constant for the reaction of
containing of platinum electrode immersed in
the cell is
2.0 M Fe2+ and 0.02 MFe3+.
o
(a) 1025
(b) 1020
Given : EFe3+ /Fe2+ = 0.771 V.
15
(c) 10
(d) 1030
Fe3+ + e– → Fe2+
Karnataka CET- 2010
Objective Chemistry Volume-II
177
YCT
Ans. (d) : Given, Eo = 0.59 V, n = 3, Kc = ?
0.0591
E ocell =
log K c
n
0.0591
0.59 =
log K c
3
0.59 × 3
= log K c
0.059
∴
log Kc = 30
Kc = Antilog 30
= 1030
200. An acidified solution of 0.1M CuSO4 is
electrolysed using platinum electrodes. What is
the reaction that occurs at the anode?
(a) Cu → Cu+2 + 2e–
(b) Cu → Cu+ + e–
(c) 2H2O → +4H+ + O2↑+ 4e–
(d) None of the above
SCRA-2010
Ans. (c) : Reaction occur at anode and cathode are
given belowAt anode:- 2H2O → O2(g)+4H+ (aq) + 4e–
At cathode :- Cu2+ (aq) + 2e–→ Cu(s)
201. If the solution of copper sulphate in which a
copper rod is immersed, is diluted 100 times,
what is the change in electrode potential
(Reduction)?
(a) –29.5 mV
(b) 29.5 mV
(c) –59.0 mV
(d) 59.0 mV
AP- EAMCET(Medical) -2010
Ans. (d): According to Nernst equation–
0.059
1
E Cu 2+ / Cu = E oCu 2+ / Cu −
log n +
2
[M ]
0.059
1
E Cu 2+ / Cu = E oCu 2+ / Cu −
log
2
[0.01]
0.059
E Cu 2+ / Cu = E oCu 2+ / Cu −
log100
(∵ log100 = 2)
2
= E oCu 2+ / Cu − 0.059 V
Where,
E = Electrode potential
Eº = Standard electrode potential = – 0.25 V
n = No. of electrons transferred
Kc = equilibrium constant
2.303RT
[Ni]
E cell = E ocell −
log
nF
[Ni 2+ ]
0.06
1
log
(∵n = 2)
2
0.1
= − 0.25 − 0.03 × 1
Ecell = −0.25 −
E cell
E cell = −0.28V
203. Which of the following is not correct regarding
the electrolytic preparation of H2O2 ?
(a) Lead is used as cathode
(b) 50% H2SO4 is used
(c) Hydrogen is liberated at anode
(d) Sulphuric acid undergoes oxidation
JIPMER-2009
Ans. (c) : Hydrogen is discharged at cathode
204. An aqueous solution containing 6.5 g of NaCl
of 90% purity was subjected to electrolysis.
After the complete electrolysis, the solution was
evaporated to get solid NaOH. The volume of
1M acetic acid required to neutralize NaOH
obtained above is
(a) 1000 cm3
(b) 2000 cm3
3
(c) 100 cm
(d) 200 cm3
Karnataka CET-2009
Ans. (c) : Given that,
Weight of pure NaCl = 6.5 × 0.9 = 5.85 g
5.85
= 0.1
No. of equivalence of NaCl =
o
58.5
= E Cu 2+ / Cu − 59.0 mV
No. of equivalence of NaOH obtained = 0.1 × volume
Thus, electrode potential decreased by 59.0 mV
of 1M acetic acid required for the neutralisation of
202. What is the electrode potential (in V) of the
0.1× 1000
NaOH =
= 100 cm3
following electrode at 25o C ?
1
Ni2+(0.1M)|Ni(s) (Standard reaction potential of
205. The standard electrode potential for the half2.303 RT
2+
cell reactions are
Ni |Ni is –0.25V,
= 0.06)
F
Zn2+ + 2e– → Zn; Eo = – 0.76 V
(a) –0.28 V
(b) –0.34 V
Fe2+ + 2e– → Fe; Eo = – 0.44 V
(c) –0.82 V
(d) – 0.22 V
The emf of the cell reaction,
JIPMER-2009
Fe2+ + Zn → Zn2+ + Fe is
+2
–
Ans. (a) : Ni + 2e → Ni
(a) – 0.32 V
(b) – 1.20 V
For Nernst equation,
(c) +1.20 V
(d) + 0.32 V
2.303RT
0.059
Karnataka-CET, 2009
E = Eº −
log K c = Eº −
log K c
2+
–
o
F
n
Ans. (d) : Zn + 2e → Zn; E = – 0.76 V
Fe2+ + 2e– → Fe; Eo = – 0.44 V
Objective Chemistry Volume-II
178
YCT
Half cell reaction is
Fe2+ + Zn → Zn2+ + Fe
Ecell = Ecathode – Eanode
= – 0.44 – (– 0.76)
= –0.44 + 0.76
= 0.32 V
206. The standard reduction potential for Mg2+ /Mg
is –2.37 V and for Cu2+ /Cu is 0.337. The
°
Ecell
for the following reaction is
So, here order of reducing power is,
Mg > Zn > Fe
So, Mg can reduce both Zn2+ & Fe2+. Zn can only
reduce Fe2+ but not Mg2+. Similarly, Fe2+ can oxidise Zn
and Mg both.
209. The reaction that takes place at the anode
during electrolysis is
(a) reduction
(b) oxidation
(c) hydrolysis
(d) redox reaction.
SRMJEEE – 2009
Mg + Cu 2+
→ Mg 2+ + Cu
Ans. (b) : As we know the loss of electrons is known as
(a) + 2.03 V
(b) – 2.03 V
oxidation. The electrode where electrons are released or
(c) – 2.7 V
(d) + 2.7 V
the process of oxidation takes place is known as anode.
MHT CET-2009 210. During the charging of a lead-acid storage
°
battery, the cathode reaction is
= 0.337V ,
Ans. (d) : Given, E Cu
2+
/ Cu
(a) formation of PbSO4
E °Mg2+ / Mg = −2.37V
(b) reduction of Pb2+ to Pb
(c) formation of PbO2
Then, standard Potential ( E ocell )
(d) oxidation of Pb to Pb2+
E °cell = E °cathode − E °anode
AP - EAMCET(MEDICAL) - 2009
= E °Cu 2+ / Cu − E°Mg2+ / Mg
Ans. (c) : The following electrode reaction take place–
At cathode–
= 0.337 − (−2.37)V
PbSO 4 + 2H 2 O → PbO 2 + SO 42− + 2e −
= 2.7V
At anode–
207. Given:
2+
–
0
(i) Cu + 2e → Cu, E = 0.337 V
Pb 2+ + 2e − → Pb
2+
–
+ 0
(ii) Cu + e → Cu ,E = 0.153 V
211. If 0.1 M solutions of each electrolyte are taken
Electrode potential, E0 for the reaction
and if all electrolytes are completely
Cu+ + e– → Cu, will be
dissociated, then whose boiling point will be
highest?
(a) 0.90 V
(b) 0.30 V
(c) 0.38 V
(d) 0.52 V
(a) Glucose
(b) KCl
(AIPMT -2009)
(c) BaCl2
(d) K4[Fe(CN)6]
Ans. (d) : Given,
CG PET -2009
Cu 2+ + 2e−
→ Cu; E1° = 0.337V
Ans. (d) : Elevation in boiling point is a colligative
property i.e depends upon the number of particles.
Cu 2+ + e−
→ Cu + ; E °2 = 0.153V
Hence, the electrolytes which give largest number of
The required reaction is
particles
in the solution, has the highest boiling point.
Cu + + e−
→ Cu; E 3° = ?
Since K 4 Fe (CN )6 gives largest number of particles,
Applying, ∆G ° = −nFE ° , ∆G 3° = ∆G1° −∆G °2
hence it has the highest boiling point.
−(n 3 FE °3 ) = −(n1FE1° ) − (−n 2 FE °2 )
212. In Daniell cell, anode and cathode are
respectively
E °3 = 2× E1° − E °2
(a) Zn Zn 2 + and Cu 2 + Cu
E o = (2 × 0.337) − (0.153)
3
(b) Cu Cu 2 + and Zn 2 + Zn
= 0.52V
Eo values for Mg 2+ /Mg = −2.37 V, Zn 2+ / Zn =
– 0.76 V and Fe2+/Fe = – 0.44 V. Which
statement is correct?
(a) Zn reduces Fe2+
(b) Zn reduces Mg2+
(c) Mg oxidises Fe
(d) Zn oxidises Fe
UP CPMT-2009
Ans. (a): Given,
E oMg2+ / Mg = −2.73V, E oZn 2+ / Zn = −0.76V,E oFe2+ / Fe = −0.44 V
208.
Reducing power ∝
Objective Chemistry Volume-II
1
E oreduction
(c) Fe Fe 2 + and Cu 2 + Cu
(d) Cu Cu 2 + and Fe 2 + Fe
CG PET -2009
Ans. (a) : In Daniell cell, oxidation occurs at anode and
reduction occurs at cathode thus, the element, which has
higher negative value of reduction potential, is used as
anode and that with lower negative value of reduction
potential is used as cathode.
Reduction potential of Zn is more negative than Cu.
Zn | Zn2+ is anode and Cu2+ | Cu is cathode.
179
YCT
213. The standard reduction electrode potentials of
the three electrodes P, Q
and R are
respectively –1.76 V, 0.34 V and 0.8 V. Then
(a) metal Q will displace the cation of P from its
aqueous solution and deposit the metal P
(b) both metals Q and R will displace the cation
of P from its aqueous solution and deposit the
metal P
(c) metal R will displace the cation of P from its
aqueous solution and deposit the metal P
(d) metal P will displace the cation of R from its
aqueous solution and deposit the metal R.
J & K CET-(2009)
Ans. (d) : The metal having low standard reduction
electrode potential displace the metal from its salt's
solution which has higher value of standard reduction
potential.
214. Which of the following electrolytes will have
maximum flocculation value for Fe(OH)3 sol ?
(a) NaCl
(b) Na2S
(c) (NH4)3PO4
(d) K2SO4
JCECE - 2009
Ans. (a) : Fe(OH)3 is a positively charge solution.
Comparing the charge on negative ions, Cl– has least
charge and hence least coagulating power and
maximum coagulation values.
1
Flocculation value ∝
Coagulating power
Among the given electrolytes, NaCl has lowest
coagulation power, so its flocculation value will be
maximum.
215. Kohlrausch's law states that at
(a) Infinite dilution, each ion makes definite
contribution to conductance of an electrolyte
whatever be the nature of the other ion of the
electrolyte
(b) Infinite dilution, each ion makes definite
contribution to equivalent conductance of an
electrolyte, whatever be the nature of the
other ion of the electrolyte
(c) Finite dilution, each ion makes definite
contribution to equivalent conductance of an
electrolyte, whatever be the nature of the
other ion of the electrolyte
(d) Infinite dilution each ion makes definite
contribution to equivalent conductance of an
electrolyte depending on the nature of the
other ion of the electrolyte
(AIPMT -2008)
Ans. (b) : At infinite dilution, when the dissociation of
electrolyte is complete, each ion makes a definite
contribution towards the molar conductivity irrespective
of the nature of the other ion with which it is associated.
According to Kohlrausch’s law the molar conductivity
of an electrolyte at infinite dilution can be expressed as
the sum of the contributions from its individual ions.
λ∞ = λa + λc
Where, λ a and λ c are know ionic conductance of anion
and cation at infinite dilution respectively.
Objective Chemistry Volume-II
216. Standard free energies or formation (in kJ/
mol) at 298 K are – 237.2, – 394.4 and –8.2 for
H2O(l), CO2(g) and pentane(g) respectively.
The value of E0Cell for the pentane-oxygen fuel
cell is
(a) 1.0968 V
(b) 0.0968 V
(c) 1.968 V
(d) 2.0968 V
(AIPMT -2008)
Ans. (a) : C5 H12 + 8O 2
→ 5CO 2 + 6H 2 O, n = 32
∆G ° = [5∆F G ° (O 2 ) + 6∆FG ° (H 2 O)] −
[∆F G ° (C5 H12 ) + 8∆FG ° (O 2 )]
= [5(−394.5) + 6(−237.2)] − [−8.2) + 0]
= − 3387.5 kJ Mol−1
∆FG ° = −n × F× E °Cell
∴
−3387500 = −32×96500× E °Cell
E °Cell = 1.0968 V
217. What is the electrode potential (in V) of the
following electrode at 25oC?
Ni2+ (0.1 M)|Ni(s) (Standard reduction potential
2.303 RT
of Ni2+|Ni is –0.25V,
= 0.06 )
F
(a) –0.28 V
(b) –0.34 V
(c) –0.82 V
(d) –0.22 V
UP CPMT-2008
2+
–
Ans. (a): Ni + 2e → Ni
From Nernst equation,
[ Ni ]
2.303RT
log
nF
Ni 2 +
0.06
1
= −0.25 −
log
(∵n = 2)
2
0.1
= –0.25 – 0.03 × 1
E cell = – 0.28V
E cell = E ocell –
218. The potential of the following cell is 0.34 volt at
25ºC. Calculate the standard reduction
potential of the Copper half cell.
Pt H 2 (1atm ) Cu 2+ (1M) Cu
(a) + 0.34 volt
(c) + 3.4 volt
(b) – 0.34 volt
(d) – 3.4 volt
GUJCET-2008
Ans. (a) : Given, E ocell = 0.34V
The half reaction take place in anode and cathode is–
Anode: H 2
→ 2H + + 2e○−
Cathode: Cu 2+ + 2e○−
→ Cu
From the Nernst equation we get–
2
E cell
180
H +
0.0592
= E cell −
log
2
Cu 2+
º
YCT
0.0592
(1)
log
2
1
0.0592
or 0.34 = ( E º cathode − 0 ) −
×0
2
E º Cu 2+ Cu = +0.34V
2
0.34 = E º cathode − E º anode −
219. The standard reduction potential of Cu2+/Cu
and Cu2+/Cu+ are 0.337 V and 0.153 V
respectively. The standard electrode potential
of Cu+/Cu half-life is
(a) 0.184 V
(b) 0.827 V
(c) 0.521 V
(d) 0.490 V.
AMU– 2008
Ans. (c) : Applying ∆G = –nFE°,
Cu2++ 2e– → Cu ..........(i); ∆G = – 2 × F × 0.377
= – (0.674)F
Cu2++ e– → Cu+.........(ii); ∆G = –1× F× 0.153
= – (0.153)F
Subtracting equation (ii) from (i)
Cu++ e– → Cu; ∆G = [–0.674– (–0.153)]F
= –0.674 + 0.153. = – (0.521)F
∆G
−0.521F
E° for the reaction=
= −
= 0.521V
nF
1× F
220. For a cell reaction involving two electron
change, the standard EMF of the cell is 0.295 V
at 25°C. The equilibrium constant of the
reaction at 25°C will be
(a) 29.5×10–2
(b) 10
(d) 2.95×10–10
(c) 1×1010
AMU– 2008
Ans. (c) : According to Nernst equation,
0.059
E = E°–
log Q
at 25°C
n
At equilibrium E = 0, Q = K
0.059
∴ 0 = E° –
log K
n
0.059
or E° =
log K
n
E ×n
log K =
0.059
0.295 × 2
log K =
0.059
0.295 × 2
K = antilog
0.059
K = 1× 1010
221. The relationship between Gibb’s free energy
change (∆G) and emf (E) of a reversible
electrochemical cell is given by
(a) ∆G = nFE
(b) ∆G = nF/E
(c) ∆G = –nFE
(d) ∆G = E/nF
J & K CET-(2008)
Ans. (c) : If the free energy change (∆G) is negative
then any redox reaction would occur spontaneously.
∆G = –nFEcell
Objective Chemistry Volume-II
where,
n = No. of electrons
F = value of Faraday
E = The emf of cell
∆G can be negative if E is positive .
222. The standard electrode potentials of the halfcells are given as below:
Zn → Zn2+ + 2e–, E° = 0.76V
Fe → Fe2+ + 2e– , E° = 0.44V
The E.M.F of the cell reaction:
Zn + Fe2+ → Zn2+ + Fe is
(a) – 0.32 V
(b) + 0.32 V
(c) + 1.20 V
(d) – 1.20 V
MPPET-2008
Ans. (b) : Given that, standard electrode potentials of
Half – cell–
Zn → Zn 2+ + 2e – ..........(i), E ° = 0.76V
Fe → Fe 2+ + 2e – ............(ii), E ° = 0.44V
Subtract Eq. (i) from Eq. (ii), we get.
Zn + Fe2+ → Zn2+ + Fe
The E.M.F of the cell,
E ocell = E cathode − E anode
E ocell = 0.76 – 0.44 = 0.32V
223. The standard redox potential (reduction
reaction) of Pt/Cr2 O -7 , Cr +3 ; Pt/Mn O 4 ,Mn2+;
Pt/Ce+4, Ce+3 in the presence of acid are 1.33 V,
1.51 V and 1.61 V respectively at 25°C. The
oxidising power of these systems decrease in
the order
2−
−
−
2−
(a) Cr2 O 7 > Mn O 4 > Ce+4
(b) Mn O 4 > Cr2 O 7 > Ce+4
−
2−
(c) Ce+4 > Mn O 4 > Cr2 O 7
−
2−
(d) Mn O 4 > Ce+4 > Cr2 O 7
SRMJEEE – 2008
Ans. (c) : The given reduction potential are:
Cr2 O72–
→ Cr 3+ E° = 1.33V
MnO 4–
→ Mn 2+ E° = 1.51V
Ce +4
→ Ce3+ E° = 1.61V
As we know that more the reduction potential of the
substance, stronger is the oxidising agent. Thus, the
oxidising power of the given system decrease in the
–
2−
order – Ce+4 > MnO 4 > Cr2 O 7
224. The standard electrode potential of hydrogen
electrode at 1 M concentration and hydrogen
gas at 1 atmosphere pressure is
(a) 1 Volt
(b) 6 Volt
(c) 8 Volt
(d) 0 Volt
J & K CET-(2007)
181
YCT
0.059
1
log 2 = 0V
2
(1)
Standard electrode potential of hydrogen at 1M
concentration and 1atm pressure is Zero Volts.
225. The standard oxidation potentials of Zn, Cu,
Ag and Ni electrodes are +0.76, 0.34, –0.80 and
+0.25V respectively. Which of the following
reaction will provide maximum voltage?
(a) Cu + 2Ag+ (aq) → Cu2+ (aq) + 2Ag
(b) Zn + 2Ag+ (aq)→ Zn2+(aq) + 2Ag
(c) H2 + Ni2+ (aq) → 2H+ (aq) + Ni
(d) Zn + Cu2+ (aq) → Zn2+ (aq) + Cu
(e) Zn + 2H+ (aq) → Zn2+ (aq) + H2
Kerala-CEE-2007
Ans. (b) : Oxidation potentials are given in the
question. The reduction potentials are obtained by
changing the sign, i.e. the reduction potentials of Zn,
Cu, Ag and Ni are, – 0.76, – 0.34, + 0.80 and – 0.25V
respectively.
E ocell = E oAg − E oZn
= 0.80 – (– 0.76) = 1.56V
Ans. (d) : E = 0 −
229. Zn2+ → Zn(s); Eo = – 0.76 V
Cu2+ → Cu(s); Eo = – 0.34 V
Which of the following is spontaneous?
(a) Zn2+ + Cu → Zn+Cu2+
(b) Cu2++Zn → Cu+Zn2+
(c) Zn2+ +Cu2+ → Zn+Cu
(d) None of the above
UP CPMT-2007
Ans. (b) : Electrode potential of cell must be +ve for
spontaneous reaction.
Zn2+ → Zn; Eo = – 0.76 V
Cu2+ → Cu; Eo= – 0.34 V
Redox reaction is
Zn → Zn2+ +2e– (oxidation)
Cu2+ +2e– → Cu (reduction)
Zn + Cu 2 + → Zn 2 + + Cu
Ecell = Eocathode – Eoanode
= – 0.34 – (– 0.76)
= + 0.42 V
Ecell is positive so above reaction is feasible.
230. For the redox reaction
°
226. The standard Ered
values of A, B and C are
Zn(s) + Cu2+ (0.1M) → Zn2+ (1M) + Cu(s)
+0.68 V, – 2.54 V, – 0.50 V respectively. The
Taking place in a cell, Eocell is 1.10 V. Ecell for
order of their reducing power is
RT
(a) A > B > C
(b) A > C > B
the cell will be 2.303
= 0.0591
(c) C > B > A
(d) B > C > A
F
MHT CET-2007
(a) 2.14 V
(b) 1.80 V
Ans. (d) : Greater in the negative value of standard
(c) 1.07 V
(d) 0.82 V
reduction potential, stronger is the reducing character.
UPTU/UPSEE-2007
Thus, order of reducing power–
Ans. (c) : From Nernst equation,
B(−2.54V) > C(−0.50V) > A(+0.68V)
2.303RT
[Zn 2+ ]
°
E cell = E cell
−
log10
227. The equilibrium constant of the reaction:
nF
[Cu 2+ ]
Cu(s) + 2Ag+(aq) → Cu2+(aq) +2Ag(s);
For the given cell reaction n = 2.
E0 = 0.46 V at 298K is
0.0591
1
10
10
(a) 2.0 × 10
(b) 4.0 × 10
E cell = 1.10 −
log10
15
10
2
0.1
(c) 4.0 × 10
(d) 2.4 × 10
0.0591
(AIPMT -2007)
Ecell = 1.10 −
= 1.10 – 0.02955
Ans. (c) : For a cell reaction in equilibrium at 298K,
2
Ecell =1.07V
E o = 0.46V and n= 2
0.0591
231. When 3.86 A current is passed through an
E °Cell =
× log10 K c
electrolyte for 50 min, 2.4 g of a divalent metal
n
is deposited. The gram atomic weight of the
0.0591
metal (in grams) is
0.46 =
× log10 K c
n
(a) 24
(b) 12
2× 0.46
(c)
64
(d) 40
log10 K c =
= 15.57
0.0591
AP - EAMCET (Medical) - 2007
K c = 3.7 ×1015 ≃ 4 ×1015
Ans. (d) : Given data–
228. In electrolytic cell, cathode acts as an/a
Current = 3.86 A, t = 50 min = 50 × 60 sec = 3000 sec
(a) oxidising agent
(b) reducing agent
Now, W = Zit
(c) either of the two
(d) Neither (a) nor (b)
Atomic weight
W=
×i× t
UP CPMT-2007
Valency × 96500
Ans. (b) : In electrolytic cell, cathode acts as a reducing
2.4 × valency × 96500
agent.
∴
Atomic weight =
+
–
i× t
At cathode,
Na + e → Na (reduction)
2.4 × 2 × 96500
At anode,
Cl– → Cl + e– (oxidation)
Atomic weight =
3.86 × 3000
Na + + Cl− → Na + Cl
Cathode Anode
Atomic weight = 40g
Objective Chemistry Volume-II
182
YCT
232. Calculate the emf of the cell
Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)
Given
o
= +0.34 V, EoAg+ /Ag = 0.80 V
ECu
2+
/ Cu
0.0591
1
log
(∵ log10=1)
2
0.1
Ecell = – 0.25 – 0.0295
Ecell = – 0.279 V ≈ – 0.28 V
236. Given the following half cell reactions and the
(a) +0.46 V
(b) +1.14 V
corresponding electrode potentials
(c) +0.57 V
(d) –0.46 V
(I) A + e ⇌ A– ; Eo = – 0.24 V
AP EAMCET (Engg.) -2007
(II) B– +e ⇌ B2– ; Eo = + 1.32 V
Ans. (a) : Given that, E oCu 2+ / Cu = 0.34V (Anode)
(III) C–+2e ⇌ C3– ; Eo = – 1.32 V
E o + = 0.80V (Cathode)
Ecell = – 0.25 –
Ag / Ag
We know that,
o
o
o
E cell
= E (cathode)
− E (anode)
= 0.80 − 0.34 = 0.46V
233. If K < 1.0, what will be the value of ∆G 0 of the
following?
(a) 1.0
(b) Zero
(c) Negative
(d) Positive
GUJCET-2007
Ans. (d) : If K<1.0 i.e. lnK has the negative value.
Now, from the relation between free energy and rate
constant.
We get,
∆G º = − RT lnK
∵ K has negative value.
∴
∆G ° = −RT ln ( K )
∆G ° = RT lnK
So, value of ∆G ° is positive.
234. If the standard electrode potential for the cell
Zn Zn 2+ ( aq ) Cu 2+ ( aq ) Cu is 1.10 V then the
(IV) D +2e ⇌ D2– ; Eo = + 0.65 V
which combination of the two half cells would
result in a cell with the largest EMF?
(a) (i) and (ii)
(b) (i) and (iii)
(c) (i) and (iv)
(d) (ii) and (iii)
AMU–2006
Ans. (d) : The two half cells with the greatest difference
in potentials would give the cell with largest EMF.
237. The metal that does not displace hydrogen
from an acid is :
(a) Ca
(b) Al
(c) Zn
(d) Hg
BCECE-2006
Ans. (d) : Al,Ca and Zn are placed above hydrogen
hence these metals can displace hydrogen, whereas, Hg
placed below the hydrogen. So, it does not displace
hydrogen from an acid.
238. Two electrochemical cell Fe|Fe2+ || Cu2+ | Cu
and Zn|Zn2+ || Cu2+ | Cu are connected in series.
(Fe2+/ Fe = –0.41V, Zn2+/Zn = –0.076V, Cu2+/
Cu = + 0.34V). The net emf of this cell is
(a) 1.10V
(b) 0.75V
(c) 0.35V
(d) 1.85V
[BITSAT – 2006]
Ans. (d) : For Fe & Cu cell, Fe is anode and Cu is
Cathode, E cell = 0.34 – (– 0.41) = 0.75V and for Zn &
Cu cell Zn is anode and Cu is Cathode,
E cell = 0.34 – (– 0.76) = 1.10V.
So, net emf = 0.75V + 1.10V = +1.85V
239. For the redox reaction
Zn ( s ) + Cu 2+ ( 0.1M ) → Zn 2+ (1M ) + Cu ( s )
maximum work done by this cell will be
(a) −106.15kJ
(b) −212.30kJ
(c) −318.45kJ
(d) −424.60kJ
CG PET -2007
Ans. (b) : W=–nEF= –2×1.1×96500
= –212.3 kJ
235. What is the electrode potential (in V) of the
taking place in a cell, E º cell is 1.10 V E cell for the
following electrode at 25 oC?
2+
Ni (0.1M)| Ni (s)
RT
cell will be 2 ⋅ 303
= 0.0591
(Standard reduction potential of Ni2+| Ni is)
F
2.303 RT
(a) 2.14 V
(b) 1.80 V
– 0.25V,
= 0.06
F
(c) 0.82 V
(d) 1.07 V
(a) – 0.28 V
(b) – 0.34 V
CG PET -2006
(c) – 0.82 V
(d) – 0.22 V
Zn 2+
AP-EAMCET (Medical), 2006
2.303RT
°
log10 2+
Ans. (d) : E cell = E cell −
Ans. (a) : The given cell is:
nF
Cu
Ni 2 + ( 0.1M ) | Ni(s)
For the given cell reaction n = 2
°
E Ni2+ / Ni = −0.25V
0.0591
1
E cell = 1.10 −
log10
2
0.1
The Nernst equation for the given cell0.0591
0.0591
1
⇒ 1.10 −
= 1.07V
Ecell = E °Ni2+ / Ni −
log
2+
2
n
[Ni ]
Objective Chemistry Volume-II
183
YCT
240.
Zn Zn 2 + Cu 2 + Cu
c =1 c =1
243. Given the electrode potentials
Fe3+ + e–→Fe2+, E o = 0.771 volts
I 2 + 2e − → 2I − , E o = 0.536 volts
If the standard reduction potential of zinc
E ocell for the cell reaction
electrode and copper half cell is-0.76V and 0.34
2Fe3++2I–→ 2Fe2+ + I2, is
V respectively then the emf will be
(a) 1.006 V
(b) 0.503V
(a) 1.1 V
(b) 1.4 V
(c) 0.235V
(d) –0.235V
(c) 1.34 V
(d) None of these
AMU-2005
CG PET -2006
Ans.
(c)
:
We
have
Ans. (a) : Given,
2Fe3+ + 2I
→ 2Fe + + I2
E oZn 2+ / Zn = − 0.76 V
E°cell = E ocathode − E oanode
E oCu 2+ / Cu = 0.34 V
= (0.771 − 0.536)
EMF = E oR − E oL
= 0.235V
o
o
−
EMF = E Cu
E
2+
2+
244.
What
is the emf of the cell, whose half cells are
/ Cu
Zn / Zn
given below Mg 2+ + 2e – → Mg ( s ) E = –2.37V
= 0.34 + 0.76
Cu 2+ + 2e – → Cu ( s ) E = + 0.34V
=1.1V
(a) −2.30V
(b) 1.336V
EoFe2+ /Fe = –0.441V and EoFe3+ /Fe2+ = 0.771 V, and
(c)
2.71V
(d)
2.03V
standard EMF of the reaction Fe + 2Fe3+→
CG PET -2005
3Fe2+ will be
Ans. (c) : These are oxidation potentials. Reduction
(a) 0.111 V
(b) 0.330 V
(c) 1.653 V
(d) 1.212 V
potentials are equal and opposite.
(AIPMT -2006) So,
E ocell = E ooxi. (Mg) + E ored (Cu)
o
Ans. (d) : Given, E Fe2+ / Fe = −0.441V ,
= 2.37 + 0.34 = 2.71V
245. For the feasibility of a redox reaction in a cell,
E oFe3+ / Fe2+ = 0.771V
the emf should be
For the cell reaction
2+
–
(a) positive
(b) fixed
Fe → Fe + 2e ….(i)
(c)
zero
(d)
negative
Anode reaction;
2+
J & K CET-(2005)
Fe → Fe +2e
Cathode reaction;
Ans. (a) : Any redox reaction would occur
2Fe3++2e- → 2Fe2+
spontaneously if the free energy change (∆G) is
E ocell = E oCathode − E oAnode
negative. ∆G ° = −nFE °
o
Where n is the number of electrons involved, F is the
E = 0.771 − ( −0.441)
value of Faraday and E° is the cell emf.
E ocell = 1.212V
∆G ° can be negative if E° is positive.
242. Find out emf of cell,
246. What is potential of platinum wire dipped into
Zn;Zn 2+ (1M) || Cu 2+ Cu (1M); Eo for
a solution of 0.1 M in Sn2+ and 0.01 in Sn4+?
2+
o
2+
(a) Eo
(b) Eo + 0.059
Zn /Zn = –0.76;E for Cu /Cu = +0.34
(a) +1.10V
(b) –1.10V
0.059
0.059
(d) Eo =
(c) Eo +
(c) –0.76
(d) –0.42
2
2
UP CPMT-2006
JCECE - 2005
Ans. (a) :
Ans. (c) : For a oxidation half-cell
Eocell = Eocathode – Eoanode
Sn2+ → Sn4+ + 2e–
Oxidation half reaction = Zn
→ Zn 2+ +2e –
RT [Sn 4+ ]
0.059
0.01
E = Eº −
ln
= Eº −
log
Reduction half reaction = Cu 2+ + 2e –
→ Cu
2+
nF [Sn ]
2
0.1
Cell reaction = Zn + Cu 2+
→ Zn 2+ +Cu
0.059
1
E = Eº −
log
Eoright – Eoleft = Eocell
2
10
= 0.34 – (– 0.76)
0.059
= 0.34 + 0.76
E = Eº +
2
= +1.10 V
241.
Objective Chemistry Volume-II
184
YCT
247. The standard electrode potentials of Ag+/Ag is
+0.80 V and Cu+/Cu is +0.34 V. These
electrodes are connected through a salt bridge
and if :
(a) copper electrode acts as a cathode then E ocell
is +0.46 V
(b) silver electrode acts as anode then E ocell is
–0.34V
(c) copper electrode acts as anode then E ocell is
+0.46 V
(d) silver electrode acts as a cathode then E ocell is
–0.34V
(e) silver electrode acts as anode and E ocell is
+1.14 V
Kerala-CEE-2005
Ans. (c) : The cell reaction is
Cu(s) + 2Ag+
→ Cu2+ + 2Ag
The two half cell reactions are
Cu
→ Cu2+ + 2e–
Oxidation (anode)
+
–
Ag + e
→ Ag
Reduction (cathode)
Ecell = Ecathode – Eanode
= 0.80 – 0.34
= + 0.46V
248. 4.5 g of aluminium (at. mass 27 amu) is
deposited at cathode from Al3+ solution by a
certain quantity of electric charge. The volume
of hydrogen produced at STP from H+ ions in
solution by the same quantity of electric charge
will be
(a) 44.8 L
(b) 22.4 L
(c) 11.2 L
(d) 5.6 L
(AIPMT -2005)
Ans. (d) : According to Faraday’s law of electrolysis
equivalent of different substance deposited or liberated
at electrodes during electrolysis are always equal.
m Al E Al
=
m = mass of the substance
mH
EH
E = equivalent weight of the substance
27
4.5
= 3
mH
1
⇒
m H = 0.5g
Volume at 2g H2 at STP = 22.4L
22.4× 0.5
∴ Volume of 0.5g H2 at STP =
L
2
= 5.6 L
249. The electrode potentials for
Cu 2+ ( aq ) + e − → Cu + ( aq )
Ans. (b) :
Cu 2+ + e − → Cu + ∆G1o ,
Cu + + e − → Cu ∆G o2
Cu 2+ + 2e − → Cu , ∆G o = ∆G1o + ∆G o2
− nFE o = − n1FE1o + ( − n 2 FE o2 )
0.15 ×1 + 1× 0.50
2
0.15 + 0.50
=
2
= 0.325 V
250. The emf of a particular voltaic cell with the cell
+
reaction Hg 2+
2 + H 2 ⇌ 2Hg + 2H is 0.65 V. The
maximum electrical work of this cell when 0.5 g
of H2 is consumed.
(a) –3.12 × 104 J
(b) –1.25×105 J
6
(d) None
(c) 25.0×10 J
VITEEE-2016
Ans. (a) : Given cell reaction,
Hg 22+ + H 2 ⇌ 2Hg + 2H + , E = 0.65V
Eo =
∴
Number of moles in 0.5gm H2 =
=
0.5
= 0.25mole
2
Wmax = −1.25 × 105 × 0.25
= −3.12 × 104 J
251. The standard reduction potential for Cu2+ /Cu
is + 0.34. Calculate the reduction potential at
pH = 14 for the above couple. (Ksp Cu (OH)2 =
1 × 10–19)
(a) –0.22 V
(b) +0.22 V
(c) –0.44 V
(d) +0.44 V
VITEEE-2013
Ans. (a) : Given that,
pH = 14, [H+] = 10–14
K w 10−14 m
[OH–]=
=
=1
[H + ] 10−14 m
(b) 0.325 V
(d) 0.150 V
VITEEE-2015
Objective Chemistry Volume-II
Weight
Molecular weight
Hence,
and Cu + ( aq ) + e − → Cu ( s )
are + 0.15 V and + 0.50 V, respectively. The
o
value of ECu
will be:
2+
/Cu
(a) 0.500 V
(c) 0.650 V
Wmax = −nFE
= −2 × 96500 × 0.65
= −1.25 ×105 J
185
2
K sp = Cu 2+ OH − = 10−19
Cu +2 =
10−19
= 10−19
2
OH −
The half cell reaction
Cu 2+ + 2e − → Cu
0.059
1
E = E° −
log
2
Cu 2+
∴
YCT
0.059
1
so,
E °cell = E °ox + E °red
log −19
2
10
E °cell = −E °Co+2 / Co + E°Ce4+ / Ce+3
0.059
E= 0.34–
× 19
1.89 = − ( − 0.28) + E °Ce+4 / Ce+3
2
E = – 0.22 V
∴
E °Ce+4 / Ce+3 = 1.89 − 0.28 = 1.61V
252. For hydrogen-oxygen fuel cell, the cell reaction
255. Which of the following reaction is possible at
is 2H 2 (g) + O 2 (g) → 2H 2O(l)
anode?
If ∆G of (H2O) = –237.2 kJ mol–1, then emf of
(a) Fe2 + 2e– → 2F–
this cell is
1
(b) 2H+ + O2 + 2e– → H2O
(a) + 2.46 V
(b) – 2.46 V
2
(c) + 1.23 V
(d) –1.23 V
(c) 2Cr23+ + 7H 2 O → Cr2 O72 − + 14H + + 6e −
VITEEE-2016
(d) None of the above
Ans. (c) : Given cell reaction is
VITEEE-2016
2H 2 ( g ) + O 2 ( g )
→ 2H 2 O ( l )
3
6+
–
Ans. (c) : Cr → Cr +3e
Oxidation State of Cr changes from +3 to +6 by
∆G °f ( H 2 O ) = −237.2 kJ.mol−1
loss of electrons.
E° = ?
At anode oxidation takes place.
∴
∆G° = – nFE°
So this reaction is possible at anode.
∆G °
∴ E° = −
256.
nF
− ( −237.2 ) × 1000
=
= 1.23V
[∵ n = 2]
2 × 96500
E = 0.34 −
253.
MnO 4− + 8H + + 5e − → Mn 2+ + 4H 2O; E° = 1.51
V
MnO 2 + 4H + + 2e − → Mn 2+ + 2H 2O; E°= 1.23 V
E° MnO- |MnO is
4
2
(a) 1.70 V
(c) 1.37 V
(b) 0.91 V
(d) 0.548 V
VITEEE-2011
Ans. (a) : Given
MnO 4− + 8H + + 5e − → Mn 2+ + 4H 2 O;E° = 1.51V …(i)
MnO 2 + 4H + + 2e − → Mn 2+ + 2H 2 O; E° = 1.23V
or Mn 2+ + 2H 2 O → MnO 2 + 4H + + 2e− ; E° = −1.23V ..(ii)
From equation (i) and (ii) we get
MnO 4 − + 4H + + 3e − → MnO 2 + 2H 2 O; E 3 = ?
EoCu2+ |Cu = + 0.34V
EoZn2+ |Zn = – 0.76V
1.51× 5 + (−1.23) × 2
E3 =
3
E3 = 1.70V
254. For the cell reaction 2Ce4+ + Co → 2Ce3+ +
Co2+; Eocell is 1.89 V. If ECo2+ /Co is – 0.28 V, what
Identify the incorrect statement from the
options below for the above cell.
(a) If Eext > 1.1V, Zn dissolves at Zn electrode
and Cu deposits at Cu electrode.
(b) If Eext = 1.1V, no flow of electrons or current
occurs.
(c) If Eext > 1.1V, electrons flows from Cu to Zn.
(d) If Eext < 1.1V, Zn dissolves at anode and Cu
deposits at cathode.
JEE Main 11 Sep 2020
is the value of EoCe4+ /Ce3+ ?
(a) 0.28 V
(c) 2.17 V
(b) 1.61 V
(d) 5.29 V
VITEEE-2017
Ans. (b) : Given cell reaction
2Ce +4 +Co → 2Ce3+ +Co 2+ ; E °cell = 1.89
and E Co2+ / Co = −0.28
Objective Chemistry Volume-II
Ans. (a) : B2 H 6 + 3O 2
→ B2 O3 + 3H 2O
27.66g of B2H6 (1mole) requires 3 moles of oxygen
(O2) for complete burning.
186
YCT
2.
Faraday's Law
257. Given below are half cell reactions
MnO4− +8H+ +5e−→Mn2+ +4H2O,
EoMn2+ /MnO- = –1.510 V
∴
4
1
O 2 + 2H + + 2e- → H 2O
2
EoO2 /H2O = 1.223 V
∴
Will the permanganate ion, MnO4− liberate O2
from water in the presence of an acid?
(a) No, because E ocell = −2.733 V
∴
Equivalent mass of Au (E2) = ?
We know that –
m1 E1
=
m2 E 2
2.15 107.9
=
1.31
E2
107.9×1.31
E2 =
= 65.7
2.15
Atomic mass
Equivalent weight of Gold =
valency
Atomic mass 197
Valency =
=
=3
E2
65.7
(b) Yes, because E ocell = +0.287 V
260. If 5 ampere of current is passed for 193 seconds
through a solution containing Copper salt, 0.32
(d) Yes, because E = +2.733 V
g of copper is deposited. What is the oxidation
state of the Cu in the salt?
NEET-17.06.2022
(a) + 2
(b) + 1
Ans. (b) : Reduction
3
Cathode :
(c) + 3
(d) +
2
→2Mn 2 + + 8H 2 O; E oRP =1.510V
2MnO −4 + 16H + + 10e −
AP-EAMCET-08.07.2022, Shift-I
Anode :
Ans. (a) : Given that,
I = 5A,
t = 193 second
5
+
−
o
→ O 2 + 10H + 10e ; E OP =− 1.223V
5H 2 O
W
=
0.32g,
let oxidation state = n
Cu
2
By
faraday’s
lawCell reaction:
W
I× t
5
Equation of solution is given by =
=
2MnO −4 + 6H +
→2Mn 2+ + O 2 + 3H 2 O
eq.wt 96500
2
After putting the values we get
E o = (SRP)
− (SRP)
(c) No, because E ocell = −0.287 V
o
cell
cell
cathode
Anode
E ocell = 1.510 V − 1.223 V
0.32
5
×
193
atomic
weight
= 0.287 V
∵ eq.wt =
=
63.54 96500
oxidation state
Yes, the given cell reaction is possible.
n
258. The amount of charge in F (Faraday) required
to obtain one mole of iron from Fe3O4 is____
n=+2
(Nearest Integer)
261. A certain quantity of electricity is passed
JEE Main 26.07.2022, Shift-I
through aqueous Al2(SO4)3 and CuSO4
Ans. (3) : Given that,
solutions connected in series. 0.09 g of Al is
deposited on cathode during electrolysis. The
+ 8e −
Reaction
Fe3 O 4
→ 3Fe
amount of copper deposited on cathode in
To obtain one moles of Fe we consider charge of 1 mole
grams is
8
(At. wt. of Al = 27; Cu = 63.6)
Fe = F
(a) 0.318
(b) 31.8
3
(c) 0.636
(d) 3.18
= 2.67 F
AP EAMCET (Medical) - 2013
259. In two separate experiments, the same quantity
of electricity was passed through silver and Ans. (a) : The deposition reaction of aluminium (Al) is
gold solutions [Assume 't' constant] The
Al3+ + 3e −
→ Al
amounts of Ag and Au deposited are 2.15 and
∴
0.09
g
of
Al
will
be deposited by =
1.31 g, respectively. The valency of gold is
3× F
[Atomic mass of Ag = 107.9; Au = 197]
× 0.09
(a) 1
(b) 2
molecular weight of Al
(c) 3
(d) 4
3 × 96500
=
× 0.09
TS-EAMCET-19.07.2022, Shift-II
27
Ans. (c) : Given that –
= 965C
Mass of silver (m1) = 2.15 g
The deposition reaction of copper (Cu) is Mass of gold (m2) = 1.31 g
Cu 2 + + 2e −
→ Cu
Equivalent mass of Ag (E1) = 107.9
Objective Chemistry Volume-II
187
YCT
Thus, 2F = 2×96500C, deposit Cu = 63.6g
63.6 × 965
∴
965C will deposit Cu =
= 0.318g
2 × 96500
262. When 9.65 C of electricity is passed through a
solution of AgNO3 (atomic weight of silver 108),
the amount of silver deposited is :
(a) 10.8 mg
(b) 5.4 mg
(c) 16.2 mg
(d) 21.2 mg
COMEDK-2007
Ans. (a) : According to Faraday's Law of electrolysis :
i× t ×M
W=
nF
Where,
W = Weight deposited,
i = current, t = time
M = Molar mass,
n = number of electron.
Given that, Molar mass of silver (M) = 108 g/mole
n = 1, it = 9.65 C
9.65 ×108
∴
W=
1× 96500
W = 0.0108 g
or
W = 10.8 mg.
263. What is the approximate quantity of electricity
(in coulomb) required to deposit all the silver
from 250 mL of 1 M AgNO3 aqueous solution?
(a) 96500
(b) 24125
(c) 48250
(d) 12062.5
AP-EAMCET-2000, 05
Ans. (b) : Moles of silver =
250
1
× 1 = mole
1000
4
Ag+ + e– → Ag
1 mole Ag requires 1 mole e–
1
1
∴
mole Ag requires = mole electron
4
4
1
∴ Total charge = × 96500
4
= 24125 C.
264. What is the time (in sec) required for
depositing all the silver present in 125 mL of 1
M AgNO3 solution by passing a current of
241.25 A?
(1F = 96500 C)
(a) 10
(b) 50
(c) 1000
(d) 100
AP-EAMCET-2006
Ans. (b) : Given that, Current (i) = 241.25 A
∵
125 mL of AgNO3 solution contain
108 × 125
g Ag
=
1000
= 13.5 g Ag
96500
∴
13.5 g of Ag is deposited by =
× 13.5
108
= 12062.5 C
Objective Chemistry Volume-II
Since,
Q = i.t
Q 12062.5
t = =
i
241.25
t = 50 sec.
265. Assertion (A) : A current of 96.5 A is passed into
aqueous AgNO3 solution for 100 s. The weight of
silver deposited is 10.89 g (at.wt. of Ag = 108).
Reason (R) : The mass of a substance deposited
during the electrolysis of an electrolyte is
inversely proportional to the quantity of
electricity passing through the electrolyte.
(a) Both (A) and (R) are true and (R) is the
correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the
correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
AP-EAMCET-2006
Ans. (c) : Given that, Current (i) = 96.5 A, t = 100 sec.
Atomic weight of Ag = 108
∵
Q = it
= 96.5 × 100 = 9650 C
108 × 9650
∴ 9650 C deposited weight of Ag =
96500
= 10.8 gm.
266. When same quantity of electricity is passed
through aqueous AgNO3 and H2SO4 solutions
connected in series, 5.04 × 10–2g of H2 is
liberated. What is the mass of silver (in gram)
deposited? (Eq. wts. of hydrogen = 1.008, silver
= 108)
(a) 54
(b) 0.54
(c) 5.4
(d) 10.8
AP-EAMCET-2008
Ans. (c) : Given that :
Weight of hydrogen liberated = 5.04 × 10–2 gm.
Eq. wt. of hydrogen = 1.008
Eq. wt. of silver = 108
Mass of silver deposited (m) = ?
According to Faraday's second law of electrolysis–
Mass of silver deposited
Eq. wt. of silver
=
Weight of hydrogen liberated
Eq. wt. of H 2
m
108
=
5.04 ×10−2 1.008
108 × 5.04 × 10−2
m=
1.008
m = 5.4 gm.
267. When electric current is passed through
acidified water for 1930 s, 1120 mL of H2 gas is
collected (at STP) at the cathode. What is the
current passed in ampere?
(a) 0.05
(b) 0.50
(c) 5.0
(d) 50
AP-EAMCET-2008
Ans. (c) : Given that : t = 1930 sec.
1120 × 10−3
No. of moles of hydrogen collected =
mole
22.4
= 0.05 mole
188
YCT
271. Salts of A, B and C were electrolyzed under
identical conditions using the same quantity of
electricity. It was found that when 2.1 g of A
was deposited. The weights of B and C
deposited were 2.7 g and 9.6 g respectively. If
the atomic mass of A, B and C respectively are
7, 27 and 64 respectively are ____.
(a) 3, 1, 2
(b) 1, 3, 2
(c) 3, 1, 3
(d) 2, 3, 2
∵ 1 mole of H2 is liberated = 96500 C
2 × 0.05 mole of H2 will be liberate
= 96500 × 2 × 0.05
= 9650 electricity
∴ Q = i.t
Q 9650
i= =
= 5A
t 1930
268. For a hypothetical reaction
A→C
AP-EAMCET 25-08-2021 Shift - I
k
A ↽ k 1 ⇀ B[Fast]
Ans. (b) :
2
(c) ∝ [A]2 [B]2
(d) ∝ [A]2 [B]
TS-EAMCET (Engg.), 07.08.2021 Shift-II
Ans. (a) : Rate of reaction depend on slow step. Then,
K3
A + B
→ C (slow) will determine rate.
Rate law ∝ [A][B]
269. An electric current of 0.965 A is passed for
2000 seconds through a solution containing
[Cu(CH3CN)4]+ and metallic copper is
deposited at the cathode. The amount of copper
deposited is
(a) 0.005 mol
(b) 0.01 mol
(c) 0.02 mol
(d) 0.04 mol
SCRA-2014
Ans. (c) : Equivalent weight of [Cu(CH3CN)4]+ = 227/1
= 227
Equivalent weight × i × t
∴ Mass deposited on cathode =
96500
227 × 0.965 × 2000
=
96500
= 4.45 gm
4.45
The amount of Cu deposit =
= 0.01960 ≃ 0.02 mole
227
270. If the resistance of 0.1 M KCl solution in a
conductance cell is 300 Ω and conductivity is
0.013Scm–1, then the value of cell constant
is_______
(a) 3.9 cm–1
(b) 39 m–1
–1
(c) 3.9 m
(d) 0.39 cm–1
AP EAMCET 23-08-2021 Shift-I
Ans. (a) : Given that,
Conductivity (k) = 0.013 S.cm–1
Resistance (R) = 300Ω
Cell constant = ?
Cell constant
∴ Conductivity (k) =
Resistance(R)
Cell constant
0.013 =
300
⇒ Cell constant = 300×0.013
= 3.9 cm–1
Objective Chemistry Volume-II
EQ
(Faraday's Law of electrolysis)
96500
Where, E = Equivalent weight
M (atomic weight)
E=
x ( Valency factor )
For A (Atomic mass = 7)
( 7 / x ) Q = 7 ⋅ Q ⇒ x = 7 = 1 = 10
2.1 =
96500
x 96500
2.1 0.3 3
For B(Atomic mass = 27)
( 27 / y ) Q = 27 ⋅ Q ⇒ y = 27 = 1 = 10
2.7 =
96500
y 96500
2.7 0.1 1
For C(Atomic mass = 64)
( 64 / z ) Q = 64 ⋅ Q ⇒ z = 64 = 4 = 2 = 20
9.6 =
96500
z 96500
9.6 0.6 .3 3
∴
x : y : z = 10 : 30 : 20
W = ZQ =
k3
A + B
→ C(Slow)
Rate law for this reaction is
(b) ∝ [A]2
(a) ∝ [A][B]
x : y : z = 1: 3 : 2
Where x, y and z are valency factor of A, B and C.
272. A constant current of 30 A is passed through
an aqueous solution of NaCl for a time of 1.00
h. What is the volume of Cl2 gas at STP
produced?
(a) 30.00 L
(b) 25.08 L
(c) 12.54 L
(d) 1.12 L
VITEEE-2013
Electrolysis
Ans. (c) : 2NaCl
→ 2Na + + 2Cl −
At anode 2Cl- → Cl2 +2e-
At cathode 2Na + + 2e − → 2Na
1 mole = 22.4 litres of Cl2 at STP
Amount of current passed = 30 × 60 × 60 = 108000
coulomb
108000 coulomb will liberate–
108000
= 0.5596 mole Cl2
96500× 2
So, the chlorine liberated at STP = 0.5596 × 22.4
= 12.54 litres
273. The number of Faradays needed to reduce 4 g
equivalents of Cu2+ to Cu metal will be
(a) 1
(b) 2
1
(c)
(d) 4
2
VITEEE- 2010
189
YCT
Ans. (d) :
Cu 2+ + 2e − → Cu
1mol
2F
1mol
1
1F
1
mol
mol
2
2
1g eq.
1F
1g eq.
Thus, to reduce 4 equivalent of Cu2+ into Cu Four
Faradays are required.
274. The amount of electricity required to produce
one mole of copper from copper sulphate
solution will be
(a) 1 Faraday
(b) 2.33 Faraday
(c) 2 Faraday
(d) 1.33 Faraday
VITEEE- 2008
Ans. (c) : In CuSO4 solution, oxidation state of Cu is
+2. Hence one mole of copper sulphate will required
charge equal to two moles of electrons to form metallic
Cu. Mole charge = 1 F. Hence 2 Faraday is required.
Cu2+ +2e–
→ Cu(s)
275. How long (in hours) must a current of 5.0
amperes be maintained to electroplate 60g of
calcium from molten CaCl2?
(a) 27 hours
(b) 8.3 hours
(c) 11 hours
(d) 16 hours
VITEEE- 2007
Ans. (d) : Given,
W = 60 g, i = 5 amp.
Mol.mass 40
Eq. mass of Ca 2+ =
=
= 20
2
2
Eq.mass of Ca 2+
20
Z=
=
96500
96500
Faraday’s first law of electrolysis,
W = zit
20
or 60 =
×5× t
96500
96500 × 60
∴t =
= 57900 sec = 16 h.
20 × 5
276. When a quantity of electricity is passed
through CuSO4 solution, 0.16 g of copper gets
deposited. If the same quantity of electricity is
passed through acidulated water, then the
volume of H2 liberated at STP will be
(Given : At. wt. of Cu = 64)
(a) 4.0 cm3
(b) 56 cm3
3
(c) 604 cm
(d) 8.0 cm3
SRMJEEE-2014
Ans. (b) : Given –
Deposited amount of Cu = 0.16g
Liberated amount of H2 = ?
Now,
Weight of Cu deposited Equivalent weight of Cu
=
Weight of H 2 produced
Equivalent weight of H
WH2
22400
× 5 × 10−3 cc
2
= 56 cm3
277. Which of the following conversions involves
gain of 5 electrons per ion?
(a) MnO −4 → Mn2+
(b) CrO 24− → Cr3+
∴ Volume of H2 librated at STP =
(c) MnO 24− → MnO2
(d) Cr2 O72− → 2Cr3+
SRMJEEE-2013, 2012
Ans. (a) :
−
+5e−
(a) Mn O 4
→ Mn2+
O.S. of Mn–
x–8=–1
x=+2
x=+7
Hence, 5e– gain the Mn and convert in + 2 oxidation
state.
−
+3e
(b) CrO 24 −
→ Cr 3+
O.S. of Cr–
x–8=–2
x = +3
x=+6
−
+2e
(c) MnO 24 −
→ MnO 2
O.S. of Mn –
x–8=–2
x=+4
x=+6
−
+3e
(d) Cr2 O 27 −
→ 2Cr 3+
O.S. of Cr –
2x – 14 = – 2
2x = + 6
x=+6
x=+3
278. What is the electrochemical equivalent (in gCoulomb–1) of silver? (Ag = 108; F = Faraday)
108
(a) 108 F
(b)
F
F
1
(c)
(d)
108
108F
A P - EA M C ET ( En g g . ) - 2 0 0 5
Ans. (b) : For silver atom–
Equivalent weight of Ag = 108
E
∴ Electrochemical equivalent (Z) =
F
108
–1
or
Z=
g. coulomb
F
279. A 200 W, 100 V bulb is connected in series with
an electrolytic cell. If an aqueous solution of an
Sn-salt is electrolysed for 5 hrs, 11.1 g of Sn
gets deposited. The chemical formula of the
compound is ......... (Given atomic weight of Sn
is 118.7 g mol–1)
(a) SnO
(b) SnCℓ2
0.16 32
=
WH2
1
Objective Chemistry Volume-II
0.16
32
= 5 × 10−3 g
or WH2 =
(c) SnCℓ4
(d) SnO2
AP EAMCET (Engg.) 18.09.2020, Shift-I
190
YCT
Ans. (c) : Given,
A = atomic weight of Sn = 118.7 g mol–1
200W
i = current =
= 2A
100V
t = time = 5hrs (5 × 3600 s)
A
×i× t
Eit n
∴ W=
=
F
F
A × i × t 118.7 × 2 × ( 5 × 3600 )
⇒ n=
=
= 3.989
W×F
11.1× 96500
A
E = =equavalent weight of Snn+
n
n = valency of Sn in the salt
W = mass of Sn deposited = 11.1 g
Here, SnO and SnO2 are non-electrolytes (aqueous). So,
valency of Sn will be 4 in SnCℓ4 (Sn4+ + 4Cℓ–).
280.
How much current is required to produce H2
gas at the rate of 1 cc/sec under STP?
(a) 33 A
(b) 6.0 A
(c) 7.9 A
(d) 8.61 A
AP EAMCET (Engg.) 21.09.2020, Shift-II
Ans. (d) : 2H+ + 2e– → H2
1 mole of H2 = 22400 cc at NTP
2 moles of electron required = 2 × 96500
2 × 96500
1 cc at NTP =
= 8.616
22400
∵
Q = It
8.616
∴
I=
= 8.616 A
1
Hence, option (d) is correct.
281. When a current of 10 A is passes through
molten AlCl3 for 1.608 minutes. The mass of Al
deposited will be–––[Atomic mass of Al = 27g]
(a) 0.09 g
(b) 0.81 g
(c) 1.35 g
(d) 0.27 g
AP EAPCET 19-08-2021 Shift-I
Ans. (a) : Given:- I = 10A t = 1.608 minutes
t = 96.5 sec
=?
E
Formulam=
i.t.
nF
Where
m = mass of Al
e = atomic mass of Al
i = current
t = time
n = oxidation number (3)
27 × 10 × 96.5
Or
m=
3 × 96500
m = 0.09g
282. An electric current is passed through silver and
water voltmeters connected in series. The
cathode of silver voltmeter weighed 0.054 g
more after the electrolysis. The volume of O2
liberated is
(a) 5.6 cm3
(b) 11.2 cm3
3
(c) 22.4 cm
(d) 2.8 cm3
COMEDK-2012
Objective Chemistry Volume-II
Ans. (d) : For deposition of Ag, reaction is;
Ag + + e –
→ Ag
Thus, 108 gm of Ag is deposited by 1F.
∴ 0.054 of Ag will be deposited by
1
× 0.054
108
= 5 × 10 –4 F
For 1 mole of H2O, the reaction is:
1
H 2 O
→ 2H + + O 2 + 2e –
2
1
Thus, 2F = mol or 11200 cm3 of O2
2
11200
Therefore, 5 ×10 –4 F =
× 5 ×10 –4
2
= 2.8 cm3 of O 2
=
283. The volume of H2 obtained at S.T.P when Mg
obtained by passing a current of 0.5 amp
through molten MgCl2 for 32.2 minutes is
treated with excess dilute HCl is approximately
[Eq.mass of Mg = 12]
(a) 56 cm3
(b) 28 cm3
3
(c) 5.6 cm
(d) 112 cm3
COMEDK-2012
Ans. (d) :Given that,
i = 0.5 amp
t = 32.2 ×60 sec
∴ Mass of Mg depositied = Z it
12
=
× 0.5 × 32.2 × 60
96500
= 0.12012gm
Now,
Mg + 2HCl
→ MgCl2 + H 2
24gm
→ 22400 cm3
22400
0.12012 gm
→
× 0.12012
24
3
= 112cm
284. When the same quantity of current passed
through CuSO4 and AgNO3 solution 2.7 g of
silver is deposited. The amount of copper
deposited is
(a) 0.4 g
(b) 3.2 g
(c) 1.6 g
(d) 0.8 g
COMEDK-2012
Ans. (d) : According to Faraday’s second law:
Weight.of Cu deposited Eq.wt.of Cu
=
Weight.of Ag deposited Eq.wt.of Ag
Weight.of Cu deposited 63.5 / 2
=
2.7gm
108
63.5 × 2.7
∴ Wt. of Cu deposited=
2 × 108
= 0.793 gm
≈ 0.80gm
191
YCT
285. How many gram of cobalt metal will be
deposited, when a solution of cobalt(II)
chloride is electrolyzed with a current of 10
amperes for 109 minutes (1 Faraday =96,500
C; Atomic mass of Co= 59 u)
(a) 4.0
(b) 20.0
(c) 40.0
(d) 0.66
Karanataka NEET-2013
Ans. (b): Given that,
i=10 amp
t= 109 min =109×60 sec
E=59/2
From faraday’s first law
Eit
10 × 109 × 60 × 59
W=
=
96500
2 × 96500
W=20
286. When 965 coulombs of electricity is passed
through a solution of silver nitrate, the amount
of silver deposited is
(a) 10.8g
(b) 2.16g
(c) 1.08g
(d) 0.54g
COMEDK-2019
Ans. (c) : Weight of Ag deposited
E Ag × 8
WAg =
96500
gram atomic wt 108
E Ag =
=
= 108
valency
1
108 × 965
WAg =
= 1.08gm
96500
287. How many Faradays are needed for reduction
of 2.5 mole of Cr2O 7 2- into Cr 3+?
(a) 15
(b) 12
(c) 6
(d) 3
GUJCET-2007
Ans. (a) : The reduction reaction is:
( +6 )
( +3 )
Cr2 O72−
→ Cr 3+
n factor = 2 × ( 3) = 6
∴ Number of equivalent = n factor × no. of moles
= 6 × 2.5
= 15
288. A metal plate of dimension of (1 × 2 cm2) has to
be coated on both the sides by Cu metal. How
long does it take to deposit Cu of 0.01 cm
thickness, if 1.5A current is used?
[Electrochemical equivalence of Cu is 0.0003
g/c and the density of Cu is 9 g/cm3]
(a) 400s
(b) 800s
(c) 120s
(d) 160s
TS-EAMCET (Engg.), 06.08.2021
Ans. (a): Given that
I = 1.5amp , density of Cu is 9g/cm
m = ρV
m=ρ×A×t
First law, m = z × it
Objective Chemistry Volume-II
9 × 2 × 0.01 = 0.0003 × 1.5 × t
t=
18 × 0.01
= 400sec.
0.0003 × 1.5
289. When the same quantity of electricity is passed
through the aqueous solutions of the given
electrolytes for the same amount of time, which
metal will be deposited in maximum amount on
the cathode?
(a) ZnSO4
(b) FeCl3
(c) AgNO3
(d) NiCl2
TS EAMCET 05.08.2021, Shift-I
Ans. (c) : When the same quantity of electricity is
passed through the aqueous solution of the given
electrolytes for the same amount of time, AgNO3 metal
will be deposited in maximum amount on the cathode.
290. MnO 4– + 8H + + 5e – → Mn 2+ + 4H 2O,E° = 1.51V.
The quantity of electricity required in Faraday
to reduce five moles of MnO 4− is _______.
JEE Main 26.02.2021, Shift-I
Ans. (25) : Given reaction,
MnO 4− + 8H + + 5e – → Mn 2+ + 4H 2 O
1 mole of MnO 4− require of 5 Faraday charge
∴ 5 mole of MnO −4 will require 25 Faraday charge
291. A current of 0.5 amperes is passed for 30
minute through a voltmeter containing CuSO4
solution. Find the Weight of Cu deposited.
(a) 3.18 g
(b) 0.318 g
(c) 0.296 g
(d) 0.150 g
AMU-2010
Ans. (c) : Given ,
current passed = 0.5 A,
time = 30 min = 30 × 60 s
we know that,
W = ZIt
E
Mol.wt./ 2
Z=
=
[ For Copper, E = Mol.wt/2]
96500
96500
63.5
1
W=
×
× 0.5 × 30 × 60 = 0.296g
2
96500
292. When one Faraday of electricity is passed
through CuSO4 solution, number of atoms
formed at cathode will be
(a) 6.02 × 1023
(b) 3.01 × 1023
(c) 2
(d) 6.02 × 10–23
Assam CEE-2018
2+
2−
Ans. (b): CuSO4 → Cu + SO 4
Anode
SO 24 − → 2e– + SO2 + O2
Cathode
Cu2 + 2e– → Cu
2F → 1mole of Cu
1
1F → mole of Cu
2
No. of atom = Anode × NA
1
= × 6.02 × 1023
2
= 3.01 × 1023
192
YCT
293. The charge required to liberate one gram
equivalent of an element is :
(a) 96500 F
(b) 1 F
(c) 1 C
(d) none of these
BCECE-2006
Ans. (b) : The charge required to liberate one gram
equivalent of an element is always equal to 1F.
1 Faraday = 96500 coulomb
294. Which one of the following can be considered
as a weak electrolyte?
(a) NaCl
(b) HCl
(d) K2SO4
(c) CH3COOH
BCECE-2011
Ans. (c) : CH3COOH we know that Acetic acid is a
weak acid because it won't liberate Hydrogen ion very
easily in a polar solvent. Means it is weak acid and also
called a weak electrolyte.
295. 1-faraday of charge is passed through a
solution of CuSO4, the amount of copper
deposit will be equal to its–
(a) gram equivalent weight
(b) gram molecular weight
(c) atomic weight
(d) electrochemical weight
BCECE-2016
Ans. (a) : 1 Faraday of
charge will deposit the
substance equal to its equivalent weight. So, 1F charge
gives copper equal to charge gram equivalent weight.
296. How long a current of 3 amperes has to be
passed through a solution of AgNO3 to coat a
metal surface of 80 cm2 and 0.005 mm thick
layer. Density of Ag is 10.5 g cm–3
(a) 125.1 seconds
(b) 12.5 seconds
(c) 155.2 seconds
(d) 200 seconds
[BITSAT-2008]
Ans. (a) : Given that,
i = 3amp
Density = 10.5g cm-3
Volume = Area × thickness
Mass = volume × density
∴
Mass of Ag to be deposited
80 × 0.005
=
× 10.5 = 0.42g
10
Eq.wt × i × t
Amount deposited =
96500
108 × 3 × t
0.42 =
96500
t = 125.1 seconds.
297. The number of coulombs required for the
deposition of 108 g of silver is
(a) 96500
(b) 48250
(c) 193000
(d) 10000
[BITSAT – 2011]
+
–
Ans. (a) : Ag + e → Ag(s)
Eq.wt × Q
Amt. deposited =
96500
Objective Chemistry Volume-II
107.870
×Q
96500
Q = 96500 C
298. A current of 2.0 A passed 5 hours through a
molten metal salt deposits 22.2 g of metal (At
wt. = 177). The oxidation state of the metal in
the metal salt is
(a) +1
(b) +2
(c) +3
(d) +4
[BITSAT – 2013]
Ans. (c) : Given that
T = 5hours
i = 2 amp
Q = i.t
= 5 × 2 × 60 × 60 = 36000C
Eq ⋅ wt × Q
m=
96500
m × 96500
Eq ⋅ wt =
Q
22.2 × 96500
=
36000
= 59.50
At. wt 177
Oxidation state =
=
= 2.97 ≈ 3
Eq.wt 59.50
299. The mass of the substance deposited when one
Faraday of charge is passed through its
solution is equal to
(a) relative equivalent weight
(b) gram equivalent weight
(c) specific equivalent weight
(d) None of the above
[BITSAT – 2015]
Ans. (b) : The mass of the substance deposited when
1Faraday of charge is passed through its solution is
equal to gram equivalent weight.
300. On passing a current of 1.0 ampere for 16 min
and 5 sec through one litre solution of CuCl2,
all copper of the solution was deposited at
cathode. The strength of CuCl2 solution was
(Molar mass of Cu = 63.5; Faraday constant
= 96,500 Cmol-1)
(a) 0.01N
(b) 0.01 M
(c) 0.02 M
(d) 0.2 N
[BITSAT – 2018]
Ans. (a) : Given that, i = 1amp
t = 16min.
= 16 × 60 + 5 = 965sec.
107.870 =
W
q
=
E 96500
(Where, q = it = charge of ion)
W
it
1× 965
1
=
=
=
E 96500 96500 100
By Faraday's Ist Law,
1
No. of equivalent 100
Normality =
=
Volume (in litre)
1
= 0.01N
193
YCT
301. A 100.0 mL dilute solution of Ag+ is
electrolysed for 15.0 minutes with a current of
1.25 mA and the silver is removed completely.
What was the initial [Ag+]?
(a) 2.32 × 10–1
(b) 2.32 × 10–4
–3
(c) 2.32 × 10
(d) 1.16 × 10–4
[BITSAT – 2018]
Ans. (d) : Given, V = 100 mL, t = 15 min, I = 1.25 mA
Q = I × t = 1.25 × 10–3 × 15 × 60 = 1.125 C
Cathodic reaction,
Ag+ + e– → Ag
1F → 108 g
∴
96500C → 108 g
108×1.125
1.125C →
96500
= 1.259 × 10–3 g
1.259×10−3 ×103
Conc. [Ag+] =
= 1.16×10−4
108×100
302. One coulomb of charge passes through a
solution of AgNO3 and CuSO4 connected in
series and the concentration of two solutions
being in the ratio 1:2. The ratio of amount of
Ag and Cu deposited on Pt electrode is
(a) 107.9:63.54
(b) 54:31.77
(c) 107.9:31.77
(d) 54:63.54
CG PET- 2016
+
–
Ans. (c) : Ag + 1e → Ag(s)
Cu2+ + 2e– → Cu(s)
As Faraday's IInd law states that ratio of the weight
deposited on the electron is equal to the ratio of the
equivalent weight of the substance.
w1 E1
Hence,
=
w 2 E2
w Ag
=
E Ag
w Cu E Cu
107.9
2
107.9
×
=
1
63.54 31.77
w Ag 107.9
=
w Cu 31.77
303. How much Coulomb needed to convert 1 mole
of MnO4 into Mn2+?
(a) 482500C
(b) 193000C
(c) 96500C
(d) 36500 C
CG PET-2013
−
−
2+
Ans. (a) : MnO 4 + 5e → Mn
Here, 5 moles of electrons are needed for reduction of 1
mole of MnO −4 to Mn2+
5 moles of electron = 5 Faraday
⇒ Quantity of charge needed = 5 × 96500
= 482500C
304. 4.5 g of aluminum (atomic mass 27 amu) is
deposited at cathode from Al3+ solution by a
certain quantity of electric charge. The volume
of hydrogen produced at STP from H+ ions in
solution by the same quantity of electric charge
will be
Objective Chemistry Volume-II
(a) 44.8 L
(c) 22.4 L
(b) 11.2 L
(d) 5.6 L
CG PET-2007
eq.wt × Q
96485
4.5 × 96485 m × 96485
=
27 / 3
1
4.5 × 3
=m
27
m = 0.5
0.5
Moles of H 2 =
= 0.25
2
Volume = 0.25 × 22.4
= 5.6L
305. The weight of Cu deposited, when 2 Faraday of
electricity is passed will be
(a) 31.75 g
(b) 23.85 g
(c) 63.5 g
(d) 125.67 g
CG PET-2018
Ans. (c) : Cu+2 + 2e– → Cu
2F = 1 mole Cu
= 63.5 gm
306. One litre of 1 M CuSO4 solution is electrolysed.
After passing 2F charge, the molarity of CuSO4
solution will be
M
(a)
(b) zero
2
M
(d) M
(c)
4
CG PET-2017
Ans. (b) : Given,
Molarity of CuSO4 solution = 1M, i.e. 1 mol/L
∴ 1 mol of Cu 2+ ions require 2F of charge for complete
deposition by electrolysis.
On passing 2F charge, all Cu 2+ ions deposited and no
Cu 2+ left in CuSO4 solution. So, the molarity of CuSO4
solution will be zero.
307. 2×10−5 C charge on Zn2+ ions corresponds to
(a) 2×10−5 mol of Zn2+ ions
(b) 6.22× 107 Zn2+ ions
(c) 0.64 mg of Zn2+ ions
(d) 1×10−10 mol Zn2+ ions
CG PET-2017
Ans. (c) : ∵ 2F amount of charge give 1 mol of Zn2+
ions.
or, 2 × 96500 C amount of charge give 1 mole of Zn2+
ions.
2 ×10−5
∴ 2 × 10−5 C charge will give =
2 × 96500
(mole of Zn2+ ions)
2+
−5
Amount of Zn (n) ions = 1 × 10
w
Also, n =
M
Ans. (d) : m =
194
YCT
Where,
n= number of moles
w= mass of Zn2+ ions
M= molar mass of Zn2+
∴ w = n × M = 10−5 × 65.5 (M molar Zn2+ =65.5)
= 65.5×10−5g
= 65.5 × 10 −5 × 103 milligram.
= 0.65 ≈ 0.64 milligram of Zn2+ions.
308. On passing 0.5 Faraday of electricity through
NaCl, the amount of Cl deposited on cathode is
(a) 35.5g
(b) 17.75g
(c) 71 g
(d) 142 g
CG PET-2015
Ans. (b) : Equivalent of Cl deposited = No. of Faraday
deposited wt. of Cl = 0.5 × Eq. wt
= 0.5 × 35.5
= 17.75 gm
309. How many cc of oxygen will be liberated by 2A
current flowing for 3 min 13s through acidulated
water?
(a) 11.2 cc
(b) 33.6 cc
(c) 44.8 cc
(d) 22.4 cc
CG PET-2010
Ans. (d) : Given,
i = 2A
t = 3 min 13 sec
= 3 × 60 + 13
t = 193 sec
∵
Q = it
Q = 2 × 193
= 386 C
2 F = 2 × 96500 C gives ½ mole O2
= 11200 cc of O2
11200 × 386
∴ 386 C will give O 2 =
2 × 96500
= 22.4 cc
310. How much electricity in terms of Faraday is
required for reduction of 2 mole Cr2O 72– into
Cr3+ in acidic medium ?
(a) 12 F
(b) 3 F
(c) 6 F
(d) 9 F
GUJCET-2022
WB-JEE-2017
Ans. (c) The reduction reaction in the acidic medium is
Cr2 O72− + 14H + + 6e − → 2Cr 3+ + 7H 2 O
Thus, six Faraday of charge is required.
311. For how much time, 10 ampere electric current
should be passed through a dilute aqueous
NiSO4 solution during electrolysis using inert
electrode, in order to get 5.85 gm Nickel?
[At. mass of Ni = 58.5 gm]
(a) 965 sec
(b) 3860 sec
(c) 1930 sec
(d) 9650 sec
GUJCET-2018
Objective Chemistry Volume-II
Ans.(c): Given that, i = 10amp
W = 5.85gm
According to Faraday's 1st law
W = zit
58.5
5.85 =
×10 × t
2 × 96500
t = 1930 sec
312. Two electrolytic cells containing molten
solutions of Nickel chloride & Aluminium
chloride are connected in series. If same
amount of electric current is passed through
them, what will be the weight of Nickel
obtained when 18 gm of Aluminum is
obtained? (Al=27 gm/mole, Ni=58.5 gm/mole)
(a) 58.5 gm
(b) 117 gm
(c) 29.25 gm
(d) 5.58 gm
GUJCET-2016, 2015
Ans. (a) : We know that, NiCl2 → Ni2+ + 2Cl–
AlCl3 → Al3+ + 3Cl–
According to Faraday's IInd law;
W∝Z
∵
Z∝E
∴
W∝E
W
∴
= Constant
E
Where, Z = Eelectrochemical equivalent
W = Mass liberated at electrode
E = Equivalent mass
WNi E Ni
=
WAl E Al
=
∴
M Ni / n Ni
M Al / n Al
∵
E=
M
n
WNi 58.5g / mol / 2
=
18g
27g / mol / 3
M = At. mass
WNi = 58.5g
n = Valency
313. These are physical properties of an element.
A. Sublimation enthalpy
B. Ionisation enthalpy
C. Hydration enthalpy
D. Electron gain enthalpy
The total number of above properties that
affect the reduction potential is........
(Integer answer)
[JEE Main-2021, 26 Aug Shift-I]
Ans. (3) : Sublimation enthalpy, Ionisation enthalpy
and hydration enthalpy affect the reduction potential.
The total number of above properties that affect the
reduction potential is 3.
314. Consider the cell at 25ºC
Zn|Zn2+ (aq), (1M)||Fe3+ (aq), Fe2+ (aq)|Pt(s)
The fraction of total iron present as Fe3+ ion at
the cell potential of 1.500 V is X × 10–2. The
value of X is .......(Nearest integer)
(Given, E oFe3+ / Fe2+ = 0.77V, E oZn 2+ / Zn = −0.76V)
195
[JEE Main-2021, 25 July Shift-I]
YCT
Ans. (24) : Cell reaction,
Fe3+ + e– → Fe2+
Zn → Zn2+ + 2e–
∴
E ocell = 0.77 + 0.76
= 1.53 V
E cell = E ocell −
∆H = –825.2 kJ mol–1
∆S° = ?
We know that,
∆G = –nF E ocell
And, ∆G = ∆H – T∆S
∴
–nF E ocell = ∆H – T∆S
E° = 0.77 V
E° = +0.76 V
0.0591 [Zn 2+ ][Fe 2+ ]2
log
2
[Fe3+ ]2
1.5 = 1.53 −
0.0591
[Fe ]
log
2
[Fe3+ ]2
2
∴
Fe3+
2+ = 10
Fe
Fe3+
1
=
Fe 2+
10
Fraction of Fe3+ ion =
Fe3+
0.316
=
3+
Fe + Fe
1.316
= 0.24
= 24 × 10–2
2+
∴
X = 24
315. Assume a cell with the following reaction
Cu(s) + 2Ag+ (1×10–3M)→ Cu2+ (0.250M) +
2Ag(s),
E ocell = 2.97V
E cell for the above reaction is..........V.
(Nearest integer)
[Given : log 2.5 = 0.3979, T = 298 K]
[JEE Main-2021, 22 July Shift-II]
Ans. (3) : Cu → Cu2+ + 2e–
2Ag+ + 2e– → 2Ag
∴
n=2
By Nernst equation,
0.0592
[Cu 2+ ]
log
2
[Ag+ ]2
0.250
0.0592
E cell = 2.97 −
log
(1×10−3 )2
2
Ecell = 2.89 V
E cell ≈ 3V
E cell = E ocell −
∴
∴
316. Consider the following cell reaction,
9
Cd ( s ) + Hg 2SO4 ( s ) + H 2O ( l ) ⇌
5
9
CdSO 4 ⋅ H 2O ( s ) + 2Hg ( l )
5
The value of E ocell is 4.315 V at 25ºC. If ∆ H o = –
825.2 kJ mol–1, the standard entropy change
∆ So in JK–1 is ..........(Nearest integer) [Given,
Farady constant = 96487 C mol–1]
[JEE Main-2021, 31 Aug Shift-I]
Ans. (25): Given,
E °cell = 4.315
T = 25°C
Objective Chemistry Volume-II
−nFE ocell −∆H
T
−2×96487 × 4.315 + 825.2×103
−∆S =
298
= –25.11
∆S ≈ 25 JK–1
317. Aluminium oxide may be electrolysed at
1000ºC to furnish aluminium metal (atomic
mass = 27 amu; 1 Faraday = 96,500 C). The
cathode reaction is
Al3+ + 3e– → Al0
To prepare 5.12 kg of aluminium metal by this
method would require
(a) 5.49×101 C of electricity
(b) 5.49×104 C of electricity
(c) 1.83×107 C of electricity
(d) 5.49×107 C of electricity
[AIEEE-2005]
Ans. (d) : Aluminium oxide electrolysed at 1000°C
Atomic mass of aluminium metal = 27 amu
1 Faraday = 96500 coulomb
Then we know that,
EQ
W=
96500
27 × Q
5.12 × 103 =
3 × 96500
512 × 103 × 3 × 965
Q=
27
= 5.49 × 107 C of electron
318. Two Faraday of electricity is passed through a
solution of CuSO4. The mass of copper
deposited at the cathode is (At. mass of Cu =
63.5 u)
(a) 0g
(b) 63.5g
(c) 2g
(d) 127g
[JEE Main-2015]
Ans. (b) : Give that,
Q = 2F
Valancy of the metal, Z = 2
Atomic mass of Cu = 63.59 u
CuSO4 → Cu2+ + SO 24−
Cu2+ + 2e– → Cu
E
Alternatively, W = ZQ = .2F = 2E
F
2 × 63.5
=
2
= 63.5
∴
2+ 2
196
−∆S =
YCT
Equivalent mass of copper =
Atomic mass
Valency
63.5
= 31.75
2
Amount of Copper deposited by 2 Faraday = 2 × 31. 75
= 63.5g
319. Potassium chlorate is prepared by the
electrolysis of KCl in basic solution
6OH − + Cl − → ClO 3− + 3H 2O + 6e − . If only 60%
of the current is utilized in the reaction, the
time (rounded to the nearest hour) required to
produce 10 g of KClO3 using a current of 2 A
is........
(Given : F = 96,500 C mol–1; molar mass of
KClO3 = 122 g mol–1)
[JEE Main-2020, 6 Sep Shift-I]
Ans. (11) : Cl − + 6OH − → ClO3− + 3H 2 O + 6e −
Given, i = 2 amp
∵
60% of current is used
=
60
= 1.2A
100
Zit
∵
W=
96500
Z
W=
× 1.2 × t
96500
122
1
10 =
×
× 1.2 × t
6 96500
122
∵ Equi.wt.of KClO3 = 6
t = 10.94 hr
t ≈ 11 hr
320. An acidic solution of dichromate is electrolysed
for 8 minutes using 2A current. As per the
following equation
Cr2 O 72- + 14H + + 6e- → 2Cr 3+ + 7H 2 O
The amount of Cr3+ obtained was 0.104 g. The
efficiency of the process (in%) is
(Take: F = 96000 C, atomic mass of chromium
= 52)..............
[JEE Main-2020, 3 Sep Shift-II]
Ans. (60) :
Actualobtained amount
% efficiency =
× 100
Theoreticalobtained amount
Actual amount = 0.104 g
We need to find theoretical obtained amount,
8 × 60 × 2
Moles of Cr3+ × 3 =
96000
52
3+
Moles of Cr =
g
300
0.104
% efficiency =
×100
52 / 300
30 × 104
=
= 60%
52
So,
i = 2×
Objective Chemistry Volume-II
321. Consider the following reaction,
MnO −4 + 8H + + 5e − → Mn2+ + 4H2O, Eº = 1.51V.
The quantity of electricity required in Faraday
to reduce five moles of MnO −4 is........................
[JEE Main-2021, 26 Feb Shift-I]
Ans. (25) :
1 mole of MnO −4 = 5 mole of electron required
=5×1F
(NA × e– = 1 F)
=5F
−
5 mole of MnO 4 = 5 × 5
= 25 F
The quantity of electricity required = 25 Faraday.
322. Amongst the following, the form of water with
the lowest ionic conductance at 298 K is
(a) saline water used for intravenous injection
(b) distilled water
(c) water from a well
(d) sea water
JEE Main-2020
Ans. (b) : Water from well or sea contains mineral ions
They have ionic conductance. Saline water is an
electrolyte and contains ions. It also has ionic
conductance.
Distilled water have lowest ionic conductance.
323. Three faradays of electricity are passed
through molten Al2O3, aqueous solution of
CuSO4 and molten NaCl taken in three
different electrolytic cells. Then the mole ratio
of Al, Cu and Na deposited on the cathode will
be
(a) 3 : 4 : 6
(b) 2 : 1 : 6
(c) 3 : 2 : 1
(d) 2 : 3 : 6
J & K CET-(2011)
Kerala-CEE-2010
J & K CET-(1998)
Ans. (d) : Mn+ + ne– → M
nF = 1 mole of M
Al3+ + 3e– → Al
Cu2+ + 2e– → Cu
Na+ + e– → Na
The mole ratio of Al, Cu and Na deposited at the
1 1
respective cathode is : :1
3 2
= 2 : 3: 6
324. Faraday’s constant is defined as
(a) charged carried by 1 electron
(b) charge carried by one mole of electrons
(c) charge required to deposit one mole of
substance
(d) charge carried by two moles of electrons.
J & K CET-2010
197
YCT
Ans. (b) : Faraday constant (F) is defined as charge
carried by one mole of electrons. Charge carried by
6.023 × 1023 electron (1 mol electrons)
= 1.6 × 10–19 × 6.023 × 1023
= 96368 = 1F
325. A certain current liberates 0.504 g of hydrogen
in 2 hours. How many grams of oxygen can be
liberated by the same current in same time?
(a) 2.0 g
(b) 0.4 g
(c) 4.0 g
(d) 8.0 g
J & K CET-2010
Ans. (c) : From Faraday's law, we get
W
= Constant
E
WO WH
=
EO
EH
Here,
WH = 0.504 (weight of hydrogen)
Eo = 8 (equivalent weight of oxygen)
EH = 1(equivalent weight of hydrogen)
0.504×8
∴
WO =
= 4.032
1
WO ≈ 4g
326. The highest electrical conductivity among the
following aqueous solutions, is of
(a) 0.1 M difluoroacetic acid
(b) 0.1 M fluoroacetic acid
(c) 0.1 M chloroacetic acid
(d) 0.1 M acetic acid
JCECE-2008
Ans. (a) : Fluoro group causes negative inductive
effect, thus increasing ionisation, hence 0.1 M
difluoroacetic acid has highest electrical conductivity,
H O
| ||
F−C−C−O−H
|
F
327. Aluminium oxide may be electrolysed at
1000ºC to furnish aluminium metal (atomic
mass = 27 amu; 1 F = 96,500 C). The cathode
reaction is
Al3+ + 3e- → Al
To prepare 5.12 kg of aluminium metal by this
method would require
(a) 5.49 × 101 C of electricity
(b) 5.49 × 104 C of electricity
(c) 1.83 × 107 C of electricity
(d) 5.49 × 107 C of electricity
JCECE-2008
Ans. (d):
Al3+ + 3e– → Al
W = ZQ
Where W = amount of metal
= 5.12 Kg
= 5.12 × 103g
Z = Electrochemical equivalent
Objective Chemistry Volume-II
Equivalent weight
96500
27
W=
3 × 96500
27
5.12 × 103 =
×Q
3 × 96500
5.12 × 103 × 3 × 96500
Q=
C
27
= 5.49 × 107C
328. Which of the following does not conduct
electricity?
(a) Fused NaCl
(b) Solid NaCl
(c) Brine solution
(d) Copper
JCECE-2009
Ans. (b) : Sodium Chloride does not have any free
mobile ion in its solid-state, which are responsible for
the conduction of electricity but in fused or aqueous
state. It conducts electricity as it has free mobile ions in
that state.
329. What is the time (in sec) required for
depositing all the silver present in 125 mL of 1
M AgNO3 solution by passing a current of
241.25 A? (1F = 96500 C)
(a) 10
(b) 50
(c) 1000
(d) 100
JCECE-2009
Ans. (b): 125 mL of AgNO3 solution will contain
108 × 125
= 13.5 Ag
=
1000
Ag+ + e– → Ag
(1F = 96500) 108 g
108 g of Ag is deposited by 96500 C
13.5 g of Ag is deposited by
96500
× 13.5 = 12062.5 C
=
108
∵ Q = it
=
Q 12062.5
=
= 50
i
241.25
330. Number of Faraday required to get 1 g atom of
Mg from MgCl2 is
(a) 0.0035 F
(b) 2 F
(c) 1 F
(d) 0.2 F
JCECE-2016
Ans. (b) : Mg2+ + 2e– → Mg
One gram atom means one mole magnesium from
MgCl2
MgCl2 → Mg2+ + 2Cl–
Hence 1 mole of MgCl2 required 2 Faradays of charge
1 g atom of Mg = 2g eq Mg = 2 Faraday
331. How many Faradays are required to reduce 1
mol of Br O3– to Br–.
(a) 3
(b) 5
(c) 6
(d) 4
JIPMER-2017
∴t=
198
YCT
Ans. (c) : BrO3− + 6H+ + 6e– → Br– + 3H2O
From the above reaction It is cleared that reduction of
BrO3– to Br– required to 6e– . Hence 6 Faradays is
required to 1 mole of BrO3– to Br–.
332. The charge required for the reduction of 1 mol
of MnO −4 to MnO2 is
(a) 1F
(b) 3F
(c) 5F
(d) 7F
Karnataka-CET-2018
AIIMS-2006
Ans. (d) : In purification process of copper impure
copper is anode and pure Copper is cathode. So increase
in the mass of cathode is due to Cu deposited on it.
Amount of impurity
= 22.26 – 22.011
= 0.2491 g
Amount of pure Cu deposited
E
W = Zit =
× it (Z = E/F, 1F = 96500 C)
96500
63.54
=
× 140 × 482.5
+7
2 × 96500
+4
−
Ans. (b) : MnO 4
→ MnO 2
= 22.239 g
From the above reaction reduction of MnO4– to MnO2 But increased mass of cathode = 22.011g
So, the amount of impurities (Fe)
require 3 mole of e– . so it required 3F charge.
333. The amount of the current in Faraday, is = 22.239 – 22.011
required for the reduction of 1 mole of Cr2 O 27 − = 0.228 g
Faraday's second law & Electrolysis
ions to Cr3+ is
= W 1 W 2 = E 1E 2
(a) 1 F
(b) 2 F
27.75 × 0.228
(c) 6 F
(d) 4 F
∴ wt.of Fe =
= 0.199
31.77
Karnataka-CET-2016
∴ % of iron in impure carbon
+6
+3
Ans. (c) : Cr2 O72−
→ Cr 3+
0.199
=
× 100
Reduction of Cr2 O72− to Cr 3+ change in oxidation
22.26
number ie change of 3 mole of electron per mole of Cr– = 0.89 ≈ 0.90
atom.
336. The approximate time duration in hours to
electroplate 30g of calcium from molten
∴ Cr2 o 72− has 2 moles of Cr atom. Hence total 6 mole of
calcium chloride using a current of 5 A is
electron required ie 6F of Charge (current) is required
2−
[At. mass of Ca = 40]
for reduction of 1 mole of Cr2 O7 ion.
(a) 80
(b) 10
334. How many Coulombs of electricity are
(c) 16
(d) 8
required for the oxidation of one mole of water
Karnataka-CET-2012
of dioxygen?
4
4
Ans.
(d):
Given,
W
=
30g,
I
=
5A
(a) 9.65 × 10 C
(b) 1.93 × 10 C
W=Z×I×t
(c) 1.93 × 105C
(d) 19.3 × 105C
Where,
Karnataka-CET-2015
Z = Electrochemical Equivalent
1
+
–
Eq ⋅ wt
Ans. (c) : H2O → 2H + O2 + 2e
30 =
×5× t
2
96500
1
H2O → O2
40
Eq. wt =
2
2
No. of faraday required for O = 2
40
Therefore 2 moles of electricity is required for oxidation
30 =
×5× t
of 1 mol of water.
2 × 96500
= 2 × 96500 C
(1F = 96500 C)
30 × 2 × 96500
t=
= 193000C
40 × 5
= 1.93 × 105C
t = 289505 second
335. Impure copper containing Fe, Au, Ag as
28950
impurities is electrolytically refined. A current
=
≈ 8 hours
3600
of 140 A for 482.5 s decreased the mass of the
anode by 22.26 g and increased the mass of 337. 9.65 C of electric current is passed through
cathode by 22.011 g. Percentage of iron in
fused alhydrous MgCl2. The magnesium metal
impure copper is
thus obtained is completely converted into a
(Given molar mass Fe = 55.5 g mol–1, molar
Grignard reagent. The number of moles of
mass Cu = 63.54 g mol–1)
Grignard reagent obtained is
(a) 0.95
(b) 0.85
(a) 5 × 10–4
(b) 1 × 10–4
–5
(c) 0.97
(d) 0.90
(c) 5 × 10
(d) 1 × 10–5
Karnataka-CET-2014
Karnataka-CET-2011
Objective Chemistry Volume-II
199
YCT
Ans. (c) :
96500 Coulombs of electric = 12 g of magnesium.
9.65 × 12
9.65 Coulombs of electric =
96500
= 1.2 × 10–3 g of magnesium
∴ The moles of Grignard reagent obtained is
1.2 × 10 –3
=
= 0.05 × 10−3 = 5 ×10−5
24
338. In the electrolysis of acidulated water, it is
desired to obtain 1.12 cc of hydrogen per
second under STP condition. The current to be
passed is
(a) 1.93A
(b) 9.65 A
(c) 19.3 A
(d) 0.965 A
Karnataka-CET, 2009
1.12
Ans. (b) : No. of moles of H2 =
22400
No. of equivalence of hydrogen
1.12 × 2
=
= 10−4
22400
No. of Faradays required = 10–4
∴ Current to be passed in one second
= 96500 × 10–4
= 9.65 A
339 How many moles of platinum will be deposited
on the cathode when 0.40 F of electricity is
passed through a 1.0 M solution of Pt4+?
(a) 0.60 mol
(b) 1.0 mol
(c) 0.40 mol
(d) 0.45 mol
(e) 0.10 mol
Kerala-CEE-2020
Ans. (e) : Faraday’s first law of Electrolysis state that
“the chemical deposition due to the same flow of
current through an electrolyte is directly proportional to
the quantity of electricity passed through it.
Pt4+ + 4e– → Pt
So, 4 moles of electricity or 4F of electricity is required
to deposit 1 mol of Pt.
1
0.40 F of electricity will deposit = × 0.40 = 0.10 mol
4
340. 1 C electricity deposits :
(a) half of electrochemical equivalent of Ag
(b) electrochemical equivalent of Ag
(c) 96500 g of Ag
(d) 10.8 g of Ag
Manipal-2016, MHT CET-2010
Ans. (b) : Faraday's first law,
W=Z×Q
Q=1C
W= Z
Z = electrochemical equivalent.
341. The number of moles of electrons passed when
current of 2A is passed through a solution of
electrolyte for 20 minutes is
(a) 4.1×10–4 mol e–
(b) 1.24×10–2 mol e–
–2
–
(c) 2.487×10 mol e
(d) 2.487×10–1 mol e–
MHT CET-2018
Objective Chemistry Volume-II
Ans. (c) : Q = I × t = 2A × 20 × 60 Sec = 2400C
1 mole of electrons = 96500C
Q
Numbers of mole of electron =
96500
2400
=
= 0.02487
96500
= 2.487 × 10–2 mol e–
342. How many Faradays of electricity are required
to deposit 10 g of calcium from molten calcium
chloride using inert electrodes?
(Molar mass of calcium = 40 g mol–1)
(a) 0.5 F
(b) 1 F
(c) 0.25 F
(d) 2 F
MHT CET-2016
Ans. (a) : Ca2+ + 2e– → Ca
Number of moles of
10g
Ca =
= 0.25 mol
40g / mol
1 mol Ca requires 2 moles of electrons = 2 faradays of
electricity.
0.25 mole Ca = 2 × 0.25 = 0.50 moles of electrons = 0.5
faraday's of electricity
343. According to Faraday's first law
96500 × E
E× I× t
(b) w =
(a) w =
I× t
96500
I × t × 96500
I× w
(d) E =
(c) E =
w
t × 96500
MHT CET-2011
Ans. (b) : According to faraday's first law
E× I× t
W = ZIt =
96500
The amount of substance that undergoes oxidation or
reduction at each electrodes during electrolysis is
directly proportional to the amount of electricity that
passes through the cell.
344. The amount of silver deposited on passing 2 F
of electricity through aqueous solution of
AgNO3
(a) 54 g
(b) 108 g
(c) 216 g
(d) 324 g
MHT CET-2007
Ans. (c) : Number of electron transferred = 1
Electricity passed = 2F
Number of moles = 2 moles
Amount of Ag deposited = moles × molecular weight
of Ag
= 2 × 108 = 216 g
345. The number of Faradays (F) required to
produce 20g of calcium from molten CaCl2
(Atomic mass of Ca = 40 mol-1) is
(a) 1
(b) 2
(c) 3
(d) 4
NEET-2020
200
YCT
Ans. (a) : Number of required moles ––
Mass
=
Molar Mass
20
= 0.5 mol
40
Electricity required to produce 1 mole of calcium = 2F
The electricity required to produce 0.5 mol of Ca = 0.5
× 2F
=1F
346. During the electrolysis of molten sodium
chloride, the time required to produce 0.10 mol
of chlorine gas using a current of 3 amperes is
(a) 55 minutes
(b) 110 minutes
(c) 220 minutes
(d) 330 minutes
NEET II-2016
→ Na + + Cl –
Ans. (b): NaCl
=
349. The weight of silver (at. wt. = 108) displaced by
a quantity of electricity which displaces 5600
mL of O2 at STP will be
(a) 5.4 g
(b) 10.8 g
(c) 54.0 g
(d) 108.0 g
(NEET-2014)
Ans. (d) : At STP 1 mole of oxygen occupies 22400
mL.
Hence, number of moles of oxygen corresponding to
5600
1
5600 mL =
mol O2= mol O2
4
22400
1
Weight of oxygen = × 32 = 8g
4
Equivalents of Ag = Equivalents of O2
Weight of Ag
Weight of O 2
=
Equivalent Weight of Ag Equivalent Weight of O 2
2Cl –
→1Cl 2 + 2e –
electrolyse 1mole
WAg
2F
M Ag
Q = It
Q
t=
I
For 1 mole of Cl2. produce = 2F = 2×96500C
Charge for 0.1 mole of Cl2 = 2 × 96500 × 0.1
Given I = 3A
2 × 96500 × 0.10
∴t =
3
t = 6433 sec or t = 107.22 min ≈ 110 min
WAg =
347. The number of electrons delivered at the
cathode during electrolysis by a current of 1
ampere in 60 seconds is (charge on electron =
1.60 × 10-19C)
(a) 6 × 1023
(b) 6 × 1020
20
(c) 3.75 × 10
(d) 7.48 × 1023
(NEET-II 2016)
Ans. (c): Q = I × t
Q = 1 × 60 = 60C
Now 1.60 × 10–19 C = 1 electron
60
∴ 60C =
1.6 × 10−19
= 37.5 × 1019
= 3.75 × 1020 electrons
348. When 0.1 mol MnO 42− is oxidized, the quantity
of electricity required to completely oxidise
MnO 42− to MnO 4− is
(a) 96500 C
(b) 2 × 96500 C
(c) 9650 C
(d) 96.50 C
(NEET-2014)
Ans. (c) : The oxidation reaction
MnO 24 −
→ MnO 4− + e −
Q = 0.1 × F = 0.1× 96500C
= 9650C
The Quantity of electricity require to oxidizing MnO 24 −
completely will be 0.1 faraday i.e 9650 C.
Objective Chemistry Volume-II
WO2
M O2
=
WO2
M O2
× M Ag
8
WAg = × 4 × 108
32
WAg = 108 gm
350. How many grams of cobalt metal will be
deposited when a solution of cobalt (II)
chloride is electrolyzed with a current of 10
amperes for 109 minutes? (1 Faraday = 96, 500
C;) Atomic mass of Co = 59 u)
(a) 4.0
(b) 20.0
(c) 40.0
(d) 0.66
(Karnataka NEET 2013)
EIt
Ans. (b) : W = ZIt =
96500
59
Equivalent weight of cobalt (II) =
u
2
E × It
W= B
96500
59 10 × 109 × 60
M B 59
W=
×
∴ E B = z = 2
2
96500
W = 19.9 ≈ 20
351. Al2O3 is reduced by electrolysis at low
potentials and high currents. If 4.0 × 104
amperes of current is passed through molten
Al2O3 for 6 hours, what mass of aluminium is
produced? (Assume 100% current efficiency,
at. mass of Al = 27 g mol–1)
(a) 8.1 × 104 g
(b) 2.4 × 105 g
4
(c) 1.3 × 10 g
(d) 9.0 × 103 g
(AIPMT -2009)
Ans. (a) : Al2O3 → 2Al3+ + 3O2–
At Cathode
Al3+ + 3e– → Al
First we calculate the equivalent weight
201
YCT
E=
Mole wt
Valency
E=
27
3
27
= Z × 96500
3
9
Z=
96500
Now faraday's first law. W = Z × I × t
9
W=
× 4 × 104 × 6 × 3600
96500
9 × 4.0 ×104 × 6 × 3600
=
96500
7776 ×104
=
965
= 8.1 × 104 g
352. For the cell reaction:
2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(aq)
E0cell = 0.24 V at 298 K. The standard Gibbs'
energy (∆rG0) of the cell reaction is
[Given that Faraday constant, F=96500 C mol–1]
(a) 23.16 kJ mol–1
(b) –46.32 kJ mol–1
–1
(c) –23.16 kJ mol
(d) 46.32 kJ mol–1
(NEET – 2019)
o
Ans. (b) : The relation between E cell and ∆ r G o is as
follows:
o
∆ r G o = −nFE cell
For the cell reaction,
2Fe3+ + 2I– → 2Fe2+ + I2
n=6–4
=2
E ocell = 0.24 V
o
∆ r G o = −nFE cell
= –2 × 96500 × 0.24 J mol–1
= –46.32 × 103 J mol–1
= –46.32 kJ mol–1
353. The equilibrium constant of the reaction,
Cu(s) + 2Ag + (aq )
→ Cu 2+ (aq) + 2Ag(s)
0
Ecell
= 0.46V at 289 K is
(a) 2.4 × 1010
(b) 2.0 × 1010
10
(c) 4.0 × 10
(d) 4.0 × 1015
UP CPMT-2011
Ans. (d) : Cu(s) + 2Ag + (aq)
→ Cu 2 + (aq) + 2Ag(s)
E 0cell = 0.46 V at 298 K
0.059
E 0cell =
log K c
2
0.059
0.46 =
log K c
2
0.46
log K c =
= 15.59
0.0295
Kc = Antilog 15.59
Kc = 3.89 ×1015 = 4 × 1015
Objective Chemistry Volume-II
354. Aluminium oxide may be electrolysed at
o
1000 C to furnish aluminum metal (atomic
mass = 27 amu; 1F = 96,500 C). The cathode
reactions is Al3+ + 3e– → Al
To prepare 5.12 kg of aluminium metal by this
method would enquire
(a) 5. 49 × 101 C of electricity
(b) 5. 49 × 104 C of electricity
(c) 1.83 × 107 C of electricity
(d) 5.49 × 107 C of electricity
UPTU/UPSEE-2007
3+
–
Ans. (d): Al + 3e → Al
1g of Al is obtained by passing a Current of 3 ×
96500
C
27
∴ 5.12 × 103 g of Al is obtained by passing
mole of Al
1
Mole Ratio =
=
moles of electrons 3
Charge 1
Moles of Al =
×
96500 3
5.12 × 103
Q
1
=
×
27
96500 3
5.12 × 103 × 3 × 96500
Q=
= 5.49 × 107 C
27
355. In acidic medium MnO4 is converted to Mn2+.
The quantity of electricity in faraday required
to reduce 0.5 mole of MnO4 to Mn2+ would be:
(a) 2.5
(b) 5
(c) 1
(d) 0.5
UPTU/UPSEE-2006
Ans. (a) : In MnO −4 the oxidation number of Mn is +7
Mn+5 + 5e– → Mn+2
The quantity of electricity needed
= (1F mol–1) × (5 × 0.05 mol) = 0.25F
356. In electrochemical reaction of which set of
reactants, the metal displacement will not take
place ?
(a) Mg + Cu2+
(b) Pb + Ag+
2+
(c) Zn + Cu
(d) Cu + Mg2+
UPTU/UPSEE-2018
Ans. (d): In electrochemical reaction of Cu + Mg2+, the
metal displacement will not take place.
Displacement reaction occurs when a metal form the
electrochemical series is mixed with the ions of metal
lower down in the series. In the electrochemical series
Mg is placed above Cu.
Hence metal displacement reaction won't occur in Cu +
Mg2+. In Pb + Ag+ and Zn + Cu2+, lead and Zinc are
more reactive than silver and copper, hence they can
Displace silver and copper respectively from their salt
solution.
357. One faraday of current was passed through the
electrolytic cells placed in series containing
solution of Ag+, Ni2+ and Cr3+ respectively. The
ratio of amounts of Ag, Ni and Cr deposited
will be
(At. wt. of Ag = 108, Ni = 59, Cr = 52)
202
YCT
(b) 17.4 : 29.5 : 108
Zn2+ + 2e– → Zn ; 2F
Al3+ + 3e– → Al ; 3F
(d) 108 : 59 : 52
Ag+ + e– → Ag ; 1F
UPTU/UPSEE-2013
Ans. (a) : When 1 Faraday of current was passed The quantity of electricity separately requirement is 2 :
through the electrolytic cell then amount deposited will 3 : 1
360. What amount of electricity can deposit 1 mole
be
of Al metal at cathode when passed through
+
–
Ag + e → Ag
molten AlCl3?
108
(a) 0.3 F
(b) 1 F
For Ag, w =
= 108g
1
(c) 3 F
(d) 1/3 F
WB-JEE-2018
(i) Ni2+ + 2e– → Ni
3+
–
Ans. (c) : AlCl3 → Al + 3Cl
59
w=
= 29.5g
At Cathode : Al3+ + 3e– → Al
2
1 mole electron = Faraday charge
(iii) Cr3+ + 3e– → Cr
3F → 1 mol
52
3 moles electrons = 3 Faraday Charge (3F)
w=
= 17.3
3
It means the conversion of every aluminium ion to
So, the ratio of amount of Ag : Ni : Cr =108 : 29.5 : aluminium atom requires three electrons.
17.5
361. Assertion : One mole of silver deposits by 1
Faraday charge.
358. On passing ‘C ampere of current for time ‘t’
Reason
: Faraday charge required depends upon
sec through 1 L of 2 (M) CuSO4 solution
no
of
electron.
(atomic weight of Cu=63.5), the amount ‘m’ of
(a) If both Assertion and Reason are correct and
Cu (in gram) deposited on cathode will be
Reason is the correct explanation of
(a) m = Ct/(63.5 × 96500)
Assertion.
(b) m = Ct/(31.25 × 96500)
(b) If both Assertion and Reason are correct, but
(c) m = (C × 96500)/(31.25×t)
Reason is not the correct explanation of
Assertion.
(d) m = (31.75 × C × t)/96500
(c) If Assertion is correct but Reason is incorrect.
WB-JEE-2012
(d) If both the Assertion and Reason are
Ans. (d) : Faraday's law of electrolysis m ∝ Ct, M =
incorrect.
ZCt
AIIMS 25 May 2019 (Morning)
C = Current
Ans. (a) : 1 Faraday charge deposits one gram
t = time
equivalent of any substance.
There for the formula Z
Ag + + e −
→ Ag (s)
1F
Equivalent weight of metal
1mol
1mol(Deposited)
Z=
The quantities of substances liberated at electrodes
96500
depend upon charge on the ions being deposited.
63.5
Eq. Wt. of Cu
362.
Assertion : Galvanised iron does not rust.
2
Reason: Zinc has a more negative electrode
63.5
potential than iron.
Z=
2 × 96500
(a) If both Assertion and Reason are correct and
63.5 × C × t 31.75 × C × t
the Reason is a correct explanation of the
m=
=
Assertion.
2 × 96500
96500
(b) If both Assertion and Reason are correct but
359. The quantity of electricity needed to separately
Reason is not a correct explanation of the
electrolyze 1 M solution of ZnSO4, AlCl3 and
Assertion.
AgNO3 completely is in the ratio of
(c) If the Assertion is correct but Reason is
(a) 2 : 3 : 1
(b) 2 : 1 : 1
incorrect.
(c) 2 : 1 : 3
(d) 2 : 2 : 1
(d) If both the Assertion and Reason are
incorrect.
WB-JEE-2014
nd
(e)
If the Assertion is incorrect but the Reason is
Ans. (a) : From the Faradya's 2 law,
correct.
E
[AIIMS-2005]
W ∝ ZIt ∝
It
96500
Ans. (a) : Galvanised iron does not rust easily because
zinc (which is present in the form of coating) has more
1
1
negative electrode potential (–0.76V) than iron
It ∝ or electricty ∝
E
E
( –0.41V). Hence Zn is less reactive than Fe.
(a) 108 : 29.5 : 17.3
(c) 1 : 2 : 3
Objective Chemistry Volume-II
203
YCT
363. Time taken to completely (in hrs) decompose
36 g water by passing 3A current is:
(a) 35.8 hrs
(b) 40 hrs
(c) 51.8 hrs
(d) 22.5 hrs
[AIIMS-27, May, 2018 (E)]
Ans. (a) : Completely decompose = 36 g
Current = 3A
2H2O → 2H2 + O2
2 moles of H2O = 4 moles of e–
So, 4F of charge needed for completely decompostion.
Now, Q = I × t
Q 4 × 96500
t=
=
I
3
386,000
=
3
= 128666.6 sec
128666
=
3600
= 35.74 ≈ 35.8 hrs.
3.
Ans. (a): Given –
*
C = 0.01N, R = 220ohm, G = 0.88cm −1 = Λ eq = ?
Now,
*
G = κ⋅R
*
Where- G = Cellconstant
κ = conductivity
∴
R = resistance
0.88 −1
κ=
cm ohm −1
220
κ = 0.004 cm −1ohm −1
∴
Λ eq =
κ × 1000
molarity
0.004 × 1000
Λ eq =
0.01
Λ eq = 4 × 100
Λ eq = 400ohm cm 2 g eq −1
366. The EMF of the cell,
Mg|Mg2+ (0.01M)|| Sn2+ (0.1 M) | Sn at 298K is
(E
Cell Constant
°
Mg 2+ /Mg
°
= −2.34V,ESn
= − 0.14V
2+
/Sn
)
(a) 2.17 V
(b) 2.23 V
364. The cell potential for Zn|Zn2+ (aq)||Snx+|Sn is
(c) 2.51 V
(d) 2.45 V
0.801 V at 298 K. The reaction quotient for the
VITEEE- 2011
above reaction is 10−12. The number of
electrons involved in the given electrochemical Ans. (b) : The given half cell reaction is :
cell reaction is ……. .
→ Mg 2+ + 2e − ;E °Mg2+ / Mg = −2.34 V
Mg
oxidation
0
0
(Given
EZn2+ |Zn = –0.763V,ESnx+ |Sn = +0.008V
Sn 2+ + 2e −
→ Sn; E °Sn 2+ /Sn = −0.14 V
reduction
2.303RT
and
= 0.06V )
E cell = E°Sn − E °Mg
F
= − 0.14 − ( −2.34 )
JEE Main 25.07.2022, Shift-I
Ans. (4) : Given that,
= 2.2V
Cell potential (E) of Zn Zn 2 + Sn x + Sn = 0.801V
Form the Nernst equationReaction quotient (Q) = 10−12
E oZn 2+ / Zn = 0.763 V
E cell = E cell −
o
ESn
= 0.008 V
x+
/ Sn
2.303RT
log Q
nF
0.06
0.801V = 0.771 −
log10−12
n
n=4
The total number of electrons involve in this
electrochemical cell reaction is 4.
365. The resistance of 0.01 N solution of an
electrolyte was found to be 220 ohm at 298 K
using a conductivity cell with a cell constant of
0.88cm-1. The value of equivalent conductance
of solution is –
(a) 400 ohm cm2 g eq-1 (b) 295 ohm cm2 g eq-1
(c) 419 ohm cm2 g eq-1 (d) 425 ohm cm2 g eq-1
JEE Main-2019
E = Eo =
Objective Chemistry Volume-II
E cell
Mg 2+
0.0591
log 2+ ( where n = 2 )
n
Sn
0.0591
[ 0.01]
log
= 2.2 −
2
[ 0.1]
Given − Mg 2+ = 0.01
Sn 2+ = 0.1
0.0591
1
log
2
10
Ecell = 2.2 + 0.029
or
Ecell = 2.229
or
Ecell ≈ 2.23 V
367. In the reversible reaction.
k
2NO 2 ↽ 1 ⇀ N 2O 4
E cell = 2.2 −
204
k2
the rate of disappearance of NO2 is equal to
2k1
2
(a)
[ NO 2 ]
k2
(b) 2k1 [ NO 2 ] − 2k 2 [ N 2 O 4 ]
2
YCT
(c) 2k1 [ NO 2 ] − k 2 [ N 2 O 4 ]
Ans. (a) : ∆G° = –115 kJ at 298K.
Now, ∆G° = –2.303 RT log kp
R = 8.314 JK–1 mol–1 & T = 298K.
VITEEE- 2011
∆G° = –2.303 × 8.314 × 298 × log kp
2
(d)
( 2k1 − k2 )[ NO2 ]
Ans. (c) : In the reversible reaction
k
2NO 2 ↽ 1 ⇀ N 2O 4
−115 ×103
= 20.155
−2.303 × 8.314 × 298
log kp ≈ 20.16
370. The
emf
of
the
Daniel
Cell
Zn ZnSO4 (0.01M) CuSO 4 (1M) Cu at 298 K is
E1. When concentration of ZnSO4 is changed to
1M and that of CuSO4 is changed to 0.01 M.
the emf changed to E2. Then find the
relationship between E1 and E2.
(a) E1>E2
(b) E1<E2
(c) E1=E2
(d) E2 =0 ≠ E1
AP EAPCET-6 Sep. 2021, Shift-II
Ans. (a): According to Nernst equation:–
log k p =
k2
The concentration of NO2 is decrease in the
forward reaction and concentration increase in
backward reaction.
The rate of disappearance of NO2 in the forward
d[NO 2 ]
reaction is −
= 2k1[NO2 ]2
dt
Rate of formation of NO2 in the reversible reaction is
d[NO 2 ]
−
= k 2 [N 2O 4 ]
dt
∴ Rate of disappearance of
NO 2 = 2k1 [ NO2 ] − k2 [ N2O4 ]
2
368. For the following cell reaction,
Ag | Ag + | AgCl | Cl – | Cl 2 , Pt
Ecell = E οcell –
∆G °f ( AgCl ) = −109kJ/mol
0.0591
[Zn 2+ ]
log
n
[Cu 2+ ]
0.0591
[0.01]
log
2
1
0.0591
∆G f° ( Ag + ) = 78kJ/mol
= 1.1 −
× log[1×10−2 ]
2
E° of the cell is
0.0591
(a) – 0.60 V
(b) 0.60 V
= 1.1 +
×2
2
(c) 6.0 V
(d) None of these
= 1.16 V
VITEEE- 2009
Again, when concentration changed,
Ans. (a) : For the given cell,
Ag|Ag+|AgCl|Cl– |Cl2,Pt
0.0591
1
Ecell = E οcell –
log
The cell reactions are as follows
2
0.01
+
−
At anode :
Ag
→ Ag + e
0.0591
Ecell= 1.16 –
log102
−
At cathode : AgCl + e
→ Ag(s) + Cl−
2
Ecell= 1.16 – 0.0591
Net cell reaction: AgCl
→ Ag + + Cl−
°
°
°
Ecell= 1.041 V
∴ ∆G
= ∑ ∆G − ∑ ∆G
= 1.1 −
∆G °f ( Cl − ) = −129kJ/mol
reaction
p
R
So, E1 >E 2
= ( 78 − 129 ) − ( −109 )
= –51 + 109
= + 58 kJ / mol
∆G° = –nFE°
58 × 103 J = −1 × 96500 × E°cell
371. Calculate the maximum work that can be
obtained from the cell,
Zn|Zn2+(1 M)| |Ag+(1M)|Ag
Where E° Zn2+ |Zn = − 0.76 V and E° Ag + |Ag = 0.80 V
−58 × 1000
96500
−58,000
°
E cell =
96500
= – 0.6010 V ≈ –0.60V
369. The standard free energy change of a reaction
is ∆G° = –115kJ at 298 K. Calculate the
equilibrium
constant
kp
in
log
kp
−1
−1
( R = 8.314 Jk mol ) .
E °cell =
(a) 20.16
(c) 2.016
(a) −301.080 kJ
(c) 112.830 kJ
(b) 201.830 kJ
(d) 212.630 kJ
AMU-2013
Ans. (a) : E °cell = E °cathode − E °anode
= 0.80 – (– 0. 76)
= + 1.56 V
Zn|Zn2+ (1M) || Ag+ (1M) | Ag
Zn + e– → Zn–
Wmax = ∆G° = – nFE°
= – 2 × 96500 × 1.56
(b) 2.303
= – 301080.00 J
(d) 13.83
VITEEE- 2008
= – 301.080 kJ
Objective Chemistry Volume-II
205
YCT
372. For a cell reaction involving a two electron
change, the standard emf of the cell is found to
be 0.295 V at 25ºC. The equilibrium constant of
the reaction at 25ºC will be
(a) 1×10–10
(b) 29.5×10–2
(c) 10
(d) 1×1010
COMEDK-2010
0.059
log K
Ans. (d) : E °cell =
n
0.059
log K
∴0.295 =
n
0.295 × 2
log K =
= 10
0.059
10
K = 1 × 10
373. Given the data at 25ºC
Ag + I– → AgI + e–; Eº = 0.152 V
Ag → Ag+ +e– ; Eº = – 0.8 V
What is the value of log Ksp for AgI?
RT
= 0.059V
2.303
F
(a) –8.12
(b) +8.612
(c) –37.83
(d) –16.13
[AIEEE 2006]
+
–
Ans. (d) : Ag → Ag + e ; E° = – 0.8 V
Ag → Ag+ + I ; E° = – 0.952
0.059
E °cell =
log K sp
n
0.0952
– 0.952 =
log K sp
1
0.952
log Ksp =
0.059
= – 16.135
375. The standard electrode potential Eº and its
dEo
temperature coefficient
for a cell are 2V
dT
and –5 × 10–4 VK–1 at 300 K respectively. The
cell reaction is Zn(s) + Cu2+ (aq)→ Zn2+ (aq) +
Cu (s). The standard reaction enthalpy (∆r H Θ )
at 300 K in kJ mol–1 is, [Use, R = 8 JK–1 mol–1
and F = 96,000 C mol–1]
(a) – 412.8
(b) – 384.0
(c) 206.4
(d) 192.0
AIEEE-2007
Ans. (a) : According to Gibbs – Helmholtz equation
∆G = ∆H – T ∆S
….(i)
∆G = – nFE °cell
....(ii)
On substituting the given values in equation
∆G = – 2 × 96000 C mol–2 × 2 V
= – 4 × 96000 J mol–1
= – 384000 J mol–1
dE°
∆S = nF
dT
∆S = 2 × 96000 C mol –1 × (–5 ×10–4 VK–1)
= – 96 JK–1 mol–1
∆G = ∆H – 300 K × (– 96 JK–1 mol–1)
∆H = – 384000 – 28800 J mol–1
= – 4128000 J mol–1
= – 412.800 kJ mol–1
376. Calculate the standard cell potential (in V) of
the cell in which following reaction takes place
Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag(s)
Given that, EºAg+/Ag = x V,
Eo Fe2+ /Fe = yV
Eo Fe3+ /Fe = z V
o
374. Given, ECr
= 0.72 V,
3+
/Cr
(a) x + 2y – 3z
(b) x – y
(c) x + y – z
(d) x – z
[JEE Main-2019, 8 April Shift-II]
o
EFe
= – 0.42 V
2+
/Fe
The potential for the cell Cr|Cr3+ (0.1M)|| Fe2+
(0.01M)| Fe is
(a) 0.26 V
(b) 0.399 V
(c) – 0.339 V
(d) – 0.26 V
[AIEEE 2008]
Ans. (a) : Cr → [Cr3+ + 3e–] × 2
Reduction half – cell
Fe2+ + 2e– → Fe] × 3
Net cell reaction
2Cr + 3Fe2+ → 2Cr3+ + 3Fe
n=6
°
°
°
E cell = E oxi + E red = 0.72 – 0.42 = 0.30 V
0.0591
10−2
log ⋅ −3
10
n
0.059
4
= 0.30 –
log10
6
Ecell = 0.2606 V
Ecell = E °cell –
Objective Chemistry Volume-II
Ans. (a) : Fe2+ + 2e– → Fe; E °Fe2+ / Fe = yV
∆G1° = – 2Fy
Fe → Fe ; E
3+
……(i)
°
Fe 2+ / Fe
∆G °2 = – 3Fz
= zV
……(ii)
Ag + e → Ag ; E° = + xV
∆G °3 = – Fx
…… (iii)
+
–
On adding equation (i) & (iii) and subtracting from (ii)
we get.
Fe2+ + Ag+ → Ag + Fe3+
∆G = ∆G1° + ∆G °3 – ∆G °2
– FE °cell = – 2Fy – Fx – (–3Fz)
– E °cell = – 2y – x + 3z
E °cell = x + 2y – 3z
206
YCT
377. The standard Gibbs energy for the given cell
reaction in kJ mol-1 at 298K is
Zn(s) + Cu2+ (aq) → Zn2+ (aq)
+Cu(s), Eº = 2 V at 298 K
(Faraday's constant, F = 96000 C mol–1)
(a) 384
(b) 192
(c) –384
(d) –192
[JEE Main-2019, 9 April Shift-I]
2+
Ans. (c) : Zn + Cu → Zn2+ + Cu
Zn + 2e– → Zn2+
Cu + 2e– → Cu2+
∆G = – nFE °cell
= – 2 × 96000 × 2 = – 384000 J
= – 384 kJ
378. For an electrochemical cell Sn(s)|Sn2+ (aq,
Sn 2+
1M)||Pb2+ (aq, 1 M)|Pb (s) the ratio 2+
Pb
when this cell attains equilibrium is ....... .
Ans. (d) : ∆G = – nFEcell
∆G is negative if Ecell is positive
Cu → Cu2+ (C1) + 2e– : E°
Cu2+ (C2) + 2e– → Cu (s) : E° .
Cu2+ (C2) → Cu+2 (C1) E °cell = 0
2.303RT
log Q
nF
C
2.30 RT
Ecell = 0 –
log 1
nF
C2
Since Ecell > 0
C1
< 1 = C1 < C 2 .
C2
381. Given that the standard potentials (Eº) of Cu2+
/Cu and Cu+/ Cu are 0.34 V and 0.522 V
respectively, the Eº of Cu2+/Cu+ is
(a) – 0.158 V
(b) + 0.158 V
(c) – 0.182 V
(d) 0.182 V
[JEE Main-2020, 7 Jan Shift-I]
Given : EoSn2+ |Sn = − 0.14V,
Ans.
(b)
:
2.303RT
o
Cu2+ + e– → Cu+; ∆G1 = – nFE1° = – 1× F × E1°
= 0.06
EPb2+ |Pb = − 0.13V,
F
Cu+ + e– → Cu; ∆G2 = – 1 × F× 0.522
[JEE Main-2020, 8 Jan Shift-II] Cu2+ + 2e– → Cu ; ∆G = –2 × F × 0.34
3
Ans. (2.1544) : E °Sn 2+ /Sn = − 0.14V
Now ∆G3 = ∆G1 + ∆G2
– 2 × F × 0.34 = – FE1° – 0.522F
E°
= − 0.13V
Pb 2+ / Pb
E1° = 0.158 V .
2.303RT
For cell reaction =
= 0.06
F
Sn 2+
0.06
E = E° –
log 2+
n
Pb
0.01× 2
Sn 2 +
=
= log 2 +
Pb
0.06
382. For the reaction,
2Fe3+ (aq)+ 2I– (aq) → 2Fe2+ (aq) + I2 (s)
The magnitude of the standard molar Gibbs
free energy change, ∆r G om = – ...........kJ (Round
off to the nearest integer).
1
Sn 2+
= 2+ = 10 3 = 2.1544
Pb
379. An oxidation - reduction reaction in which 4
electrons are transferred has a ∆Gº of 17.37 kJ
mol–1 at 25 ºC. The value of Eocell (in V) is.....×
10–2. (1F = 96, 500 C mol–1)
[JEE Main-2020, 5 Sep Shift-I]
–2
Ans. (–6×10 ) : We Know that, ∆G° = – nFE
17.37 × 1000 = – 3 × 96500 × E°
E° = 17370/ (– 3 × 96500)
E° = – 579/9650 Volt
= – 0.06
= – 6 × 10–2 Volt
380. For the given cell;
Cu(s)|Cu2+ (C1M)||Cu2+ (C2M)|Cu(s)
change in Gibbs energy (∆G) is negative, if
C
(a) C1 = C2
(b) C 2 = 1
2
(c) C1 = 2C2
(d) C 2 = 2C1
[JEE Main-2020, 6 Sep Shift-II]
Objective Chemistry Volume-II
Ecell = E °cell –
EoFe2+ /Fe( s) = - 0.440V;
Eo - = 0.539V;
l 2 /2l
Eo 3+
= - 0.036V
Fe /Fe( s )
F = 96500C
[JEE Main-2021, 18 March Shift-I]
Ans. (45) : Given that,
E oFe2+ / Fe = −0.440V
E oI / 2I− = 0.539V
2
E
o
Fe3+ / Fe
= −0.036V
By given equation–
2Fe3+ + 2I − → 2Fe 2+ + I 2
Since,
Now, nE1o + nE o2 = nE3o
(Where, n = no. of e– transferred)
E oFe3+ / Fe2+ + 2E oFe2+ / Fe = 3E oFe3+ / Fe
207
YCT
Ans. (3776) : Given that,
E2 – E1 = x × 10–4 V
2.303RT
= 0.059
F
According to question, reaction is–
MnO −4 + 8H + + 5e − → Mn 2+ + 4H 2 O
By Nernst equation–
E oFe3+ / Fe2+ + 2 × (− 0.440) = 3 × (− 0.036)
E oFe3+ / Fe2+ = −108 + 0.880
= + 0.772
E ocell = E ocathode − E oanode
= 0.772 – 0.539
E ocell = 0.233
Then by Gibbs free energy–
∆G = − nFE ocell
o
−
E1 = E cell
∆G = – 2 × 0.233 × 96500
∆G = 44,969 J
∆G = 45 kJ
383. Emf of the following cell at 298 K in V is x ×10–
2
Zn|Zn2+ (0.1M)|| Ag+ (0.01 M)|Ag
The value of x is.............
(Rounded off to the nearest integer).
[Given, EoZn2+ /Zn = -0.76V,
1
[Mn +2 ]
2.303RT
log + 8 ×
−
nF
[H ] [MnO 4 ]
0.059
1
log 8
n
1
0.059
o
E1 = E cell
−
log1
n
0.059
o
E1 = E cell
−
×0
[∵ log1 = 0]
n
o
E1 = E cell
−0
o
E1 = E cell
−
o
E1 = E cell
EoAg+ /Ag = +0.80V,
1
[Mn +2 ]
2.303RT
o
And E 2 = E cell
−
log −4 8 ×
2.303RT
−
= 0.059 ]
nF
[10 ] [MnO 4 ]
F
[JEE Main-2021, 26 Feb Shift-II] E = E o − 0.059 log1032
2
cell
n
Ans. (147.15×10–2) : Zn → Zn2+ + 2e–
+
–
0.059
× 32
2Ag → 2Ag + 2e
E 2 = E1 −
5
Zn + 2Ag +
→ Zn 2+ + 2Ag
0.059 × 32
E 2 − E1 = −
E °cell = E °Ag+ / Ag − E °Zn 2+ / Zn
5
0.059 × 32
We know that
E 2 − E1 =
5
E °cell = 0.80 – (– 0.76)
E
−
E
=
0.3776
2
1
E °cell = 1.56
−4
E
−
E
2
1 = 3776 × 10 V
According to nearest integer
x × 10 −4 V = 3776 × 10 −4 V
Zn 2+
0.059
Ecell = 1.56 –
log
So, x = 3776
2
2
[ Ag + ]
Ecell = 1.56 –
0.059
0.1
log
2
2
( 0.01)
385. The Gibbs energy change (in J) for the given
reaction at
[Cu2+] = [Sn2+] = 1M and 298 K is Cu(s) + Sn2+
(aq)→Cu2+ (aq) + Sn (s);
EoSn2+ /Sn = -0.16V,
0.059
×3
2
Ecell = 1.56 – 0.0885
EoCu2+ /Cu = 0.34V,
Ecell = 1.4715
Ecell = 147.15 × 10–2
Take, F = 96500 C mol–1
[JEE Main-2020, 2 Sep Shift-I]
384. The magnitude of the change in oxidising
power of the MnO -4 /Mn2+ couples is x × 10–4 V, Ans. (96500) : Given that,
E °Sn 2+ |Sn = − 0.16 V
if the H+ concentration is decreased from 1 M
–4
to 10 M at 25 ºC. (Assume concentration of
E °Cu 2+ |Cu = − 0.34V
2+
+
MnO 4 and Mn to be same on change in H
F = 96500 C mol–1
concentration). The value of x is .........
At anode : − Cu → Cu 2+ + 2e −
(Rounded off to the nearest integer).
At cathode: − Sn 2+ + 2e− → Sn
2.303RT
= 0.059
Cu + Sn 2+ → Cu 2+ + Sn
Given,
F
°
E cell = E °cathode − E°anode
[JEE Main-2021, 24 Feb Shift-II]
Ecell = 1.56 –
Objective Chemistry Volume-II
208
YCT
2.303 × 0.025 E °cell = – 0.16 – 0.34
E °cell = – 0.50
Then,
Now, change in Gibbs Energy ––
∆G = – nF E °cell
κ
1.07 × 106
⇒
1
G
0.243
G * = 0.243 × 1.07 × 106
G * = 0.26001 × 10 ≅ 26 × 104 m–1
G* =
∆G = – 2 × 96500 × – 0.50 = 96500 J
388. According to the expression ∆G o = – nFEo, the
386. For the disproportionation reaction
cell reaction is spontaneous when
2Cu + ( aq ) ⇌ Cu ( s ) + Cu 2+ ( aq ) at 298 K, ln K
(Notations and symbols carry their usual
meanings)
(where is the equilibrium constant) is............
(a) ∆G o is positive
(b) ∆G o is zero
×10-1
o
(c) E is negative
(d) Eo is positive
Given :
J & K CET-(2017)
(EoCu2+ /Cu+ = 0.16V,
Ans. (d) : ∆G° = – nFE°
RT
We get the –
EoCu+ /Cu = 0.52Vand
= 0.025)
F
For a spontaneous process, Gibb's free energy change
[JEE Main-2020, 2 Sep Shift-II] must be negative. So, E° should be positive.
389. The resistance of 0.01 m KCl solution at 298 K
Ans. (144×10-1) : Given that,
°
is 1500 Ω. If the conductivity of 0.01 m KCl is
E Cu 2+ |Cu + = 0.16 V
solution at 298 K is 0.146 × 10-3 S cm-1. The cell
°
E Cu + |Cu = 0.52V
constant of the conductivity cell in cm-1 is
RT
(a) 0.219
(b) 0.291
= 0.025
F
(c) 0.301
(d) 0.194
Ecell = 0
Kerala-CEE-29.08.2021
Then,
Ans. (a) : Given that,
°
°
°
R = 1500 Ω
E cell = E Cu + |Cu − E Cu 2+ |Cu +
κ = 0.146 × 10-3 Scm–1
= 0.52 – 0.16 V
Cell constant (G*) = ?
o
E cell = 0.36V
Now, G* = R κ
G* = 1500 × 0.146 × 10-3
2.303RT
o
E cell = E cell
−
log K
G* = 0.219 cm–1
nF
390. The standard reduction potentials for two half0.025
0 = 0.36 −
log K
cell reactions are given below.
1
Cd 2+ (aq) + 2e – → Cd(s); Eº = –0.40 V
2.303 × 0.025 log K = 0.36
Ag + (aq) + e – → Ag(s); Eº = 0.80 V
0.36
ln K =
The standard free energy change for the
0.025
reaction
ln K = 14.4
2Ag + (aq) + Cd(s) → 2Ag(s) + Cd 2+ (aq)
ln K = 144 × 10–1
is given by :
387. If the conductivity of mercury at 0ºC is
(a) 115.8 kJ
(b) – 115.8 kJ
1.07×106 S m–1 and the resistance of a cell
(c)
–
231.6
kJ
(d)
231.6 kJ
containing mercury is 0.243 Ω, then the cell
4
–1
Manipal-2018
constant of the cell is x × 10 m . The value of x
+
–
Ans. (c) : 2Ag + 2e → 2Ag ; E° = 0.80 V
................. (Nearest integer)
2Cd + 2e– → 2Cd2+ ; E° = 0.400 V
[JEE Main-2021, 1 Sep Shift-II]
2Ag+ + Cd → Cd2+ + 2Ag
Ans. (26) : Given that,
E° = 1.20 V
Conductivity κ = 1.07 × 106 S m–1
∆G° = – nE°F
Resistance R = 0.243 Ω
= – 2 × 1.20 × 96500
G* = x × 104 m–1
= – 231.6 kJ
We know –
391. Which among the following solutions is not
1
used in determination of the cell constant?
G=
& κ = G × G*
R
(a) 10–2 M KCl
(b) 10–1 M KCl
1
(c) 1 M KCl
(d) Saturated KCl
G=
Ω −1
MHT CET-2015
0.243
Objective Chemistry Volume-II
209
YCT
Ans. (d) : Saturated KCl solution is not used in the
determination of the cell constant.
While its equivalent conductance is inversely
proportional to its normality for a given volume Vcm3.
So, cell constant will also change with the change in
molarity or normality of KCl solution.
392. For the reduction of silver ions with copper
metal, the standard cell potential was found to
be + 0.46 V at 25 oC. The value of standard
Gibbs energy, ∆Go will be (F = 96500 C mol-1)
(a) – 89.0 kJ
(b) + 89.0 kJ
(c) – 44.5 kJ
(d) –98.0 kJ
(AIPMT -2010)
Ans. (a) : Cu + 2Ag+ → Cu2+ + 2Ag
We know that,
∆G° = – nFE °cell
= – 2 × 96500 × 0.46
= – 88780 J
= – 88.780kJ
≃ −89kJ
393. The Gibbs' energy for the decomposition of
o
Al2O3
at
500
C
is
as
follows
2
4
Al O → Al + O 2 , ∆rG = + 966 kJ mol–1
3 2 3
3
The potential difference needed for the
electrolytic reduction of aluminium oxide
(Al2O3) at 500 oC is at least
(a) 4.5 V
(b) 3.0 V
(c) 2.5 V
(d) 5.0 V
(AIPMT -Mains- 2012)
2
4
→ Al + O 2
Ans. (c) : Al2 O3
3
3
∆r G = + 966 kJ mol–1
The half cell reactions for the given reaction can be
written as,
2
Anode O32−
→ O 2 + 4e −
3
2
4
Cathode Al 2 O3 + 4e −
→ Al
3
3
∆rG = – nFE °cell
When ∆rG is change in gibbs free energy, n is number
of electrons transferred, F is Faraday constant, E °cell is
cell potential.
966 × 103 = –(4 × 96500 × E °cell )
Ans. (a): ∆G° = Work done by cell
E °cell < 0
∆G° = – n × F × E °cell
∆G° = n × F × E °cell
∆G° > 0
Nernst Equation
RT
Ecell = E °cell –
log K eq
nF
Ecell = 0
RT
E °cell = −
ln K eq
nF
RT ln K = –n × F × E °cell
RT ln K = – ∆G°
If
Keq < 1
If
Keq < 0
395. A hypothetical electrochemical cell is shown
below:
A|A+ (x M)|| B+(y M) | B
The emf measured is + 0.20V. The cell reaction
is
(a) A+ B+ → A+ + B
(b) A++ B → A + B+
(c) A+ + e– → A; B+ + e– → B
(d) The cell reaction cannot be predicted.
(AIPMT -2006)
Ans. (a): Electrochemical Cell –The EMF of cell is +
0.20 V. So cell reaction is possible. The half cell
reactions are given as follows.
At positive pole
B+ + e– | | → B (reduction)
At negative pole.
A | | → A+ + e– (Oxidation)
Cell reaction is
A + B+ | | → A+ + B;
°
E cell = + 0.20V
396. Standard electrode potential for Sn4+/Sn2+
couple is +0.15 V and that for the Cr3+/Cr
couple is –0.74 . These two couples in their
standard state are connected to make a cell.
The cell potential will be
(a) +1.19V
(b) + 0.89V
(c) + 0.18V
(d) + 0.83 V
(AIPMT -2011)
4+
–
2+
°
Ans. (b) : Sn + 2e → Sn ; E RP = 0.15 V
966 × 103
Cr3+ + 3e– → Cr
; E °RP = – 0.74V
4 × 96500
= – 2.5 V
Eocell = E °RPSn – E °POCr
The potential difference needed for the electrolytic
Eocell = 0.74 + 0.15
reduction of aluminium oxide = 2.5 V
Eocell = 0.89 V
394. If the E0cell for a given reaction has a negative
value, which of the following gives the correct 397. A button cell used in watches function as
following :
relationship for the values of ∆G0 and Keq?
Zn(s)+Ag2O(s)+H2O(l)
(a) ∆G0 > 0; Keq <1
(b) ∆G0 > 0; Keq > 1
2+
–
(c) ∆G0 < 0; Keq > 1
(d) ∆G0 < 0; Keq < 1
↽ ⇀ 2Ag(s)+Zn (aq)+ 2OH (aq)
(NEET- II 2016, 2011)
If half cell potentials are
E °cell = −
Objective Chemistry Volume-II
210
YCT
Zn2+(aq) + 2e- → 2Ag(s) +;E0 = -0.76V
0.059
log K
Ans. (a) : E °cell =
Ag2O(s)+H2O(l)+2e- → 2Ag(s)+2OH-(aq);E0 =
2
0.34
0.059
The cell potential will be
E °B+ / B − E °A2+ / A =
log k
2
(a) 0.84V
(b) 1.34V
(c) 1.10V
(d) 0.42V
0.059
E °B+ / B – 0.34 =
×15.6
(NEET-2013)
2
Ans. (c): The formula of standard cell potential is
E °B+ / B = 0.80V.
E °cell = E °right − E°left
401. Consider the following cell reaction:
= 0.344 – (– 0.76)
→
2Fe(s) + O 2 (g) + 4H + (aq)
E °cell = 0.344 + 0.076
2+
0
2Fe (aq) + 2H 2 O(I);E = 1.67V
E ° = + 1.10 V
cell
398. Assertion : The cell potential of mercury cell is
1.35 V, which remains constant.
Reason : In mercury cell, the electrolyte is
paste of KOH and ZnO.
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is in
correct.
(d) If both the Assertion and Reason are
incorrect.
[AIIMS-2008]
Ans. (b) : The cell potential remains constant during its
life as the overall reaction does not involve any ion in
the solution whose concentration may change during its
life time.
Zn + HgO → ZnO + Hg
399. The standard EMF for the cell reaction,
Zn + Cu 2+
→ Cu + Zn 2+ is1.1 volt at 25 º C .
The EMF for the cell reaction, when 0.1 M
Cu2+ and 0.1 M Zn2+ solutions are used, at 25
ºC is
(a) 1.10V
(b) 1.10V
(c) –1.10V
(d) –0.110V
[AIIMS-27, May, 2018 (M)]
Ans. (a) : Zn + Cu2+ → Cu + Zn2+
At[Fe 2+ ] = 10-3 M,p(O 2 ) = 0.1 atm and pH =3,
the cell potential at 25 ºC is
(a) 1.47 V
(b) 1.77V
(c) 1.87 V
(d) 1.57 V
[AIIMS-2017]
+
2+
Ans. (d) : 2Fe + O2 + 4H → 2Fe + 2H2O
The half cell reactions are
2Fe → 2Fe2+ + 4e–
O2 + 4H+ + 4e– → 2H2O
Given that,
[Fe2+] = 10–3 M, p (O2) = 0.1 atm, pH = 3,
0.059
(10−3 ) 2
log −3 4
4
(10 ) (0.1)
= 1.67 – 0.10
= 1.57 V
402. The cell constant of a given cell is 0.47 cm-1.
The resistance of solution placed in this cell is
measured to be 31.6 ohm. The conductivity of
the solution (in S cm-1 where S has usual
meaning) is
(a) 0.15
(b) 1.5
(c) 0.015
(d) 150
[AIIMS-2012]
Ans. (c) : The conductivity ( κ ) is the product of
reciprocal resistance.
l 1
1
κ = × =
× 0.47 cm −1
R a 31.6 Ohm
= 0.015 S cm–1
403. The products formed when an aqueous solution
of NaBr is electrolysed in a cell having inert
electrode are :
(a) Na and Br2
(b) Na and O2
(c) H2 Br2 and NaOH
(d) H2 and O2
[AIIMS-2006]
Ans. (c) : NaBr ↽ ⇀ Na+ + Br–
E = E° −
Cu 2+
0.059
log 2+
2
Zn
Ecell
= 1.10 + 0
Ecell
= 1.10 V
400. Cell equation:
A + 2B 2+ → A 2+ + 2B
A 2+ + 2e- → A
E 0 = +0.34V
and log10 K = 15.6 at 300K for cell reactions.
Na+ + OH– → NaOH
Find E0 for B + + e - → B
At Cathode, 2H2O + 2e– → H2 + 2OH–
2.303 RT
Given
= 0.059 at 300K
At Anode,
Br– + e– → Br
F
Br +Br
→ Br2
(a) 0.80
(b) 1.26
So the products are H2 and NaOH (at Cathode) and Br2
(c) –0.54
(d) +0.94
[AIIMS-26, May, 2018 (M)] (At Anode).
Ecell = E °cell +
Objective Chemistry Volume-II
211
YCT
406. Copper reduces NO 3− into NO and NO2
4. Nernst Equation
depending upon the concentration of HNO3 in
solution. (Assuming fixed [Cu2+] and
404. At 298 K, the standard electrode potentials of
PNO = PNO2 ), the HNO3 concentration at which
Cu2+/Cu, Zn2+/Zn, Fe2+/Fe and Ag+/Ag are 0.34
the thermodynamic tendency for reduction of
V,−0.76 V, −0.44 V and 0.80 V respectively.
NO 3− into NO and NO2 by copper is same is 10x
On the basis of standard electrode potential,
predict which of the following reaction can not
(Rounded-off to the
M. The value of 2x is
occur?
nearest integer)
(a) 2CuSO4(aq) +2Ag(s) → 2Cu(s) +Ag2SO4(aq)
[Given,
°
(b) CuSO4(aq) +Zn(s) ZnSO4(aq) + Cu(s)
ECu
= 0.34 V, E°NO /NO = 0.96 V, E°NO /NO = 0.79 V
/Cu
(c) CuSO4(aq) +Fe(s) → FeSO4(aq) + Cu(s)
RT
(d) FeSO4(aq) + Zn(s) →ZnSO4(aq) + Fe(s)
and at 298 K,
( 2.303 ) = 0.059]
F
NEET-17.06.2022
JEE Main 25.02.2021, Shift-II
Ans. (a) : Reactivity order : Zn > Fe > Cu > Ag
−
3
2+
−
3
2
Ans. : (4) Given, E °Cu 2+ / Cu = 0.34V , T = 298 K
SRP: E oZn 2+ / Zn < E oFe2+ / Fe < E oCu 2+ / Cu < E oAg+ / Ag
More reactive metals (lower SRP) can displace less
reactive metals (higher SRP) from their salt solution, in
case of displacement reaction.
CuSO4 (aq) + Ag(s) → Cu(s) + AgSO4 (aq)
So, this reaction is not possible because Ag is less
reactive metal as compare to Cu.
405. For
yields
Cr2O72- + 14H + + 6e –
→ 2Cr 3+ + 7H 2O.E°
E °NO − / NO = 0.96V ,
3
RT
( 2.303) = 0.0591
F
E °NO − / NO = 0.79V
3
2
Reaction is given below:–
Anode: Cu → Cu 2+ + 2e −
…..(i)
−
+
−
Cathode: 3e + 4H + NO3 → NO + 2H 2 O …(ii)
Multiply in eq. (i) by 3 and eq. (ii) by 2, we get –
= 1.33 Vat Cr2O72- = 4.5 millimole. [Cr3+] =
3Cu → 3Cu 2+ + 6e −
....(iii)
1.5 millimole and E = 1.067 V. Then calculate
+
−
−
8H + 2NO3 + 6e → 2NO + 4H 2 O ....(iv)
the pH of the solution.
8H + + 3Cu ( s ) + 2NO3− → 3Cu 2+ + 2NO + 4H 2 O
(a) 2
(b) 3
(c) 2.5
(d) 1.5
By adding equation (iii) and (iv),
AP EAPCET 19-08-2021, Shift-II
°
∵E °cell = E ox
+ E °red
Ans. (a) : Given data, E ° = 1.33V
= 0.34 + 0.96V
Cr2 O72− = 4.5millimole
°
E cell = 1.30 V
Ecell = 1.067V
3
2
pH = ?
Cu 2+ [ NO ]
0.0591
2−
+
−
3+
(E cell )1 = 1.30 −
log
…(v)
Now, for Cr2 O7 + 14H + 6e → 2Cr + 7H 2 O
8
2
6
H + NO3−
From Nernst equation we get–
Cu 2+ [ NO ]
Q1 =
8
2
H + NO3−
3
(15 ×10 )
0.059
1.067 = 1.33 −
log
6
( 4.5 ×10 )( H )
(15 ×10 ) = 0.263 × 6
log
−3 2
−3
−3 2
or
or
or
or
(4.5)(H + )14
0.059
50 ×10−3
log
= 26.74
H + 14
log(50 × 10–3) – log(H+)14 = 26.74
–1.3010 – 14log[H+] = 26.74
–14log[H+] = 28.041
28.041
pH =
14
pH = 20
Objective Chemistry Volume-II
+ 14
2
Anode :– Cu → Cu 2+ + 2e − ….(vi)
Cathode :– 2H + + NO3− + e− → NO 2 + H 2 O …(vii)
Multiply by 2 in eqn (vii), we get4H + + 2NO −3 + 2e − → 2NO 2 + 2H 2 O ....(viii)
Cu + 4H + + 2NO3− → 2NO 2 + 2H 2 O + Cu 2+
By adding eq n (vi) and(viii)
E °cell = 1.13 V
Cu 2+ [ NO 2 ]
Q2 =
4
2
H + NO3 −
2
Cu 2+ [ NO 2 ]
0.0591
(E cell ) 2 = 1.13 −
log
.....(ix)
4
2
2
H + NO3−
2
212
YCT
Equating equation (v) and (ix), we get (E cell )1 = (E cell ) 2
c
d
(a) E(cell) = E °cell − RT log [C] a[D]b
nF
[A] [B]
c
RT
[C]
[D]d
(b) E(cell) = E °cell −
ln
nF [A]a [B]b
0.0591
0.0591
log Q1 = 1.13 −
log Q 2
6
2
0.0591
0.17 =
[log Q1 − 3log Q2 ]
6
3
6
12
2
Cu 2+ [ NO ] NO3− H +
0.0591
0.17 =
log
2
8
3
6
6
NO3− H + Cu 2+ [ NO 2 ]
1.3 −
(c) E(cell) = E °cell +
(d) E(cell) = E °cell +
4
4
NO3− H +
0.0591
0.17 =
log
4
6
[ NO 2 ]
0.0591
0.17 =
× 8 log ( HNO3 )
6
log ( HNO3 ) = 2.16
RT
[C]c [D]d
log
[A]a [B]b
nF
RT [C]c [D]d
ln
nF [A]a [B]b
J & K CET-(2019)
Ans. (b) :
∆G = −nFE cell
∆G ° = −nFE °cell
∆G = ∆G ° + RT ln Q
−nFE cell = −nFE °cell + RT ln Q
HNO3 = 10 2.16 M = 10 x M
x = 2.16
2x = 4.32 ≈ 4
RT
ln Q
nF
Product [C]c [D]d
º
=
Q=
407. H 2 (g) + 2AgCl(g) ⇌ 2Ag(s) + 2HCl(aq),Ecell
at
Reactant [A]a [B]b
25ºC for the cell is 0.22V. The equilibrium
RT [C]c [D]d
°
constant at 25ºC is
−
E cell = E cell
ln
7
8
nF [A]a [B]b
(a) 2.8 × 10
(b) 5.2 × 10
(c) 2.8 × 105
(d) 5.2 × 104
410. Standard cell voltage for the cell
Kerala-CEE-29.08.2021
Pb |Pb2+ ||Sn2+ | Sn is – 0.01 V. If the cell is to
o
exhibit Ecell = 0, the value of
Ans. (a) : Given that, E cell = 0.22V
[Sn2+]/[Pb2+] should be antilog of –
Equilibrium constant (Kc) = ?
(a) + 0.3
(b) 0.5
0.0592
(c) 1.5
(d) – 0.5
∴ We know that, E ocell =
log K c
n
VITEEE-2018
0.0592
Ans.
(a)
:
We
use
the
Nernst
equation
–
0.22 =
log K c
2
0.059 [ Reduction ]
E cell = E o −
ln
log K c = 7.43
2
[ Oxidation ]
Kc = 107.43
Given, Ecell =0,
E° = – 0.01V
or
Kc = 2.8×107
Now, Oxidation and reduction of the cell is–
408. For the reaction,
2NH 3 ( s ) + CO 2 ( g ) ⇌
Pb → Pb 2+ + 2e − − (Oxidation)
E cell = E °cell −
NH 2CONH 2 ( aq ) + H 2O ( ℓ ) find the value of
equilibrium constant at 295 K. Given, standard
Gibbs energy change at the given temperature
is 13.9 kJ mol–1.
(a) 2.88 × 102
(b) 2.58 × 102
2
(c) 2.40 × 10
(d) 2.65 × 102
AP EAMCET (Engg.) 21.09.2020, Shift-II
Ans. (a) : ∆Gº = –2.303 nRT log Kc
–13.9 × 103 J = –2.303 × 1 × 8.314 × 295 log Kc
−13.9 ×103
⇒
log K c =
= 2.46
−2.303 × 1× 8.314 × 295
2.46
2
⇒
Kc = 10 ⇒ 288.4 = 2.88 × 10
Hence, option (a) is correct.
409. For n-electron redox reactions of the type
aA+bB→cC +dD, the cell potential can be
expressed as
Objective Chemistry Volume-II
Sn 2+ + 2e − → Sn − (Reduction)
Pb + Sn 2+ → Sn + Pb 2+ (Full reaction)
∴
or
∴
0 = −0.01 +
Sn 2 +
0.059
log 2 +
2
Pb
Sn 2+ 0.02
ln 2+ =
Pb 0.059
Taking antilog both side –
Sn 2+
= Antilog ( 0.3)
Pb 2+
411. Which of the following relation represents
correct relation between standard electrode
potential and equilibrium constant?
213
YCT
I. log K =
II. K = e
nFEo
2.303 RT
nFEo
RT
III. log K =
-nFEo
2.303 RT
nFEo
RT
Choose the correct statement (s).
(a) I, II and III are correct
(b) II and III are correct
(c) I, II and IV are correct
(d) I and IV are correct
[BITSAT – 2017]
°
Ans. (c) : ∆G = – 2.303 RT log K
IV. log K = 0.4342
– nFE ° = – 2.303 RT log K
T = 300K
∆G° = –RT ln K
– 690. 9R = – R × 300 × ln K
690.9
ln K =
= 2.303
300
2.303 log K = 2.303
log K = 1
K = 101 = 10
Unit of K = (atm)∆n = (atm)2–3
Unit of K = (atm)–1
K = 10 atm–1
414. If for the cell reaction, Zn + Cu 2+ ⇌ Cu + Zn 2+
Entropy change ∆S° is 96.5 J mol–1K-1, then
temperature coefficient of the emf of a cell is
(a) 5 × 10-4 VK-1
(b) 1 × 10-3 VK-1
-3
-1
(c) 2 × 10 VK
(d) 9.65 × 10-4 VK-1
VITEEE-2013
Ans. (a) : We know that
dE
∆G = ∆H − nFT
….(i)
dT P
and from Gibbs free energy
∆G = ∆H – T∆S
….(ii)
from (1) and (2) we get
∆S
dE
=
dT P nF
nFE °
1
×
…… (i)
2.303 RT
nFE °
log K = 0.4342
…… (ii)
RT
nFE °
lnK =
RT
nFE°
K = e RT
…… (iii)
96.5
dE
Hence, equation (I), (II) and (IV) are correct.
=
dT P 2 × 96500
412. λ° m for NH4Cl, NaOH and NaCl are 130, 248
–1
2
–1
= 5 ×10 −4 VK −1
and 126.5 ohm cm mol respectively. The
λ° m of NH4OH will be
415. The degree of dissociation (α) of a weak
(a) 251.5
(b) 244.5
electrolyte, AxBy is related to van't Hoff factor
(c) 130
(d) 504.5
(i) by the expression
COMEDK 2015
i +1
i −1
(a) α =
(b) α =
Ans. (a) : Given:
x
+
y
−
1
(x
+
y − 1)
λ°m( NH4Cl) = 130ohm –1cm 2 mol –1
x + y −1
x + y +1
(c) α =
(d) α =
λ°m( NaOH) = 248ohm –1cm 2 mol –1
i −1
i −1
SRMJEEE – 2008
λ°m( NaCl) = 126.5ohm –1cm 2 mol –1
Ans. (b) : The dissociation of weak electrolyte is given
λ°m( NH4OH) = λ°m( NH4Cl) + λ°m( Na OH ) – λ°m( NaCl)
as :
λ°m( NH4OH) = 130+248–126.5
A B ⇌ xA+ + yB –
log K =
x
λ°m( NH4OH) = 251.5ohm –1cm 2 mol –1
lnitially
y
L
0
0
At equi 1 − α xα yα
413. For the reaction,
2SO 2 ( g ) + O 2 ( g ) ⇌ 2SO 3 ( g ) at 300 K, the Total moles = 1 – α + xα + yα = 1 + α(x + y – 1)
value of Go is – 690.9 R. The equilibrium ∴ Vant 's Hoff factor
constant value for the reaction at that
Observed valueof colligative property
=
temperature is (R is gas constant)
Calculated colligative property
(a) 10 atm-1
(b) 10 atm
1
+
α
(x
+
y
− 1)
(c) 10
(d) 1
i=
1
WB-JEE-2015
i = 1 + α (x + y − 1)
Ans. (a) : For the reaction
or
2SO2 + O2 ⇌ 2SO3
i −1
Given that,
α=
( x + y − 1)
∆G° = – 690.9R,
Objective Chemistry Volume-II
214
YCT
5.
Ans. (c) : Given that,
2
–1
Λ °m NaCl = 126.5 S cm mol
Conductance and Conductor
416. Resistance of a conductivity cell filled with 0.1
mol L-1 NaCl is 100 Ohm. If the resistance of
the same cell when filled with 0.02 mol L-1 NaCl
solution is 258 Ohm. The conductivity of
0.02mol L-1 NaCl solution is (Conductivity of
0.1 L-1 NaCl is 1.29 Sm-1)
(a) 1.0 S m-1
(b) 0.2 S m-1
-1
(c) 2.0 S m
(d) 0.5 S m-1
AP-EAMCET-05.07.2022, Shift-II
Ans. (d) : For 0.1 mol L–1 NaCl solution,
Conductivity (K) = 1.29 S m–1, Resistance (R) = 100 Ω
∵ Cell constant = Conductivity × Resistance
= 1.29 × 100 = 129
For,
0.02 mol L–1 NaCl solution
Resistance = 258 Ω
Cell constant
Conductivity (K) =
Resistance
129
=
258
= 0.5 S m−1
Λ °m HCl = 426.16 S cm mol
2
–1
Λ °m CH3COONa = 91.0 S cm mol
2
–1
Λ °m CH3COOH = Λ °m CH3COONa+ Λ °m HCl – Λ °m NaCl
= 91.0 + 426.16 –126.5
= 390.71 S cm2 mol–1
419. Unit of equivalent conductance is :
(a) ohm −1cm 2 (g − eq) −1 (b) ohm cm(g − eq)
(c) ohm cm2 (g − eq)−1
(d) ohm −1 cm(g − eq)−1
SRM JEE 2014
BCECE-2005,
J & K CET-(2007, 2006, 2003)
Ans. (a) : The equivalent conductance is calculated by
multiplying the specific conductivity and the volume of
the solution having one gram equivalent of the
electrolyte.
1000
Equivalent conductivity = κ ×
Normality
ohm −1cm
417. At 25°C, the molar conductances at infinite
=
dilution for the strong electrolytes NaOH, NaCl
g equivalent cm −1
–4
–4
and BaCl2 are 248 × 10 , 126 × 10 and 280 ×
= ohm–1 cm2 (g equivalent)–1
10–4 Sm2 mol–1 respectively, λ m° Ba(OH)2 in Sm2
= ohm −1cm 2 (g − eq) −1
mol–1 is
420. The highest electrical conductivity of the
(a) 52.4 × 10–4
(b) 524 × 10–4
–4
following aqueous solutions is of
(c) 402 × 10
(d) 262 × 10–4
(a) 0.1 M difluoroacetic acid
BITSAT – 2021, COMEDK 2014
VITEEE- 2009
(b) 0.1 M fluoroacetic acid
AP EAMCET (Engg.)-2009
(c) 0.1M choloroacetic acid
Ans. (b) : Given that,
(d) 0.1 M acetic acid
BaCl2 = 280 × 10–4
UPTU/UPSEE-2007
NaOH = 248 × 10–4
AIEEE-2005
NaCl = 126 × 10–4
Ans. (a) : 0.1 M difluoroacetic acid has highest number
BaCl2 + 2NaOH
→ Ba(OH) 2 + 2NaCl
of ions in solution because of higher degree of
dissociation.
λ ∞m Ba(OH)2 = λ ∞m BaCl2 + 2λ ∞m NaOH − 2λ ∞m NaCl
Di fluoro acetic acid is strong acid due to inonization
= 280 × 10 −4 + 2 × 248 × 10 −4 −2 × 126 × 10−4
enthalpy (IE) of fluorine atom.
= ( 280 + 496 − 252 ) × 10−4
= 524 × 10−4 Sm 2 mol−1
418. The molar conductance of NaCl, HCl and
CH3COONa at infinite dilution are 126.45,
426.16 and 91.0 S cm2 mol–1 respectively. The
molar conductance of CH3COOH at infinte
2+
–
dilution is Choose the right option for your 421. Equivalent conductance's of Ba and Cl ions
–1
–1
–1
are 127 and 76 ohm cm eq respectively.
answer.
Equivalent conductance of BaCl2 at infinite
(a) 540.48 S cm2mol–1 (b) 201.28 Scm2 mol–1
2
–1
2
–1
dilution is
(c) 390.71 S cm mol
(d) 698.28 Scm mol
(a) 139.5
(b) 101.5
(NEET-2021)
(c)
203
(d) 279
MHT CET-03.05.2019, SHIFT-I
VITEEE- 2011
UP CPMT-2013
(AIPMT -Mains 2012, 1997)
(AIPMT -2000)
Objective Chemistry Volume-II
215
YCT
Ans. (a) : Chemical equation––
κ
∵
Λm =
BaCl2 → Ba2+ + 2Cl–
C ×1000
From the Kohlrausch law ––
0.0715 Sm −1
1 ∞ 1 ∞
Λm =
λ∞ = λ+ + λ−
n
n
5.0 ×10−3 mol L−1 × 1000
1 ∞
1 ∞
Λ m = 14.3 mSm2 mol–1
λ ∞ (BaCl2 ) = × λ Ba 2+ + × λ Cl−
2
1
424. A KCl solution of conductivity 0.14 S m–1
127
shows a resistance of 4.19 Ω in a conductivity
=
+ 76
cell.
If the same cell is filled with an HCl
2
solution, the resistance drops to 1.03Ω. The
= 63.5 + 76
conductivity of the HCl solution is............×10–2
= 139.5 ohm–1 cm–1 eq–1
S m–1 (Round off to the nearest integer)
422. If an electrolytic solution has specific resistance
[JEE Main 2021, 17 March Shift-II]
x and y is the molarity of that solution, then
Ans.
(57)
:
Given
that
molar conductance ( λ m ) of that solution will be
Conductivity ( κ ) = 0.14Sm-1
1000y
1000x
Resistance R = 4.19 ohm
(b)
(a)
For KCl solution
x
y
xy
1000
1 l
(d)
(c)
R= ×
xy
1000
κ A
Tripura JEE-2022 l
= R×κ
Ans. (c): Molar Conductivity λm
A
κ ×1000
= 4.19× 0.14
Λm =
M
= 0.5866 ohm
κ = Specific conductance
For HCl solution
y = Molarity of solution
1 l
x = Specific resistant
R =
κ A
1
Specific resistant =
l/A
Specific conductance
κ=
R
1
x=
0.5866
κ
κ=
1.03
1
κ=
κ
=
56.95
×10−2 Sm−1
x
425. The resistance of a conductivity cell with cell
1 1000
constant 1.14 cm–1, containing 0.001 M KCl at
Λm = ×
298 K is 1500 Ω. The molar conductivity of
x
y
0.001
M KCl solution at 298 K in S cm2 mol–1
1000
is..........
(Integer answer)
=
xy
[JEE Main 2021, 27 Aug Shift-II]
423. A 5.0 m mol dm–3 aqueous solution of KCl has Ans. (760) : Given data
a conductance of 0.55 mS when measured in a
Cell constant = 1.14cm–1
cell of cell constant 1.3 cm–1. The molar
= 1.14 × 103m–1
conductivity of this solution is --------- mSm2
Resistance = 1500Ω
mol–1. (Round off to the Nearest Integer).
Molar concentration (C) = 0.001M
JEE Main 16.03.2021, Shift-II
Cell constant = Resistance × Conductivity ( κ )
Ans. (14.3 mSm2 mol–1)
Cellconstant
1.14×103 m−1
Given:κ =
=
–3
–3
Concentration(C) of KCl= 5.0×10 mol dm
Resistance
1500
Conductance (G) = 0.55mS
κ
Molar conductivity =
Cell constant = 1.3cm–1
C
Λm = ?
1.14×103
= 760Scm 2 mol−1
*
l
1500× 0.001
–1
∵ Conductivity = G × = 0.55(mS)×1.3(cm )
426. The conductivity of a weak acid HA of
A
concentration 0.001 mol L–1 is 2.0×10–5 S cm–1.
−3
κ = 0.55 × 10 × 1.3 × 100 S / m
If Aºm (HA) = 190 S cm2 mol–1, the ionisation
constant (Ka) of HA is equal to .........×10–6.
κ = 0.55 × 10−3 × 1.3 ×100 S / m
(Round off to the nearest integer)
κ = 0.715×10–1Sm–1
[JEE Main 2021, 27 July Shift-I]
(
Objective Chemistry Volume-II
216
)
YCT
Ans. (12) : Given that C = 0.001 mol L–1
κ = 2 × 10–5 S cm–1
Conductivity ( κ)
Molar conductivity =
Concentration (M )
2×10−5
×1000
0.001
= 20 Scm 2 mol−1
=
Degree of dissociation (α) =
=
∧m
∧∞
m
20
2
=
190 19
HA ↽ ⇀ H + + A−
0.001α
0.001α
0.001(1−α)
α 2
K a = 0.001
1− α
2
2
0.001×
19
=
= 12.3×10−6
2
1−
19
427. Given below are two statements.
Statement I : The limiting molar conductivity
of KCl (strong electrolyte) is higher compared
to that of CH3 COOH (weak electrolyte)
Statement II : Molar conductivity decreases
with decrease in concentration of electrolyte. In
the light of the above statements, choose the
most appropriate answer from the options
given below
(a) Statement I is true but statement II is false.
(b) Statement I is false but statement II is true.
(c) Both statement I and statement II are true.
(d) Both statement I and statement II are false.
[JEE Main 2021, 26 Aug Shift-I]
Ans. (d) : Limiting molar conductivity of KCl
Λ ∞mKCI = Λ ∞m K + + Λ ∞m CI−
( )
( )
= 73.55 S cm2/mol + 76.3 S cm2/mol
=149.3 S cm2/mol
Limiting molar conductivity of CH3COOH
∞
∞
Λ∞
CH3COOH = Λ CH COO− + Λ H+
3
2
= 349.8Scm / mol + 40.9Scm 2 / mol
= 390.7Scm 2 / mol
Limiting molar conductivity of CH3COOH is more than
KCl. Hence Statement I is false
Also, as the concentration decreases, dilution of the
electrolyte increases which increase the degree of
dissociation of weak electrolyte. Thus the number of
ions in the solution increases and hence, the molar
conductance of electrolyte increases. Therefore, we can
say that with decreases in concentration of electrolytes
molar conductance increase.
Hence, Statement II is also false.
Objective Chemistry Volume-II
428. The resistance of 0.01m KCl solution at 298 K
is 1500Ω. It is the conductivity of 0.01 m KCl
solution at 298 K is 0.1466×10–3 S cm–1. The cell
constant of the conductivity cell in cm–1 is
(a) 0.219
(b) 0.291
(c) 0.301
(d) 0.194
Karnataka-CET-2021
Ans. (a) : Given that,
Resistant (R) = 1500 Ω
Conduitivity ( κ ) = 0.146 × 10–3 S cm–1
Relation between resistance and cell constant,
1
κ = × cellcons tant ( G * )
R
G* = κ.R
= 0.146 × 10–3 × 1500 = 219 × 103
= 0.219 cm–1
429. Zeta potential is
(a) Potential required to bring about coagulation
of a colloidal sol.
(b) Potential required to give the particle a speed
of 1 cm s–1
(c) Potential difference between fixed charged
layer and the diffused layer having opposite
charges
(d) Potential energy of the colloidal particles.
Kerala-CEE-29.08.2021
Ans. (c) : The charges of opposite signs on the fixed
and diffused part of the double layer results in a
difference in potential between these layers in the same
manner as potential difference is developed in a
capacitor. This potential difference between the fixed
layer and the diffused layer of opposite charges is called
the zeta potential.
430. The molar conductivity of 0.007 M acetic acid
is 20 S cm2 mol–1 . What is the dissociation
constant of acetic acid? Choose the correct
option.
Λ ºH + = 350S cm 2 mol -1
º
2
-1
Λ CH 3COO- = 50 S cm mol
(a) 2.50 × 105 mol L–1 (b) 1.75 ×10–4 mol L–1
(c) 2.50 ×10–4 mol L–1 (d) 1.75 × 10–5 mol L–1
(NEET- 2021)
Ans. (d) :
º
Given, Λ ºH+ = 350Scm 2 mol-1 , Λ CH
= 50 Scm 2 mol-1
COO3
Λ
217
°
m
( CH3COOH ) = Λ
°
m
( H ) + Λ ( CH COO )
+
°
m
−
3
= 350 + 50
= 400S cm2 mol–1
α=
Λ cm
Λ °m
YCT
20
= 5×10−2
α=
400
K a (CH 3COOH ) = Cα 2
2
= 0.007 ×(5×10−2 )
= 1.75×10−5 mol L−1
431. Given
λ° 2+ = 106 S cm2mole–1 ,λ°
SO42–
Mg
The value of λ °
MgSO 4
= 160 S cm2mole–1 .
(in S cm2 mole–1) is
Ans. (57) : Given that : κ = 0.14 Sm–1, R= 4.19 Ω
The relation between conductivity and cell constant is:
G* = κ .R
For same conductivity cell, the cell are constant.
For KCl solution–
G * = 0.14 × 4.19 = 0.5866
......(i)
For HCl solution–
G* = κ ×1.03
From equation (i) put G* = 0.5866
0.5866
or κ =
= 0.5695
1.03
κ HCl = 56.9 × 10 –2 Sm –1
(a) 271.6
(b) 266
434. The molar conductivities at infinite dilution of
(c) 390
(d) 126
barium chloride,
sulphuric
acid
and
TS-EAMCET (Engg.), 05.08.2021 Shift-II
hydrochloric acid are 280, 860 and 426 S cm2
Ans. (b) : Given that,
mol–1 respectively. The molar conductivity at
°
2
−1
infinite dilution of barium sulphate is......S cm2
λ 2+ = 106Scm mole
Mg
mol–1 (Round off to the nearest Integer).
°
2
–1
[JEE Main 2021, 18 March Shift-II]
λ 2– = 160 S cm mole
SO4
Ans. (288) : Given that, Molar conductivity = λ ∞
m
λ°
=?
= 280 S cm2 mol–1
MgSO 4
We know that –
(Λ∞m )H2SO4 = 860 S cm2 mol–1
MgSO 4 → Mg 2 + + SO 24 −
(Λ∞m )HCl = 426 S cm2 mol–1
°
°
°
λ[ MgSO ] = λ 2+ + λ 2−
4
[Mg ]
[SO 4 ]
(Λ∞m )BaSO4 = (Λ∞m )Ba 2+ + (Λ∞m )SO24−
= 106 + 160
∞
∞
= (Λ ∞
°
m )BaCl + (Λ m )H SO – 2 (Λ m )HCl
λ MgSO4 = 266Scm 2 mole−1
2
2
4
= 280 + 860 – (2 × 426)
432. The molar conductivities of KCl. NaCl and
= 288 S cm2 mol–1
2
–1
KNO3 are 100, 120 and 90 cm . mol .
respectively. The molar conductivity of NaNO3 435. The molar conductivity of AgNO3, NaCl, and
NaNO3 at infinite dilution are 116.5, 110.3, and
would be ______ cm2. mol–1.
105.2 mho cm2 mol–1 respectively. In the same
(a) 110
(b) 290
unit, the molar conductivity of AgCl is
(c) 310
(d) 120
(a) 121.6
(b) 111.4
AP EAPCET 24.08.2021, Shift-I
(c) 130.6
(d) 150.2
Ans. (a) : Given that,
Assam CEE-2020
°
–1
–1
Ans.
(a)
:
Given
that,
∧ m ( KCl ) = 100Scm cm
The molar conductivity of solutions at infinite dilution
∧ °m ( NaCl ) = 120Scm 2 cm –1
Λ ∞AgNO3 = 116.5 mho cm 2 mol −1
∧ °m ( KNO3 ) = 90Scm 2 cm –1
Λ ∞NaCl = 110.3mho cm 2 mol−1
°
∧ m ( NaNO3 ) = ?
Λ ∞NaNO3 = 105.2mho cm 2 mol −1
Hence,
∞
Λ ∞AgCl = Λ AgNO
+ Λ ∞NaCl − Λ ∞NaNO3
3
NaCl + KNO3 → NaNO3 + KCl
= 116.5 + 110.3 – 105.2
∴ ∧ °m ( NaNO3 ) = ∧°m ( NaCl ) + ∧ °m ( KNO3 ) – ∧ °m ( KCl )
= 121.6 mho cm2 mol–1
= 120 + 90 – 100
436. The variation of molar conductivity with
= 210 – 100
concentration of an electrolyte (X) in aqueous
2
–1
= 110 S cm cm
solution is shown in the given figure.
433. A KCl solution of conductivity 0.14 S m–1
shows a resistance of 4.19 Ω in a conductivity
cell. If the same cell is filled with an HCI
solution, the resistance drops to 1.03 Ω. The
conductivity of the HCl solution is ________
×10–2 Sm–1. (Round off to the Nearest Integer).
JEE Main 17.03.2021, Shift-II
Objective Chemistry Volume-II
218
YCT
The electrolyte X is
(a) HCl
(b) NaCl
(c) KNO3
(d) CH3COOH
[JEE Main 2020, 5 Sep Shift=II]
Ans. (d) : Molar conductivity, Λ m is defined as the
conducting power of all the ion produced by dissolving
one gram mole of on electrolyte in solution. It is a
function of the ionic strength of a solution.
The molar conductivity of both weak and strong
electrolytes increases with decrease in concentration.
Molar conductivity is the conductivity affected by one
mole of ions. Increased dilution results in the
dissociation of more electrolytes into ions and
effectively increasing the number of active ions in the
solution.
These active ions impart more conductivity.
437. Let CNaCl and CBaSO4 be the conductances (in S)
measured for saturated aqueous solutions of
NaCl and BaSO4, respectively, at a
temperature T. Which of the following is false?
(a) Ionic mobilities of ions from both salts
increase with T.
(b) CBaSO4 (T2) > CBaSO4 (T1) for T2>T1
(c) CNaCl (T2)> CNaCl(T1) for T2>T1
(d) CNaCl >> CBaSO4 at a given T
[JEE Main 2020, 3 Sep Shift-I]
Ans. (d) : Because at a given temperature these will
have equal conductance release two ions in aqueous
both of the compounds release two ions in aqueous
solution
CNaCl >> CBaSO4 at a given T
438. The pair of electrolytes that possess same value
for the constant (A) in the Debye-HuckelOnsager equation, Λ m =Λ om – A C is
(a) MgSO4, NaSO4
(b) NH4Cl, NaBr
(c) NaBr, MgSO4
(d) NaCl, CaCl2
Karnataka-CET-2020
Ans. (b) : NH4Cl and NaBr , the charge on the ions are
the same Hence the pair of electrolyte that will possess
the same value for the constant will be NH4Cl , NaBr
439. Which among the following statements is
incorrect for interstitial compounds ?
(1) They are very hard and rigid
(2) They have higher melting point than pure
metal
(3) They do not show conductivity
(4) They are chemically reactive
(a) 1
(b) 2
(c) 3
(d) 4
AP EAMCET (Engg.) 18.9.2020 Shift-I
Objective Chemistry Volume-II
Ans. (d) : Statement (3) is incorrect because interstitial
compounds of d- or f- block metals show conductivity,
which is similar to their parent metals.
440. The specific conductivity of 0.1 N KCl solution
is 0.0129 ohm–1 cm–1. The resistance of the
solution in the cell is 100 ohm. The cell constant
of the cell will be :
(a) 1.10
(b) 1.29
(c) 0.56
(d) 2.80
Manipal-2019
Ans. (b) : Given,
specific conductivity ( κ ) = 0.0129 ohm–1 cm–1 and
resistance (R) = 100 ohm
We know,
1
Specific conductivity ( κ ) = × Cellconstant
R
Cell constant = κ × R
= 0.0129×100
= 1.29
1
441. The resistance of
M solution is 2.5×103 ohm.
10
What is the molar conductivity of solution?
(cell constant = 1.25 cm–1)
(a) 3.5 ohm–1 cm2 mol–1
(b) 5.0 ohm–1 cm2 mol–1
(c) 2.5 ohm–1 cm2 mol–1
(d) 2.0 ohm–1 cm2 mol–1
MHT CET-02.05.2019, SHIFT-III
Ans. (b) : Given, resistance (R) 2.5 × 103 ohm
Cell constant (G*) = 1.25 cm–1
We know,
G*
Specific conductivity ( κ ) =
R
1.25
=
= 5 × 10−4 ohm −1cm −1
2.5 × 103
κ × 1000
Molar conductivity ( Λ m ) =
m
Λ m = 5 × 10–4 × 1000 × 10
Λ m = 5 ohm–1cm2 mol–1
442. The conductivity of an electrolytic solution
decreases on dilution due to
(a) decrease in number of ions per unit volume
(b) increase in ionic mobility of ions
(c) increase in percentage ionisation
(d) increase in number of ions per unit volume
MHT CET-02.05.2019, SHIFT-III
Ans. (a) : Conductivity of an electrolytic concentration
of the electrolyte is decreases due to the number of ions
per unit volume carrying the current decreases on
dilution, So conductivity always decreases with
decrease in concentration.
443. Consider the statements S1 and S2:
S1 : Conductivity always increases with decrease in
the concentration of electrolyte.
S2 : Molar conductivity always increase with
decrease in the concentration of electrolyte.
219
YCT
The correct option among the following is
(a) S1 is correct and S2 is wrong
(b) S1 is wrong and S2 is correct
(c) Both S1 and S2 are wrong
(d) Both S1 and S2 are correct
[JEE Main 2019, 10 April Shift-I]
Ans. (b) : On dilution number of ions per unit volume
decrease so conductivity decreases with dilution and
hence, S1 is wrong.
Conductivity ( κ ) ∝ Concentration
1
Molar conductivity (C) ∝
Concentration
On dilution (C) and ( κ ) both decrease but the effect of
(C) is more dominating so Λ m increases hence S2 is
right.
1000 × κ
Λm =
C
444. Which one of the following graphs between
molar conductivity ( Λ m) versus C is correct?
(a)
(b)
(c)
(d)
[JEE Main 2019, 10 April Shift-II]
Ans. (c) : The graphs and molar conductivity (Λ m )
versus both NaCl and KCl are strong. Electrolytes, but
Na+ is more hydrated with respect to K+. KCl
electrolyte have higher Λ m with respect to NaCl.
Λ om for NaCl, HCl and NaA are 126.4, 425.9
and 100.5 S cm2 mol–1, respectively. If the
conductivity of 0.001 M HA is 5×10–5 S cm–1,
degree of dissociation of HA is
(a) 0.25
(b) 0.50
(c) 0.75
(d) 0.125
[JEE Main 2019, 12 Jan Shift-II]
Ans. (d) : Given that,
Λ m NaCl = 126.45 cm2 mol–1
…. (i)
445.
Λ °m HCl = 425. 95 cm2 mol–1
°
m
…..(ii)
Λ NaA = 100. 55 cm 55 cm mol–1 …(iii)
According to Formula–
Λ °m HA = Λ °m HCl + Λ °m NaA – Λ °m NaCl
= 425.9 + 100.5 – 126.4
= 400 S cm2mol–1
Λ °m =
2
1000 κ 1000 × 5 × 10−5
=
M
10−3
Objective Chemistry Volume-II
2
1000
= 50 S cm2 mol–1
0.001
Λ
50
a= m =
= 0.125
Λ m 400
= 5 × 10–5 ×
446. The decreasing order of electrical conductivity
of the following aqueous solution is
0.1 M formic acid (A),
0.1 M acetic acid (B),
0.1 M benzoic acid (C).
(a) A > C > B
(b) C > B > A
(c) A > B > C
(d) C > A > B
[JEE Main 2019, 12 April Shift-II]
Ans. (a) : Electrical conductivity of the aqueous
solution depends on the degree of ionisation. Degree of
ionisation is directly proportional to the acidic strength.
HCOOH > C6H5COOH > CH3COOH.
A
B
C
As the acidic strength decreases rate of dissociation
decreases and hence conductivity decreases.
Order of acidic strength
A>C>B
Acidic strength ↑ = degree of ionization ↑
447. Addition of excess of AgNO3 to an aqueous
solution of 1 mole of PdCl2·4NH3 gives 2 moles
of AgCl. The conductivity of this solution
corresponds to
(a) 1 : 1 electrolyte
(b) 1 : 3 electrolyte
(c) 1 : 2 electrolyte
(d) 1 : 4 electrolyte
Karnataka-CET-2019
Ans. (c) : Formation of moles of AgCl suggests that the
complex is.
2+
Pd ( NH 3 ) 4 Cl2
→ Pd ( NH 3 ) 4 + 2Cl –
2 mole of AgClget
Precipitated.
It is 1 : 2 electrolyte .
448. The aqueous solution of which of the following
complex has the least conductivity under
identical conditions.
(a) Penta aqua chlorido chromium (III) chloride
(b) Tetra aqua dichlorido chromium (III) chloride
(c) Hexa aqua chromium (III) chloride
(d) Tri aqua trichlorido chromium (III)
GUJCET-2018
Ans. (d) :
Compound
Ions
[Cr(H2O)5Cl]Cl2
3
[Cr(H2O)4Cl2]Cl
2
[Cr(H2O)6]Cl3
4
Cr(H2O)3Cl3
0
conductivity order :– D < B < A < C
449. When during electrolysis of solution of AgNO3,
9650 C of charge pass through the
electroplating bath, the mass of silver deposited
on the cathode will be
220
YCT
(a) 1.08 g
(c) 21.6 g
(b) 10.8 g
Ans. (a) : Given that
2
–1
(d) 108 g
Λ m = 240 Scm mol …..(i)
JEE Main 2018 23 Oct.
2
–1
Λ ∞m = 420 Scm mol …….(ii)
Ans. (b) : Ag+ + e– → Ag
Kb = 0.52 K. Kg mol–1….(iii)
9650
1
Number of moles of Ag =
= moles
Λ
96500 10
Degree of ionization (α) = ∞m
+
−
Λm
Ag + e
→ Ag
240Scm 2 mol –1
9650 C = 0.1 F = 0.1
α=
equivalent Ag = 0.1 mol Ag
420Scm 2 mol –1
∴ mass of Ag produced = 0.1 × 108 = 10.8 g
= 0.57
450. If molar conductivity of Ca2+ and Cl − ions are Van't Hoff factor
119 and 71 S cm2mol–1 respectively, then the i = 1 + (n – 1)α
[ for ionization]
molar conductivity of CaCl2 at infinite dilution i = 1 + (2 – 1) 0.57 = 1.57
[ For HCl, n = 2]
is
Elevation in boiling
(a) 215 S cm2mol–1
(b) 340 S cm2mol–1
∆Tb = iKb ×m = 1.57 × 0.52 × 3 = 2.45 K
(c) 126 S cm2mol–1
(d) 261 S cm2mol–1
Tb = Tb° +∆Tb = 373.15 + 2.45 = 375.6 K
JIPMER-2018
453. 0.1 mole, per litre solution is present in a
Ans. (d) : Given that,
conductivity cell where electrode of 100 cm2
°
2
–1
Λ Ca 2+ = 119 S cm mol
area are placed at 1cm apart and resistance
observed is 5×103 Ohm, what is molar
conductivity of solution?
Λ °CaCl2 = Λ °Ca 2+ + 2 × Λ °Cl−
(a) 5×102 S cm2 mol-1 (b) 2×104 S cm2 mol-1
(c) 200 S cm2 mol-1
(d) 0.02 S cm2 mol-1
Λ °CaCl2 = 119 + 2 × 71
[AIIMS-26, May, 2018 (E)]
Λ °CaCl2 = 261 S cm2 mol–1
Ans. (d) : Given that,
Molarity = 0.1 mol/L
451. At a particular temperature, the ratio of molar
Resistance = 5× 103 Ω
conductance of specific conductance of 0.01 M
We know that
NaCl solution is
(a) 105 cm3 mol–1
(b) 103 cm3 mol–1
Molar conductivity of solution ( Λ m )
(c) 10 cm3 mol–1
(d) 105 cm3 mol–1
k ×1000
Karnataka-CET-2018
=
M
κ × 1000
Where k = specific conductivity
Ans. (a) : molar conductance ( Λ m )=
M
1
k = × Cell constant
Λ m ( ohm –1cm 2 mol –1 ) 1000
R
=
1 l
M
κ ( ohm –1cm –1 )
= ×
[l = 1cm (given)]
R A
Λ m 1000
=
1
1cm
=
×
0.01
κ
3
5×10 Ω 100cm 2
Λm
5
3
–1
= 10 cm mol
= 2×10−6 Ω−1cm−1
κ
452. Consider
the
galvanic
cell,
Pt
(s) Then, (Λ m ) = k ×1000
M
H 2 (1bar) HCl (aq) (1M) Cl 2 (1bar) Pt (s).
−6
2×10 ×1000
After running the cell for sometime, the
=
concentration of the electrolyte is automatically
0.1
raised to 3M HCl. Molar conductivity of the
2
=
or 0.02 Scm 2 mol−1
3M HCl is about 240 S cm2 mol-1 and limiting
100
2
molar conductivity of HCl is about 420 S cm
of 0.1 MHA is
mol-1. If Kb of water is 0.52 K kg mol-1, 454. Specific –4 conductance
–1
–1
3.75
×
10
ohm
cm
.
If
calculate the boiling point of the electrolyte at
∞
the end of the experiment
Λ (HA) = 250 ohm–1 cm2 mol–1, the
(a) 375.6 K
(b) 376.3 K
dissociation constant Ka of HA is:
(c) 378.1 K
(d) 380.3 K
(a) 1.0 × 10–5
(b) 2.25 × 10–4
–5
(e) 381.6 K
(c) 2.25 × 10
(d) 2.25 × 10–13
[BITSAT – 2017]
Kerala-CEE-2018
Λ °Cl− = 71 S cm2 mol–1
Objective Chemistry Volume-II
221
YCT
Ans. (b) : We know that,
Molar conductivity Λ m = Λ °m − A C
Where, Λ m molar conductivity
Ans. (c) : Given that,
Specific conductance at certain concentration
= 3.75 × 10–4
Specific conductance at infinite dilution
= 250 ohm–1 cm2 mol–1
Λ °m is limiting molar conductivity
A is constant
C is concentration is electrolyte
This equation is in the form of the state line equation
y = mx + c
1000 κ 1000 × 3.75 ×10−4
= 3.75
=
0.1
0.1
Λ
3.75
α = m∞ =
= 1.5 × 10–2
Λ m 250
Λm =
Where α is a Degree of dissociation,
Ka = Cα2 = 0.1 × (1.5 × 10–2)2
= 2.25 × 10–5
So,
455. The graph of C → Λ m for an aqueous solution
of which substance is not obtained as a straight
line?
(a) HCl
(b) NaCN
(c) NaCl
(d) HCN
GUJCET-2017
Ans. (d) : HCN is weak acid and with dilution molar
conductivity increases rapidly. Where as in case of
strong electrolyte HCl and NaOH and also for NaCl the
change almost linear.
456. Which of the following electrolytic solutions
has the least specific conductance?
(a) 0.002 N
(b) 0.1 N
(c) 0.2 N
(d) 2 N
JCECE - 2017
Ans. (a): Specific conductance of a solution is directly
proportional to concentration of electrolytes.
Hence, specific conductance is least for the solution of
0.002N concentration.
457. What is the SI unit of conductivity?
(a) Sm
(b) Sm–1
2
(c) Sm
(d) Sm–2
MHT CET-2017
Ans. (b) : The reciprocal of specific resistance or
resistivity is called conductivity. The SI unit of
conductivity is Sm–1
458. Which one of the following is corrected plot of
Λm (in S cm2 mol–1) and C [in (mol/L)1/2] for
KCl solution? (Y = Λm; x = C )
(a)
(c)
C is increases then Λm is decreases
Where, Y = Λm and X = C
459. The molar conductivity of a 0.5 mol/cm3
solution
of
AgNO3
with
electrolytic
conductivity of 5.76 × 10–3 S cm–1 at 298 K is
(a) 2.88 S cm2/mol
(b) 11.52 S cm2/mol
2
(c) 0.086 S cm /mol
(d) 28.8 S cm2/mol
(NEET-II 2016)
Ans. (b) : Given,
Molar conductivity ( κ ) = 7.76 × 10–3 S cm–1
molar mass (m) = 0.5 mole cm–3
κ ×1000
Λ=
(m)
5.76 × 10−3 × 1000
0.5
= 11.52 S cm2 mol–1
460. The order of equivalent conductances at
infinite dilution for LiCl NaCl and KCl is
(a) LiCl > NaCl > KCl (b) KCl > NaCl > LiCl
(c) NaCl > KCl > LiCl (d) LiCl > KCl > NaCl
WB-JEE-2016
Ans. (b) : The order of equivalent conductance at
infinite dilution for LiCl, NaCl and KCl is KCl > NaCl
> LiCl. Li+ , due to its high polarizing power gets
hydrated and its ionic mobility is reduced.
In LiCl, NaCl and KCl, anion are same. Cations have
same charge but different size. Smaller cations are
more heavily hydrated in aqueous solution giving
larger hydrated radius and thus smaller ionic speeds and
equivalent conductance.
1
conductance ∝
(b)
radii
Thus, equivalent conductance: KCl>NaCl>LiCl.
461. Consider the following statements:
Statement I : The conductance depends on the
number of ions and ion mobility. The equivalent
conductance increases with increase in dilution,
(d)
the specific conductance diminishes.
Statement II : The total number of ions increases
on account of increased ionization due to dilution,
TS-EAMCET-2016
but the number of ions per unit volume decreases.
Objective Chemistry Volume-II
222
=
YCT
Which one of the following is correct in respect
of the above statements?
(a) Both the statements are true and statement II
is the correct explanation of statement I
(b) Both the statements are true but statement II is
not the correct explanation of statement I
(c) Statement I is true, but statement II is false
(d) Statement I is false, but statement II is true
SCRA-2015
Ans. (a): The conductance depend on the numbers of
ions and ion mobility. The equivalent conductance
increases with increase in dilution, the specific
conductance diminishes.
The total number of ions increases on account of
increased ionization due to dilution, but the number of
ions per unit volume decreases.
1 A
A
Conductance (G) = =
=κ
R δ.l
l
Where, R = Resistance
δ = Resistivity
A = cross section area
l = length
κ = Conductivity
462. The value of electrical resistance at super
conductivity state is
(a) 100
(b) 0
(c) low
(d) high.
SRMJEEE – 2015
Ans. (b) : For the super conductivity state, the electrical
resistance will be 0 by which current flow fastly.
463. Pure silicon doped with phosphorus is :
(a) amorphous
(b) p-type semiconductor
(c) n-type semiconductor
(d) insulator
AP-EAMCET (Engg.) 2015
Ans. (c) : Doping of silicon (group 14) with phosphorus
element (group 15) to give excess electron are called ntype semiconductors. An n-type material by itself has
mainly negative charge carriers (electrons) which are
able to move freely, but it is still neutral.
464. Silicon doped with arsenic is
(a) p-type semiconductor
(b) n-type semiconductor
(c) intrinsic semiconductor
(d) insulator.
COMEDK 2015
Ans. (b) : Silicon doped with arsenic is n-type
semiconductor because silicon has valence electrons
and arsenic has 5 valence electron to form a four
covalent bond and fifth electron remain free and
increase conductivity.
465. The conductivity of 0.001028 mol L–1 acetic
acid is 4.95×10–5 S cm–1, Find out its
dissociation constant if Λ m for acetic acid is
390.5 S cm–1 mol–1.
(a) 2.18×10–5 mol–1 L–1 (b) 1.78×10–5 mol L–1
(c) 3.72×10–4 mol–1 L–1 (d) 2.37×10–4 mol L–1
JIPMER-2015
Objective Chemistry Volume-II
κ 4.95 ×10 –5 Scm –1 1000cm3
=
×
C 0.001028mol L–1
L
= 48.15 S cm2 mol–1
κ × 1000
Λm =
C
Λ m 48.15
α= ° =
= 0.123
Λ m 390.5
Ans. (b) : Λ m =
0.001028 × ( 0.123)
Cα 2
=
(1 – 0.123)
(1 – α )
= 1.78 × 10–5 mol L–1
466. At a particular temperature, the ratio of
equivalent conductance to specific conductance
of a 0.01 N NaCl solution is
(a) 105cm3
(b) 103cm3
3
(c) 10 cm
(d) 105cm2
WB-JEE-2015
Ans. (a) : As we know that,
Equivalent conductance ( Λ )
Specific conductance ( κ)
=
Concentration
κ ×1000
Λ =
,
0.01
(Given in question Concentration= 0.01)
2
Kα =
⇒
Λ 1000 Ω−1 cm 2 eq−1
=
κ
0.01 Ω−1 cm−1
⇒
Λ
= 105 cm3 eq−1
κ
***
467. The equivalent conductivity of a solution
containing 2.54g of CuSO4 per L is 91.0 W–1cm2
eq–1. Its conductivity would be
(a) 2.9 × 10–3 Ω–1 cm–1 (b) 1.8 × 10–2 Ω–1 cm–1
(c) 2.4 × 10–4 Ω–1 cm–1 (d) 3.6 × 10–3 Ω–1 cm–1
VITEEE-2008
Ans. (a) : Given that
Λ eq = 91.0 W −1cm 2 eq −1
We know that
K = Λ eq .C
2.54
= 91.0 ×
159 ×1000
2
−3
−1
−1
= 2.9 × 10 Ω cm
468. The equivalent conductance of silver nitrate
solution at 250°C for an infinite dilution was
found to be 133.3Ω–1cm2equiv–1. The transport
number of Ag+ ions in very dilute solution of
AgNO3 is 0.464. Equivalent conductances of
Ag+ and NO 3– (in Ω–1 cm2 equiv–1) at infinite
dilution are respectively
223
YCT
(b) 61.9, 71.4
(a) 101 S cm2mol–1
(b) 87 S cm2mol–1
2
–1
(d) 133.3, 195.2
(c) –101 S cm mol
(d) –391 S cm2mol–1
VITEEE-2013
VITEEE- 2008
Ans. (b) : Since NaNO3 is formed by the reaction
Ans. (b) : Given, transport no. of Ag+ = 0.464
NaCl + KNO3 → NaNO3 + KCl
Equivalence conductance 133.3 Ω–1
∞
+
Hence, using Kohlrausch's law
So, λ (Ag ) = Transport number of Ag + × A (∞AgNO3 )
Λ mo NaNO = Λ oNaCl + Λ oKNO3 − Λ oKCl
3
= 0.464 × 133.3
= 128 + 111 – 152 = 87 S cm2 mol–1
= 61.9 Ω–1 cm2 equiv–1
472. If the molar conductance values of Ca2+ and Cl–
According to Kohl rausch's low
at infinite dilution are respectively 118.88×10–4
Λ ∞( AgNO ) = λ ∞Ag+ + λ ∞NO−
3
( ) ( 3)
m2 mho mol–1 and 77.33×10–4 m2 mho mol–1
∞
∞
∞
then that of CaCl2 is (in m2 mho mol–1)
∴
λ − =Λ ( AgNO ) − λ Ag+
(
)
3
( NO3 )
(a) 118.88 × 10–4
(b) 154.66 × 10–4
–4
= 133.3 − 61.9
(c) 273.54 × 10
(d) 196.21 × 10–4
−1
2
−1
VITEEE- 2007
= 71.4Ω cm equiv
Ans.
(c)
:
Molar
conductance
of
CaCl
2
469. Which of the following expressions correctly
= Molar conductance of Ca2+ + 2 × (molar
represents the equivalent conductance at
Conductance of Cl–)
infinite dilution of Al2(SO4)3 ? Given that
–4
°
°
= 118.88 × 10 + (77.33 × 10–4)×2
Λ Al 3+ and
the
equivalent
Λ SO2− are
4
= 273.54 × 10–4 m2 mho mol–1
conductance's at infinite dilution of the
473. For strong electrolytes the plot of molar
respective ions?
°
°
°
°
conductance vs C is
(a) 2Λ Al3+ + 3ΛSO2−
(b) Λ Al3+ + ΛSO2−
4
4
(a) parabolic
(b) linear
1 °
1 °
°
(c)
sinusoidal
(d)
circular
(c) Λ °Al3+ + 3Λ SO
×
6
(d)
Λ
+
Λ
2−
3+
2−
4
3 Al
2 SO4
VITEEE- 2007
VITEEE- 2012 Ans. (b) : According to Debye – Huckel – Onsagar
Ans. (b) : Al 2 (SO 4 )3 ↽ ⇀ 2Al3+ + 2SO 24−
equation Λ m = Λ ° m − ( A + B Λ ° m ) C
Kohlrausch's law states that the equivalent conductivity where A and B are the Debye – Huckel constants. If we
of an electrolyte at infinite dilution is equal to the sum plot a graph between molar conductance (Λm) against
of the conductance's of the anions and cations.
C a straight
At infinite dilution, when dissociation is complete, each the square roots of the concentration
ion makes a definite contribution towards molar line is obtained
conductance of the electrolyte.
Λ ∞eq Al2 (SO4 ) = Λ°Al3+ + Λ°SO 2−
(a) 195.2, 133.3
(c) 71.4, 61.9
(
)
( )
3
4
470. The equivalent conductance at infinite dilution
of a weak acid such as HF
(a) can be determined by extrapolation of
measurements on dilute solutions of HCl,
HBr and HI
(b) can be determined by measurement on very
dilute HF solutions
(c) can best be determined from measurements
on dilute solutions of NaF, NaCl and HCl
(d) is an undefined quantity
VITEEE- 2010
Ans. (c) : We know that according to Kohlrausch law,
at infinite dilution, the limiting molar conductivity is the
sum of the limiting molar conductivity of its
conductivity of its constitution.
HF → H+ + F–
H+ + F– = Na+ + F– – (Na– + Cl–) + Cl–
∴ Equivalent conductance of HF = equivalent
conductance of NaF – NaCl + HCl.
471. The molar conductivities of KCl, NaCl and
KNO3 are 152, 128 and 111 S cm2 mol–1
respectively. What is the molar conductivity of
NaNO3 ?
Objective Chemistry Volume-II
474. The charge carriers in p-type semiconductors
are
(a) electrons
(b) protons
(c) neutrons
(d) positive holes.
SRMJEEE – 2009
Ans. (d) : In the p-type semiconductor the mobile
charge carriers i.e. holes are present in majority and
electrons are the minority charge carriers.
475. The electrical conductivity of metal decreases
with increase of temperature, this is due to
(a) enhanced vibration of metal ions
(b) movement of electrons
(c) chemical energy
(d) increase of thermal energy.
SRMJEEE – 2011
224
YCT
Ans. (a) : As we know the band gap in the metal is 478. At a certain temperature and at infinite
dilution, the equivalent conductances of sodium
small and thus the electrons can easily take a lap to
benzoate, hydrochloric acid and sodium
conduction band and conduct electricity but with
chloride are 240, 349 and 229 ohm-1 cm2 equiv-1
increase in temperature, the thermal motion makes the
respectively. The equivalent conductance of
electrons to collide and disturb the free flow. Thus
benzoic acid in ohm–1 cm2 equiv–1 at the same
conductivity is decreased.
conditions is
476. At 298 K the molar conductivities at infinite
(a) 80
(b) 328
(c) 360
(d) 408
dilution ( Λ °m ) of NH4Cl, KOH and KCl are
AP-EAMCET- (Engg.) - 2010
152.8, 272.6 and 149.8 S cm2 mol–1 respectively.
The λ om of NH4OH in S cm2 mol–1 and % Ans. (c) : Given that
∞
–1
2
–1
dissociation of 0.01 M NH OH with Λ = 25.1 S ∧ C6H5COONa = 240Ω cm equiv ----(i)
4
m
cm2 mol–1 at the same temperature are
∧ ∞ HCl = 349Ω −1cm 2 equiv –1 ----(ii)
(a) 275.6, 0.91
(b) 275.6, 9.1
∧ ∞ NaCl = 229Ω −1cm 2 equiv –1 ----(iii)
(c) 266.6, 9.6
(d) 30, 84
AP-EAMCET (Engg.) - 2014 From equation (i), (ii) and (iii), we get
∧ ∞ C6H5COOH = ∧ ∞ C6H5COONa + ∧ ∞ HCl − ∧ ∞ NaCl
Ans. (b) : Given that
= 240 + 349 − 229
Λ ° (NH4Cl) = 152.8 S cm2 mol–1
m
= 360Ω −1cm 2equiv−1
°
m
Λ (KOH) = 272.6 S cm2 mol–1
479. A solution of concentration C g equiv/L has a
specific resistance R. The equivalent
conductance of the solution is
R
C
(a)
(b)
C
R
1000
1000R
(c)
(d)
RC
C
AP-EAMCET- (Engg.) - 2010
Ans. (c) :
Λ °m (KCl) = 149.8 S cm2 mol–1
Λ °m (NH4OH) =
Λ °m (NH4Cl) + Λ °m (KOH) – Λ °m (KCl)
= 152.8 + 272.6 – 149.8
= 275.6 S cm2 mol–1
Therefore, degree of dissociation (α) =
Λm
Λ °m
25.1
= 0.091
275.6
Thus, % of degree of dissociation = 9.1%
=
Equivalent conductance =
477. If the values of Λ ∞ of NH 4Cl, NaOH and NaCl
are 130, 217 and 109 ohm-1 cm2 equiv-1
respectively, the Λ ∞ of NH4OH in ohm-1 cm2
equiv-1 is
(a) 238
(b) 196
(c) 22
(d) 456
AP-EAMCET- (Engg.)-2011
Ans. (a) : Given that
Λ ∞ NH 4 Cl = 130 Ohm -1 cm 2 equiv −1
.....(I)
Λ ∞ NaOH = 217 Ohm-1 cm 2 equiv-1
.....(II)
Λ ∞ NaCl = 109 Ohm cm equiv
.....(III)
-1
2
-1
Λ ∞ NH 4 OH =347–109
= 238 Ohm-1cm 2 equiv−1
Objective Chemistry Volume-II
⇒
⇒
K × 1000
Concentration
1
K
1
K=
R
R=
1000
.
RC
480. Which among the following expressions is not
correct?
(a) µ ∞ = γ + λ +∞ + γ − λ −∞
Equivalent conductance =
1 ∞
1
λ + + − λ −∞
+
n
n
∞
(c) λ ∞cation = µ cation
× faraday
(b) λ ∞ =
On adding equation (I) and (II) and then subtracting
equation (III), we get.
Λ ∞ NH 4 OH =Λ ∞ NH 4 Cl+Λ ∞ NaOH − Λ ∞ NaCl
Λ ∞ NH 4 OH =130+217–109
=
specific conductance ×1000
concentration ( in gm equiv/L )
∞
(d) λ ∞anion = µcation
× faraday
AMU-2011
Ans. (d) : The value of conductance and number of
ions, respectively, superscript infinity denotes the
condition of infinite dilution and subscript. + and –
indicate cation and anion respectively.
225
YCT
Ionic mobility and conductance are related as
Conductance = mobility × faraday
µ∞ = γ + λ +∞ + γ − λ ∞−
Where, γ is the activity coefficient.
481. A 0.5 M NaOH solution offers a resistance of
31.6 ohm in a conductivity cell at room
temperature. What shall be the approximate
molar conductance of this NaOH solution if cell
constant of the cell is 0.367 cm–1?
(a) 234 S cm2 mol–1
(b) 23.2 S cm2 mol–1
2
–1
(c) 4645 S cm mol
(d) 5464 S cm2 mol–1
AMU – 2008
Ans. (b) : The molar conductance is defined as the
conductance of all the ions produced by ionization of 1
g mole of an electrolyte present in V ml of solution. It is
denoted by Λm.
Given that: R= 31.6Ω
1
1
Conductance = =
R 31.6
= 0.0316 Ohm–1
Specific conductance = Conductance × cell constant
= 0.0316×0.367
= 0.0116 ohm–1
cm–1
Now molar concentration = 0.5 M.
= 0.5×10–3 mol cm–3
κ
Molar Conductance =
C
0.0116
=
0.5×10−3
= 23.2 cm2 mol–1
482. The conductivity of a saturated solution of
BaSO4 is 3.06 × 10–6 ohm–1 cm–1 and its
equivalent conductance is 1.53 ohm-1 cm2
equiv–1. The Ksp for BaSO4 will be
(a) 4 × 10–12
(b) 2.5 × 10–9
–13
(c) 2.5 × 10
(d) 4 × 10–6
[BITSAT – 2009]
Ans. (d) : Given that,
Equivalent conductance = 1.53 ohm–1 cm2
κ × 1000
Solubility =
Λ eq
Ans. (c) : Kohlrausch law:– At infinite dilution, when
dissociation is complete, each ion makes a definite
contribution towards equivalent conductance of the
electrolyte irrespective of the nature of the ion with
which it is associated and the equivalent conductance at
infinite dilution for any electrolyte is the sum of the
contribution of its constitution ions i.e. anions and
cations.
Considering that,
ionic conductance of Ba+2 = 127 S cm2 mole–1
ionic conductance of Ca– = 76 S cm2 mole–1
The equivalent weight,
(λ )
= ( λ ∞m )
(λ )
1
127
= ( λ ∞m ) 2+ + ( λ ∞m ) − =
+ 76 = 139.5ohm −1cm 2 eq −1
Ba
Cl
2
2
∞
m BaCl
2
∞
eq BaCl
2
Ba 2 +
+ 2 ( λ ∞m )
Cl −
484. By diluting a weak electrolyte, specific
conductivity (Kc) and equivalent conductivity
(λc) changes as
(a) both increase
(b) Kc increase, λc decreases
(c) Kc decreases, λc increases
(d) both decrease
CG PET -2009
Ans. (c) : When the solution of a weak electrolyte is
diluted, the volume of the solution increases, hence
equivalent conductivity (λc) increases. However , during
this process, the number of current carrying particles
per per unit volume decreases, hence specific
conductivity (K)c decreases.
485. A cell, with cell constant 0.4 cm−1, has the
resistance of 40 ohm of a 0.01 M solution of an
electrolyte, then the molar conductivity in
ohm−1 cm2 mol−1 will be
(a) 104
(b) 103
2
(c) 10
(d) 10
CG PET -2007
Ans. (b) : Given that,
Cell constant = 0.4 cm–1
Resistance (R) = 40 Ohm
1
Conductivity ( κ ) =
R
3.06 ×10−6 × 1000
1
=
=
Ohm−1
1⋅ 53
40
= 2 × 10–3 mol L–1
Concentration of solution (molarities) = 0.01 M
BaSO4 is AB type of electrolyte,
Specific conductivity×1000
2
Ksp = S
molar conductivity =
–3 2
Concentration of Solution
= (2×10 )
= 4 × 10–6 mol2 L–2
specific conductivity = conductivity × cell constant
–
1
483. The ionic conductance of Ba2+ and Cl are
=
× 0.4 ohm–1 cm–1
respectively 127 and 76Ω –1cm 2 mol–1 at infinite
40
dilution. The equivalent conductance of BaCl2
= 0.01 ohm–1 cm–1
at infinite dilution will be
0.01×1000
Molar conductivity =
ohm−1cm 2 mol−1
(a) 330Ω –1cm 2
(b) 203Ω –1cm 2
0.01
(c) 139Ω –1cm 2
(d) 51Ω –1cm 2
= 1000 ohm–1 cm2 mol–1
[BITSAT – 2012]
= 103 ohm–1 cm2 mol–1
Objective Chemistry Volume-II
226
YCT
486. Resistance of 0.2 M solution of an electrolyte is
50Ω. The specific conductance of the solution is
1.4Sm–1. The ressitance of 0.5M solution of the
same electrolyte is 280 Ω. The molar
conductivity of 0.5 M solution of the electrolyte
in S m2 mol–1 is
(a) 5×10–4
(b) 5×10–3
3
(c) 5×10
(d) 5×102
[JEE Main 2014]
Ans. (a) : Given that,
For 0.2M solution,
R = 50 Ω
κ = 1.4 S m–1 = 1.4 × 10–2 S cm–1
In is know that,
l R
l
R 1
= ⇒ = = 50×1.4×10−2 cm
R= ρ ⇒
A
ρ a
a ρ
For 0.5 M Solution,
R = 28Ω
κ=?
l
= 50 × 1.4 × 10–2 cm
a
The specific conductance of the solution is given by
1 1 l
1
κ = = × =
× 50 × 1.4 × 10–2
280
ρ R a
1
=
× 70 × 10–2
280
= 2.5 × 10–3 S cm–1
κ × 1000 2.5 ×10−3 × 1000
Λm =
=
M
0.5
= 5 S cm2mol–1
= 5 × 10–4 S m2 mol–1
487. The equivalent conductance of NaCl at
concentration C and at infinite dilution are λc
and λ∞ is given as
(a) λc = λ∞ + (B) C
(b) λc = λ∞ – (B) C
(c) λc = λ∞ – (B) C
(d) λc = λ∞ + (B) C
[JEE Main 2014]
Ans. (c) : The equivalent conductance of NaCl at
concentration C and at infinite dilution are λc and λ∞ is
given as
(a)
(b)
(c)
(d)
6250 S m2 mol–1
6.25×10–4 S m2 mol–1
625×10–4 S m2 mol–1
62.5 S m2 mol–1
[AIEEE 2011]
Ans. (b) : Given that,
Specific conductance = 1.3 S m–1
Resistance = 50Ω
For 0.2M solution,
Specific conductance ( κ )
= conductance (1/R) × Cell constant (l/a)
1 l
l
1.3 = × ⇒ = 1.3×50×10−2 cm−1
50 a
a
For 0.4 M solution, R = 260 Ω
l
1
1 l
R = ρ ⇒ =κ= ×
a
ρ
R a
1.3×50×10−2
Scm−1
260
κ × 1000
Molar conductivity (Λ m ) =
Molarity
κ =
1.3 × 50 × 10−2 × 1000
260 × 4
= 6.25 S cm2 mol–1
= 6.25 × 10–4 S m2 mol–1
489. The equivalent conductance of two strong
electrolytes at infinite dilution in H2O (where
ions move freely through a solution) at 25ºC
are given below
Λ o CH3COONa = 91.0 S cm2/ equiv
=
Λ o HCl = 426.2 S cm2/ equiv
What additional information/quantity one
needs to calculate Λ o of an aqueous solution of
acetic acid?
(a) Λ o of NaCl
(b) Λ o of CH3COOK
(c) The limiting equivalent conductance of H+
(λoH+)
(d) Λ o of chloroacetic acid (ClCH2COOH)
[AIEEE 2007]
Ans. (a) : According to Kohlrausch's law, molar
conductivity of weak electrolyte acetic acid
(CH3COOH)can be calculated as follows:
Λ°CH3 COOH = Λ°CH3 COONa + Λ°HCl − Λ°NaCl
λc = λ∞ – B C
Where,
λc = molar conductivity of the solution at certain Hence to calculate Λ °CH3COOH we need Λ °NaCl
concentration.
490. The limting molar conductivities Λ °NaOAc and
λ∞ = Limiting molar conductivity
Λ °HCl at infinite dilution in water at 25ºC are
C = concentration
91.0 and 426.2S cm2/mol, respectively. To
B = constant that depends on temperature, charges on
calculate Λ °HOAc the additional value required
the ion.
is
488. Resistance of 0.2 M solution of an electrolyte is
50Ω. The specific conductance of the solution is
(a) Λ o H2 O
(b) Λ o KCl
–1
1.3 S m . If resistance of the 0.4M solution of
(c) Λ o NaOH
(d) Λ o NaCl
the same electrolyte is 260Ω, its molar
[AIEEE 2006]
conductivity is
Objective Chemistry Volume-II
227
YCT
Ans. (d) : Λ °NaOAc = Λ °Na + + Λ °AcO−
…..(i)
Λ°HCl = Λ°H+ + Λ°Cl−
…..(ii)
Λ°HOAc = Λ°H+ + Λ°AcO−
…..(iii)
On adding equation (i) & (ii) and equating with
equation (iii) we get
Λ°HOAc = Λ°NaOAc + Λ°HCl − Λ°NaCl
491.
Electrolyte KCl KNO3 HCl NaOAc NaCl
126.5
149.9 145.0 426.2 91.0
Λ∞ (S
cm2mol–1)
Calculate Λ ∞HOAc using
appropriate
molar
conductances of the electrolytes listed above at
infinite dilution in H2O at 25ºC.
(a) 217.5
(b) 390.7
(c) 552.7
(d) 517.2
[AIEEE 2005]
Ans. (b) : Given that,
Λ ∞HCl = 426.92 S cm2 mol–1
…(i)
Λ ∞NaOAc = 91.0 S cm2 mol–1
Λ
∞
NaCl
∞
AcOH
2
…(ii)
–1
= 126.5 S cm mol ...(iii)
∞
Λ
= Λ AcONa
+ Λ ∞HCl − Λ ∞NaCl
= 91.0 + 426.2 – 126.5
= 390.75 S cm2 mol–1
492. Molar conductivity decreases with decrease in
concentration
(a) for strong electrolytes
(b) for weak electrolytes
(c) both for strong and weak electrolytes
(d) for non-electrolytes.
J & K CET-(2012)
Ans. (c) : Specific conductivity decreases
with
decreases in concentration for weak as well as strong
electrolytes.
493. What is conductance?
(a) Inverse of resistance
(b) Proportional of resistance
(c) Equal of resistivity
(d) Equal of resistance
J & K CET-(2014)
Ans. (a) : The express of the ease of the passing of the
electrons inverse of resistance called conductance.
Conductance is the reciprocal or inverse of resistance.
1
C∝
R
494. When an electrolytic solution conducts
electricity the current is carried by
(a) the electrons
(b) cations and anions
(c) neutral molecules
(d) the atoms of the electrolyte
J & K CET-(2013)
Objective Chemistry Volume-II
Ans. (b) : Because there are rearrangement of passage
of current in electrolytic solution is due to migration of
ions towards opposite electrodes. The current is carried
by cations and anions.
495. Which one of the following is correct?
(a) Equivalent conductance decreases with
dilution.
(b) Specific conductance increases with dilution.
(c) Specific conductance decreases with dilution.
(d) Equivalent conductance increases with
increasing concentration.
J & K CET-(2011)
Ans. (c) : On dilution, it is the equivalent and molar
conductance
which
increases
while
specific
conductance decreases. It is because concentration of
ions per CC decreases on dilution, thus specific
conductance decreases with dilution.
496. The correct expression in SI system relating the
equivalent conductance (Λ c ), specific
conductance ( κ ) and equivalent concentration
(C) is
κ
κ ×1000
(a) Λc =
(b) Λc =
C
C
κ ×10 –6
C
J & K CET-(2009)
Ans. (b) : Equivalent conductance is related to specific
conductance κ as
κ × 1000
Molar conductivity Λ c =
C
Where, C is the number of gram equivalents of solute
dissolved in one litre of the solution.
497. Which of the following conducts electricity?
(a) Crystal NaCl
(b) Diamond
(c) Molten KBr
(d) Sulphur
JCECE - 2006
Ans. (c) : Only molten or aqueous solution of ionic
crystals conduct electricity. Ionic compounds in the
solid state do not conducts electricity.
Hence, molten KBr conducts electricity.
498. The resistance of 1 N solution of acetic acid is
250 Ω, when measured in a cell having a cell
constant of 1.15 cm–1. The equivalent
conductance (in Ω–1 cm2 equiv–1) of 1 N acetic
acid is
(a) 2.3
(b) 4.6
(c) 9.2
(d) 18.4
JCECE - 2012
κ × 1000
Ans. (b) : Equivalent conductivity (Λeq) =
C
cell constant 1.15
κ=
=
S cm−1
resistance
250
1.15 ×1000
Λeq =
= 4.6Ω−1 cm2 equiv–1
250 × 1
228
(c) Λc =
κ ×10 –3
C
(d) Λc =
YCT
499. When 0.1 mol CoCl3(NH3)5 is treated with
excess of AgNO3, 0.2 mole of AgCl are
obtained. The conductivity of solution will
correspond to
(a) 1 : 3 electrolyte
(b) 1 : 2 electrolyte
(c) 1 : 1 electrolyte
(d) 3 : 1 electrolyte
JCECE - 2013
Ans. (b) : Formation of 0.2 mole of AgCl from 0.1
mole of complex means that there are two ionizable Cl.
Hence, formula is [Co (NH3)5 Cl ]Cl2 i.e., 1 : 2 type
electrolyte.
500. Which one of the following has the highest
molar conductivity?
(a) Diaminedichloroplatinum (II)
(b) Tetraaminedichlorocobalt (III) chloride
(c) Potassium hexacyanoferrate (II)
(d) Hexaaquochromium (III) bromide
(e) Pentacarbonyl iron (0)
Kerala-CEE-2006
Ans. (c) : K 4 Fe ( CN )6
→ 4K + + Fe ( CN )6
The molar conductivity of Potassium hexacyanoferrate
(II) i.e K4 [Fe (CN)6] is highest because it gives
maximum number of ions on ionisation.
K4[Fe(CN6)] gives 5 ions in the solution which is
maximum out of the given complexes.
501. A weak electrolyte having the limiting
equivalent conductance of 400 S cm2 g equiv–1at
298 K is 2% ionized in its 0.1 N solution. The
resistance of this solution (in ohm) in an
electrolytic cell of cell constant 0.4 cm–1 at this
temperature is
(a) 200
(b) 300
(c) 400
(d) 500
(e) 600
Kerala-CEE-2012
Ans. (d) : The weak electrolyte is 2% ionized.
C=α.N
2
=
× 0.1
100
= 2 × 10–3 g equiv. L–1
Limiting equivalent conductance
κ × 1000
Λ eq =
C
Λm
α=
Λ∞
Λ ∞ = 400 Scm2g eq–1
2
Λ m = Λ ∞ × α = 400 ×
=8
100
κ × 1000
1
Λm =
and. κ = × Cell constant
C
R
1
1000
Λ ∞ = × Cell constant ×
R
C
Cellconstant
× 1000
R=
Λm × C
0.4 × 1000
R=
= 500 ohm
8 × 0.1
4–
Objective Chemistry Volume-II
502. The values of limiting ionic conductance of H+
and HCOO– ions are respectively 347 and 53 S
cm2mol–1 the dissociation constant of methanoic
acid at 298 K is
(a) 1×10-5
(b) 2 × 10-5
-4
(c) 1.5×10
(d) 2.5×10-5
-4
(e) 2.5 × 10
Kerala-CEE-2014
Ans. (e) : Λ °m (HCOOH) = Λ° H + + Λ° HCOO–
= (347 + 53) Scm2 mol–1 = 400 Scm2 mol–1
Dissociating constant Ka = Cα2 (for weak electrolytes)
Degree of dissociation α =
Λ cm
°
Λm
=
40
= 0.1
400
Ka = Cα2
= 0.025 × (0.1)2
= 2.5 × 10–4
503. How is electrical conductance of a conductor
related with length and area of cross section of
the conductor?
(a) G = l.a.k–1
(b) G = k.l.a–1
–1
(c) G = k.a.l
(d) G = k.l.a–2
MHT CET-2014
1
Ans. (c) : Specific conductance, ( κ ) = G ×
a
G = electrical conductance
a
G = k×
l
Electrical conductance is given by
G= k.al–1
504. Which of the following complexes has lowest
molar conductance?
(a) CoCl3.3NH3
(b) CoCl3.4NH3
(c) CoCl3.5NH3
(d) CoCl3.6NH3
MHT CET-2014
Ans. (a) : Molar conductance is the conducting power
of all the ions produced by dissolving 1g mole of an
electrolyte.
Therefore,
Molar conductance ∝ Number of ions produced.
Werner
ModernIonisation
complex
notation
CoCl3.3NH3 [Co(NH3)3Cl3]
[Co(NH3)3Cl3]
(No ionisation)
CoCl3.4NH3 [Co(NH3)4Cl2]Cl
[Co(NH3)4Cl3]+ +
Cl–
(2 ions)
CoCl3.5NH3 [Co(NH3)5Cl]Cl2
[Co(NH3)5Cl]2+ +
2Cl– (3 ions)
CoCl3.6NH3 [Co(NH3)6]Cl3
[Co(NH3)6]3+ +
3Cl– (4 ions)
From above table it is clear that, molar conduction of
the complexes will be in the following order a < b < c
< d.
229
YCT
505. At 25 ºC molar conductance of 0.1 molar
aqueous solution of ammonium hydroxide is
9.54 ohm–1 cm2 mol–1 and at infinite dilution its
molar conductance is 238 ohm–1 cm2 mol–1. The
degree of ionisation of ammonium hydroxide at
the same concentration and temperature is
(a) 4.008%
(b) 40.800%
(c) 2.080%
(d) 20.800%
(NEET-2013)
Ans. (a) : Given,
Molar conductance of 0.1m solution
λ cm = 9.54 ohm–1 cm2 mol–1
Ans. (b) : Al2 (SO 4 )3 ↽ ⇀ 2Al3+ + 3SO 24−
At infinite dilution dissociation of electrolytic solution
is complete and each ion makes a definite contribution
toward molar conductance of each electrolyte.
We can calculate the equivalent conductance only ions,
so the equivalent conductance at infinite dilution.
°
∧°Al2 (SO4 ) = ∧°Al3+ + ∧SO
2−
3
°
eq
°
Al3+
∧ =∧
4
°
SO24−
+∧
509. The equivalent conductance of M/32 solution of
a weak monobasic acid is 8.0 mho cm-2 and at
infinite dilution is 400 mho cm2. The
At infinite dilution = λ ∞m = 238 ohm–1 cm2
dissociation constant of this acid is
λ cm
(a) 1.25 × 10–6
(b) 6.25 × 10–4
Degree of ionization (a) = ∞
–4
(c) 1.25 × 10
(d) 1.25 × 10–5
λm
(AIPMT -2009)
9.54
=
× 100
Ans.
(d)
:
Given
that
238
M 1
=4.008%
Concentration, C =
= M
32
32
506. Limiting molar conductivity of NH4OH
c
Λ m = 8.0 mho cm 2
i.e., Λ º
is equal to
m ( NH 4OH )
Λ om = 400 mho cm 2
º
(a) Λ m( NH4Cl ) + Λ ºm( Na 4Cl ) − Λ ºm( NaOH )
Degree of dissociation
(b) Λ mº ( NaOH ) + Λ mº ( NaCl ) − Λ mº ( NH4Cl )
Λ cm
8
(
α
)
=
=
= 0.02
o
º
º
º
Λ m 400
(c) Λ m( NH OH ) + Λ m( NH Cl ) − Λ m( HCl )
4
(d) Λ
º
m ( NH 4 Cl )
4
+Λ
º
m( NaOH )
− Λ ºm( NaCl )
Kc = C
α2
1
= × 0.02 × 0.02
1 − α 32
1 – α =1 – 0.02
=.98 ≃ 1
(AIPMT -2012)
Ans. (d) : According to Kohlrausch law, limiting molar
Kc = 1.25×10–5
conductivity of NH4OH
510. Equivalent conductivity at infinite dilution for
Λ mº ( NH4OH ) = Λ ºm( NH4Cl ) + Λ ºm( NaOH ) − Λ ºm( NaCl )
sodium-potassium oxalate [(COO-)2Na+K+] will
507. An increase in equivalent conductance of a
be [given molar conductivities of oxalate, K+
strong electrolyte with dilution is mainly due to
and Na+ ions at infinite dilution are 148.2, 50.1,
(a) increase in ionic mobility of ions
73.5 S cm2 mol-1 respectively]
(b) 100% ionisation of electrolyte at normal
(a) 271.8 S cm2 eq-1
(b) 67.95 S cm2 eq-1
2
-1
dilution
(c) 543.6 S cm eq
(d) 135.9 S cm2 eq-1
(c) increase in both i.e., number of ions and ionic
WB-JEE-2013
mobility of ions
∞
∞
∞
+
Ans. (d) : λ m = λ m (oxalate) + λ m (Na ) + λ ∞m (K + )
(d) increase in number of ions
(AIPMT -2010)
λ ∞m = (148.2 + 50.1 + 73.5)Scm 2 mol−1
Ans. (a) : Strong electrolytes are completely ionised at
λ ∞m = 270.8Scm 2 mol–1
all concentrations. On increasing dilution the no. of ions
271.8
remains the same but the ionic mobility increases and
λ ∞Eq =
= 135.9Scm 2 eq −1
the equivalent conduction increases.
2
508. Which of the following expressions correctly 511. A conductivity cell has been calibrated with a
represents the equivalent conductance at
0.01M 1 : 1 electrolyte solution (specific
infinite dilution of Al2(SO4)3 ?. Given that
conductance, k = 1.25 × 10–3 S cm–1) in the cell
º
and the measured resistance was 800 Ω at 25oC.
Λ Al 3+ and Λ SO2− are the equivalent conductances
4
The cell constant will be
at infinite dilution of the respective ions.
(b) 0.102 cm-1
(a) 1.02 cm-1
º
º
(a) 2 Λ ºAl3+ + 3Λ SO
(b) Λ ºAl3+ + Λ SO
-1
2−
2−
(c) 1.00 cm
(d) 0.5 cm-1
4
4
WB-JEE-2013
1
1 º
º
(c) ( Λ ºAl3+ + Λ SO
(d) Λ ºAl3+ + Λ SO
2−
2− ) × 6
–3
4
4
Ans. (c) : Given that κ = 1.25 × 10 s/cm
3
2
(AIPMT -Mains 2010)
R = 800Ω
Objective Chemistry Volume-II
230
YCT
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is in
correct.
(d) If both the Assertion and Reason are
incorrect.
[AIIMS-2011]
Ans. (a) : According to Kohlrausch law limiting molar
conductivity of an electrolyte can be represented as the
sum of the individual contributions of the anion and
cation of the electrolyte.
515. Molar conductances of BaCl2, H2SO4 and HCl
at infinite dilutions are x1, x2 and x3
respectively. Equivalent conductance of
BaSO4 at infinite dilution will be:
(a) (x1+x2–x3)/2
(b) x1+x2–2x3
(c) (x1–x2–x3)/2
(d) (x1+x2–2x3)/2
[AIIMS-2011]
Ans. (d) :
Λ m .BaCl2 = Λ mBe2+ + 2Λ mCl− x1
We know,
l
1
and ρ =
A
κ
1
l
800 =
×
1.25 × 10−3 A
l
= cell constant
A
l
= 800×1.25×10–3
A
= 1 cm–1
512. The correct order of equivalent conductance’s
at infinite dilution in water at room
temperature for H+, K+, CH3COO– and HO–
ions is
(a) HO– > H+ > K+ > CH3COO–
(b) H+ > HO– > K+ > CH3COO–
(c) H+ > K+ > HO- > CH3COO–
(d) H+ > K+ > CH3COO– > HO–
WB-JEE-2013
Ans. (b) : Equivalent conductance is the conductivity
power of all the ions produce by one gram equaivalent
of an electrolyte in the given solution.
Eq. conductance of cation > Eq. conductance of anion
Λ m .H 2SO 4 = 2Λ mH+ + Λ mSO− x 2
4
(It is due to size of ion)
+
–
+
–
Λ
Λ
=
+
x3 ]× 2
Λ
+
−
mHCl
H > HO > K > CH3COO
mH
mCl
o
513. At 25 C, the molar conductance of 0.007 M ∴ Λm BaSO 4 = ( x1 + x 2 − 2x 3 )
hydrofluoric acid is 150 mho cm2mol–1 and it’s
o
Λ m .BaSO 4
A m = 500 mho cm2 mol-1The value of the Λeq BaSO 4 =
Total
charge
(cation / anion )
dissociation constant of the acid at the given
concentration at 25oC is
[ x + x 2 − 2x 3 ]
(a) 7 × 10–4 M
(b) 7 × 10–5 M
Λeq BaSO 4 = 1
2
–3
–4
(c) 9 × 10 M
(d) 9 × 10 M
516. Equivalent conductance of an electrolyte
WB-JEE-2014
containing NaF at infinite dilution is
Ans. (d) : Given that Molar conductance (C) = 0.007M
90.1 Ohm–1 cm2. If NaF is replaced by KF
what is the value of equivalent conductance?
λ
α = om
(a) 90.1 Ohm-1 cm2
(b) 111.2 Ohm–1 cm2
λm
(c) 0
(d) 222.4 Ohm-1 cm2
150
[AIIMS-2009]
α=
= 0.3
500
Ans. (a) : Because at infinity dilution the equivalent
conductance of strong electrolytes furnishing same
HF ⇌ H + + F−
number of ions is same. In this case both NaF and KF
2
2
α
0.007 × (0.3)
are strong electrolyte.
K=C
=
1− α
1 − 0.3
Thus, they have same equivalent conductance. Hence,
the equivalent conductance of FF is also 90.1 Ohm–1.
0.007 × 0.09
=
0.7
6. Type of Cell
= 9×10–4 M.
R =ρ
°
514. Assertion : If λ °Na+ and λ Cl
are molar limiting
-
517. Match list I with list II
List-I
list-II
(a) Li
(i)
absorbent
for
carbon
dioxide
(b) Na
(ii)
electrochemical cells
(c) KOH
(iii)
Coolant in fast breeder
reactor
Reason : This is according to Kohlrausch law
(d) Cs
(iv)
photoelectric cell
of independent migration of ions.
conductivity of sodium and chloride ions
respectively, then the limiting
molar
conductivity for sodium chloride is given by the
equation:
°
λ °NaCl = λ °Na + λ Cl
-
Objective Chemistry Volume-II
231
YCT
Choose the correct answer from the options
given below:
(a) (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
(b) (a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)
(c) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
(d) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
NEET-17.06.2022
Ans. (a) : Li – Electrochemical cells or bearing for
motor engines
Na – coolant in fast breeder reactors
KOH – Absorbent for CO2
Cs – Photoelectric cell which converts energy of light
directly into electrical energy (electricity)
518. The energy conversion involved in a galvanic
cell is
(a) Chemical energy to mechanical energy
(b) Chemical energy to electrical energy
(c) Electrical energy to chemical energy
(d) Electrical energy to thermal energy
AP-EAMCET-07.07.2022, Shift-I
Ans. (b) : The energy conversion involved in a galvanic
cell is chemical energy into electrical energy.
Galvanic cell consists of two segment box called half
cell. The half cell contains electrode and electrolyte
solution that is connected in the circuit.
Two dissimilar metals [eg → copper and zinc] are
immersed in electrolyte. These metals are connected by
external circuit then one metal (copper) is reduced and
other metal [zinc] is oxidized.
519. E1, E2, E3 are the emf values of the three
galvanic cells respectively.
(i)
520. For a galvanic cell Cr3+/Cr3+//Cd2/Cd.
calculated ∆ G0 for its all reaction will be
[E 0Cr3+ / cr = −0.74V, E 0Cd2 / Cd = −0.40V]
1
+M +M +.
u
u
C u C
|
|
C
|
+M
M
1 +M 21
+.
2 0 2 1u
u C
u
|
C |
C |
|
|
|
M
21 20
2
1
n
n
n
Z |
Z |
Z
|
n n n
Z Z Z
(a) –28.95 kJ/mol
(b) –125.4 kJ/mol
(c) –196.8 KJ/mol
(d) –87.6 kJ/mol
AP- EAPCET- 07-09-2021, Shift-I
Ans. (c) : Given that,
Ecathode = – 0.40
Eanode = – 0.74
Ecell = Ecathode – Eanode
= – 0.40 – ( – 0.74)
= 0.34
∆G° = –nf Ecell
= –6 × 96500 × 0.34
= – 196869 J/mol
= – 196.869 kJ mol–1.
521. Among the following, number of metal/s which
can be used as electrodes in the photoelectric
. (Integer answer)
cell is
(a) Li
(b) Na
(c) Rb
(d) Cs
JEE Main 25.02.2021, Shift-II
Ans. (d) : Cesium is used as an electrode in
photoelectric cells because its ionization potential is
low. Due to which its valence shell electrons can be
easily ejected out by the energy of light.
• Li is used to make electrochemical cells.
• Sodium is used to make a Na/Pb alloy needed to
make PbEt4 and PbMe4.
• Liquid sodium metal is used as a coolant in fast
(ii)
breeder nuclear reactors.
(iii)
Hence, Cs is only metal which used as electrodes in
the photoelectric cell.
Which one of the following is true?
(a) E2 > E3 > E1
(b) E3 > E2 > E1
522. Match the entries from Column I and Column
(c) E1 > E2 > E3
(d) E1 > E3 > E2
II and choose the correct order.
BCECE-2014, Karnataka-CET, 2009
Column I
Column II
Ans. (b) : For the given cell,
A. Leclanche
1. Converts energy of
cell
combustion
into
0.0591
[Zn 2 + ]
E cell = E ocell −
log
2+
electrical
energy
[Cu ]
2
B. Fuel cell
2. Rechargeable cell
0.0591
1
(i)
E1 = E ocell −
log
C. Ni-Cd Cell 3. At anode, Zn→Zn+2 +
2
0.1
Ze–
0.0591
o
Codes
= E cell −
2
A
B
C
0.0591
1
o
(a)
3
2
1
E 2 = E cell −
log
(ii)
2
1
(b) 1
2
3
(c)
3
1
2
0.0591
= E ocell −
×0
(d)
2
1
3
2
AP EAMCET (Engg.) 18.9.2020 Shift-I
= E ocell
Ans. (c) : The correct matching between Column I and
0.0591
0.1
Column II is
E 3 = E ocell −
log
(iii)
2
1
(A) Leclanche cell-anode reaction :
0.0591
o
Anode : Zn ( s ) → Zn +2 + 2e −
= E cell +
2
Cathode : MnO 2 + NH +4 + e − → MnO ( OH ) + NH 3
E3 > E2 > E1
Objective Chemistry Volume-II
232
YCT
(B) Fuel cell produced energy due to combustion
2H2 + O2 → H2O + Energy
(gets converted into electrical energy).
(C) Ni-Cd cell is rechargeable , so it has more life time.
523. The hydrogen ion concentration in a standard
hydrogen electrode is
(a) 1 mol/L
(b) 1g/L
(c) 1kg/L
(d) 1mol/mL
J & K CET-(2019)
Ans. (a) : Hydrogen ion concentration in a standard
hydrogen electrode is 1 mol/L.
524. A galvanic cell transforms the energy released
by a spontaneous redox reaction into electrical
energy that can be used to perform work. In a
galvanic cell, electrons flow from
(a) anode to cathode through the external circuit
(b) cathode to anode through the external circuit
(c) anode to cathode through the electrolyte
(d) cathode to anode through the electrolyte.
J & K CET-(2019)
Ans. (a) : A galvanic cell transforms the energy
released by a spontaneous redox reaction into electrical
energy from negative pole (anode) to the positive pole
(cathode) through the external circuit.
525. Which of the following is not true regarding the
usage of hydrogen as a fuel?
(a) High calorific value
(b) The combustible energy of hydrogen can be
directly converted to electrical energy in a
fuel cell
(c) Combustion product is ecofriendly
(d) Hydrogen gas can be easily liquefied and
stored
Karnataka-CET-2019
Ans. (d) : It is true about hydrogen to be used as a fuel.
Hydrogen is highly inflammable and hence storage is
difficult. Hydrogen has low critical temperature and
hence not easily liquefied.
It is true that the combustible energy can be directly
converted to electrical energy is a fuel cell.
526. A fuel cell operates at constant current, with H2
fuel (1 bar) and O2 oxidant (1 bar). The
electrolyte used is 0.001 M HCl and the
product(s) of the reaction are confined inside
the fuel cell. Which of the following is true
about the electrolyte?
(a) Boiling point of the electrolyte decrease with
increase in the duration of fuel cell operation
(b) Boiling point of the electrolyte increases with
increase in the duration of fuel cell operation
(c) Open circuit voltage of the fuel cell remains
constant with increase in duration of
operation
(d) Open circuit voltage of the fuel cell increase
with increase in duration of operation
(e) Both (A) and (C)
Kerala-CEE-2019
Objective Chemistry Volume-II
Ans. (a) : A fuel cell operates at constant current, with
H2 fuel (1 bar) and O2 oxidant. The electrolyte used is
0.001 M HCl and the products of the reaction are
confined inside the fuel of boiling point at the
electrolyte decreases with increase in the duration of
fuel cell operation.
527. Which of the following acts as oxidising agent
in hydrogen-oxygen fuel cell ?
(a) H2
(b) O2
(c) KOH
(d) C
MHT CET-02.05.2019, SHIFT-II
Ans. (b) : A hydrogen-oxygen fuel cell is an
electrochemical cell that uses a pair of redox reactions
to turn the chemical energy of hydrogen which serves as
a fuel, and oxygen as an oxidizing agent into electricity.
2H 2 + 4OH −
→ 4H 2 O + 4e −
→ 4OH −
O 2 + 2H 2 O + 4e −
2H 2 + O 2
→ 2H 2 O
528. Electrolyte used in Ni-Cd cell
(a) KOH
(b) H2SO4
(c) LiOH
(d) Al2O3
JCECE - 2018
Ans. (a) : Electrolyte used in Ni-Cd cell is KOH
(potassium hydroxide). It is a type of secondary cell. It
consists of a cadmium anode and metal grid containing
NiO2 as cathode and KOH solution is electrolyte. It has
longer life than the lead electrolyte. It has longer life
than the lead storage cell but more expensive to
manufacture.
529. For a fuel cell, which statement is correct?
(a) Hydrogen is oxidized to H2O at cathode
(b) Oxygen is reduced to OH− at anode
(c) Hydrogen is oxidized to H2O at anode
(d) Hydrogen is oxidized to H+ at anode
CG PET -2017
Ans. (c) : Galvanic cells which use energy for
combustion of fuels like H 2 ,CH 4 ,CH 3OH etc. as the
source to produce electrical energy are called fuel cells.
Fuel cell uses the reaction of H2 and O2 in gaseous state
to form water.
At anode, hydrogen is oxidized to H2O
(i) 2H 2 → 4H + + 4e −
(ii) 4H + + 4OH − → 4H 2 O
At cathode, oxygen is reduced to OH– ions
O 2 (g) + 2H 2 O(l ) + 4e − → 4OH − ( aq )
530. When a lead storage battery is discharged:
(a) SO2 is evolved
(b) lead sulphate is consumed
(c) lead is formed
(d) sulphuric acid is consumed
JIPMER-2017
Ans. (d) 2H 2SO 4 (aq) → 4H + ( aq ) + 2SO 24− (aq)
233
Pb(s) + SO 24− (aq) → PbSO 4 (s) + 2e − ( anode )
PbO 2 (s) + 4H + (aq) + SO 24− (aq) + 2e −
→ PbSO 4 (s) + H 2 O(l )(cathod)
YCT
Pb(s) + PbO 2 (s) + 2H 2SO 4 (aq) → 2PbSO 4 (s)
+2H 2 O(l ) (complete reaction)
Sulphuric acid is consumed during discharging of lead
strong battery.
531. Consider a fuel cell supplied with 1 mol of H2
gas and 10 moles of O2 gas. If fuel cell is
operated at 96.5 mA current, how long will it
deliver power?
(Assume 1F = 96500 C/mole of electrons)
(b) 0.5 × 106S
(a) 1 × 106S
6
(c) 2 × 10 S
(d) 4 × 106S
6
(e) 5 × 10 S
Kerala-CEE-2017
At.wt
Ans. (c) : ∵ w = zit and Z =
nF
Since the cell is supplied with 1 mole of H2 gas and 10
mole of O2. H2 is the limiting reagent, 1 mole of H2
reacts with 10 mole of O2.
1 mole of H2 required 0.5 mol of O2
charge requirement = 2F
H2 → 2H++2e–
Q
I = , I = 96.5 mA
t
Q 2 × 96500
t= =
= 2×106 S
I 9.65 × 10−2
532. For car battery which one is correct statement?
(a) Cathode is lead dioxide (PbO2) and anode is
copper (Cu)
(b) Cathode is copper (Cu) and anode is lead
dioxide (PbO2)
(c) Cathode is copper (Cu) and anode is lead (Pb)
(d) Cathode is lead dioxide (PbO2)and anode is
lead (Pb)
UPTU/UPSEE-2017
Ans. (d) : The anode is a positively charged electrode
within the battery which allows the electrons to travel
outside of the battery .
The negatively charged electrode through which the
electrons re-enter.
Car battery is a lead storage battery in which cathode is
made of PbO2 and anode is made of Pb–plate.
PbO2 (Pb4+) reduce to Pb2+ at cathode while Pb oxidise
to Pb2+ at anode.
533. The space in the dry cell is filled with
(a) paste of KOH and ZnO
(b) MnO2, ZnCl2, a filter
(c) MnO2, ZnCl2, NH4Cl and a filter
(d) MnCl2, ZnCl2, NH4Cl and a filter
COMEDK 2016
Ans. (c) : A dry cell is a type of electric battery which
has two electrodes. There is a mixture of MnO2, ZnCl2,
NH4Cl and a filter. It contains an electrolyte that is
contained within a paste or other moist medium.
534. Which of the following is incorrect in a
galvanic cell?
(a) Oxidation occurs at anode
(b) Reduction occurs at cathode
Objective Chemistry Volume-II
(c) The electrode at which electrons are gained is
called cathode
(d) The electrode at which electrons are lost is
called cathode
Karnataka-CET-2016
Ans. (d) : A galvanic cell is an electrochemical cell in
which oxidation occurs at anode containing negative
charge and reduction occur at cathode containing a
positive charge.
At the electrode at which electrons are lost is called anode.
The anode is where the oxidation reaction take place. In
other words this is where the metal loses electrons.
535. A secondary cell is one
(a) can be recharged
(b) can be recharged by passing current through it
in the same direction
(c) can be recharged by passing current through it
in the opposite direction
(d) cannot be recharged
Karnataka-CET-2016
Ans. (c) : A secondary cell is a type of cell which can
be recharged by passing current through it in the
opposite direction once the cell is discharged.
They can be recharged by passing current through it in
the opposite direction.
536. In dry cell, what acts as a negative electrode?
(a) Zinc
(b) Graphite
(c) Ammonium chloride (d) Manganese dioxide
MHT CET-2016
Ans. (a) : In a dry cell zinc acts as negative electrode.
A zinc carbon battery is a dry cell battery between a
zinc metal electrode and a carbon rod from an
electrochemical reaction between zinc and manganese
dioxide mediated by a suitable electrolyte.
Zinc acts as a negative electrode as it has more
negative potential and acts as the anode.
537. Zinc can be coated on iron to produce
galvanized iron but the reverse is not possible.
It is because
(a) zinc is lighter than iron
(b) zinc has lower melting point than iron
(c) zinc has lower negative electrode potential
than iron
(d) zinc has higher negative electrode potential
than iron
(NEET-II 2016)
Ans. (d) : Reduction potential values of
E oZn 2+ / Zn = −0.76V
E oFe2+ / Fe = −0.44V
Zinc can be coated on iron to produce galvanized iron
but the reverse is not possible it is because Zinc has
higher negative electrode potential than iron.
538. What will happen if a cell is placed into 0.4%
(mass/volume) NaCl solution?
(a) There will be no change in cell volume
(b) Cell will dissolve
(c) Cell will swell
(d) Cell will shrink
UPTU/UPSEE-2016
234
YCT
Ans. (c) : Cell will swell due to osmotic pressure when 543. In H2-O2 fuel cell the reaction occurring at
cathode is
the blood cells are placed in 0.4% sodium chloride
solution, water flows into the cells and the cells swell.
(a) 2H2(g) + O2(g) → 2H2O (l)
539. The standard emf of Zn-Cu voltaic cell is
(b) O2(g) + 2H2O(l) + 4e–
→ 4OH − (aq)
(a) 2.1 V
(b) 2.8 V
1
(c) 1.2 V
(d) 1.1 V
→ H2
(c) H+ + e–
2
SRMJEEE – 2015
(d) H + (aq) + OH − (aq)
Ans. (d) : The standard electromotive force (EMF) of
→ H 2 O(l )
Zn-Cu voltaic cell is 1.1V at room temperature.
Karnataka-CET-2015
Zn 2+
0.059
log 2+
Cu
2
0.059
0.1
log
= 1.10 −
2
0.1
0.059
E cell = 1.10 −
log1
2
= 1.10 – 0
= 1.10
540. The standard electrode potential for Daniell
cell is 1.1 volt. What is the standard Gibbs
energy for the reaction?
(a) 212. 3 kJ mol–1
(b) –212.3 kJ mol–1
–1
(c) 106.15 kJ mol
(d) –106.15 kJ mol–1
COMEDK 2015
Ans. (b) : Given that,
E°= 1.1 volt.
The standard electrode potential for Daniell cell is _
Zn ( s ) + Cu 2+ ( aq )
→ Zn 2+ ( aq ) + Cu ( s ) , n=2
E cell = E cell −
∴ ∆G o = –nFE°
∆G ° = –2 × 96500 ×1.1
∆G ° = –212.3KJ / mole
541. The cathode reaction in the dry cell will be
(a) Zn (s)
→ Zn 2 + + 2e –
(b) MnO 2 + NH +4 + e –
→ MnO(OH) + NH 3
(c) Zn(Hg) + 2OH –
→ ZnO(s) + H 2 O + 2e –
(d) MnO(OH) + NH 3
→ MnO 2 + NH +4 + 2e –
COMEDK 2015
Ans. (b) : The cathode reaction in the dry cell will beMnO 2 + NH +4 + e –
→ MnO(OH) + NH 3
542. In the lead-acid battery during charging, the
cathode reaction is
(a) formation of PbO2
(b) formation of PbSO4
(c) reduction of Pb2+ to Pb
(d) decomposition of Pb at the anode.
AMU-2015
Ans. (c) : During charging of lead–acid battery Pb2+
ions of PbSO4 are reduced to Pb on cathode.
At
Anode
PbSO4 (s) + 2H 2O → PbO2 (s) + SO42− (aq) + 4H + (aq) + 2e−
At Cathode PbSO 4 (s) + 2e − → Pb(s) + SO 42− (aq)
Thus, Pb2+ ions of PbSO4 are reduced to Pb on the
cathode while PbSO4 is oxidized to PbO2 at anode.
Objective Chemistry Volume-II
Ans. (b) : In a fuel cell cathode reaction can be written
as
O2 + 2H2O + 4e– → 4OH–
Anode reaction can be written as
H2 + 2OH– → 2H2O + 2e–
Net cell reaction ––
2H2 + O2 → 2H2O
544. While charging the lead storage battery,
(a) PbSO4 on anode is reduced to Pb
(b) PbSO4 on cathode is reduced to Pb
(c) PbSO4 on cathode is oxidised to Pb
(d) PbSO4 on anode is reduced to PbO2
Karnataka-CET-2015
Ans. (a) : During charging PbO2 is formed at anode
from PbSO4 (oxidation of Pb ion) Pb is formed at
cathode from PbSO4 (reduction of Pb ion).
545. A device that converts energy of combustion of
fuels like hydrogen and methane, directly into
electrical energy is known as
(a) dynamo
(b) Ni-Cd cell
(c) fuel cell
(d) electrolytic cell
(NEET-2015, Cancelled)
Ans. (c) : Fuel cell converts energy of combustion of
fuels like H2, CH4 etc. directly into electrical energy.
Dynamo produce direct current by using a commutator.
Ni–Cd cell is a type of rechargeable battery which
consists of a cadmium anode and a metal grid
containing NiO2 acting as a cathode.
546. Which one of the following cells can convert
chemical energy of H2 and O2 directly into
electrical energy?
(a) Mercury cell
(b) Daniel cell
(c) Fuel cell
(d) Lead storage cell
VITEEE- 2010
Ans. (c) : A fuel cell is a device that convert chemical
potential energy to electrical energy. In fuel cell the
energy produced by the combustion of fuels like, H2,
O2, CH4, etc is converted into electric energy, eg H2–O2
fuel cell.
547. The electrochemical cell stops working after
sometime because :
(a) electrode potential of both the electrodes
becomes zero
(b) electrode potential of both the electrodes
becomes equal
(c) one of the electrodes is eaten away
(d) the cell reaction gets reversed
VITEEE- 2008
235
YCT
Ans. (b) : As we know that,
∆Ecell = Ecathode – Eanode
when Ecathode = Eanode
∆Ecell = 0
If ∆Ecell = 0
When electrode potential of both the electrodes
becomes equal then the electrochemical cell stops
working after sometime.
548. The extent of charge of lead accumulator is
determined by
(a) amount of PbSO4 in the battery
(b) amount of PbO2 in the battery
(c) specific gravity of H2SO4 of the battery
(d) amount of Pb in the battery
AP-EAMCET (Engg.) 2013
Ans. (c) : The lead accumulator refers to a secondary
cell as the electrical energy does not generate itself
inside the cell, but it is stored prior to an external
source.
The extent of charge of a lead accumulator is
determined by the specific gravity of H2SO4 of the
battery.
Pb ( s ) + PbO 2 ( s ) + 4H + ( aq ) + 2SO 4 2− ( aq )
→
2PbSO 4 ( l ) + 2H 2 O
The fully charged lead accumulator is 1.30 g/ml and
when it drops below 1.20 g/ml, the lead accumulator
need recharging.
549. In a electrochemical cell, the reaction will be
feasible when
(a) ∆G = –ve, E = + ve (b) ∆G = + ve, E = –ve
(c) ∆G = 0, E = –ve
(d) ∆G = 0, E = 0
COMEDK 2014
Ans. (a) :For a spontaneous reaction, ∆G should be –ve
which is possible only when E is +ve as ∆G = –nFE.
then ∆G = –ve and E= +ve.
550. Which metal is protected from corrosion by a
layer of its own oxide?
(a) Tl
(b) Ag
(c) Al
(d) Au
AMU-2013
Ans. (c) : Aluminum reacts with air to form Al2O3
which protects the metal from further corrosion.
Al is oxidized to form Al2O3 which forms a protective
coating on the surface of the metal.
551. During charging of lead storage battery, the
reaction occurring at the cathode is
→ Pb
(a) Pb2++2e-
2+
(b) Pb
→ Pb +2e(c) PbSO4+2H2O
→ 2PbO2+4H++ SO 42 − +2e–
(d) Pb2++ SO 42 −
→ PbSO4
AMU-2012
Ans. (c) : A lead storage battery is a secondary cell so
we can charge the battery by passing our current
directly so, it can be relies.
Objective Chemistry Volume-II
At anode
Pb + SO 24 − → PbSO 4 + 2e−
At cathode
PbO 2 + SO 42 − + 4H + + 2e − → PbSO 4 + 2H 2 O
552. Other things being equal, the EMF of a Daniel
cell may be increased by
(a) keeping low temperature
(b) using large copper electrodes
(c) using large zinc electrodes
(d) decreasing concentration of Cu2+ ions
BCECE-2011
Ans. (a) : For Daniel cell–
Zn + Cu2+ → Zn2+ + Cu
RT [Zn 2 + ]
o
E cell = E cell
−
ln
nF [Cu 2 + ]
The decrease in temperature and increase in
[Cu2+] will increase the EMF.
Thus, low temperature will increases EMF of Daniel cell.
553. Match the electrode (in column I) with its
general name (in Column II) and choose the
correct option given below.
Column I
Column II
Calomel
A
1 Reference
Glass
B
2 Redox
Hydrogen
C
3 Membrane
Quinhydrone
D
4 Gas
A
B
C
D
(a)
1
3
4
2
(b)
3
4
2
2
(c)
3
2
4
1
(d)
2
4
3
1
BCECE-2013
Ans. (a) : Calomel
→ Reference
Glass
→ Membrane
Hydrogen
→ Gas
Quinhydrone
→ Redox
554. In a Daniell cell constructed in the laboratory,
the voltage observed was 0.9V instead of 1.1V
of the standard cell. A possible explanation is(a) [Zn2+] > [Cu2+]
(b) [Zn2+] < [Cu2+]
(c) Zn electrode has twice the surface of Cu
electrode
(d) mol ratio of Zn2+ : Cu2+ is 2 : 1
BCECE-2013
Ans. (a) :
Zn 2+
0.0591
°
E cell = E cell −
log 2+
2
Cu
Zn 2+
0.0591
= 1.1−
log 2+
Cu
2
Zn 2+
E cell < 1.1V
if 2+ > 1
Cu
Thus, option (a) is correct.
236
YCT
555. In electrolysis of NaCl when Pt electrode is
taken then H2 is liberated at cathode while with
Hg electrode if forms sodium amalgam
(a) Hg is more inert that Pt
(b) more voltage is required to reduce H+ at Hg
than at Pt
(c) Na is dissolved in Hg while it does not
dissolved in Pt
(d) concentration of H+ ions is larger when Pt
electrode is taken
CG PET -2007
Ans. (b) : Sodium chloride in water dissociates as
NaCl → Na++ Cl–
H2O → H+ + OH–
When electric current is passed through this
solution using platinum electrodes Na+ and H+ move
towards cathode.
More voltage is required to reduce H+ at Hg than at Pt.
556. On the basis of the information available from
4
2
the reaction Al + O 2 → Al 2O3 , ∆G = − 827 kJ
3
3
mol−1 of O2, the minimum emf required to
carry out an electrolysis of Al2O3 is (F=96500 C
mol−1)
(a) 2.14 V
(b) 4.28 V
(c) 6.42V
(d) 8.56 V
CG PET -2007
Ans. (a) : Given that,
∆G = 827 kJ mol–1
4
2
and Al + O3 →
Al2O3
3
3
For O2 ∆G = –nFE° n=4
– 827 × 103 J = – 4 × E° × 96500
∆G 827 × 103
E=
=
= 4.28
nF 4 × 96500
4.28
E =
= 2.14V
2
557. The standard emf of a galvanic cell involving
cell reaction with n = 2 is found to be 0.295 V at
25ºC. The equilibrium constant of the reaction
would be
(a) 2.0× 1011
(b) 4.0× 1012
2
(c) 1.0× 10
(d) 1.0× 1010
(Given, F=96500 C mol −1
R=8.314 JK−1mol−1)
CG PET -2007
Ans. (d) : Given that,
E °cell = 0.295 V
R = 8.314 Jk–1
T = 25°C + 273 = 298 k
F = 96500 C mol–1 and n = 2
By Nernst equation ,
2.303RT
Ecell = E °cell –
log10 k eq
nF
At equilibrium Ecell = 0
Objective Chemistry Volume-II
2.303RT
log k eq
nF
2.303 × 8.314 × 298
0.295 =
log k eq
2 × 96500
2 × 96500 × 0.295
log k eq =
= 9.97 ≈ 10
2.303 × 8.314 × 298
10
Keq= antilog 10 =1×10
558. In galvanic cell, the salt bridge is used to
(a) complete the circuit
(b) reduce the electric resistance in the cell
(c) separated cathode from anode
(d) carry salts for the chemical reaction
CG PET -2006
Ans. (a) : In galvanic cell the salt bridge used to
complete the circuit as it provides an electrical contact
between the two solution.
It maintains electrical neutrality in both the solutions by
a flow of ions.
559. Anode reaction of a fuel cell is
(a) Zn(Hg) + 2OH − → ZnO(s) + H 2 O + 2e−
E ocell −
(b) Pb(s) + SO 24− (aq)
→ PbSO 4 (s) + 2e −
→ 4H 2 O (l) + 4e−
(c) 2H 2 (g) + 4OH − (aq)
(d) 2Fe(s)
→ 2Fe 2+ + 4e −
J & K CET-(2012)
Ans. (c) Fuel cells is an electrochemical cell that
generates electrical
energy from fuel via an
electrochemical reaction.
Steps involve in fuel cell reaction at anode ––
• Fuel transported to the anode of the cell.
• Fuel undergoes the anode reaction.
• Anode reaction splits the fuel into H+ and e–
2H 2 (g) + 4OH − (aq)
→ 4H 2 O (l) + 4e−
560. A cell is constructed by coupling the two
electrodes Sn/Sn2+ and Cu/Cu2+. If Eo (Sn2+,
Sn), Eo (Cu2+, Cu), and Eocell are – 0.14 V, 0.34
V and 0.48 V respectively, the correct
representation of the cell is
(a) Sn (s) Sn 2+ (0.1M) Cu 2+ (1.0M) Cu (s)
(b) Sn (s) Sn 2+ (1.0M) Cu + (1.0M) Cu (s)
(c) Sn (s) Sn 2+ (1.0M) Cu 2+ (1.0M) Cu (s)
(d) Cu (s) Cu 2+ (1.0M) Sn 2+ (1.0M) Sn (s)
Ans. (c) : The value of E
o
cell
J & K CET-(2011)
in (c) option is
E ocell = E ocathode − E oanode
= 0.34 – (–0.14)
= 0.34 + (0.14)
= 0.48 V
Half cell reaction
Cathode (reduction) Cu2++ 2e– → Cu
Anode (oxidation) Sn → Sn2+2e–
237
YCT
In the representation of the cell, anode keeps on the left
side and cathode on the right side,
Hence, the correct representation the cell ––
Sn (s) Sn 2+ (1.0M) Cu 2+ (1.0M) Cu (s)
561. In the electrolysis of aqueous solution of CuSO4
using copper electrode, the process that takes
place at the anode is
–
(a) SO 2–
4 → SO 4 + 2e
(b) Cu → Cu + + e –
(c) 2OH – → H 2 O + 1/ 2O 2 + e –
(d) Cu → Cu 2+ + 2e –
J & K CET-(2009)
Ans. (d) : When aqueous solution of CuSO4 is
subjected to electrolysis, the following reactions takes
place at cathode and anode.
Cathode
Cu2+ + 2e– → Cu
Anode
Cu (s) → Cu2+ + 2e–
Therefore, copper will get deposited at cathode and
copper will get dissolved at anode.
562. Galvanic cell is a device in which :
(a) chemical energy is converted into electrical
energy
(b) electrical energy is converted into chemical
energy
(c) chemical energy is seen in the form of heat
(d) thermal energy from an outside source is used
to drive the cell reaction
J & K CET-(2008)
Ans. (a) : Galvanic cell is an electrochemical cell that
converts the chemical energy of spontaneous redox
reactions into electrical energy .
563. What is the cell reaction occurring in Daniel
cell (Galvanic cell)?
(a) Cu(s) + ZnSO4(aq) → CuSO4(aq) + Zn(s)
(b) Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4 (aq)
(c) Ni(s) + ZnSO4(aq) → NiSO4(aq) + Zn(s)
(d) 2Na(s) + CdSO4(aq) → Na2SO4(aq) + Cd(s)
J & K CET-(2006)
Ans. (b) : Daniell cell consists of two electrodes of
dissimilar metal, Zn and Cu each electrode is in contact
with a solution of its own i.e. Zinc sulphate and copper
sulphate respectively.
The redox reaction of the Daniel cell is
Zn(s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq)
564. Which of the following statements (or equation)
is correct?
(a) The unit of cell emf is V cm–1
nF
(b) ∆G = –
E
(c) In galvanic cell, chemical energy is
transformed into electrical energy
(d) Oxidation state of Mn in potassium
permanganate is + 6
J & K CET-(2006)
Objective Chemistry Volume-II
Ans. (c) : In a galvanic cell the electrical energy is
produced from chemical reactions (i.e. chemical energy
is transformed into electrical energy).
• The unit of cell e.m.f is volt.
• ∆G = – nFEcell
• In KMnO4 oxidation state of Mn is +7.
565. Cell reaction is spontaneous, when
(a) E °red is negative
(b) E °red is positive
(c) ∆G° is negative
(d) ∆G° is positive
MHT CET-2008
Ans. (c) : For a cell reaction to be spontaneous ∆Go
should be negative.
As ∆Go= –nFEo.
So, the value will be –ve only when E is +ve.
n = no. of electrons involved
F = value of Faraday’s constant
When ∆G < 0 the cell reaction is spontaneous
∆G must be negative for spontaneity of cell.
566. The efficiency of a fuel cell is given by
(a) ∆G/∆S
(b) ∆G/∆H
(c) ∆S/∆G
(d) ∆H/∆G
(AIPMT -2007)
Ans. (b) : Thermodynamic efficiency of fuel cell is given
as the ratio of the Gibb's function change to the Enthalpy
change. Gibb's Function measures the electrical work
whereas the enthalpy change is a measure of the heating
value of the fuel. This is given as∆G
Efficiency =
∆H
567. When an acid cell is charged, then
(a) voltage of cell increase
(b) resistance of cell increases
(c) electrolyte of cell dilutes
(d) All of the above
UP CPMT-2014
Ans. (a) : The voltage of the cell increases when an acid
cell is charged. During charging of an acid cell the
external source supplies current due to emf increases
and also the relative density of cell increases. So, the
voltage of cell increases.
568. Anode in the galvanic cell is
(a) negative electrode (b) positive electrode
(c) neutral electrode
(d) None of the above
UP CPMT-2005
Ans. (a) : Anode in galvanic cells is negative electrode
where oxidation take place.
In galvanic cell the current flows from cathode to anode
in external circuit.
∴ Anode is negative electrode in galvanic cell.
569. The electrolyte in lead storage battery is dilute
sulphuric acid. The concentration of sulphuric
acid in a lead-storage battery must be between
4.8 M and 5.3 M for most efficient functioning
A 5 mL sulphuric acid sample of a particular
battery requires 50 mL of 1.0 M NaOH for
complete neutralisation. Which of the following
statements about the functioning of battery is
the most appropriate?
238
YCT
(a) The acid concentration in the battery is not in and,
the most effective range
0.0591
1
°
(b) The acid concentration in the battery is in the E 2 = E − 2 log 0.01
most effective range
0.0591
°
2
(c) The acid concentration in the battery is hardly E 2 = E − 2 log10
in the most effective range
0.0591
°
(d) Only a good mechanic can tell whether or not E 2 = E −
2
the acid concentration in the battery is in the
Thus E1 > E2
most effective range
UPTU/UPSEE-2011 572. The chemical reaction,
2AgCl(s) + H 2 (g) → 2HCl(aq ) + 2Ag (s)
Ans. (b) : Sample of sulphuric acid volume - 5 ml
taking place in a galvanic cell is represented by
Volume of HaOH for neutrolisation = 50 ml
the notation:
Normality of naOH required for neutralization = 1.0M =
1.0N
(a) Pt (s) H 2(g ) ,1bar 1M KCl(aq) AgCl(S) Ag (s)
∵ N 1V 1 = N 2V 2
(b) Pt (s) H 2(g) ,1bar 1M HCl(aq) 1M Ag +(aq) Ag (s)
1.0 × 50
N1 =
= 10 N
(c) Pt (s) H 2(g) ,1bar 1M HCl(aq) AgCl(S) Ag (s)
5
(d) Pt (s) H 2(g) ,1bar 1M HCl(aq) Ag (s) AgCl(S)
normality 10
∴ Molarity =
=
= 5M
valence
2
[AIIMS-2005]
Hence the acid concentration in lead storage battery in Ans. (b) :
the most effective range (between 4.8 and 5.3 M).
2AgCl(s) + H 2 (g) → 2HCl(aq) + 2Ag(s)
570. What flows in the internal circuit of a galvanic
The activities of solids and liquids are taken as unity
cell?
and at low concentrations the activity of a solute is
(a) Ions
(b) Electrons
approximated to its molarity.
(c) Electricity
(d) Atoms
The cell reaction will be.
UPTU/UPSEE-2008
Pt(s) | H 2 (g)(1 bar) | 1M HCl(aq) | 1M Ag + (aq) | Ag(s)
Anode
Cathode
Ans. (a) : Ions flow in the internal circuit of a galvanic
cell. The internal circuit of a galvanic cell, ions flow
where as in the external circuit the electrons flow from
7. Molar conductance
one electrode to another electrode.
571. The emf of Daniell cell at 298 K is E1
573. Specific conductance of 0.1 M HNO 3 is
Zn ZnSO4 (0.01M) CuSO4 (1.0M) Cu.
6.3x10-2 ohm -1 cm -1 . The molar conductance of
When the concentration of ZnSO4 is 1.0M and
the solution is
that of CuSO4 is 0.01 M, the emf changed to E2
(a) 6.300 ohm −1cm 2 mol−1
What is the relation between E1 and E2?
(b) E2=0≠E2
(a) E1= E2
(b) 63.0 ohm −1cm 2 mol −1
(c) E1>E2
(d) E1<E2
(c) 630ohm −1cm 2 mol −1
[AIIMS-2008]
(d) 315ohm −1cm 2 mol −1
Ans. (c) :
Karnataka CET-17.06.2022, Shift-II
Zn 2+
0.0591
Ecell = E °cell −
log 2+
Ans. (c) : Given that,
Cu
n
Concentration of nitric acid = 0.1 M
Substituting the given value in two cases.
Specific conductance, K = 6.3× 10−2 ohm−1 cm−1
0.0591
0.01
K × 1000
E1 = E ° −
log
Molar conductance, λ =
2
1.0
C
−2
0.0591
°
−2
6.3
×
10
× 1000
E1 = E −
log10
=
2
0.1
0.0591
°
E1 = E +
×2
= 630ohm −1 cm −1mol −1
2
Objective Chemistry Volume-II
239
YCT
03.
Chemical Kinetics
3.
1.
1.
Rate of Chemical Reaction
For a cell, Cu ( s ) Cu 2+ (0.001M Ag+ ( 0.01M) Ag ( s )
the cell potential is found to be 0.43 V at 298 K.
The magnitude of standard electrode potential
for Cu 2+ / Cu is ____× 10 −2 V.
It has been found that for a chemical reaction
with rise in temperature by 9 K the rate
constant gets double. Assuming a reaction to be
occurring at 300 K, the value of activation
energy is found to be _____ kJ mol–1 [nearest
integer]
(Given ln 10 = 2.3, R = 8.3 J K–1 mol–1, log 2 =
0.30)
JEE Main-27.06.2022, Shift-II
2.303RT
Θ
= 0.06V
K
Ea 1
1
Given: E Ag+ / Ag = 0.80V and
−
Ans. (59) : log10 2 =
F
K1 2.303R 300 309
JEE Main 29.07.2022, Shift-II
Ea
9
Ans. (34) : At anode Cu → Cu2+ +2e–
0.3 =
+
–
2.303 × 8.3 300 × 309
At cathode 2Ag + 2e → 2Ag
0.3 × 2.303 × 8.3 × 300 × 309
Cell reaction →
Ea =
Cu + 2Ag+ → Cu2+ +2Ag
9
2+
=
59065.4
J
Cu
2.303RT
log
Ecell = E 0cell −
E a = 59.06KJ
2
F
Ag +
4.
The equation
12 -1 -26000K/T
2+
K=(6.5×10
s )e
is followed for the
Cu
0.06
log
Ecell = E 0cell −
decomposition
of compound A. The activation
2
2
Ag +
energy for the reaction is _____ kJ mol-1.
[nearest integer]
( 0.001)
0.06
(Given: R = 8.314 JK-1 mol-1]
0.43 = E 0cell −
log
2
2
JEE Main-29.06.2022, Shift-II
( 0.01)
0
Ans.
(216)
:
Given,
E cell = 0.46
K=(6.5×1012s-1)e-26000K/T
0
0
0
E cell = E Ag+ / Ag − E Cu + / Cu
We know,
0
K = Ae − Ea / RT
0.46 = 0.80 – E Cu + / Cu
Then
0
E Cu
= 0.34 volt
−26000K
+
/ Cu
K= (6.5×1012) e T
0
–2
E Cu + / Cu = 34 × 10
Ea
= 26000
2.
Among the following which are mismatched:
8.314
Order
Unit of k
Ea= 216.164 kJ/mol.
(A) Zero order
–
mol L–1 time–1
–1
5.
Catalyst
A reduces the activation energy for a
(B) First order
–
mol L
reaction by 10 KJ mol–1 at 300 K. The ratio of
–1
–1
(C) second order
–
L mol time
K T Catalysed
(D) Third order
–
L2 mol–1 time–1
rate constants,
is ex. The value
th
1–n n–1
–1
K
Uncatalysed
(E) n order
–
mol
L
time
T
of x is ––––––[nearest integer]
(a) (A) and (C)
(b) (B) and (D)
[Assume that the pre-exponential factor is
(c) (A) and (E)
(d) (C) and (E)
same in both the cases given R = 8.31 J K–1
CG PET-22.05.2022
mol–1]
Ans. (b) :
JEE Main-26.06.2022, Shift-II
Order
Unit of k
Ans. (4) : We know
(i) Zero order
mol L–1 time–1
–1
K = Ae − Ea / RT
(ii) First order
sec
For calalysed
(iii) Second order
L mol–1 time–1
(iv) Third order
(v) nth order
L2 mol–2 time–1
mol1 – n Ln – 1 time–1
Objective Chemistry Volume-II
1
K cat = Ae− Ea / RT
And For Uncatalysed
240
YCT
8.
K uncat = Ae − Ea / RT
1
Ea − Ea
K cat
= e RT
K uncat
10×1000
= e 8.31×300
X
e = e4.009
x=4
6.
The activation energy of one of the reaction in
a biochemical process is 532611 J mol–1. When
the temperature falls from 310 K to 300 K, the
change in rate constant observed is
k300= x ×10–3k310. The value of x is ––––––.
[Given : ln 10 = 2.3
R = 8.3 j K–1mol–1]
JEE Main-29.06.2022, Shift-I
Ans. (1) : We know,
K = Ae − Ea / RT
K E 1 1
ln 310 = a −
K 300 R T1 T2
K 532611
10
ln 310 =
×
K
8.3
310
×
300
300
The rate constants for decomposition of
acetaldehyde have been measured over the
temperature range 700 -1000K. The data has
103
graph.
been analysed by plotting in k vs
T
The value of activation energy for the reaction
is _______ kJ mol–1. (Nearest integer)
(Given: R = 8.31 J K–1mol–1)
JEE Main-24.06.2022, Shift-I
Ans. (154) :
From Arrhenius equation–
K = Ae − Ea / RT
Taking log both side–
E
We get, log K = log A − a
K 310
RT
ln
= 6.9
Ea
K 300
Slope of the graph = −
= −18.5
= 3 × ln10
R × 103
Ea= 18.5 ×8.31×1000=154kJ mol–1
K
ln 310 = ln103
9.
In a reaction 3A→ products, the concentration
K 300
of A decreases from 0.4 mol L–1 to 0.1 mol L–1
3
in 20 minutes at 300K. The rate of decrease in
K310 = K300 × 10
–3
[A] during this interval (in mol L–1 min–1) at
K300 = K310 × 10
–3
300 K is
K300= x ×10 K310
(a) 0.005
(b) 0.015
So,
x=1
(c)
0.001
(d) 0.15
7.
The rate constant for a first order reaction is
(e)
0.05
given by the following equation
Kerala CEE -03.07.2022
2.0×104
Ans. (a) : We know that
Ink = 33.24 −
T
1 [A ] − [A]
The activation energy for the reaction is given
Rate = × o
-1
3
t
by ______ kJ mol . (In nearest integer)
(Given : R = 8.3 J K-1 mol -1)
1 [0.4] − [0.1]
Rate = ×
JEE Main-27.06.2022, Shift-I
3
20
Ans. (166) : Given
1 [0.3]
Rate = ×
2 × 104
3 20
lnk = 33.24 −
…….(i)
T
1 3
Rate = ×
We know,
3 200
E
Rate = 0.005 mol L–1 min–1
lnk = lnA − A
…….(ii)
RT
10. The half-life period of a first order reaction at
comparing both equation
298K is 20 minutes. The time (in min.) required
EA
4
for 99.9% completion of the reaction at the
= 2.0×10
R
same temperature, is
EA= 2.0×104×R
(a) 100
(b) 200
EA= 2.0×104×8.3
(c) 15
(d) 250
EA= 16.6×104 Jule
(e) 300
Kerala CEE -03.07.2022
EA= 166 kJ.
Objective Chemistry Volume-II
241
YCT
Ans. (b) : For first order reaction
We know,
2.303
a
t=
log
K
a−x
Given that,
X= 0.999a for t99.9%
And X= 0.5a for t50%
2.303
a
log
t 99.9%
K
a
−
0.999a
=
2.303
a
t 50%
log
K
a − 0.5a
1
log
t 99.9%
0.001
=
1
20
log
0.5
t 99.9% − ( −3 )
−(−3)
=
=
20
log 2 0.0301
Given that,
X= 0.9a for t90%
X= 0.5a for t50%
And t90% = X t50%
2.303
a
log
t 90%
a − 0.9a
= K
2.303
a
t 50%
log
K
a − 0.5a
Xt 50%
log10
=
log 2
t 50%
13.
t99.9% = 10×20
t99.9% = 200 min.
11. For a first order reaction A → B, the rate
constant, k = 5.5 × 10–14 s–1. The time required for
67%
completion
of
reaction
is
X × 10–1 times the half life of reaction. The
value of x is –––––– (Nearest integer).
JEE Main-28.06.2022, Shift-I
Ans. (16) : For first order reaction
We know,
2.303
a
t=
log
K
a−x
Given that,
X= 0.67a for t67%
X= 0.5a for t50%
And t67% = X×10–1 t50%
2.303
a
log
t 67%
a − 0.67a
= K
2.303
a
t 50%
log
K
a − 0.5a
X × 10 −1 t 50%
log 3
=
t 50%
log 2
X ×10 −1 = 1.585
X×10–1 = 15.84 ×10–1
X ≈16
12. For a first order reaction, the time required for
completion of 90% reaction is ‘X’ times the
half life of the reaction. The value of ‘X’ is
(Given: In 10= 2.303 and log 2 = 0.3010)
(a) 1.12
(b) 2.43
(c) 3.32
(d) 33.31
JEE Main-24.06.2022, Shift-II
Ans. (c) : For first order reaction
We know,
2.303
a
t=
log
K
a−x
Objective Chemistry Volume-II
2.303
0.3010 × 2.303
X= 3.32
At 345 K, the half life for the decomposition of
a sample of a gaseous compound initially at
55.5 kPa was 340 s. When the pressure was 27.8
kPa, the half life was found to be 170 s. The
order of the reaction is–––.[integer answer]
JEE Main-25.06.2022, Shift-II
X=
Ans. (0) : t1/ 2 ×
t1
t2
1
[ Po ]
n −1
(P )
= 2 n −1
( P1 )
n −1
340 27.8
=
170 55.5
1
2 = n −1
( 2)
n=0
14.
n −1
For the reaction 2A
→ 4B + C, if.
d[A ]
= k1 [A]
dt
then,
(a) 3k1 = k 3
–
(c) 2k1 = k 3
d [B ]
dt
= k 2 [A]
d [C]
dt
= k 3 [A]
(b) k1 = 2k 3
(d) k1 = 3k 3
Assam-CEE-31.07.2022
Ans. (b) : 2A→4B+C
From law of chemical equilibrium,
1 dA 1 dB dC
−
=
=
2 dt 4 dt
dt
1
1
∴ K1[A] = K 2 [A] = 1K 3 [A]
2
4
1
∴ K1[A] = 1K 3 [A]
2
∴ K1[A] = 2K 3 [A]
15.
242
+d(B)
is equals to
dt
2 d(A)
(b) –
3 dt
3A → 2B, rate of reaction
(a) –
3 d(A)
2 dt
YCT
1 d(A)
d(A)
(d) +2
3 dt
dt
COMEDK-2018, Karnataka-CET-2016
WB-JEE-2008, UP CPMT-2006
(AIPMT -2002)
Ans. (b) : Given the reaction is 3A→2B
Rate of reaction is 1 d[A] 1 d[B]
=–
=
3 dt
2 dt
d[B]
2 d[A]
or
=–
dt
3 dt
16. Assertion: The kinetics of the reaction
mA + nB + pC → m'X + n'Y + p'Z
dX
obey the rate expression as
= k[A]m [B]n
dt
Reason: The rate of the reaction does not
depend upon the concentration of C.
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are
incorrect.
[AIIMS-2011, 2017]
Ans. (a): The rate of reaction can be determined
experimentally by using rate law,
dX
m
n
= k [ A ] . [ B]
rate
dt
∴ The rate of reaction is independent of concentration
of c. hence option (a) is correct.
17. 1L of 2 M CH3COOH is mixed with 1 L of 3 M
C2H5OH to form an ester. The rate of the
reaction with respect to the initial rate when
each solution is diluted with an equal volume of
water will be
(a) 0.25 times
(b) 2 times
(c) 0.5 times
(d) 4 times
Karnataka-CET-2015, 2019
Ans. (a) :
CH 3COOH + C 2 H 5OH
→ CH 3COOC 2 H 5 + H 2 O
(c) –
t=0
2
At
time t
2
=1
2
3
3
= 1.5
2
Initial rate = k [ CH 3 COOH ][ C 2 H 5 OH ]
= k × 2×3
ri = 6k
∴
rate = k × 1× 1.5
1
= − × 6k = 0.25 × initial rate
4
Objective Chemistry Volume-II
For the reaction : H 2 + I 2
→ 2HI , the
differential rate law is
d [H2 ]
d [ I2 ]
d [ HI ]
(a) −
=−
=2
dt
dt
dt
d [H2 ]
d [ I 2 ] d [ HI ]
(b) −2
= −2
=
dt
dt
dt
d [H2 ]
d [ I2 ] d [ HI ]
(c) −
=−
=
dt
dt
dt
d [H2 ]
d [ I2 ] d [ HI ]
=−
=
(d) −
dt
dt
dt
MHT CET-2008, AIPMT -1997
Ans. (b) : Reaction: H 2 + I 2
→ 2HI
18.
∴ Reaction rate
∴
19.
−2
−d [ H 2 ]
d [H2 ]
dt
= −2
=
−d [ I 2 ]
d [ I2 ]
dt
=+
=+
1 d [ HI]
2 dt
d [ HI]
dt
dt
dt
The rate of a gaseous reaction is given by the
expression k[A][B]2. If the volume of vessel is
reduced to one half of the initial volume, the
reaction rate as compared to original rate is
1
16
(c) 8
1
8
(d) 16
Kerala-CEE-29.08.2021
Karnataka-CET-2021
Ans. (c) : Given that, the rate of gaseous reaction –
r1 = k [A] [B]2
.....(i)
∵
C ∝ 1/V
When the volume of vessel is reduced to one half of the
initial volumer2 = k [A/2] [B/2]2
[A][B]2
r2 = k
.....(ii)
8
Dividing equation (i) by (ii)(a)
(b)
r1
k[A][B]2
=
r2 k[A][B]2 / 8
r1
=8
r2
∴
r1 = 8r2
So, the reaction rate as compared to original rate in 8
times.
20. For a chemical reaction A→B, it was found
that concentration of B is increased by
0.2 mol L–1 in 30 min. The average rate of the
reaction is .......×10–1 mol L–1 h–1. (Nearest
integer)
[JEE Main 2021, 25 July Shift-II]
−1
−1 −1
Ans. ( 4 × 10 molL h ) : Given, A
→B
d [A]
d [ B]
∴
reaction rate, = −
=+
dt
dt
243
YCT
reaction rate =
23.
0.2
= 2 × 0.2
130
= 0.4 molL−1h −1
= 4 × 10−1 molL−1h −1
Rate of the reaction, xA + yB → zC is given
by r = k[A]x[B]y. If the concentration of A is
tripled, rate of reaction increased by 27 times
over the initial. Then choose the correct plot
for variation of half-life (t1/2 on y-axis) of the
reaction w.r.t. total initial concentration of
reactants (on x - axis) is ........
Rate of reaction is 4 × 10 −1 molL−1h −1
21. The reaction that occurs in a breath analyser, a
device used to determine the alcohol level in a
person's blood stream is
2K2Cr2O7+8H2SO4+3C2H6O→
2Cr2(SO4)3+3C2H4O2+2K2SO4+11H2O
If the rate of appearance of Cr2(SO4)3 is 2.67
mol min–1 at a particular time, the rate of
disappearance of C2H6O at the same time is
AP EAMCET (Engg.) 18.9.2020 Shift-I
....... mol min–1. (Nearest integer)
Ans.
(a)
:
In
this question, we have to assume that
[JEE Main 2021, 27 Aug Shift-I]
order of the reaction xA + yB → zC with respect to B is
−1
−1
Ans. ( 4.005 molL min ) : Given that,
zero, i.e., y = 0.
2K 2 Cr2 O7 + 8H 2SO 4 + 3C 2 H 6 O
So, the rate expression becomes,
r = k [ A ] [ B] = k [ A ] [ B] = k [ A ]
….. (i)
Now, according to questionIf the concentration of A is tripled, rate of reaction
increased by 27 times over the initial.
According to available data
(27r) =k[3A]x
…. (ii)
Divide equation (ii) by equation (i), we get
→ 2Cr2 ( SO 4 )3 + 3C 2 H 4O 2 + 2K 2SO 4 + 11H 2 O
x
Rate of appearance of
d Cr2 ( SO 4 )3
= 2.67mol min −1
Cr2 ( SO 4 )3 .
dt
∴ reaction rate
1 d [ K 2 Cr2 O 7 ]
1 d [ H 2SO 4 ]
1 d [ C2 H6 O]
=−
=−
=−
2
dt
8
dt
3
dt
=
∴
y
x
0
x
27r k [3A ]
=
x
r
k [A]
x
1 d Cr2 ( SO 4 )3 1 d [ C 2 H 4 O 2 ]
1 d [ K 2SO 4 ]
=
=−
2
dt
3
dt
2
dt
or 27 = ( 3) or ( 3) = ( 3)
x=3
Thus, order of reaction is 3
Now, for a third order reaction half-life will be
3
1
t1/ 2 =
or t1/ 2 ∝ 2
2
2k 3a 0
a0
x
1 d [ C 2 H 6 O ] 1 d Cr2 ( SO 4 )3
=
3
dt
2
dt
d [ C2 H6 O] 3
−
= × 2.67
dt
2
= 4.005 molL−1 min −1
3
x
∴ rate of disappearance of C 2 H 6 O = 4molL−1 min −1
22.
For a chemical reaction
k
k2
A + B ↽k 1 ⇀ C
→ D . The rate
−1
dc
is
dt
(a) k1[A][B] – k –1[C] – k 2 [C]
24.
(b) –k1[A][B] + ( k –1 + k 2 )[C]
(c) k1[A][B]
(d)
[ k1 – k 2 – k –1 ][C]
TS-EAMCET 09.08.2021, Shift-I
Ans. (a) :
k1 →
k2 D
A + B ←
C →
k −1
dc
The rate is,
= k1 [ A ][ B] – k –1 [ C ] – k 2 [ C ]
dt
Objective Chemistry Volume-II
244
3
For the reaction 2A + 3B + C → 3P, which
2
statement is correct?
dn A 2 dn B 3 dn C
(a)
=
=
dt
3 dt
4 dt
dn A 3 dn B 3 dn C
(b)
=
=
dt
2 dt
4 dt
dn A dn B dn C
(c)
=
=
dt
dt
dt
dn A 2 dn B 4 dn C
(d)
=
=
dt
3 dt
3 dt
[JEE Main 2020, 3 Sep Shift-II]
YCT
3
Ans. (d) : Given reaction : 2A + 3B + C
→ 3P
2
Rate of reaction
1 d [A]
1 d [ B]
2 d [ C]
1 d [P]
=
=−
=−
=+
2 dt
3 dt
3 dt
3 dt
∴ [A ] ∝ n A
∴ Rate of reaction
1 dn A
1 dn B
2 dn C 1 dn P
=−
=−
=−
=
2 dt
3 dt
3 dt
3 dt
dn A 2 dn B 4 dn C
∴
=
=
dt
3 dt
3 dt
25. For the reaction
2H2(g)+2NO(g)→N2(g)+2H2O(g)
the observed rate expression is,
rate = kf[NO]2[H2]. The rate expression for the
reverse reaction is
(a) kb[N2][H2O]2/[H2] (b) kb[N2][H2O]
(c) kb[N2][H2O]2/[NO] (d) kb[N2][H2O]2
[JEE Main 2020, 7 Jan Shift-II]
Ans. (a) : Given that,
2H 2 ( g ) + 2NO ( g )
→ N 2 ( g ) + 2H 2 O ( g )
Forward reaction rf = k f [ H 2 ][ NO ]
2
[ N ][ H O]
At equilibrium, k eq = 2 2 2 2
[ H 2 ] [ NO]
2
k f [ N 2 ][ H 2 O ]
=
k b [ H 2 ]2 [ NO ]2
2
N 2 ][ H 2 O ]
[
2
k f [ H 2 ][ NO ] = k b
[H2 ]
2
[ N ][ H 2 O]
rf = K b 2
[H2 ]
2
In the chemical reaction A → B, what is the
order of the reaction? Given that, the rate of
reaction doubles if the concentration of A is
increased four times.
(a) 2
(b) 15
(c) 0.5
(d) 1
AP EAMCET (Engg.) 21.09.2020, Shift-I
Ans. (c) : We know that,
27.
r1 = k [ A ]
n
r2 = 2r1 = k [ 4A ]
n
Divide equation (ii) by (i).
( 2) = ( 4)
1
= ( 2)
2n
∴
2n = 1
1
⇒
n = = 0.5
2
Hence, the correct option is (c).
28. For a reaction of the type aA + bB → Products,
d[A]
the –
is equal to :
dt
d[B]
1 d[B]
(a) –
(b) – ×
dt
b dt
a d[B]
b d[B]
(c) – ×
(d) – ×
b dt
a dt
Manipal-2019
Ans. (c) : Reaction : aA + bB → products
1 d [A ]
1 d [ B]
∴ rate = −
=−
a dt
b dt
d [A]
a d [ B]
∴
−
=−
dt
b dt
29. For the elementary reaction
2SO 2 (g) + O 2 (g)
→ 2SO 3 (g) , identify
correct among the following relations
d[O 2 ( g )]
d[SO 2 (g)
(a) −
=−
dt
dt
+1 d[SO3 ( g )] d[SO 2 ( g )]
(b)
=
2
dt
dt
At equilibrium, rate of backward reaction
[ N 2 ][ H 2O]
[H2 ]
2
K b [ N 2 ][ H 2 O ] / [ H 2 ]
2
rb = rf = K b
r b=
n
26.
the
The rate of a reaction doubles when the initial
concentration of the reactant is made four-fold.
d[SO3 ( g )]
2d[O 2 ( g )]
If the initial concentration is made 400 fold,
(c) +
=−
then the rate will become :
dt
dt
(a) 400 times
(b) 200 times
d[SO 2 ( g )]
d[O 2 ( g )]
(c) 40 times
(d) 20 times
=−
(d) +
Manipal-2020
dt
dt
Ans. (d) :When reactant concentration is four-fold the
MHT CET-02.05.2019, SHIFT-III
reaction rate is double i.e.
Ans. (c) :
1/ 2
rate = k [ A ]
Reaction : 2SO 2 ( g ) + O 2 ( g )
→ 2SO3 ( g )
When the concentration becomes 400 times then
d [O2 ]
1 d [SO 2 ]
1 d [SO3 ]
1/ 2
rate
=−
=−
=+
rate = k [ 400A ]
2 dt
dt
2 dt
1/ 2
d [SO3 ]
2.d [ O 2 ]
= 20 k [ A ]
=−
dt
dt
The rate will becomes 20 times.
Objective Chemistry Volume-II
245
YCT
30.
For the elementary reaction,
3 H2(g) + N2(g) → 2NH3(g) identify the correct ∵
relation among the following relations:
d [ B]
dt
=0
= k1 [ A ] − k 2 [ B] = 0
k
∴ [ B] = 1 [ A ]
k2
33. For an elementary chemical reaction,
−2 d H 2 ( g ) d NH3 ( g )
d[A]
k
(b)
=
A 2 ↽k 1 ⇀ 2A , the expression for
is
–1
3
dt
dt
dt
2
(b) k1[A2]–k–1[A]2
(a) 2k1[A2]–k–1[A]
d NH 3 ( g ) −1 d H 2 ( g )
2
(c) 2k1[A2]–2k–1[A]
(d) k1[A2]+k–1[A]2
(c)
=
[JEE Main 2019, 10 Jan Shift-II]
dt
3
dt
dx
1/ 2
−d H 2 ( g ) d NH 3 ( g )
Ans. (c) :
= k [ H 2 ][ 4Br2 ]
(d)
=
dt
dt
dt
k
MHT CET-03.05.2019, SHIFT-I Given reaction: A 2 ↽ 1 ⇀ 2A
k −1
(AIPMT -2006), (A-P EAMCET-2003) for forward reaction
for backward reaction
Ans. (b) :
k1
k −1
A 2 → 2A
2A →
A2
Reaction : 3H 2 ( g ) + N 2 ( g )
→ 2NH 3 ( g )
2
rate = k1 [ A 2 ]
rate = k −1 [ A ]
d [ N2 ]
1 d [H2 ]
1 d [ NH 3 ]
For equation (i)
∴ rate = −
=−
=+
3 dt
dt
2 dt
1 d [A]
2
= k1 [ A 2 ] − k −1 [ A ]
∴ rate = +
2 d H 2 ( g ) d NH 3 ( g )
2 dt
∴
−
=
3
dt
dt
d [A]
2
= 2.k1 [ A 2 ] − 2k −1 [ A ]
31. For the chemical reaction,
dt
N 2 (g) + 3H 2 (g) ↽ ⇀ 2NH 3 (g)
34. NO2 required for a reaction is produced by the
decomposition of N2O5 in CCl4 as per the
the correct option is
equation, 2N2O5(g)→4NO2(g)+O2(g) the initial
d [H2 ]
d [ NH 3 ]
concentration
of N2O5 is 3.00 mol L–1 and it is
(a) 3
=2
–1
2.75
mol
L
after 30 minutes. The rate of
dt
dt
formation of NO2 is
1 d [H 2 ]
1 d [ NH 3 ]
(a) 4.167×10–3 mol L–1 min–1
(b) −
=−
3 dt
2 dt
(b) 1.667×10–2 mol L–1 min–1
(c) 8.333×10–3 mol L–1 min–1
d[ N2 ]
d [ NH 3 ]
(c) −
=2
(d) 2.083×10–3 mol L–1 min–1
dt
dt
[JEE Main 2019, 12 April Shift-II]
d [ N 2 ] 1 d [ NH 3 ]
Ans.
(b)
:
Rate
of
disappearance
of NO 2 ,
(d) −
=
dt
2 dt
d [ N 2 O5 ]
2.75 − 3.00
NEET-2019
−
= −
= 0.0083
dt
30
Ans. (d) : Reaction : N 2 ( g ) + 3H 2 ( g ) ↽ ⇀ 2NH 3 ( g )
∴ Given reaction : 2N 2O5 ( g )
→ 4NO 2 ( g ) + O 2 ( g )
d [ N2 ]
1 d [H2 ]
1 d [ NH 3 ]
∴ Rate of reaction
∴ rate = −
=−
=+
dt
3 dt
2 dt
d [O2 ]
1 d [ N 2 O5 ]
1 d [ NO 2 ]
=−
=+
=+
d [ N 2 ] 1 d [ NH 3 ]
2
dt
4 dt
dt
−
=
dt
2 dt
d [ NO 2 ]
d [ N 2 O5 ]
∴ Rate of formation of NO 2 ,
= 2×
32. For a reaction scheme,
dt
dt
k1
k2
A
→ B
→ C, if the rate of formation of
= 2 × 0.0083
B is set to be zero then the concentration of B is
= 1.667 × 10 −2 molL−1 min −1
given by
35. In the following reaction; xA→yB
k1
d [ A]
d [B]
(a) k1k 2 [A]
(b) [A]
log10 –
= log10
+ 0.3010
k2
dt
dt
(d) (k1 + k2)[A]
(c) (k1 – k2)[A]
A and B respectively can be
[JEE Main 2019, 12 April Shift-I]
(a) n-butane and iso-butane
k1
k2
(b)
C2H2 and C6H6
Ans. (b) : Given, Reaction : A → B → C
(c) C2H4 and C4H8
d [ B]
(d) N2O4 and NO2
∴
rate of formation of B,
= k1 [ A ] − k 2 [ B ]
dt
[JEE Main 2019, 12 April Shift-I]
(a)
−3 d H 2 ( g ) d NH3 ( g )
=
2
dt
dt
Objective Chemistry Volume-II
246
YCT
Ans. (c) : Given, xA
→ yB
∴
Initial pressure
Po
0
0
Pressure after time t, ( Po − P )
2P
P
∴ Initial pressure = Po
Total pressure, after time t, Pt = Po − P + 2P + P
Pt = Po + 2P
d [A ]
d [ B]
log10 −
= log10
+ 0.3010
dt
dt
d [ B]
= log10
+ log10 2
dt
d [A]
d [ B]
log10 −
= log10 2
dt
dt
d [A]
d [ B]
−
=2
dt
dt
1 d [ A ] d [ B]
⇒−
=
2 dt
dt
2A
→ B
2C 2 H 4
→ C4 H8
C 2 H 4 and C 4 H8
Pt − Po
2
∴ Po ∝ initial concentration
∴P =
( Po − P ) ∝
∴
Burning the coal is represented is
C(s)+O2(g) → CO2(g)
The rate of this reaction is increased by
(b) Decreasing the concentration of oxygen
(b) Powdering the lumps of coal
(c) Decreasing the temperature of coal
(d) Providing inert atmosphere
J & K CET-(2019)
Ans. (b) : On increasing the powdering of coal more
number of active centre are formed and there increase in
the surface area of reactant and hence the rate of
reaction increases.
36. The rate of a reaction is found to depend upon
two concentration variables. What should be
the order the reaction?
(a) 1
(b) 3
(c) 2
(d) 0
J & K CET-(2019)
Ans. (c) : When the rate depends on two concentration
variables, the possible order of reaction is 2.
37. For a first order gas phase reaction:
A(g) → 2B (g) + C (g)
P0 is initial pressure of A and Pt is the total
pressure at time 't'. Integrated rate equation is:
(b)
(c)
(d)
Ans. (b):
Po
2.303
log
t
( Po − P )
Po
Pt − Po
Po −
2
2Po
2.303
=
log
t
2Po − Pt + Po
=
35.
(a)
k=
concentration after time t.
2.303
log
t
2Po
2.303
log
t
3Po − Pt
39. Assertion: The rate of reaction is never negative.
Reason: Minus sign, used in expressing the rate,
shows that concentration of products is
decreasing.
(a) If both Assertion and Reason are correct and
Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are
incorrect.
AIIMS 26 May 2019 (Evening)
Ans. (c): The rate of reaction is always positive. Minus
sign used in expressing the rate only shows that the
concentration of the reactant is decreasing.
40. For the reaction : A + 2 B → C + D, the
expression of rate of reaction will be:
−1 d [ A ] −1 d [ B]
1 d [ A ] −1 d [ B]
(a)
=
(b)
=
1 dt
2 dt
1 dt
2 dt
Po
2.303
log
−1 d [ A ] 1 d [ B]
1 d [ A ] 1 d [ B]
t
(c)
=
(d)
=
Po − Pt
1 dt
2 dt
1 dt
2 dt
2Po
AIIMS 26 May 2019 (Evening)
2.303
log
Ans. (a):
t
3Po − Pt
d [A]
1 d [ B] d [ C ] d [ D ]
Po
2.303
Reaction rate : −
=−
=
=
log
dt
2 dt
dt
dt
t
2P
−
P
o t
d [A]
1 d [ B]
∴
−
=−
2Po
2.303
dt
2 dt
log
t
2Po − Pt
41. At what temperature rate becomes double than
AIIMS 25 May 2019 (Evening)
at 300 K?
Given reaction69 ( kJ )
Given lnk = 10 −
A ( g )
→ 2B ( g ) + C ( g )
RT
Objective Chemistry Volume-II
247
k=
YCT
(b) 307.7
k [A ]
∴
(d) 323.5
[M] = 1
k −1 [ C ]
AIIMS 26 May 2019 (Evening)
k2
M + B
→ Pr oducts.
Ans. (c): Given, T1 = 300K, T2 = ?, k1 = k, k2 = 2k
∴
rate
=
k
M
[
][ B]
69 ( kJ )
2
ln k = 10 −
Substituting the value of [M], we getRT
k k [ A ][ B]
69
rate = 1 2
∴
ln k1 = 10 −
………(i)
k −1 [ C ]
R.T1
44. Instantaneous rate of reaction 3A + 2B → 5C is
69
ln k 2 = 10 −
………(ii)
_______.
R.T2
1 d [A]
1 d [ B]
1 d [C]
Dividing equation (ii) by (i) we get=−
=+
(a) +
3 dt
2 dt
5 dt
k
69 1 1
ln 2 = − −
1 d [A ]
1 d [ B]
1 d [ C]
(b) −
=−
=+
k1
R T2 T1
3 dt
2 dt
5 dt
2k
−69 1
1
d [A ]
d [ B]
1
1
1 d [ C]
ln
=
(c) −
=+
=−
−
k 8.314 T2 300
3 dt
2 dt
5 dt
d
A
d
B
d [ C]
[
]
[
]
1
1
1
300 − T2
−69
(d) +
=−
=−
log 2 =
⇒ T2 = 292.03K
3 dt
2 dt
5 dt
8.314 × 2.303 T2 × 300
GUJCET-2019
42. The reaction is
Ans. (b) : Given reaction, 3A + 2B
→ 5C
2A+B+C→A2B+C, k=2×10–6 mol–2L2s–1. The
d
A
d
B
initial concentration of [A] = 0.05 mol/L, [B] = Rate of reaction: − 1 [ ] = − 1 [ ] = + 1 d [ C ]
0.1 mol/L and [C]=0.5 mol/L. The rate after
3 dt
2 dt
5 dt
0.04 mol/L of A has reacted, will be
45. In NH3 synthesis by Haber's process, what is
(a) 1.28×10–8
(b) 1.28×10–10
the effect on the rate of the reaction with the
addition of Mo and CO, respectively?
(c) 1.60×10–8
(d) 1.60×10–10
(a) Increases and decreases
AMU-2019
(b) Decreases and decreases
+ C
→ A2B + C
Ans. (b) : 2A + B
(c) Decreases and increases
Initial
0.05
0.1
0.5
(d) Both Mo and CO increases the rate
After (0.05–0.04) (0.1–2×0.01)
(e) Both Mo and CO does not affect the rate
= 0.01
= 0.08
Kerala-CEE-2018
2
Ans. (a) : In NH 3 synthesis by Haber's process, Mo
Rate
= k.[ A ] .[ B]
acts as catalyst whereas CO acts as catalyst poison
= 2 × 10 −6 × 0.01 × 0.08 × 0.08
hence Mo increases the formation of NH 3 and CO
= 128 × 10−12
decreases the formation of NH 3 .
= 1.28 × 10−10
46. The data given below are for the reaction of A
43. Consider the following reaction
and D2 to form product at 295 K. Find the
k
k2
correct rate expression for this reaction
A ↽ k 1 ⇀ M + C; M + B
→ products
−1
D2/mol L-1 A/mol L-1 Initial rate/mol L-1 S-1
The rate can be expressed as
0.05
0.05
1 × 10-3
k1k 2 [A][B]
(a) Rate = k 2 [A][B][M] (b) Rate =
0.15
0.05
3 × 10-3
k −1[C]
0.05
0.15
9 × 10-3
1
2
k1k 2 [A][B]
k1k −1[A][M]
(a) k[D2] [A]
(b) k[D2]2 [A]1
(c) Rate =
(d) Rate =
1
1
(c) k[D2] [A]
(d) k[D2]2 [A]2
k −1[C][M]
k 2 [B]
1
0
(e) k[D2] [A]
CG PET -2019
Kerala-CEE-2018
k
Ans. (b) : Given that, A ↽ k 1 ⇀ M + C
Ans. (a) : From the data, in experiment 1 and 2, the
−1
concentration of A is constant where that of D 2 is
Rate = k1 .[ A ]
tripled, and the rate is also be tripled, hence with respect
∴ Rate = k −1 [ M ][ C]
to D 2 , the reaction is of first order.
At equilibrium,
In experiment, 1 and 3, the concentration of D 2
∴
k1 [ A ] = k −1 [ M ][ C ]
remains constant while A is tripled, and the reaction rate
(a) 329
(c) 292.03
Objective Chemistry Volume-II
248
YCT
d [O2 ]
1 d [SO 2 ]
1 d [SO3 ]
=−
=+
2 dt
dt
2 dt
2
1
∴ rate = k [ A ] [ D 2 ]
d [SO3 ]
d [O 2 ]
∴ rate of appearance of SO3
= 2×
Order of reactions = 2 + 1 = 3
dt
dt
47. Find the unit of the rate constant of a reaction
1/2
=
2
×
2
×
10 −4
represented with a rate equation, rate = k [A]
−4
−1 −1
3/2
= 4 × 10 molL s
[B]
(a) mol-1 L s-1
(b) s-1
51. The chemical reaction 2O3→3O2, proceeds as
(c) mol L-1 s-1
(d) mol-2 L2 s-1
follows O 3 ⇌ O 2 + O (fast)
-3 3 -1
(e) mol L s
O + O 3 →2O 2 (slow) the rate law for the above
Kerala-CEE-2018
reaction can be1/ 2
3/ 2
Ans. (a) : rate = k [ A ] [ B]
(a) r = k1 [O3 ]2
(b) r = k1[O3 ]2 [O 2 ]−1
rate
(c) r = k1[O3 ][O 2 ]
(d) r = k1[O 2 ][O]
k=
1/ 2
3/ 2
[ A ] [ B]
BCECE-2017
mol / L.s
Ans.
(b)
:
Chemical
reaction:
2O
→
3O
Unit of k =
3
2
1/ 2
3/ 2
( Mol / L ) ( mol / L )
⇀
Step-I : O3 ↽ O 2 + O ( fast )
mol / L.s
=
Step-II : O + O3
→ 2O 2 ( slow )
2
( mol / L )
The rate of reaction depends on slowest step,
= mol −1L.s −1
∴
rate = k.[ O ][ O3 ]
48. A reaction is given as 2P + Q → products. If
concentration of Q is kept constant and ∴ For step-I, in equilibrium state, Keq = [ O 2 ][ O ]
concentration of P is doubled then rate of
[ O3 ]
reaction is
K eq. [ O3 ]
(a) doubled
(b) halved
∴ [O] =
(c) quadrupled
(d) remains same
[O2 ]
COMEDK-2018
K eq. [ O3 ]
Ans. (c) : Given the reaction
∴
Rate = k.
[ O3 ]
[O 2 ]
2P+Q →product
2
If the concentration of Q is const and concentration of P
O3 ]
[
2
−1
is doubled then rate of reaction is= k.K eq.
= k1 [ O 3 ] [ O 2 ]
2
[O2 ]
Rate= k[2P] [Q]
Rate= 4kP2Q
52. Which of the following statement is incorrect?
∴ The rate of reaction is quadrupled.
(a) The rate of law for any reaction cannot be
49. Faster a chemical reaction, smaller is the
determined experimentally
(a) Rate constant
(b) Complex reactions have fractional order
(b) Concentration of reactant
(c) Biomolecular reaction involved simultaneous
(c) Half-life
collision between two species.
(d) Energy
(d) Molecularity is only applicable for
JCECE - 2018
elementary
Ans. (c) : Faster a chemical reaction, smaller is the half
Karnataka-CET-2017
life. When a chemical reaction proceed at a faster rate, Ans. (a) : The rate law for any reaction can't be
then time in which the concentration, of a reactant is determined experimentally.
reduced to one-half of its initial concentration is small. Rate of reaction is pure experimental analysis, hence
Which is known as half-life of a reaction.
statement (a) is wrong.
50. For the reaction, 2SO 2 + O 2 ↽ ⇀ 2SO 3 the rate 53. Which of the following is not true about a
of disappearance of O2 is 2 × 10–4 mol L–1s–1.
catalysts.
The rate of appearance of SO3 is
(a) Mechanism of the reaction in presence and
(a) 2 × 10–4 mol L–1s–1 (b) 4 × 10–4 mol L–1s–1
absence of catalyst could be different.
(c) 1 × 10–1 mol L–1s–1 (d) 6 × 10–4 mol L–1s–1
(b) Enthalpy of the reaction does not change with
Karnataka-CET-2018
catalysts.
(c)
Catalyst enhances both forward and
⇀
Ans. (b) : Reaction, 2SO 2 + O 2 ↽ 2SO3
backwards reaction at equal rate.
d [O2 ]
(d) Catalyst participates in the reaction, but not
= 2 × 10−4 molL−1s −1
consumed in the process.
dt
will becomes I times hence w.r.t. A the reaction is of
third order.
Objective Chemistry Volume-II
∴ reaction rate : −
249
YCT
(e) Use of catalyst cannot change the order of the
reaction
Kerala-CEE-2017
Ans. (e) : The use of catalyst may change the order of
the reaction.
54. Average rate of reaction for the following
reaction. 2SO 2 ( g ) + O 2 ( g )
→ 2SO 3 ( g ) is
written as
∆ [SO 2 ]
∆ [O2 ]
(b) −
(a)
∆t
∆t
∆
SO
∆
SO
1 [ 2]
3
(c)
(d)
2 ∆t
∆t
MHT CET-2016
Ans. (b) : Given the average rate of reaction
2SO2(g) + O2(g)
→ 2SO3(g)
∆ [ O 2 ] 1 ∆ [SO3 ]
1 ∆ [SO 2 ]
Rate (r) = −
=−
=
∆t
2 ∆t
2 ∆t
∆ [O 2 ]
Hence, the average rate of reaction is −
∆t
55. For the reaction
5Br–(aq) + 6H+(aq) + BrO 3− (aq) → 3Br2(aq) +
3H2O(l)
−
∆[BrO 3 ]
If, −
= 0.01mol L−1 min −1 ,
∆t
∆[Br2 ]
inmol L−1 min −1 is
∆t
(a) 0.01
(b) 0.3
(c) 0.03
(d) 0.005
TS-EAMCET-2016
Ans. (c) : Given the reaction5Br–(aq) + 6H+(aq) + BrO3− (aq) → 3Br2(aq) + 3H2O(l)
Rate of reaction is ∆[BrO3− ] 1 ∆[Br2 ]
1 ∆[Br − ]
1 ∆[H + ]
=−
=−
=
=−
∆t
5 ∆t
6 ∆t
3 ∆t
∆[BrO3− ] 1 ∆[Br2 ]
∴ −
=
∆t
3 ∆t
∆[BrO3− ]
Given that, −
= 0.01mol L−1 min −1
∆t
∆[Br2 ]
So,
= 0.01× 3 = 0.03mol L−1 min −1
∆t
56. In a first order reaction, it takes 40.5 minutes
for the reactant to be 24% decomposed. Find
the rate of the reaction.
(a) 9.4 × 10–3 min–1
(b) 7.0 × 10–3 min–1
–3
–1
(c) 25.2 × 10 min
(d) 10.5 × 10–3 min–1
SRMJEEE – 2016
Ans. (b) : Given that, t = 40.5 minutes, k = ?
For the first order kinetics –
2.303
a
k =
log
t
a−x
Where, a = initial concentration
Objective Chemistry Volume-II
x = concentration of product at time t
2.303
100
k=
log
40.5
76
2.303
k=
(log100 − log 76)
40.5
2.303
k=
(2 − 1.880)
40.5
k = 0.5686 × 0.120 = 0.006823
k ≈ 7 × 10–3 min–1
The rate constant of reaction 7 × 10–3 min–1
57. For a chemical reaction, mA → xB, the rate
law is r = k [A]2
If the concentration of A is doubled, the
reaction rat will be
(a) Doubled
(b) Quadrupled
(c) Increases by 8 times (d) Unchanged
Karnataka-CET-2016
Ans. (b) : Reaction: mA
→ xB
∴ rate = k [ A ] = ri
2
If the concentration is doubled then rate = k [ 2A ]
2
= 4k [ A ]
2
The rate of reaction is quadrupled.
58. Which of the following does not influence the
rate of reaction?
(a) Molecularity of the reaction
(b) Temperature of the reaction
(c) Concentration of the reactants
(d) Nature of the reactant
Manipal-2016
Ans. (a) : Nature and concentration of reactants and
temperature of the reaction influence the rate of
reaction. But molecularity does not affect the rate
reaction as it includes the number of atoms, ions or
molecules that must collide with one another to result
into a chemical reaction.
59. In a reversible reaction a catalyst will affect the
rate of
(a) Forward reaction
(b) Forward and reverse reaction
(c) Reverse reaction
(d) Neither (a) nor (b)
CG PET- 2015
Ans. (b) : The catalyst will affect the rate of both
forward and reverse reaction and attains equilibrium
more quickly but the value of equilibrium constant does
not affected by the presence of catalyst.
60. Total order of reaction X + Y → XY is 3. The
order of reaction with respect X is 2. State the
differential rate equation for the reaction.
d [x]
3
0
(a) −
= k [X] [Y]
dt
d [x]
0
3
(b) −
= k [X] [Y ]
dt
250
YCT
(c) −
(d) −
d [x]
dt
d [x]
dt
Ans. (a) : The concentration of product increases
linearly with time, hence w.r.t. A, the reaction is of zero
order. Hence the rate of reaction remains constant with
time
= k [X] [Y]
2
= k [ X ][ Y ]
2
GUJCET-2015
Ans. (c) : Given the reaction, X + Y
→ XY
Total order of reaction = 3
Order of reaction w.r.t. X=2
Hence, the order of reaction w.r.t. Y =1
∴ Differential rate equation
63.
Observe the following reaction:
2A + B
→C
The rate of formation of C is
d [X]
2.2 ×10-3 mol L–1min -1 .
2
1
−
= k [X] [Y ]
d(A)
dt
What is the value of −
( in mol L–1 min –1 )
dt
61. For a reaction 2A→3B, if the rate of formation
(a) 2.2 × 10 −3
(b) 1.1× 10 −3
of B is x mol/L, the rate of consumption of A is
−3
(c) 4.4 × 10
(d) 5.5 × 10 −3
(a) x
(b) 3x/2
AP EAMCET-2005
(c) 2x/3
(d) 3x
Ans. (c) : Given the reaction,
J & K CET-(2015)
2A + B
→C
→ 3B,
Ans. (c) : Reaction 2A
d [ C]
−3
–1
−1
Rate of formation of
Given that,
= 2.2 × 10 mol L min
dt
d [ B]
−1
−1
∴
The
rate
of
reaction–
= x mol L min
dt
d [ B] d [ C ]
1 d [A]
=−
=−
=
1 d [A]
1 d [ B]
2 dt
dt
dt
∴ rate = −
=+
2 dt
3 dt
d
A
d
C
[ ]
1 [ ]
−
=
2 dt
dt
d [ A ] 2 d [ B]
Rate of consumption of A, −
=
d [A]
dt
3 dt
−
= 2 × 2.2 × 10−3
dt
2
= x mol L−1 min −1
d [A]
3
or
−
= 4.4 × 10−3 mol L–1 min –1.
dt
62. The variation of concentration of the product P 64. The concentrations of the reactant A in the
with time in the reaction, A → P is shown in
reaction A → B at different times are given
below
following graph.
Concentration (M)
Time (Minutes)
0.069
0
0.052
17
0.035
34
0.018
51
The
rate
constant
of
the
reaction
according to
−d[A]
The graph between
and time will be of
the correct order of reaction is
dt
(a) 0.001 M/min
(b) 0.001 min-1
the type
(c) 0.001 min/M
(d) 0.001 M-1 min-1
d[A]
Ans. (a) : A → B, rate = −
dt
Conc.change
0.069 − 0.052 = 0.017 M
0.052 − 0.035 = 0.017 M
0.035 − 0.018 = 0.017 M
JIPMER-2015
Objective Chemistry Volume-II
251
Rate
0.017
= 0.001
17 − 0
0.017
= 0.001
34 − 17
0.017
= 0.001
51 − 34
YCT
∵ Rate is constant. So, it is independent of 67.
concentration.
Therefore the reaction is of zero order.
Rate = k(conc.)0 = 0.001 M/min.
65. The most probable velocity of a gas molecule at
298 K is 300 m/s. Its RMS velocity in m/s is
(a) 420
(b) 245
(c) 402
(d) 367
The rate equation of a gaseous reaction is given
by, r = k[A][B]. If the volume of reaction vessel
is suddenly reduced to half of the initial
volume, the reaction rate relating to the
original rate will be
1
(a)
(b) 4
4
1
(d) 2
(c)
AP - EAMCET (Medical) - 2007
2
Ans. (d) : Given that,
COMEDK-2012
Vmp = 300 m/s, T = 298K, Vrms = ?
Ans. (b) : Given that,
We know that,
r = k[A][B]
When
volume
is reduced to half then concentration gets
2RT
Vmp =
.....(i)
doubled.
M
Thus r1 = k[2A][2B]
3RT
r1= 4k [A][B]
Vrms =
.....(ii)
M
r1= 4r
From equation (i) and (ii) we get–
Therefore, the rate relating to the original rate will be 4
2RT
68. For a reaction between A and B the order with
Vmp
respect to A is 2 and the other with respect to B
M
=
is 3. The concentrations of both A and B are
Vrms
3RT
doubled, the rate will increase by a factor of
M
(a) 12
(b) 16
(c) 32
(d) 10
300
2RT M
=
×
Karnataka
NEET-2013
Vrms
M 3RT
Ans. (c) : Rate(r1)= k[A]2[B]3
3
Rate (r2) = k[2A]2[2B]3
or
Vrms = 300 ×
2
Rate (r2) = k.4 [A]2 .8[B]3
Vrms = 367.46
Rate (r2) = k.4.8[A]2 [B]3
Rate (r2)= 32k [A]2 [B]3
or
Vrms ≈ 367
66. The concentrations of I– at the start and after Therefore the rate will increase the factor is 32
10 minutes of the reaction Cl2 + 2I–→ 2Cl– + I2 r2 = 32r1
which are 0.60 mol L–1 and 0.56 mol L–1
respectively. What are the rate of 69. The rate of a chemical reaction doubles for
every 10° rise in temperature. If the
disappearance of I– and the rate of appearance
temperature increases by 60°, the rate of
of Iodine in mol L–1 min–1 respectively ?
reaction increases
(a) 0.004 and 0.002
(b) 0.002 and 0.004
(a) 20 times
(b) 32 times
(c) 0.004 and 0.004
(d) 0.002 and 0.002
(c)
64
times
(d)
128 times
SCRA - 2009
COMEDK-2014
−∆ I −
−
Ans. (c) : The rate of a chemical reaction doubles for
Ans. (a) : Rate of disappearance of I =
every 10° rise in temperature if the temperature increase
∆t
by
60° then the rate of reaction is increase by (2)6 = 64
− ( 0.56 − 0.60 )
−
I =
times.
10
70. The rate of a chemical reaction doubles for
0.04
every 10ºC increase in temperature, if the
=
10
temperature of reaction is increased from 30ºC
= 0.004 mol L–1 min–1
to 80ºC. The rate of reaction increases:
(a) 16
(b) 32
Since, equation, Cl 2 + 2I − → 2Cl− → I 2
(c)
64
(d) 4
−
1 d I
AP-EAMCET
(Medical), 2006
Rate of appearance of I 2 = −
2 dt
Ans. (b) : The variation in temperature = 80–30 = 50ºC
1
It is given that rate of chemical reaction doubles in
= − × 0.004
every 10ºC increase in temperature.
2
–1
–1
Thus, for 50ºC increase in temperature increase then rate
= 0.002 mol L Min
–
of reaction = 25 = 32 times.
Here, - sign simply indicates disappearance of I
Objective Chemistry Volume-II
252
YCT
71.
Observe the following reaction:
A (g) + 3B(g) →2C (g)
d[B]
at t = 0 is
dt
d[A]
(a) 0.10 mol L-1 s-1
(b) 2.6×10-2mol L-1 s-1
The rate of this reaction is 3 ×10-3
-2
-1 -1
(c) 5.2×10 mol L s
(d) 6.5×10-3mol L-1 s-1
dt
AMU-2012
d[B]
mol L-1 min-1 What is the value of in
Ans.
(c)
:
dt
A + 2B
→ Pr oducts
mol L-1 min-1?
(a) 3 × 10–3
(b) 9 ×10–3
d [A]
Given that,
= 2.6 × 10−2 mol L−1s −1
(c) 10–3
(d) 1.5 ×10–3
dt
AP-EAMCET (Medical), 2006
d [A]
1 d [ B]
Ans. (b) : For the following reaction∴ Rate of reaction = −
=−
dt
2 dt
A(g) + 3B(g)→2C(g)
Given that,
d [ B]
∴
= 2 × 2.6 ×10−2
d[A]
−3
−1
−1
dt
=–
= 3 × 10 mol L min
dt
= 5.2 × 10 −2 molL−1s −1
−d[B]
74. When we increase the temperature, the rate of
=?
reaction increases because of
dt
(a) more number of collisions
−d[A] −1 d[B]
1 d[C]
=
=+
(1)
(b) decrease in mean free path
dt
3 dt
2 dt
(c) more number of energetic electrons
−d[A]
(d) less number of energetic electrons
We put the value of
in equation (1), we get
dt
BCECE-2011
−1 d[B]
-3
-1
-1
Ans.
(a)
:
On
increasing
the
temperature.
The number
3 ×10 mol L min =
of collisions increases due to increase of kinetic energy.
3 dt
And hence the rate of reaction increase.
−d[B]
or
= 9 × 10−3 mol L−1 min −1
75. For the reaction system:
dt
2NO(g) + O2(g) → 2NO2(g) volume is
72. For the reaction given below,
suddenly reduced to half of its value by
−
−
5Br (aq)+ BrO3 (aq)
→ 3Br2 (l ) + 3H 2 O(l )
increasing the pressure on it. If the reaction is
The rate of formation of Br2 is related to rate of
of first order with respect to O2 and second
consumption fo Br by the following relation?
order with respect to NO, the rate of reaction
will
5d Br −
d [ Br2 ]
(a)
=−
(a) diminish to one-fourth of its initial value
dt
3dt
(b)
diminish to one-eighth of its initial value
d Br −
d [ Br2 ]
(c) increase to eight times of its initial value
(b)
=−
(d) increase to four times of its initial value
dt
dt
[BITSAT – 2007]
−
d [ Br2 ] 5 d Br
Ans. (c) : Increase to eight times of its initial value
(c)
=
dt
3 dt
when the volume in reduced to half, the concentration is
−
doubled.
d [ Br2 ] 3 d Br
(d)
=
∴ Reaction, 2NO ( g ) + O 2 ( g )
→ 2NO 2 ( g )
dt
5 dt
Ans. (d) :
The initial rate,
AMU-2012 ∴ Rate of reaction = k [ NO ]2 [ O ] = r
2
1
After reducing the volume
5Br − ( aq ) + BrO3− ( aq ) + 6H + ( aq )
→ 3Br2 ( l ) + 3H 2 O ( l )
= k.[ 2NO ] .[ 2O 2 ] = 8.r1
2
Rate of reaction
→
∴ The reaction rate is increased by 8 times.
−
−
+
−1 d Br −d BrO3 −1 d H 1 d [ Br2 ]
–
=
=
=
76. The reduction of peroxydisulphate ion by I
5
dt
dt
6 dt
3 dt
2−
−
2−
−
ion is expressed by S 2O8 + 3I
2SO4 + I 3 . If
−
–
d [ Br2 ]
3 d Br
rate of disappearance of I is 9/2 × 10–3 mol lit–1
=−
dt
5 dt
s–1, what is the rate of formation of 2SO 42−
during same time?
d[A]
The initial rate, −
at t = 0 was found to be
(a) 3 × 10–3 mol Lit–1s–1 (b) 2 × 10–3 mol Lit–1s–1
dt
(c) 10–3 mol Lit–1s–1
(d) 4 × 10–3 mol Lit–1s–1
2.6×10-2
molL-1s-1
for
the
reaction,
[BITSAT – 2014]
A+2B→Products
∴
73.
Objective Chemistry Volume-II
253
YCT
Ans. (a) :
Reaction : S2 O82 − + 3I −
→ 2SO 24 − + I3−
(c)
−
Given :
d I 9
= × 10 −3 mol L−1s −1
dt
2
2−
8
−
(d)
2−
77.
dt
dt
d [ NO 2 ] d [ O 2 ]
=4
=
dt
dt
CG PET -2009
d SO 4
1 d I
= 2×
dt
3 dt
1 9
= 2 × × ×10−3
3 2
= 3 × 10 −3 molL−1s −1
1.0L of 2.0 M acetic acid is mixed with 1.0L of
3.0 M ethyl alcohol. The reaction is
CH 3 COOH + C 2 H 5 OH ↽ ⇀
CH 3 COOC 2 H 5 + H 2 O
If both the solutions are diluted by adding 1.0L
of water in each, the initial rate of reaction is
slow by
(a) 0.5 times
(b) 2.0 times
(c) 4.0 time
(d) 0.25 time
CG PET -2008
Ans. (d) : Given reaction,
CH 3COOH + C 2 H 5OH ↽ ⇀ CH 3COOC 2 H 5 + H 2 O
2
3
1
1
∴ Initial rate, r = k.[ CH 3 COOH ] .[ C 2 H 5 OH ]
dt
−2d [ N 2 O5 ]
d [O2 ]
Ans. (b) : 2N 2O5 ↽ ⇀ 4NO 2 + O 2
−
Initial concentration
=4
−
d S2 O
1 d I 1 d SO 4 d I3
∴−
=−
=
=
dt
3 dt
2
dt
dt
2∴ Rate of appearance of SO 4
2−
d [ NO 2 ]
1 d [ N 2O5 ] 1 d [ NO 2 ] d [ O 2 ]
− .
= .
=
2
dt
4
dt
dt
∵ Rate of reaction1 d [ N 2O5 ] 1 d [ NO 2 ]
−
=
2
dt
4 dt
Multiply by 4 both side
−2d [ N 2 O5 ] d [ NO 2 ]
=
dt
dt
79. When a biochemical reaction is carried out in
laboratory from outside of human body in the
absence of enzyme than rate of reaction
obtained is 10−6 times, then activation energy of
reaction in the presence of enzyme is
(a) 6/RT
(b) P is required
(c) different from, Ea obtained in laboratory
(d) can't say anything
∴
CG PET -2007
Ans. (c) : According to Arrhenius equation.
k = Ae − Ea / RT . The enzyme acts as catalyst hence the
rate of reaction changes and Ea changes in presence of
catalyst.
r = k. × 2 × 3 = 6k ........(i)
+d ( B )
80.
3A → 2B,rate of reaction
is equals to
When the volume of each increased by 1.0L then
dt
concentration
3 d [A ]
2 d [A ]
(a) −
(b) −
2
2 dt
3 dt
[ CH3COOH ] = = 1mol / L
2
d
A
d [A]
[
]
1
3
(c) −
(d) +2
[ C2 H 5OH ] = = 1.5 mol / L
3 dt
dt
2
CG PET -2007
3
∴ Rate of reaction. r1 = k.[ I ] ×
3A
→ 2B
Ans. (b) : Given that,
2
d
A
d
[
]
[ B]
1
1
= 1.5k. ......(ii)
Rate of reaction = −
=
3 dt
2 dt
∴ from equation (i) and (ii) we getr = 4 × r1
d [ B]
d
2 [A]
∴
=−
1
dt
3 dt
r1 = r = 0.25r.
4
81. For a reaction of the type A+B→ products, it is
observed that doubling the concentration of A
78.
2N 2 O5 ↽ ⇀ 4NO 2 + o 2
caused the reaction rate to the four times as
For the above reaction which of the following is
great, but doubling the amount of B does not
not correct about rates of reaction?
affect the rate. The rate equation is
− d [ N 2 O5 ]
d [O2 ]
2
(a) Rate = k [ A ][ B]
(b) Rate = k [ A ]
(a)
=2
dt
dt
2
2
2
(c) Rate = k [ A ] [ B]
(d) Rate = k [ A ] [ B]
−2d [ N 2 O5 ] d [ NO 2 ]
(b)
=
dt
dt
CG PET -2006
Objective Chemistry Volume-II
254
YCT
1
A→2B, rate of disappearance
2
of 'A' is related to the rate of appearance of 'B'
by the expression
d [ A ] 1 d [ B]
d [ A ] 1 d [ B]
(a) –
=
(b) –
=
dt
2 dt
dt
4 dt
d [ A ] d [ B]
d [A ]
d [ B]
(c) –
=
(d) –
=4
dt
dt
dt
dt
[AIEEE 2008]
1
Ans. (b) : Given reaction : A
→ 2B
2
d [A ]
1 d [ B]
Rate of reaction, = −2
=+
dt
2 dt
−d [ A ]
1 d [ B]
=+
∴ rate of disappearance of A,
dt
4 dt
85. Which factor has no influence on the rate of
reaction?
(a) Molecularity
(b) Temperature
(c) Concentration of reactant
(d) Nature of reactant
J & K CET-(2014)
Ans. (a) : The rate of reaction is independent of the
molecularity of the reaction.
Nature and concentration of the reactants and
temperature of the reaction influence the rate of
reaction.
86. In a catalytic experiment involving Haber’s
process
N2(g) + 3H2(g) → 2NH3(g)
The rate of reaction was measured as Rate =
[NH3] = 2.0 × 10–4 Ms–1. If there were no side
reactions, what was the rate of reaction
expressed in terms of N2?
(b) 4 × 10–4 Ms–1
(a) 1 × 10–4 Ms–1
–3
–1
(c) 5 × 10 Ms
(d) 1 × 10–3 Ms–1
J & K CET-(2010)
Ans. (b) : On doubling the concentration of A. the
reaction rate is four times then the rate w.r.t. A is 84.
second order on doubling the amount of B, the reaction
rate remains unchanged hence w.r.t. to A. Then rate
w.r.t. B is zero
∴
rate = k.[ A ] .
2
82.
For a first order reaction,
(A)→products
the concentration of A changes from 0.1 M to
0.025 M in 40 min. The rate of reaction when
the concentration of A is 0.01 M is
(a) 1.73×10–5 M/min
(b) 3.47×10–4 M/min
–5
(d) 1.73×10–4 M/min
(c) 3.47×10 M/min
[AIEEE 2012]
→ Product
Ans. (b) : Given the reaction, A
Given that, t = 0
[Ao] = 0.1M
t = 40 min
[A] = 0.025 M
[ Ao ]
2.303
∴ For a first order reaction : k =
log
t
[A ]
2.303
0.1
log
40
0.025
2.303
=
log 4
40
2 × 2.303
=
log 2
40
= 0.0347 min −1
rate = k [ A ]
=
= 3.47 × 10
−4
= 0.0347 × 0.01
molL−1 min −1
83.
For a reaction
Consider the reaction,
Cl2(aq)+H2S(aq)→S(s)+2H+(aq)+2Cl–(aq)
The rate equation for this reaction is,
rate = k[Cl2][H2S]
Which of these mechanisms is/are consistent
with this rate equation?
(I) Cl2 + H2S → H+ + Cl– + Cl+ + HS– (slow)
Cl+ + HS– → H+ + Cl– + S (fast)
d [ NH 3 ]
Ans. (a) : Given :
= 2.0 ×10−4 Ms −1
(II) H 2S ↽ ⇀ H + + HS – (fast)
dt
Cl 2 + HS – → 2Cl – + H + + S(slow)
Reaction : N 2 ( g ) + 3H 2 ( g )
→ 2NH 3 ( g )
(a) (II) only
(b) Both (I) and (II)
d [ N2 ]
1 d [H2 ]
1 d [ NH 3 ]
(c) Neither (I) nor (II) (d) (I) only
∴ rate = −
=−
=+
[AIEEE 2010]
dt
3 dt
2 dt
Ans. (d) : The rate of a complex reaction depends on
d [ N2 ] 1
∴−
= × 2.0 ×10−4
the slowest step.
dt
2
The given rate equation, rate = k [ Cl 2 ][ H 2S]
= 1.0 × 10−4 mol L−1 sec −1
(i)
Cl 2 + H 2S
→ N + + Cl− + Cl + + HS− ( slow )
87. The rate of a gaseous reaction is given by the
Cl + + HS−
→ H + + Cl − + S ( fast )
expression k[A][B]. If the volume of the
th
+
−
1
(ii)
H 2S ↽ ⇀ H + HS ( fast )
reaction vessel is suddenly reduced to
of
4
Cl 2 + HS−
→ H + + 2Cl − + S ( slow )
the initial volume, the reaction rate relating to
The given rate equation follows only (i).
original rate will be
Objective Chemistry Volume-II
255
YCT
→ products.
Ans. (a) : Reaction : A + B
Experiment (i): On doubling the concentration of A, the
rate of reaction is doubled, hence w.r.t. A the reaction is
JIPMER-2013 of first order.
(ii): On doubling the concentration of B and A both the
Ans. (d) : Rate = k [ A ][ B] = ri ( say )
rate becomes 8 times, hence w.r.t. B the reaction is of
second order, because the total order for A and B, is
1
When volume is reduced to
then the concentration third order.
4
R = k [A]m [B]n
……(i)
becomes 4 times.
m
n
2R
=
k
[2A]
[B]
……(ii)
∴ rate = k [ 4A ][ 4B]
8R = k [2A]m [2B]n
……(iii)
= 16k [ A ][ B]
From (i), (ii) & (iii)
m = 1, n = 2
rate = 16ri
So,
rate
= k [A] [B]2
∴ The rate of reaction becomes 16 times of the initial
91. The bromination of acetone that occurs in acid
rate.
solution is represented by this equation.
88. For the reaction
1
10
(c) 8
1
8
(d) 16
(a)
(b)
CH 3COCH 3 (aq) + Br2 (aq) → CH 3COCH 2 Br(aq)
2N2O5 (g)
→ 4NO2(g) + O2(g)
if the concentration of NO2 increases by 5.2 ×
10–3 M in 100s then the rate of the reaction is :
(a) 1.3 × 10–5 Ms–1
(b) 0.5 × 10–4 Ms–1
–4
–1
(c) 7.6 × 10 Ms
(d) 2 × 10–3 Ms–1
–5
–1
(e) 2.5 × 10 Ms
Kerala-CEE-2005
Ans. (a) : For the reaction,
2N 2O5 ( g )
→ 4NO 2 ( g ) + O 2 ( g )
∴ rate of reaction w.r.t. NO 2 =
d [ NO 2 ] = 5.2 × 10−3 M
dt = 100s
These kinetic data were obtained for given
reaction concentrations.
Initial concentrations, M
[CH3COCH3]
[Br2]
[H+]
0.30
0.30
0.30
0.40
1 d [ NO 2 ]
4 dt
1 5.2 ×10−3
= ×
4
100
= 1.3 × 10 −5 mol L−1s −1 or Ms −1
89. In a multistep reaction, the overall rate of
reaction is equal to the
(a) rate of slowest step
(b) rate of fastest step
(c) average rate of various step
(d) the rate of last step
MHT CET-2011
Ans. (a) : The reaction mechanism is the step by step
process by which reactants actually become products.
The overall reaction rate depends almost entirely on the
rate of the slowest step. If the first step is the slowest,
and the entire reaction must wait for it, then it is the rate
determining step.
90. For the reaction, A + B → products, it is
observed that
(i) on doubling the initial concentration of A only,
the rate of reaction is also doubled and
(ii) on doubling the initial concentration of both A
and B, there is a change by a factor of 8 in the
rate of the reaction.
The rate of this reaction is given by
(a) rate = k [ A ][ B]
(b) rate = k [ A ] [ B]
(c) rate = k [ A ][ B]
(d) rate = k [ A ] [ B]
AIPMT -2009
2
+H + (aq ) + Br(aq )
2
Objective Chemistry Volume-II
2
2
0.05
0.10
0.10
0.05
0.05
0.05
0.10
0.20
Initial rate, disappearance of Br2, M s–1
5.7 × 10 −5
5.7 × 10 −5
1.2 × 10 −4
3.1 × 10 −4
Based on these data, the rate equation is
(a) Rate = k [ CH 3COCH 3 ][ Br2 ] H +
2
(b) Rate = k [ CH 3COCH 3 ][ Br2 ] H +
(c) Rate = k [ CH 3COCH 3 ] H +
(d) Rate = k [ CH 3COCH 3 ][ Br2 ]
AIPMT -2008
Ans. (c) : Rate = k [ CH 3 COCH 3 ] H
Reaction :
+
CH 3 COCH 3 (aq) + Br2 (aq)
→ CH 3 COCH 2 Br(aq) +
H + (aq) + Br − (aq)
In experiment 1 and 2, the concentration of H + and
[ CH3COCH3 ] remains constant whereas the [ Br2 ] is
doubled, but the rate of disappearance of Br2 remains
unchanged, hence w.r.t Br2 the reaction is of zero
order.
In experiment, 2 and 3 the concentration of
[ CH3COCH3 ] and [ Br2 ] remains unchanged, while
the concentration of H + is doubled, the rate of
reaction will also be doubled hence with respect to
H + the reaction is of first order.
256
YCT
In experiment 1 and 4, the concentration of
[ Br2 ]
remains constant whereas H + becomes four times
and [ CH 3COCH 3 ] becomes 4/3 times and the net rate
also becomes 5/3 times hence the reaction is of first
order w.r.t [ CH 3COCH 3 ] .
∴ rate = [ CH 3COCH 3 ] [ Br2 ] H +
rate = [ CH 3COCH 3 ] H +
1
92.
For
the
d [ NH 3 ]
dt
–d [ H 2 ]
0
reaction,
1
N 2 + 3H 2 → 2NH 3 , if
= 2 ×10 –4 mol L–1s –1 , the
value
would be
dt
(a) 4 × 10−4 mol L−1s −1
(c) 1 × 10 −4 mol L−1s −1
of
(b) 6 × 10 −4 mol L−1s −1
(d) 3 × 10−4 mol L−1s −1
AIPMT -2009
Ans. (d) : Reaction : N 2 + 3H 2
→ 2NH 3
rate = −
Given:
∴
93.
d [ N2 ]
=−
dt
d [ NH 3 ]
dt
d [H2 ]
1 d [H2 ]
1 d [ NH 3 ]
=+
3 dt
2 dt
= 2 × 10−4 mol L−1s −1
3 d [ NH 3 ] 3
= × 2 × 10−4
dt
2 dt
2
d [H2 ]
−
= 3 ×10 −4 mol L−1s −1
dt
In the reaction,
BrO 3–( aq ) + 5Br(–aq ) + 6H (+aq ) → 3Br2( l ) + 3H 2O( l )
−
=
For the reaction, 2A + B → 3C + D, which of
the following does not express the reaction
rate?
d [A]
d [ C]
(b) −
(a) −
2dt
3dt
d [ B]
d [ D]
(c) −
(d)
dt
dt
AIPMT -2006
Ans. (b) : For the reaction2A + B
→ 3C + D
d [ B]
d [ D]
1 d [A]
1 d [ C]
rate = −
=−
=+
=+
2 dt
dt
3 dt
dt
C is the product and hence its sign should be + ve.
95. If the volume of the vessel in which the reaction
2NO+O2→2NO2 is occurring is diminished to
1/3 rd of its initial volume. The rate of the
reaction will be increased by
(a) 3 times
(b) 9 times
(c) 27 times
(d) 36 times
WB-JEE-2008
94.
→ 2NO 2
Ans. (c) : Reaction 2NO + O 2
∴ Rate = k [ NO ] [ O 2 ] = ri ( say )
2
When the volume is reduced to 1/3rd. The concentration
becomes 3 times,
Then,
rate = k [3NO ] [3O 2 ]
2
rate = 27k [NO2]2 [O2] = 27ri
The rate will become 27 times.
The rate of appearance of bromine (Br2) is 96. Consider the following reaction for 2NO2 (g) +
related to rate of disappearance of bromide
F2(g)
→ 2NO2F (g). The expression for the
ions as
rate
of
reaction
in terms of the rate of change
–
d [ Br2 ]
5 d Br
of partial pressure of reactant at and product
=−
(a)
is/ are.
dt
3 dt
–
1 dp(NO 2 )
d [ Br2 ] 5 d Br
(a) rate = −
(b)
=
2
dt
dt
3 dt
–
1 dp(NO 2 )
d [ Br2 ] 3 d Br
(b) rate =
(c)
=
2
dt
dt
3 dt
–
1
dp(NO
2 F)
d [ Br2 ]
3 d Br
(c) rate = −
(d)
=−
2
dt
dt
5 dt
1 dp(NO 2 F)
AIPMT -2009
(d) rate =
Ans. (d) : Reaction:
2
dt
BrO3− ( aq ) + 5Br − ( aq ) + 6H + ( aq )
→ 3Br2( l) + 3H 2 O( l)
WB-JEE-2013
Ans. (a&d) :
−
+
d BrO3−
1 d Br
1 d H 1 d [ Br2 ]
∴ Rate=−
=−
=−
=
Reaction : 2NO 2 ( g ) + F2 ( g )
→ 2NO 2 F ( g )
dt
5 dt
6 dt
3 dt
1 dp ( NO 2 ) dp ( F2 ) 1 dp ( NO 2 F )
∴ rate =−
=−
=+
−
2
dt
dt
d [ Br2 ]
dt 2
3 d Br
=−
Hence, a and d both are correct.
dt
5 dt
Objective Chemistry Volume-II
257
YCT
2.
(a) 66.90
(c) 22.30
Rate Law and Rate Constant
(b) 33.45
(d) 44.45
AP-EAMCET-05.07.2022, Shift-I
40% of Hl undergoes decomposition to H2 and Ans. (b) : Given,
T2 = 700 K
K2=0.2s–1
l2 at 300 K. ∆G° for this decomposition reaction
–1
T1 = 500 K
K1 =0.02s–1
at one atmosphere pressure is –––––– J mol .
[Nearest integer]
K
E a T2 − T1
(Use R = 8.31 J K–1 mol–1; log 2 = 0.3010, in 10 Now, log 2 =
K1 2.303R T1T2
= 2.3, log 3 = 0.477)
JEE Main-26.06.2022, Shift-II
Ea
0.2
700 − 500
log
=
Ans. (2735) : Given,
700 × 500
0.02
2.303
×
8.314
R = 8.30 J K–1 mol–1
Ea
200
T = 300 Kf
1=
×
2.303
8.314
700
×
× 500
We know that, ∆G° = – RT ln keq
2.303 × 8.314 × 700 × 500
1
1
Or
Ea =
HI ⇌ H 2 + I 2
200
2
2
–1
E
=
33.50
kJ
mol
a
ti 1
0
0
100. If the rate of disappearance of N2O5 in the
0.4 0.4
following reaction is 1.2×10–5 molL–1s–1, the
t eq 1 − 0.4
2
2
rate of production of NO2 (in molL–1s–1) is
1
1
2N2O5 (g) → 4NO2 (g) + O2 (g)
(0.2) 2 (0.2) 2 0.2
(a) 1.2×10–5
(b) 3.6×10–5
K eq =
=
–5
1 − 0.4
0.6
(c) 2.4×10
(d) 4.8×10–5
AP-EAMCET-07.07.2022, Shift-I
1
Ans. (c) : 2N2 O5 (g)
+NO2 (g) + O2 (g)
=
1
1
3
– d ( N 2 O5 )
d ( NO 2 )
d[O 2 ]
2
= 4
=
Substituting in eqn, we get
dt
dt
dt
Given,
1
∆G° = – 8.31 × 300×2.3× log
d [ N 2 O5 ]
3
= 1.2 × 10–5 mol L–1
∆G° = 2735 J/mol
dt
98. A flask is filled with equal moles of A and B. Then,
The half life of A and B are 100 s and 50 s
d [ NO 2 ] 4
respectively and are independent of the initial
= × 1.2 × 10 –5
dt
2
concentration. The time required for the
concentration of A to be four times that of B
d [ NO 2 ]
= 2.7 × 10 –5 mol L–1 S–1
is______s.
dt
JEE Main-26.06.2022, Shift-I
Hence rate of production of NO2 is 2.7 × 10–5 mol L–1 S–1
Ans. (200) : Given,
101. A → products, is a first order reaction. The
Moles of A = Moles of B
time required to decompose A to half its initial
Half life of A = 100 s
amount is 60 minutes. The rate constant of the
Half life of B = 50 s
reaction (in s–1) is
For first order reaction–
(a) 1.05 × 10 −2
(b) 1.15 × 10 −2
−4
0.693
(c) 1.25 × 10
(d) 1.92 × 10 −4
K=
t1/ 2
AP-EAMCET-06.07.2022, Shift-I
[A] = 4[B]
Ans. (d) : For first order reaction
2.303
a
[A]0 e − k A t = 4[B]0 e − k B t
t=
log
− kA t
− kBt
K
a
−
x
e
= 4e
2.303
a
–kAt = ln 4 – kBt
t=
log
a
K
2 × 0.693
2 ×100
a−
t=
=
= 200 sec.
2
2 −1
0.693 0.693
−
2.303
60× 60 =
log 2
100
50
K
99. The rate constant of a reaction at 500K and
700K are 0.02 s-1 and 0.2 s-1 respectively. The K = 2.303× 0.303
36×102
activation energy of the reaction (in kJ mol-1) is
-1
-1
K = 1.92 × 10–4 S–1
(R = 8.3 JK mol )
97.
Objective Chemistry Volume-II
258
YCT
102. The value of rate constant for a reaction A is Ans. (b) : Given, Equilibrium constant (Kc)= 81
twice of reaction B at the same temperature. Velocity constant for forward reaction
The difference in their energy of activation
(Kf)=162 L mol–1 s–1
A
B
(Ea – Ea ) is.
K
∴ K c = f (Where Kb = Velocity constant for
K
(a) RTln2
(b) –2.303 RT
b
(c) –RTln2
(d) 0
backward reaction.)
Assam-CEE-31.07.2022
K 162
⇒ Kb = f =
= 2 L mol –1 s –1
Ans. (c) : ∵ Ka=2Kb
K c 81
A
B
Ae − Ea / RT = 2Ae− Ea / RT
106. For the reaction 2N2O5 → 4NO2 + O2, rate and
rate constant are 1.02 × 10–4 mol lit–1 sec–1 and
Ea A
Ea B
−
= ln2 −
3.4 × 10–5 sec–1 respectively then concentration
RT
RT
of N2O5 at that time will be
E a A − Ea B
(a) 1.73M
(b) 3M
−ln2 =
5
(c)
3.4
×
10
M
(d) 10.2 × 10–4M
RT
VITEEE-2018, BCECE-2012
E a A − E a B = − RT ln 2
Ans. (b) : The given reaction is –
103. The unit of rate constant for a zero order
2 N2O5 → 4 NO2 + O2
reaction is
Given Data –
(b) L mol −1s −1
(a) mol L−1s −1
Rate (r) = 1.02×10–4mol lit–1 sec–1
Rate constant (k) = 3.4×10–5 sec–1
(d) s −1
(c) L2 mol −2s −1
–1
(AIPMT -Mains 2011) The unit of rate constant is sec which is the unit of rate
MHT CET-2010, CG PET -2004 constant of first order Kinetics.
So,
from the rate law equation–
Ans. (a) : Unit of Rate constant = (mol)1-n (1)n-1 sec–1
r = k [N2O5]
n = Order of reaction
r
for zero order reaction, n = 0
or
[ N 2 O5 ] =
k
Unit of k = mol L–1. s–1.
−1
−1
104. In the reversible reaction,
1.02 ×10−4 ( mollit sec )
N
O
=
k1
[
]
2 5
2NO 2 ↽ ⇀ N 2O4 , the rate of disappearance of
3.4 × 10−5
( sec−1 )
k2
NO2 is equal to
(M = mol.lit–1)
[ N 2O5 ] = 3M
2k1
2
(a)
107. Select the rate law that corresponds to the data
[ NO2 ]
k2
shown for the following reaction A + B → C
(b) 2k1[NO2] – 2k2[N2O4]
Expt. No.
(A)
(B)
Initial
(c) 2k1[NO2]2 – k2[N2O4]
Rate
(d) (2k1–k2)[NO2]
1
0.012
0.035
0.10
JIPMER-2018, BCECE-2018
2
0.024
0.070
0.80
AMU-2015
3
0.024
0.035
0.10
Ans. (c) :
4
0.012
0.070
0.80
k
3
4
Reaction : 2NO 2 ↽ k1 ⇀ N 2 O 4
(a)
Rate
=
k[B]
(b)
Rate
=
k
[B]
2
3
(c) Rate = k[A][B]
(d) Rate = k [A]2[B]2
Rate of formation of NO 2 in reversible reaction,
[AIIMS-2015, 2012]
d [ NO 2 ]
In
experiment
1
and
3
the
concentration of A
Ans.
(a):
−
= k 2 [ N2O4 ]
is doubled while [B] is remains constant and the rate
dt
remains unchanged. Hence w.r.t A. the reaction is of
∴ Rate of reaction in terms of disappearance of NO 2 ,
zero order.
d [ NO 2 ]
In experiment, 1 and 4, the concentration of A is
2
−
= 2k1 [ NO 2 ] − k 2 .[ N 2 O 4 ] .
constant while B is doubled and the reaction rate will
dt
becomes 8 times of initial rate. Hence w.r.t. B the
105. Equilibrium constant for the reaction,
reaction is of third order.
H 2O ( g ) + CO ( g )
→ H 2 ( g ) + CO 2 ( g ) is 81. If
o
3
∴ Rate = k.[ A ] [ B]
the velocity constant of the forward reaction is
3
162 L mol –1 s–1, what is the velocity constant
Rate = k.[ B]
–1 –1
(in L mol s ) for the backward reaction?
108. For a reaction A + B → C + D, doubling the
(a) 13122
(b) 2
concentration of both the reactants increases
(c) 261
(d) 243
the reaction rate by 8 times and doubling the
(AP-EAMCET-2001)
only B simply doubles the reaction rate. The
AP EAMCET (Engg.) 2001
rate law is given as:
Objective Chemistry Volume-II
259
YCT
(a) r = k[A]1/2[B]1/2
(b) r = [A] [B]2
111. Temperature coefficient of a reaction is 2.
2
When temperature is increased from 30oC to
(d) r = k[A] [B]
(c) r = k[A] [B]
90oC, the rate of reaction is increased by
HP CET-2018, (AIPMT -2012)
x
y
(a) 60 times
(b) 64 times
Ans. (c) : Initially, rate = k [A] [B] = r ---------(i)
(c) 150 times
(d) 400 times
When concentrations of A and B is doubled i.e., 2A and
Karnataka-CET-2018, 2013
2B then.
rate = k [2A]x [2B]y = 8r ---------------(ii)
Ans. (b) : Given:- Temperature coefficient, µ=2
When concentration of B is doubled i.e., 2B then,
T1 = 30oC , T2 = 900C ∆T = 600C
rate = k[A]x [2B]y = 2r ------------------(iii)
∆T
60
Rate of reaction = ( µ ) 10 = ( 2 ) 10 = 26
From equation (i) and (iii)
= 64 times.
k[A]x [2B]y 2r
=
=2
x
y
112.
The
following
data were obtained during the
k[A] [B]
r
first
order
decomposition
of 2A (g) B (g) +
[2B]y
C(s) at a constant volume and at a particular
=
2
temperature.
[B]y
y
1
Total pressure
S.No.
Time
2 =2
in Pascal
∴ y =1
1.
At
the
end
of
10
min
300
From equation (i) and (ii)
2.
After
completion
200
k[2A]x [2B]y 8r
–1
=
The rate constant in min is
r
k[A]x [B]y
(a) 0.0693
(b) 69.3
x y
2 .2 = 8
(c)
6.93
(d)
6.93 × 10–4
2x.21 = 8
Karnataka-CET-2011, 2009
2x = 4 = 22
Ans. (a) : 2A(g)
→ B (g)+ C(g)
x=2
2 − 2x
x
Then, rate law will be
At the end of reaction, only 1 mole of gas is present
2
1
r = k[A] [B] .
whose pressure is 200 Pascal.
109. The rate law for a reaction between the At the beginning of the reaction 2 moles of gas
substances A and B is given by rate = should have a pressure of 400 Pascal.
k[A]n[B]m. On doubling the concentration of A After time 10 min
and halving the concentration of B, the ratio of No. of moles present,
the new rate to the earlier rate of the reaction
2 × 2x + x = 2 – x
will be as
The pressure of 2 moles = 400
1
400 – x = 300
(a) m+ n
(b) ( m + n )
2
x = 100
Pressure due to 2 – 2x moles of A.
(c) ( n – m )
(d) 2( n –m )
= 400 – 200 = 200
UPTU/UPSEE-2007, BITSAT-2005
n
m
2.303
a 2.303
400
Ans. (d) :Let, r = k[A] [B] ------------(i)
k =
log
log
=
If concentration of A is doubled i.e. 2A and B is halved
t
10
a−x
200
B
2.303
0.693
i.e. then
=
log 2 =
2
10
10
m
= 0.0693 min–1
n
m n −m
n B
r' = k [2A] = k [ A ] [ B] 2 − − − −(ii)
113. For a reaction A + 2B → Products, when
2
concentration of B alone is increased half life
Dividing (ii) by (i)
remains the same. If concentration of A alone is
n
m ( n −m)
r ' k[A] [B] 2
( n −m)
doubled, rate remains the same. The unit of
=
=2
rate constant for the reaction is
r
k[A]n [B]m
(a) s-1
(b) Mol–1 s–1
110. The rate constant of a reaction depends on
–1 –1
(c) Mol L s
(d) atm–1
(a) temperature
Kerala-CEE-29.08.2021
(b) initial concentration of reactants
Karnataka-CET-2021
(c) extent of reaction
(d) time of reaction
Ans. (a) : The given reaction is :
JIPMER-2005, J & K CET-(1999)
A + 2B → Product
Ans. (a) : Rate constant of a reaction depends on It is given that when concentration of B alone is
temperature and activation energy according to increased, half life remains same. It is clear that from
Arrhenius equation. On increasing the temperature rate above statement, this condition is exist in first order
constant also increases.
kinetics. Thus, the unit of rate constant for first order
Arrhenius Equation, k = Ae–Ea/RT
kinetics is given as follow :
Objective Chemistry Volume-II
260
YCT
For,
n=1
∴
Unit of k = (mol)1–n(lit)n–1sec-1
∴
Unit of k = sec–1
114. The data for the reaction A+B→C
Ex.
[A]0
[B]0
Initial rate
1.
0.012
0.035
0.10
2.
0.024
0.070
0.80
3.
0.024
0.035
0.10
4.
0.012
0.070
0.80
The rate law corresponds to the above data is:
(a) rate = k [B]3
(b) rate = k [B]4
3
(c) rate = k [A][B]
(d) rate = k [A]2[B]2
3
(e) rate = k [A] [B]
Kerala-CEE-2004, (AIPMT -1994)
Ans. (a) For the given reaction, A+B →C
Rate, r = k[Ao]x [B]y
0.1 = k[0.012]x [0.035]y-------(i)
0.8 = k[0.024]x [0.070]y ------(ii)
0.1 = k[0.024]x [0.035]y ------(iii)
0.8= k[0.012]x [0.070]y ------(iv)
On dividing eqn (iii) by eqn(i)
1 = 2x
⇒ 2x = 20
x=0
On dividing (iv) by (i)
8 = (2)y
⇒ 23 = 2y
y=3
Overall rate = k[A]o [B]3 = k [B]3
115. If the rate constant for a first order reaction is
k, the time (t) required for the completion of
99% of the reaction is given by
4.606
2.303
(a) t =
(b) t =
k
k
0.693
6.909
(c) t =
(d) t =
k
k
Kerala-CEE-29.08.2021, NEET -2019
Ans. (a) : The mathematical expression of first order
kinetics is :
2.303
a
k=
log
t
a−x
Let us, assume that (a) = 100%
Given that, x = 99%
2.303
100
∴ k=
log
t
1
4.606
or k =
t
4.606
or t =
k
116. During the kinetic study of the reaction,
2A + B → C + D, following
results
were
obtained:
Run [A]/mol L– [B]/mol L–1 Initial rate of
1
formation
of
D/mol L–1min–1
I.
0.1
0.1
6.0 × 10–3
II. 0.3
0.2
7.2 × 10–2
Objective Chemistry Volume-II
III. 0.3
0.4
2.88 × 10–1
IV. 0.4
0.1
2.40 × 10–2
Based on the above data which one of the
following is correct?
(a) Rate = k [ A ] [ B]
2
(b) Rate = k [ A ][ B]
(c) Rate = k [ A ] [ B]
(d) Rate = k [ A ][ B]
BITSAT – 2011, AIPMT -2010, AIPMT-2010
→C + D
Ans. (d) : For the reaction, 2A + B
Rate = k [A]x [B]y
6 × 10–3 = k [0.1]x [0.1]y
––––(i)
7.2 × 10–2 = k [0.3]x [0.2]y
––––(ii)
2.88 × 10–1 = k [0.3]x [0.4]y
––––(iii)
2.40 × 10–2 = k[0.4]x [0.1]y––––(iv)
On dividing eqn. (iv) by eqn. (i)
4 = 4x
⇒ x =1
On dividing eqn (iii) by eqn(ii)
4 = 2y ⇒ 22 = 2y
⇒y=2
Rate = k [A] [B]2
117. For the reaction,
1
is
aA + bB → cC + dD, the plot of log k vs
T
given below
2
2
2
The temperature at which the rate constant of
the reaction is 10–4 s–1 ........ K
(Rounded off to the nearest integer).
[Given : The rate constant of the reaction is 10–
5 –1
s at 500 K]
[JEE Mains-2021, 25 Feb Shift- I]
Ans. (526 K) : From Arrhenius equation,
Ea
log k = log A −
2.303RT
For straight line graph, y = mx+ C
−Ea
Where, m = slope =
= −10000K
2.303R
T1 = 500K , k1 = 10−5 and
T2 = ?
, k2 = 10–4
Now,
261
log10
log10
Ea 1 1
k2
=
−
k1 2.303R T1 T2
1
10−4
1
= 104
−
−5
10
500 T2
1
1
log10 10 = 104
−
500 T2
1
1
10−4 =
−
⇒ T2 = 526K.
500 T2
YCT
118. For a certain first order reaction 32% of the
(1 + 2a ) = 10
reactant is left after 570 s. The rate constant of
–3 –1
(1 − a )
this reaction is ....... ×10 s . (Round off to the
nearest integer).
1 + 2a = 10 (1 − a ) = 10 − 10a
[Given, log10 2 = 0.301, In 10 = 2.303]
9 3
[JEE Mains-2021, 17 March Shift-I]
12a = 9 ⇒ a =
=
12 4
Ans. (2) : Given - a–x = 32% , t = 570 s
Now, concentration of C at equilibrium = (1+2a)
let a = 100%
3
For 1st order reaction
= 1 + 2 × = 2.5
4
2.303
a
k=
log10
t
(a − x)
= 25 ×10−1 M .
2.303
100
log10
570
32
2
5
−3
= 4 × 10 log10 (10 ) − log10 ( 2 )
−3
= 1.98 × 10
=
⇒
k ≈ 2 × 10−3 sec−1 .
121. PCl5(g)→PCl3(g)+Cl2(g)
In the above first order reaction, the
concentration of PCl5 reduces from initial
concentration 50 mol L–1 to 10 mol L–1 in 120
minutes at 300 K. The rate constant for the
reaction at 300 K is x × 10–2 min–1.
The value of x is ........ . [Given, log5 = 0.6989]
[JEE Main2021, 20 July Shift-II]
Ans. (1.34) : Given:- C0 = 50molL–1
Ct = 10molL–1
t = 120 min. , k=?, T= 300K.
For 1st order reaction,
C
2.303
k=
log10 0
t
Ct
1
N2O5(g)→2NO2(g)+ O2(g)
2
In the above first order reaction the initial
concentration of N2O5 is 2.40×10–2 mol L–1 at
318 K. The concentration of N2O5 after 1 hour
was 1.60×10–2 mol L–1. The rate constant of the
reaction at 318 K is ........×10–3 min–1. (Nearest
2.303
50 2.303
integer)
k=
log10
=
log10 5
120
10
120
[Given : log3 = 0.477, log5 = 0.699]
k = 1.34×10–2 min–1
[JEE Mains-2021, 22 July Shift-II] ⇒
–2
–1
122. The reaction 2A+B2→2AB is an elementary
Ans. (7) : Given:- a = 2.4010 mol L
reaction. For a certain quantity of reactants, if
After t = 1hr = 60 min, a–x = 1.6×10–2 mol L–1, T=
the volume of the reaction vessel is reduced by
318K
st
a factor of 3, the rate of the reaction increases
For 1 order reaction,
by a factor of ........... . (Round off to the nearest
2.303
a
integer).
k=
log10
t
(a − x)
[JEE Main2021, 17 March Shift-II]
−2
Rate
of
elementary reaction (r) = k [A]2 [B2]
Ans.
(27)
:
2.4 × 10
2.303
3
2.303
=
log10
=
log
10
V
60
1.6 ×10−2
60
2
If volume (V) changed to
then
–3
–1
3
k = 6.7510 min
2
⇒ k ≈ 7×10–3 min–1.
2
C A C B2
r'
=
k
V / 3 V / 3 = k [3A ] [ 3B2 ]
120. For the reaction, A+B ↽ ⇀ 2C
The value of equilibrium constant is 100 at 298 r' = 27 k[A]2 [B2] =27r
K. If the initial concentration of all the three ∴ Rate of reaction increased by 27.
species is 1 M each, then the equilibrium 123. The reaction rate for the reaction
concentration of C is x×10–1 M. The value of x
–
2–
is ....... .
[ PtCl 4 ] + H 2O ↽ ⇀ Pt ( H 2O ) Cl 3 + Cl –
(Nearest integer)
was measured as a function of concentrations
[JEE Mains-2021, 25 July Shift-I]
of different species. It was observed that
2–
2–
Ans. (25) : Given:- Keq = 100 , T = 298K
–d ( PtCl 4 ) = 4.8 ×10 –5 ( PtCl 4 )
A
+ B ↽ ⇀ 2C
1M
1M
dt
119
At t = 0
At equilibrium
1M
1− a
1− a
1+ 2a
–2.4×10 –3 {Pt ( H 2O ) Cl 3 } Cl –
Where, square brackets are used to denote
molar concentrations.
The equilibrium constant,
KC =.................. (Nearest integer)
[JEE Main2021, 26 Aug Shift-II]
–
[C]
[ A ][ B]
2
2
[1 + 2a ] = [1 + 2a ]
100 =
[1 − a ][1 − a ] [1 − a ]2
2
Keq =
Objective Chemistry Volume-II
262
YCT
2−
−d ( PtCl4 )
=0
Ans. (50) : At equilibrium
dt
∴ 4.8×10–5[(PtCl4)2–] = 2.4×10–3[{Pt (H2O)Cl3}–][Cl–]
(
( PtCl4 ) 2 −
2.4 × 10−3
=
Keq =
[Pt ( H 2 O ) Cl3 ]− Cl− 4.8 × 10−5
Keq = 50
124. Which one of the following given graphs
represents the variation of rate constant (k)
with temperature (T) for an endothermic
reaction?
(a)
(b)
(c)
(d)
For the given reaction, rate constant k is,
2.47 × 103
log10 k = 20.35 –
-------(ii)
T
On comparing equation (i) and (ii)
log10 A = 20.35
⇒ A = 1020.35
Ea
And,
= 2.47 × 103
2.303R
Activation energy, Ea =2.47×103×8.314×2.303
= 47.29×103 J mol–1
= 47 k Jmol–1.
126. The inactivation rate of a virus preparation is
proportional to the amount of virus. In the first
minutes after preparation, 10% of the virus is
inactivated. The rate constant for viral
inactivation is .......... ×10–3 min–1. (Nearest
integer)
[Use: In 10 = 2.303; log10 3 = 0.477; property of
logarithm : logxy = ylogx]
[JEE Mains-2021, 20 July Shift-I]
Ans. (106) : Given:- Let Intial concentration of virus(a)
=100%
Concentration of inactivated virus after 1min (x)=10%
Remaining concentration of virus after 1min (a–x)
=90%.
Rate constant value is given in min-1 therefore, it is first
order reaction.
For first order reaction, rate constant is
2.303
a
k=
log
t
(a − x )
[JEE Mains-2021, 1 Sep Shift-II]
Ans. (c) : From Arrhenius equation
k = Ae − E a / RT
For forward reaction, kf = A f e − ( E a )f / RT
For backward reaction, kb = A b e − ( E a )b / RT
)
2.303
100
10
log
= 2.303log
1
90
9
−( E ) / RT
k
Ae af
A − ( E ) −( E ) / RT
= 2.303[log10 – log9]
k = f = f −( E ) / RT = f e a f a b
= 2.303 [1–2×log3]
k b Abe a b
Ab
= 2.303 [1–2×.477]
−∆H / RT
= 0.1059 min–1
k = Ae
= 105.9×10–3min–1
≈ 106×10–3min–1
For endothermic reaction ∆H= +ive
127. The decomposition of formic acid on gold
∆H
On increasing the temperature,
= −ive then,
surface follows first order kinetics. If the rate
RT
constant at 300K is 1.0×10–3 s–1 and the
−∆H
activation energy, Ea = 11.488 kJ mol–1, the
= +ive
rate constant at 200 K is ........×10–5 s–1
RT
(Round off to the nearest integer).
−∆H
And slope for k v/s T is
i.e., positive. And we get
[Given, R = 8.314 J mol–1 K–1]
R
[JEE Mains-2021, 16 March Shift-I]
an exponential increasing curve.
Ans.
(10)
:
Given:T1 = 300k , k1=1×10–3S–1
125. For the reaction A→B, the rate constant k
T2 = 200k, Ea = 11.488 kJ mol–1
(in s–1) given by
k2 = ?
2.47 ×103 )
(
From
Arrhenius
equation,
log10 k = 20.35 –
. The energy of
T
Ea 1 1
k
activation in kJ mol–1 is ......... (Nearest integer)
log10 2 =
−
–1
–1
k
2.303R
[Given, R = 8.314 J K mol ]
1
T1 T2
[JEE Mains-2021, 26 Feb Shift-II]
k2
11.488 ×103 1
1
log10
=
Ans. (47) : According to Arrhenius equation,
−3
300 − 200
1
×
10
2.303
×
8.314
k = Ae − E a / RT or
log10 k2 –log10 1×10–3 = –1
Ea
log10 k2 = – 4.
log10 k = log10 A –
-------(i)
k2 = 10–4 = 10×10–5 S–1.
2.303RT
Objective Chemistry Volume-II
k=
263
YCT
128. If the activation energy of a reaction is 80.9 kJ
mol–1, the fraction of molecules at 700 k, having
enough energy to react to form products is e–x.
The value of x is .........
(Rounded off to the nearest integer)
[Use, R=8.31 JK–1 mol–1].
[JEE Mains-2021, 26 Feb Shift-II]
Ans. (14) : Given:- Ea=80.9 kJmol–1 , T=700k
From Arrhenius Equation,
k = Ae − E a / RT
Ans. (a) : For first order reaction –
t1/ 2 = 1200s
We know that, for first order reaction.
0.693
t1/ 2 =
k
0.693
0.693
∴k =
⇒
⇒ 0.000577
t1/ 2
1200
k = 5.8 × 10−4 s–1.
131. If the rate constant of a first order reaction is
4.606 × 10–3 s–1. Then find the time required for
Fraction of molecules having enough energy = e − E a / RT
400g of the reactant to reduce to 50g.
e–x = e − E a / RT
(a) 7.52 minute
(b) 0.45 minute
80.9×103
(c) 46.06 minute
(d) 15.05 minute
−
13.9
−
= e 8.31×700 = e
AP EAPCET 25.08.2021, Shift-II
∴
x ≈ 14.
Ans. (a) : We know,
–4 –1
129. A reaction has rate constant k = 2.4×10 s
2.303
a
k=
log
--------- (1)
Then find the ratio of t99.9 to t50.
t
a−x
(a) 1
(b) 5
Given that, k = 4.606 × 10–3 sec–1
(c) 10
(d) 15
Initial concentration a = 400 gm
AP EAPCET 23-08-2021 Shift-I Final concentration (a–x) = 50 gm
Putting value in eqn (1) we get
Ans. (c) : For first order reaction is2.303
400
[R ]
2.303
4.606 × 10–3 =
log
k=
log o
t
50
t
[R]
2.303
Where,
t=
log8
k = rate constant
4.606 × 10−3
t = time
2.303 × 0.9080
t=
[R o] = initial concentration
4.606 × 10−3
[R] = remaining concentration
t = 454 sec.
[R o ]
2.303
454
⇒ t=
log
t=
= 7.56 minutes.
k
[R]
60
2.303
100
132. The rate constant of a first order reaction at
t 50 =
log
25oC is 1×10-3 min–1 If the temperature
k
100 − 50
coefficient of the reaction is 2, what is the rate
2.303
100
constant (min-1) at 15oC is
t 50 =
log
k
50
(a) 2 × 10–3
(b) 2× 10–4
–4
(c) 5 × 10
(d) 4 × 10-3
2.303
2.303
t 50 =
log 2 =
× 0.3010
.....(i)
TS EAMCET 10.08.2021, Shift-I
k
k
Ans. (c) : Temperature coefficient µ = 2
Time for 99.9% completion of first order reaction.
2.303
100
Rateconsant at 25o C
t 99.9 =
log
Temperature coefficient (µ) =
k
100 − 99.9
Rate consant at15o C
2.303
100 2.303
and t99.9 =
log
=
×3
.....(ii)
Rateconstant at 25o C
Rate constant at 15o C =
k
0.1
k
Temperature coefficient
dividing (ii) by (i) we get1× 10−3
2.303
=
×3
t 99.9
3
2
k
=
=
= 9.96
2.303
t 50
0.3010
10
×
10−4
× 0.3010
=
k
2
t99.9 : t50 = 9.96 ≃ 10
= 5 ×10−4 min–1
130. For a first order reaction t1/2 is 1200s. The
133. For a reaction, K = 4.5 × 10–4 L mol–1 s–1. What
specific rate constant is s–1 is
is order of reaction?
(a) 5.8 × 10–4
(b) 5.8 × 10–5
(a) Zero
(b) Second
(c) 0.58 × 10–6
(d) 0.58 × 10–5
(c) First
(d) Third
TS-EAMCET (Engg.), 07.08.2021 Shift-II
GUJCET-2021
Objective Chemistry Volume-II
264
YCT
Ans. (b) : Given that, k = 4.5 × 10–4 L mol–1 s–1
Rate of reaction is given by–
Rate = k[A]x, where, x → order of reaction
Mole
−4
−1 −1
x
L = 4.5 ×10 L mol s [A]
2
Mole
−4 −1
x
L = 4.5 × 10 s [A]
By comparing both side x = 2
∴ The reaction is 2nd order.
134. Calculate the activation energy of a reaction,
whose rate constant doubles on raising the
temperature from 300 K to 600 K.
(a) 3.45 kJ/mol
(b) 6.90 kJ/mol
(c) 9.68 kJ/mol
(d) 19.6 kJ/mol
TS EAMCET 05.08.2021, Shift-I
Ans. (a) : Given that, k1 = k, k2 = 2k
T1 = 300 k, T2 = 600 k
k
E a T2 – T1
log 2 =
k
2.303
R T1T2
1
2k
Ea
600 – 300
log
=
k 2 ⋅ 303 × 8.314 300 × 600
E a × 300
log 2 =
2 ⋅ 303 × 8.314 × 300 × 600
2 ⋅ 303× 0 ⋅ 3010 × 8.314 × 300 × 600
Ea =
300
Ea ≈ 3·45 kJ/mol
135. Time required to decompose SO 2Cl 2 to half of
its initial amount is 40 minutes. If the
decomposition is a first order reaction, what
will be the rate constant of the reaction?
(a) 1.73 × 10 −2 s −1
(b) 2.88 × 10 −4 s −1
(c) 2.88 × 10 −2 s −1
(d) 1.73 × 10 −4 s −1
GUJCET-2020
Ans. (b) : Given that; Half-life(t1/2) = 40 minutes.
= 40×60 sec.
For first order reaction,
0.693
t1/2 =
= 40 min = 40 × 60 = 2400sec
k
0.693
⇒ Rate constant, k =
= 2.88 × 10−4 S−1 .
2400
136. Consider the following plots of rate constant
1
versus for four different reactions. Which of
T
the following orders is correct for the
activation energies of these reactions?
∴
(a) Eb>Ea>Ed>Ec
(b) Ea>Ec>Ed>Eb
(c) Eb>Ed>Ec>Ea
(d) Ec>Ea>Ed>Eb
[JEE Mains-2020, 8 Jan Shift-II]
Ans. (d) : According to Arrhenius equation
−E a
+ log A
log k =
2.303RT
Ea
1
Slope = −
, for log k v/s
graph.
2.303R
T
As we know that slope is directly proportional to
activation energy.
∴
Ec > Ea > Ed > Eb
137. For an elementary reaction 2A + 3B → 4C + D
the rate of appearance of C of time 't' is 2.8 ×
10–3 mol L–1S–1. Rate of disappearance of B at
't' t. t will be
4
(a) (2.8×10 –3 ) mol L-1S-1
3
3
(b) (2.8×10 –3 ) mol L–1S–1
4
(c) 2 (2.8 × 10–3) mol L–1S–1
1
(d) (2.8×10 –3 ) mol L–1S–1
4
Karnataka-CET-2020
Ans. (b) : For reaction, 2A+3B→4C+D
d d [ A ] −1 d [ B ] 1 d [ C ] d [ D ]
−
=
=
=
2 dt
3 dt
4 dt
dt
d [ C]
−3
−1 −1
Given:= 2.8 × 10 mol L S
dt
1 d [ B] 1 d [ C ]
−
=
3 dt
4 dt
−d [ B] 3
= × 2.8 ×10−3 mol L−1 Sec −1
dt
4
138. The rate constant of a first order reaction is
231 × 10–5 s–1. How long will 4g of this reactant
reduce to 2 g?
(a) 310 s
(b) 300 s
(c) 210 s
(d) 30.1 s
(e) 230.3 s
Kerala-CEE-2020
Ans. (b) : Given that, rate constant (k) = 231 × 10–5 s–1
0.693
For first order reaction, k =
t1/ 2
0.693
⇒
t1/ 2 =
231× 10−5
= 0.003 ×105 = 3 × 102
n
Since,
1
1
N = No × , 2 = 4 ×
2
2
n
n
∴
Objective Chemistry Volume-II
265
1 1
=
2 2
n=1
T = n × t1/2
T = 300 ×1
T = 300 sec
YCT
139. The rate constant for a first order reaction is
4.606 ×10 –3 s –1 . The time required to reduce
2.0g of the reactant to 0.2 g is
(a) 100 s
(b) 200 s
(c) 500 s
(d) 1000 s
NEET-2020
Ans. (c) : Given:k = 4.606×10–3 s–1, t = ?
Co = 2g
Ct = 0.2g
For first order reaction
2.303
C
k=
log o
t
Ct
2.303
2
log
4.606 ×10−3
0.2
= 500 log 10
= 500 S
140. For a reaction 2A+B→P, when
concentration of B alone is doubled, t1/2 does
not change and when concentrations of both A
and B is doubled rate increases by a factor of 4.
The unit of rate constant is,
(a) s–1
(b) L mol–1s–1
–1 –1
(c) mol L s
(d) L2mol–2s–1
WB-JEE-2020
Ans. (b) : 2A+B→P
When concentration of B is doubled, the half life did
not change, hence reaction of first order w.r.t.B. When
concentration of A is doubled, reaction rate is double.
Hence, overall order of reaction is 1+1=2
So, unit of rate constant mol–1 lit S–1.
141. Consider the given plot of enthalpy of the
following reaction between A and B.
A+B→C+D
Identify the incorrect statement.
t=
Activation energy for formation of A and B from C
= 20–0 = 20kJ mol–1
Activation energy for formation of A and B from D =
15–10 = 5 kJ mol–1.
Hence, formation of A and B from C has highest
enthalpy of activation.
∴ Activation enthalpy to form C is 15 kJ mol–1 greater
than to form D.
142. For the reaction, 2A + B → C, the values of
initial rate at different reactant concentrations
are given in the table below.
The rate law for the reaction is
[A]
[B]
Initial rate
(mol L–1)
(mol L–1)
(mol L–1s–1)
0.05
0.05
0.045
0.10
0.05
0.090
0.20
0.10
0.72
(a) rate = k [A] [B]2
(b) rate = k [A]2 [B]2
(c) rate = k [A] [B]
(d) rate = k [A]2 [B]
[JEE Mains-2019, 8 April Shift-I]
Ans. (a) : rate = k[A]x [B]y
0.045 = k[0.05]x [0.05]y ---------(i)
0.09 = k[0.1]x [0.05]y ----------(ii)
0.72 = k[0.2]x [0.1]y ------------(iii)
On dividing (ii) by (i)
2 = 2x
⇒ x=1
On dividing (iii) by (ii)
8 = (2)1 × 2y = 2×2y
4 = 2y
2y = 22
y=2
∴ Overall rate, r = k[A]1 [B]2
= k[A] [B]2
143. The plot of t1/2 v/s [R]0 for a reaction is a
straight-line parallel to X-axis. The unit for the
rate constant of this reaction is
(a) mol L–1s
(b) mol L–1s–1
–1 1–
(c) L mol s
(d) s–1
Karnataka-CET-2019
Ans. (d) : The plot of t1/2 vs [R]0 for a reaction is a
straight line parallel to X-axis. This type of plot is
obtained in first order reaction. The unit for the rate
constant of first order reaction is obtained as
Rate
k=
[A]x [B]y
Where, x + y = 1 (for first order reaction)
(a) D is kinetically stable product.
(b) Formation of A and B from C has highest
enthalpy of activation.
(c) C is the thermodynamically stable product.
Concentration
1
(d) Activation enthalpy to form C is 5 kJ mol–1
k=
×
n
less than that to form D.
time
( Concentration )
[JEE Mains-2019, 9 April Shift-II] So,
mol L−1
1
=
×
= sec−1
Ans. (d) : From the given graph,
1
−1
sec
–1
mol
L
Activation Energy of C, E a C = 20 – 5 = 15 kJ mol
144. For a chemical reaction rate law is, rate =
E a D = 15–5 = 10kJ mol–1
k[A]2[B]. If [A] is doubled at constant [B], the
rate of reaction
Therefore, D is kinetically stable because of low
(a) increases by a factor of 8
activation energy and C is thermodynalnically unstable
(b) increases by a factor of 4
or kinetically unstable.
(
Objective Chemistry Volume-II
266
)
YCT
(c) increases by a factor of 3
147. For a reaction, consider the plot of In k versus
1/T given in the figure. If the rate constant of
(d) increases by a factor of 2
this reaction at 400 K is 10–5 s–1, then the rate
MHT CET-02.05.2019, SHIFT-II
constant at 500 K is
Ans. (b) : Rate, = k [A]2 [B] = r
If [A] is doubled at constant [B] i.e.
[A]
→ [ 2A ]
2
Rate, r' = k [2A] [B]
= 4r
Rate increased by a factor of 4.
145. For the reaction N2 + 3H2 → 2NH3
∆[NH 3 ]
= 2 ×10−4 mol L−1s −1 , then value of
If
(a) 4×10–4 s–1
(b) 10–6 s–1
–4 –1
∆t
(c) 10 s
(d) 2×10–4 s–1
−∆[H 2 ]
[JEE Mains-2019, 12 Jan Shift-II]
would be
∆t
Ans. (c) : The Arrhenius Equation is,
(a) 1 × 10–4 mol L–1 s–1 (b) 3 × 10–4 mol L–1 s–1
E
lnk = lnA – a
(c) 4 × 10–4 mol L–1 s–1 (d) 6 × 10–4 mol L–1 s–1
RT
E
VITEEE-2019 Where,
Slope = − a = −4606 (given)
R
Ans. (b) : N2 + 3H2 → 2NH3
When T1 = 400K then k1 =10–5 S–1
−∆ [ N 2 ]
1 ∆ [ H 2 ] 1 ∆ [ NH 3 ]
and when T2 = 500K then k2 = ?
Rate =
=−
=
Now,
dt
3 dt
2 dt
∆ [ NH 3 ]
k E 1 1
Given,
= 2 ×10−4 mol L−1 s −1
ln 2 = a −
dt
k1 R T1 T2
∆
H
−1 [ 2 ] 1
k
1
1
So,
= × 2 × 10−4
2.303 log 2−5 = 4606
−
3 dt
2
10
400 500
−∆ [ H 2 ] 3 × 2 × 10−4
k
4606
100
=
×
log 2−5 = +
dt
2
10
2.303 400 × 500
−5
−∆ [ H 2 ]
−4
−1 −1
log
k
−
log10
=1
2
= 3 × 10 mol L s
dt
log k 2 − log10−5 = log10 10
146. The rate law for the reaction described by
+ log10−5
N2O2(g)→2NO(g) is first order in the log k 2 = log10 10
−4
concentration of N2O2(g). The expression for the log k 2 =−4log10
−1
time dependent behaviour of the product k 2 = 10 S .
concentration [NO] is
148. The rate constant for the reaction 2N2O5 →
4NO2 + O2 is 3.0 × 10–4s–1. If the rate is 2.4 × 10–
(a) [NO]=2[N2O2]o(1–e–kt)
5
–kt
mol–1s–1 at room temperature, then the
(b) [NO]=[N2O2]o(1–e )
–kt
concentration of N2O5 in mol L-1 is
(c) [NO]=2[N2O2]o e
(a) 1.4
(b) 1.2
(d) [NO]=2[N2O2]oe–kt
(c)
0.02
(d)
0.08
Where, [N2O2]o is the initial concentration of
Assam CEE-2018
N2O2 i.e. constant.
AMU-2019 Ans. (d) : The unit of rate constant is s −1 hence it is of
first order reaction,
Ans. (b) :
rate = k.[ N 2 O 4 ]
N O
→ 2NO
2
t=0
2
[ N 2O 2 ]o
Given : rate = 2.4 ×10 −5 mol L−1s −1
0
[ N 2O 2 ]t
[ NO]t
∴ [ N 2 O 2 ]t + [ NO ]t = [ N 2O 2 ]o
For a first order reaction, [ A ] = [ A o ].e − kt
[ N 2O2 ]t = [ N 2O2 ]o e− kt
[ N 2O2 ]o − [ NO]t = [ N 2O2 ]o .e− kt
[ NO]t = [ N 2O2 ]o .(1 − e− kt ) .
t=t
Objective Chemistry Volume-II
∴
Rate constant, k = 3.0 × 10−4 s −1
2.4 × 10−5 = 3.0 × 10−4 × [ N 2 O 4 ]
[ N 2O4 ] = 0.08mol L−1
149. At 27 o C temperature, time required for 75%
completion of a first order reaction is 20
seconds. What will be its rate constant?
(a) 0.693 sec-1 mole-1 lit.
(b) 0.0693 sec-1
267
YCT
(c) 0.693 sec-1
(d) 0.0693 sec-1 mole-1lit.
2.303
2.303
log10 ,
k=
70
70
GUJCET-2018
k=0.0329 min–1
Ans. (b) : For first order reaction,
Velocity constant of the reaction = 0.0329 min–1.
2 × 0.693
153. At 300 K, activation energy of A is higher
= 20sec
t75% = t3/4 = 2 × t1/2 =
than B by 5.75 kJ/mol in presence of catalyst.
k
k
2 × 0.693
Calculate A
⇒k =
= 0.0693 sec−1
kB
20
(a) 1
(b) 10
150. A chemical reaction was carried out at 300 K
(c) 1000
(d) 100
and 280 K. The rate constant were found to be
[AIIMS-27 May, 2018 (M)]
k1 and k2 respectively, then
(a) k2 = 4k1
(b) k2 = 0.25k1
Ans. (b): Given at T = 300K,
(c) k2 = 2k1
(d) k2 = 0.5k1
E A = higher, E A − E B = + 5.75 kJ / mol
JIPMER-2018
Ea
log10 k = log10 A −
Ans. (b) : T1= 300K , T2 = 280K
2.303RT
Rate constant becomes double for every 10° rise in
k
EA − EB
temperature. Hence, for 20°C rise in temperature, rate ∴
log10 A =
constant will become four times.
k B 2.303 × 8.314 × 300
k1 = 4k2
kA
5.75 × 1000
k2 = 0.25 k1
log10
=
k B 2.303 × 8.314 × 300
151. In the reversible reaction,
k1
k
k
2NO2 ↽ k ⇀ N2O4 the rate disappearance of
⇒ A = 101
log10 A = 1
2
kB
kB
NO2 is equal to
kA
= 10
2k1
2
(a)
[ NO2 ]
kB
k2
154. Rate of two reactions whose rate constants are
(b) 2k1[NO2]2 – 2k2[N2O4]
k1 and k2 are equal at 300K such that:
2
(c) 2k1[NO2] – k2[N2O4]
Ea2 – Ea1 = 2RT,
(d) (2k – k ) [NO ]
1
2
k=
2
JIPMER-2018
So calculate ln
k1
Ans. (b) : For the reaction, 2NO2 ↽ k ⇀ N2O4
2
−d[NO 2 ]
= 2k1[NO 2 ]2
Rate of forward reaction
dt
+d[NO 2 ]
Rate of backward reaction
= 2k 2 [N 2 O 4 ]
dt
A2
A1
(a) ln 4
(c) log 2
Ans. (b): log k = log A −
(b) 2
(d) 2-ln 2
[AIIMS-26 May, 2018 (E)]
Ea
2.303RT
Given : E a 2 − E a 1 = 2RT
Rate of disappearance of NO2,
k1 = k 2
−d [ NO 2 ]
= 2k1[NO 2 ]2 − 2k 2 [N 2 O 4 ]
E a − E a1
k
A
dt
∴
log 2 = log 2 − 2
k1
A1 2.303RT
152. 90% of a first order reaction is completed in 70
minutes. The velocity constant of the reaction is
k
A
2RT
log = log 2 −
(a) 0.0329
(b) 0.329
k
A1 2.303RT
(c) 3.29
(d) 0.0293
A
2
COMEDK 2018
log 2 =
A
2.303
Ans. (a) : For first order reaction1
A
2.303
a
2.303 × log10 2 = 2
k=
log
A1
t
a–x
Given: t90= 70 minutes.
A
ln 2 = 2
a=100
A1
x=90
155. The rate constant, the activation energy and
velocity constant of the reaction k=?
the Arrhenius parameter of a chemical reaction
then
at 25 oC are 3.0 × 10–4 s–1, 104.4 kJ mol–1 and
2.303
100
6.0×1014 s-1 respectively. The value of the rate
k=
log
constant at T → ∞ is
70
100 – 90
Objective Chemistry Volume-II
268
YCT
(a) 2.0 × 1018 s–1
(c) 3.6 × 1030 s–1
(b) 6.0 × 1014 s–1
(d) infinity
AMU-2018
Ans. (b) : Given,
k = 3 × 10 −4 s −1 , E a = 104.4kJ / mol, A = 6.0 × 1014 s −1
[ O] = k eq
[O3 ] − − − −(ii)
[O2 ]
Putting the value of (ii) in ------(i)
[O ]
2
−1
r = k k eq 3 [ O3 ] = k '[ O3 ] [ O 2 ] …(Where, k'= kkeq)
[O2 ]
Arrhenius equation k = Ae − Ea / RT
158. A reaction : A2 + B → Products, involves the
E
When T → ∞, a → 0
following mechanism :
Rt
A 2 ⇌ 2A(fast)
∴ k = A.eo
(A being the intermediate)
= A ×1
→ Products (slow). The rate law
A + B
k2
k = 6.0 × 1014 s −1
consistent to this mechanism is:
156. Two reactions R1 and R2 have identical
(a) rate = k[A2][B]
(b) rate = k[A2]2[B]
1/2
preexponential factors. Activation energy of R1
(d) rate = k[A2][B]2
(c) rate = k[A2] [B]
exceeds that of R2 by 10kJ mol–1. If k1 and k2
are rate constants for reactions R1 and R2,
VITEEE-2016
k2
Ans.
(c)
:
Given
reaction
–
respectively at 300 K, then ln is equal to
The rate of reaction depends on slowest step.
k1
A + B
→ Product ( slow )
(R = 8.314 J mol–1 K–1)
k2
(a) 8
(b) 12
..….(i)
∴ Rate = k2 [A][B]
(c) 6
(d) 4
A 2 ↽ ⇀ 2A
[JEE Main-2017]
At equilibrium
Ans. (d) :
2
Given:- A1 = A2 = A
A]
[
1/ 2
−1
4
−1
……(ii)
ke =
or [ A ] = k1/e 2 [ A ]
E a 1 − E a 2 = 10kJ mol = 10 J mol
A
[ 2]
–1
–1
T = 300k, R = 8.314 Jmol K
Substituting the value of [A] in eq. (i)
According to Arrhenius equation,
1/ 2
Rate = k 2 k1/e 2 [ A 2 ] [ B]
− E a / RT
k = Ae
1/ 2
− E / RT
Rate = k [ A 2 ] [ B].
(let k 2 k1/e 2 = k)
k1 = Ae a1 − − − −(i)
159. For the following reaction : NO2(g) + CO(g) →
− E / RT
k 2 = Ae a 2 − − − −(ii)
NO(g) + CO2(g), the rate law is: Rate =
Dividing (i) and (ii)
k[NO2]2. If 0.1 mole of gaseous carbon
− E a 2 / RT
monoxide is added at constant temperature to
k2 e
( Ea − Ea ) / RT
the reaction mixture which of the following
= − E / RT = e 1 2
k1 e a 1
statements is true?
(a) Both k and the reaction rate remain the same.
4
k2
10 / 8.314×300 )
=e (
= e4
(b)
Both k and the reaction rate increase.
k1
(c) Both k and the reaction rate decrease.
k
(d) Only k increases, the reaction rate remain the
⇒ ln 2 = 4
same.
k
1
[AIIMS-2016]
157. For the chemical reaction, 2O3 ⇌ 3O2
Ans. (a): The rate expression is independent of the
The reaction proceed as follows
concentration of [CO] hence on changing the
O3 ⇌ O2 + O (fast)
concentration of [CO] the k and reaction rate remains
unchanged.
O + O3
→ 2O2 (slow)
160.
Consider the reaction:
The rate law expression will be
2
2
–1
(a) r = k' [O3]
(b) r = k' [O3] [O2]
Cl2(g) + H2S(g) → S(s) + 2H+(aq) + 2Cl–(aq)
(c) r = k' [O3] [O2]
(d) Unpredictable
The rate equation for this reaction is
JIPMER-2017
rate = k [Cl2][H2S]
Ans. (b) : Rate of reaction depends on slowest step
Which of these mechanisms is /are consistent
i.e, rate determining step. So,
with this rate equation?
–
+
–
r = k[O] [O3] ----------(i)
A. Cl2 + H2S → H+ + Cl + Cl + HS (slow)
–
–
Overall Rate doesn't depend on intermediate therefore,
+
+
Cl + HS → H + Cl + S
(fast)
–
+
O 2 ][ O ]
[
B. H2S ⇌ H + HS (fast equilibrium)
keq =
–
–
[ O3 ]
Cl2 + HS → 2Cl + H+ + S (Slow)
Objective Chemistry Volume-II
269
YCT
(b) Both A and B
164. The rate constant of the reaction, 2N2O5 → 4NO2
+ O2 at 300 K is 3×10-5 s-1. If the rate of the
(d) A only
reaction at the same temperature is 2.4×10-5 mol
[BITSAT – 2016]
dm-3 s-1 then the molar concentration of N2O5 is
Ans. (d) : The slow step is called rate determining step,
(a) 0.4 M
(b) 0.8 M
hence the given rate law as
(c) 0.04 M
(d) 0.08 M
rate = k.[ Cl 2 ] .[ H 2S]
(e) 0.06 M
Thus, only equation A is consistent with the given rate
Kerala-CEE-2016
equation.
Ans. (b) : From the unit of rate constant, it indicates 1st
161. Reaction
3ClO − → ClO 3− + 2Cl − occurs
in order reaction
k = 3 × 10−5 s –1 ,
rate = 2.4 ×10 −5 mol dm −3s −1
following two steps.
− k1
Rate, r = k [N2O5]
(i) ClO − + ClO →
ClO 2− + Cl − (Slow step)
−
−
−k 2
−
−
2 ⋅ 4 × 10−5
(ii) ClO 2 + ClO
→ ClO3 + Cl (Fast Step)
[N2O5] =
= 0.8M
3 × 10−5
Then the rate of given reaction =
3
165. Assertion: If the activation energy of a reaction
(a) k 2 ( ClO −
(b) k1 ( ClO −
is zero, temperature will have no effect on the
rate constant.
−
−
− 2
(c) k 2 ClO 2 ClO (d) k1 ClO
Reason: Lower the activation energy, faster is
the reaction.
GUJCET-2016
(a) If both Assertion and Reason are correct and
Ans. (d) : Rate depends on slowest step i.e., rate
the Reason is the correct explanation of
determining step.
Assertion.
–
rate ∝ ClO
∴
(b) If both Assertion and Reason are correct, but
–
–
rate = kl [C1O ] [C1O ]
Reason is not the correct explanation of
⇒
rate = k1[C1O–]2
Assertion.
162. For the reaction,
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are incorrect.
2 N2O5 → 4 NO2 + O2
[AIIMS-2015]
the values of rate constant and rate of reaction
are respectively 3 × 10–5s–1 and 2.4 × 10–5 mol Ans. (b): The Arrhenius expression, k = A.e − Ea / RT
L–1s–1. The concentration of N2O5 (in mol L–1)
For zero activation energy the rate constant is
will be
independent of the temperature.
(a) 0.4
(b) 0.8
For lower value of Ea, the rate constant is high. Both the
(c) 1.2
(d) 0.6
assertion (A) and reason (R) are correct but reason is
JCECE - 2016 not the correct explanation of assertion.
Ans. (b) : Given, rate constant = 310–5S–1
166. The activation energy of a reaction is zero. The
Rate of reaction = 2.4×10–5 molL–S–1
rate constant of this reaction
Since, unit of rate constant is 3.0 × 10–5 s–1, means order
(a) increases with an increase of temperature
of reaction is one.
(b) decreased with an increase of temperature
Thus, our rate law will be
(c) decreases with decrease of temperature
r = k[N2O5]1
(d) is independent of temperature
Given,
r = 2.4 × 10–5
CG PET- 2015
k = 3 × 10–5
Ans. (d) : If Ea = 0
According to Arrhenius equation,
r 2.4 × 10−5
∴
= 0.8mol L−1
[ N2O5 ] = =
k = Ae–Ea/RT
−5
k
3 ×10
k = Aeo
{ ∵ eo = 1}
163. Two similar reactions have the same rate
k=A
constant at 25ºC, but at 35ºC, one of the
Therefore,
rate constant of this reaction is independent
reaction has a higher rate constant than the
of
temperature.
other. The appropriate reason for this is
167. The unit and value of rate constant and that of
(a) due to effective collisions
rate of reaction are same for
(b) due to different activation energies
(a) zero order
(b) first order
(c) due to different threshold energies
(c)
second
order
(d)
third order
(d) due to higher population of molecules
JCECE - 2015
JIPMER-2016
Ans. (b) : At different temperature, rate constants are Ans. (a) : For zero order reaction, e.g.
different because at higher temperature activation
A→B
energy of reacting molecules are different so collision
d [A]
0
are effective. Hence at higher temperature the value of
−
= k [A]
dt
rate constant is higher.
(a) B only
(c) Neither A nor B
Objective Chemistry Volume-II
270
YCT
d [A]
= k.
dt
Therefore, the unit and value of rate constant and that of
rate of reaction are same for zero order reaction.
168. The rate constant of the reaction A → B is
0.6 ×10 –3 mol L–1s –1 . If the concentration of A is
5M, then concentration of B after 20 minutes
is 0.6 × 10 −3
(a) 3.60 M
(b) 0.36 M
(c) 0.72 M
(d) 1.08 M
NEET-2015
Ans. (c) : Given data,
k = 0.6 × 10–3 mol L–1 s–1
[A] = 5M, t = 20 min = 1200sec, [B] = ?
It is zero order reaction by seeing the value unit of rate
constant,
For zero order reaction
[At] = [Ao] – kt
= 5 – (0.6 × 10–3 × 1200)
= 4.28 M
A
→B
t=0
5M
0
t =20min 4.28M 0.72M
169. In a reaction, A + B C, the rate expression is
R = k [A] [B]2. If the concentration of both the
reactant is doubled at constant volume, then
the rate of the reaction will be
(a) eight times
(b) double
(c) quadruple
(d) triple
UPTU/UPSEE-2015
⇒
−
10−3 10 × 10−4
=
2
2
= 5 × 10−4
K17o C =
K17o C
Hence, the rate constant at 17oC for this reaction
= 5 × 10 −4
171. The quantity of K in a rate of expression :
(a) Is independent of concentration of reactants
(b) Is called Arrhenius constant
(c) Is dimensionless
(d) Is independent of temperature
MPPET-2013
Ans. (a) : According to rate law–
A+B→C+D
r = k(A)m (B)n
Where, k = Rate constant depends on temperatures and
for different order of reaction the unit of k is different.
• k is independent of the concentrations of the reactions.
Unit of Rate constant = mol1–n Ln–1 sec –1
172. For a hypothetical reaction
A→C
k
A ↽ k 1 ⇀ B[Fast]
2
k3
A + B
→ C(Slow)
Rate law for this reaction is
(b) ∝ [A]2
(a) ∝ [A][B]
(c) ∝ [A]2 [B]2
(d) ∝ [A]2 [B]
VITEEE-2014
Ans. (a) : Rate of reaction depend on slow step. Then,
k3
A + B
→ C(slow) will determine rate.
Rate r = k3 [A] [B]
Ans. (a) : Rate = k [ A ][ B] = R
If the concentration of both the reactant is
doubled
R' = k [2A] [2B]2 = 8k[A] [B]2
R' = 8R
r ∝ [A][B]
Rate will increase by a factor of 8.
170. The rate constant of a first order reaction at 173. The half-life period of a first order reaction is
27ºC is 10–3min–1. The temperature coefficient
10 minutes. Then its rate constant is
of this reaction is 2. What is the rate constant
–1
(a) 6.93 × 102 min–1
(b) 0.693 × 10–2 min–1
(in min ) at 17ºC for this reaction?
–2
–1
(d) 69.3× 10–1 min–1
(c) 6.932 × 10 min
(a) 10 −3
(b) 5 × 10 −4
SRMJEEE – 2009
(c) 2 × 10 −3
(d) 10 −2
(AP-EAMCET-2006) Ans. (c) : Given that : For first order kinetics reaction :0.693
Ans. (b) : Given that,
k=
–3
t1/ 2
Temperature coefficient = 2 , rate constant K 27 oC = 10
min–1
t1/ 2 = 10 minutes
Temperature coefficient
k=?
Rate constant at higher temperature (at 27o C)
0.693
=
∵
k=
Rate constant at lower temperature (at17o C)
10
K 27o C
k = 6.93 × 10–2 min–1
∴ Temperature coefficient =
174. The unit of rate constant for a zero order
K17o C
reaction is
Putting the value we get(a) mol L–1s–1
(b) L mol–1s–1
10−3
2
–2 –1
(c) L mol s
(d) s–1
2=
K17o C
VITEEE- 2012
2
Objective Chemistry Volume-II
271
YCT
−E a
R
1− n
mol
−1
R
(d)
sec
lit
Ea
Put . n = 0 (for zero order reaction)
AP-EAMCET- (Engg.) - 2010
1− 0
mol
Ans. (b) : According to Arrhenius equation–
−1
sec
lit
k = Ae− Ea / RT
–1
–1
mol lit sec
Taking loge both side we get –
Or
E
o
Rate of reaction for zero order , rate = k [A] = k
logek = logeA – a
–1 –1
RT
∴ unit of k = unit of rate = mol L s
E
a
175. A chemical reaction was carried out at 320 K
lnk = lnA −
and 300 K. The rate constants were found to be
RT
k1 and k2 respectively. Then
(a) k2 = 4k1
(b) k2 = 2k1
(d) k2 = 0.5 k1
(c) k2 = 0.25 k1
VITEEE- 2010
Ans. (c) : As we know that for every 10° rise in
temperature, rate constant, k becomes doubled. Hence,
on rising the temperature 20°, the rate constant will be
four times,
1
i.e., k1 = 4k 2 ⇒ k 2 = k1 = 0.25k1
4
Or
k1
n
A graph between lnk and 1/T is a straight line with
= ( 2)
k2
−E a
slope and lnA intercept.
here ∆T = T1–T2
R
= 320 – 300 = 20
178. The time taken for the completion of 90% of a
∆T
first order reaction is t min. What is the time
∴
n=
10
(in sec) taken for the completion of 99% of the
reaction?
20
n=
=2
t
10
(a) 2t
(b)
30
k1
2
∴
= ( 2)
(c) 120t
(d) 60t
k2
AP-EAMCET (Engg.)-2005
k1 = 4k2
Ans. (a) : For the first order kineticsk2 = 0.25k1
For 90% completionWhere k1 = Rate constant at 320K
k2 = Rate constant at 300K
2.303
a
k=
log
176. The half life period of a first order reaction is 1
t
a−x
min 40 secs. Calculate its rate constant.
2.303
100
or
k=
log
(a) 6.93 × 10–3 min–1
(b) 6.93 × 10–3 sec–1
t
100 − 90
(c) 6.93 × 10–3 sec
(d) 6.93 × 103 sec
2.303
VITEEE- 2008
k=
......(i)
t
Ans. (b) : For 1st order reaction.
Now, Time (T) = ?
0.693
t1/ 2 ( half time ) =
a = 100
k
x = 99
0.693
0.693
0.693
For
99%
completionHence k =
=
=
t1/ 2
( 60 + 40 ) sec (10 )2
2.303
100
∴
k=
log
−2
−1
−3
−1
= 0.693 × 10 sec = 6.93 × 10 sec
T
100 − 99
2.303
177. What is the slope of the straight line for the
or
T=
log100
1
k
graph drawn between k and , where k is the
2.303
T
T=
× 2 − − − − − (ii)
rate constant of a reaction at temperature T?
k
Ans. (a) : The unit of rate constant can be calculated by
using following formula:–
Objective Chemistry Volume-II
272
−E a
2.303R
E
(c) a
R
(a)
(b)
YCT
Putting the value of k from equation (i) in equation (ii)
2.303
k=
× log10
we get
10
2.303
2.303
T=
×t×2
k=
×1
2.303
10
T = 2t
k = 0.2303 min–1
179. The temperature coefficient of a reaction is 2.5. 182. Unit of k for third order reaction is_______
If its rate constant at T1K is 2.5×10–3s–1, the
Litre
Mole
rate constant at T2 K in s–1 is (T2 > T1).
(b)
(a)
⋅ sec
⋅ sec
–3
–3
Mole
Litre
(a) 1.0 × 10
(b) 6.25 × 10
−1
−2
(d) 6.25 × 10–2
(c) 1.0 × 10–2
Litre
Mole
−1
−1
(c)
(d)
AP EAMCET (Medical) - 2013
⋅ sec
⋅ sec
Mole
Litre
Ans. (b) : The definition of temperature coefficient is
GUJCET-2007
that- the ratio of the specific reaction rates of a reaction
Ans. (d) : Unit of nth order reaction can be written as–
at two temperatures differing by 10°C
1− n
GivenTemperature coefficient (µ) = 2.5
mol
−1
–3 –1
Unit
of
k
=
sec
K T1 = 2.5 ×10 s
lit
for third order reaction n = 3
rateconstant at T + 10°C
∴
µ=
1−3
rateconstant at T°C
mol
−1
∴ Unit of k =
sec
lit
K T2
or
µ =
( T2 > T1 )
−2
mol
K T1
−1
or unit of k =
sec
lit
K T2 = K T1 × µ
183. Assertion: A catalyst does not alter the
–3
K T2 = 2.5 × 10 × 2.5
equilibrium constant of a reaction.
–3 –1
Reason: The catalyst forms a complex with the
Rate constant at K T2 = 6.25 × 10 s
reactants and provides an alternate path with
180. With respect to the velocity constant of a
lower energy of activation for the reaction; the
reaction, which one of the following statement
forward and the backward reactions are
is not correct?
affected to the same extent.
(a) It is a measure of the velocity of the chemical
(a) If both Assertion and Reason are correct and
reaction.
the Reason is the correct explanation of
(b) It is dependent on initial concentration of
Assertion.
reactants.
(b) If both Assertion and Reason are correct, but
(c) It depends on temperature.
Reason is not the correct explanation of
Assertion.
(d) During the progress of the reaction even
though the velocity decreases, the velocity
(c) If Assertion is correct but Reason is incorrect.
constant remains constant.
(d) If both the Assertion and Reason are
COMEDK 2011
incorrect.
[AIIMS-2010]
Ans. (b) : The value of the rate constant or velocity
constant does not depend upon the concentration of the Ans. (a): A catalyst does not affect the equilibrium
reactants.
constant because it forms a complex with the reactants
181. A (g)
→ B(g) is a first order reaction. The and provides an alternate path with lower energy of
activation for the reaction; the forward and the
initial concentration of A is 0.2 mol L–1. After backward reactions are affected to the same extent and
10 min, the concentration of B is found to be there are increase in rate of forward and backward
0.18 mol L–1. The rate constant (in min–1) for reaction. Hence, the equilibrium constant remains
the reaction is
unchanged. Hence, A and R both are correct and R is
(a) 0.2303
(b) 2.303
the correct explanation of A.
(c) 0.693
(d) 0.01
184. Assertion: The rate of the reaction is the rate of
AP-EAMCET (Medical), 2008
change of concentration of a reactant or a
Ans. (a) : Given reaction:
product.
Reason: Rate of reaction remains constant
A(g) → B(g)
during
the course of reaction.
0
At t = 0 0.2mol L−1
At t =10 0.20− 0.18
0.18mol L−1
(a) If both Assertion and Reason are correct and
Initial concentration of A(g), a = 0.2M
the Reason is the correct explanation of
Final concentration of A(g), (a–x) = 0.2– 0.18 =0.02M
Assertion.
For the first order Kinetic––
(b) If both Assertion and Reason are correct, but
2.303
0.2
Reason is not the correct explanation of
k=
log
10
0.02
Assertion.
Objective Chemistry Volume-II
273
YCT
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are
incorrect.
[AIIMS-2010]
Ans. (c): The rate of reaction is the change of
concentration of reactant or product and the
concentration of reactant changes with time hence the
rate of reaction changes during the course of reaction.
Hence the (A) is correct but (R) is incorrect.
185. Assertion: In rate law, unlike in the expression
for equilibrium constants, the exponents for
concentrations do not necessarily match the
stoichiometric coefficients.
Reason: It is the mechanism and not the
balanced chemical equation for the overall
change that governs the reaction rate.
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are
incorrect.
[AIIMS-2009]
Ans. (a): The rate law depends on the mechanism and it
is purely experimental which may or may not be equal
to stoichiometric coefficient of the reactants. Hence
both A and R are correct and R is the correct
explanation of A.
186. The rate constant for the reaction.
2N 2O 5
→ 4NO 2 + O 2 is 3.0×10-4 s-1. If start
made with 1.0 mol L-1 of N2O5, calculate the rate
of formation of NO2 at the moment of the
reaction when concentration of O2 is 0.1 mol L-1.
(a) 2.7 × 10-4 mol L-1 s-1 (b) 2.4 × 10-4 mol L-1 s-1
(c) 4.8 × 10-4 mol L-1 s-1 (d) 9.6 × 10-4 mol L-1 s-1
[AIIMS-2011]
Ans. (d):
2N 2 O5
→ 4NO 2 + O 2
t=0
1.0
0
0
1.0 − ( 2 × 0.1) = 0.8
Given: k = 3.0 × 10−4 s −1
∴
Rate = k [ N 2 O5 ]
= 3.0 × 10 −4 × 0.8
= 2.4 ×10 −4 mol L−1s −1
∴ rate of formation of NO 2 = 4 × rate of reaction
Ans. (c):
2N 2O5
→ 4NO 2 + O 2
Rate of reaction
d [O2 ]
1 d [ N 2 O5 ]
1 d [ NO 2 ]
=− .
=+ .
=+
2
dt
4
dt
dt
1 d [ NO 2 ]
∴ rate of reaction = .
4
dt
188. The decomposition of N2O5 occurs as,
2N 2 O 5
→ 4N 2 + O 2 , and follows 1st order
kinetics, hence :
(a) the reaction is unimolecular
(b) the reaction is bimolecular
(c) t1/ 2 ∝ a 0
(d) none of the above
BCECE-2006
Ans. (c) : For a first order reaction, half life
0.693
t1/ 2 =
,
i.e. independent of the initial
k
concentration of the reactant.
189. The rate constant of a reaction is 1 × 10–2 mol–2
L2s–1. The order of the reaction is
(a) 0
(b) 1
(c) 2
(d) 3
BCECE-2009
Ans. (d) : For different order of reaction, the rate
constant has different units.
Unit of k for third order reaction = mol−2 L2s −1
190. The chemical reaction 2O3 → 3O2 proceeds
as follows:
Fast
Slow
O 3 ↽ ⇀ O 2 + O; O + O 3
→ 2O 2
The rate law expression should be
(a) r = k [O3]2
(b) r = k [O3]2 [O2]–1
3
2
(c) r = k [O3] [O2]
(d) r = [O3] [O2]2
[BITSAT – 2009]
Ans. (b) : The slow step is called rate determining step,
O + O3
→ 2O 2 ( slow )
∴ rate = k.[ O ].[ O3 ]
O3 ↽ ⇀ O 2 ........... ( fast )
= 4 × 2.4 × 10 −4 = 9.6 × 10 −4 mol L−1s −1
187. For the reaction, 2N2O5 → 4NO2 + O2 the rate
of reaction is :
∴
1 d
d
(a)
(b) 2 [ N 2 O5 ]
[ N 2 O5 ]
2 dt
dt
1 d
d
(c)
(d) 4 [ NO 2 ]
[ NO2 ]
4 dt
dt
[AIIMS-2006] Rate
Objective Chemistry Volume-II
[ O 2 ] .[ O ]
[ O3 ]
keq.[ O3 ]
[O] =
[O2 ]
[O ]
rate = k.k eq 3 × [ O3 ]
[O2 ]
2
[O ]
Rate = k1. 3
[O2 ]
2
−1
= k 1 .[ O 3 ] [ O 2 ]
At equilibrium, k eq =
274
YCT
191. Velocity constant of a reaction at 290 K was
= 8 mm min–1
found to be 3.2 × 10–3. At 300 K it will be
194. A + B → Product
(a) 1.28 × 10–2
(b) 9.6 × 10–3
If concentration of A is doubled, rate increases
(c) 6.4 × 10–3
(d) 3.2 × 10–4
4 times. If concentrations of A and B both are
[BITSAT – 2013]
doubled, rate increases 8 times. The differential
rate equation of the reaction will be
Ans. (c) :
−3
dC
dC
T1 = 290 K
k1 = 3.2 × 10
(b)
(a)
= kC A × C B
= kC 2A × C3B
dt
dt
T2 = 300 K
k2 = ?
dC
dC
For each 10°C rise in temperature, the rate of reaction
(c)
= kC 2A × C B
(d)
= kC 2A × C B2
is doubled, hence k 2 = 2k1 for rise in temperature from
dt
dt
290 to 300K.
CG PET -2009
k 2 = 2 × 3.2 × 10−3
→ Pr oduct
Ans. (c) : Reaction : A + B
−3
When
concentration
of
A
is
doubled,
rate of reaction is
k 2 = 6.4 × 10
4 times hence with respect to A it is of second order.
192. 2N2O5 ⇌ 4NO2 + O2
When the concentrations of both A and B are doubled
If rate and rate constant for above reaction are simultaneously, the rate increases 8times. Hence, with
2.40 × 10–5 mol L–1 s–1 and 3 × 10–5 s–1
respect to B it is doubled and the order of reaction with
respectively, then calculate the concentration of respect to B is 1.
2
1
N 2O 5.
∴ rate = k [ A ] [ B]
(a) 1.4
(b) 1.2
= k.C 2A .C1B
(c) 0.04
(d) 0.8
[BITSAT – 2012] 195. The value of rate constant for a first order
reaction is 2.303 × 10–2 sec–1. What will be the
Ans. (d) : The unit of rate constant is s −1 hence it is of
time required to reduce the concentration to
first order.
1
∴ rate = k.[ N 2 O5 ]
th of its initial concentration?
10
Given : rate = 2.40 ×10 −5 mol L−1
(a) 10 second
(b) 100 second
(c) 2303 second
(d) 230.3 second
[ N 2 O5 ] = ?
GUJCET-2014
Rate constant, k = 3 × 10 −5 s −1
Ans. (b) : Given:- rate constant, k = 2.303×10–2 sec–1
−5
−5
∴
2.40 ×10 = 3 ×10 × [ N 2 O5 ]
Let a = 1
Then, a-x = 1/10 of an a/10
2.40 × 10−5
[ N 2 O5 ] =
−5
For first order reaction,
3 ×10
2.303
a
[ N 2O5 ] = 0.8
k=
log10
t
a−x
1
2.303
1
193. A gaseous reaction X2(g) → Y + Z (g)
t=
log10
−2
2
2.303 × 10
1/10
There is increase in pressure from 100 mm to
2.303
120 mm in 5 minutes. The rate of
=
log10 10
{∵ log10 10=1}
2.303 × 10−2
disappearance of X2 is
2
= 10 log10 10
(a) 8 mm min–1
(b) 2 mm min–1
–1
⇒
t = 100 sec.
(c) 16 mm min
(d) 4 mm min–1
[BITSAT – 2014] 196. A schematic plot of lnkeq versus inverse of
temperature for a reaction is shown below
1
Re action :
Ans. (a) : t = 0
t = 5 min,
∴ total pressure
X2 (g )
→
Y (g ) +
2
Z(g)
100
0
0
(100 − P )
P
P/2
= 100 − P + P + P / 2
120 = 100 + P / 2
P
= 120 − 100
2
P = 40 mm.
Rate of disappearance of X2 is−d [ X 2 ] 40
=
dt
5
Objective Chemistry Volume-II
The reaction must be
275
YCT
(a)
(b)
(c)
(d)
highly spontaneous at ordinary temperature
one with negligible enthalpy change
endothermic
exothermic
[AIEEE-2005]
Ans. (d) : Variation of keq with temperature t is given
by van't Hoff equation
−∆H 1 ∆So
× +
ln k eq =
R/A T R/R
Solpe of the given line is positive indicating that term A
is positive, thus ∆Ho is negative. Thus, reaction is
exothermic.
197. The rate of a chemical reaction doubles for
every 10°C rise of temperature. If the
temperature is raised by 50°C, the rate of the
reaction increases by about
(a) 10 times
(b) 24 times
(c) 32 times
(d) 64 times
[AIEEE-2011]
Ans. (c) : Rate of reaction is daubles for every 10oC rise
of temperature.
Rate of reaction becomes,
50
Rate = 2 10 = 25 = 32 times.
∴ Rate of reaction increased by 32 times.
198. Consider the reaction,
2A+B→product. When concentration of B
alone was doubled, the half-life did not change.
When the concentration of A alone was
doubled, the rate increased by two times. The
unit of rate constant for this reaction is
(a) L mol–1 s–1
(b) no unit
(c) mol L–1 s–1
(d) s–1
[AIEEE-2007]
Ans. (a) : Given:- When B alone was doubled, then t1/2
is constant. For 1st order reaction half-life doesn't
depend on concentration of reactants therefore, B follow
1st order kinetics.
When A alone was doubled, then rate increased by 2
times,
i.e. r' = 2 = k[2A]x
r = k[A]x
Then x = 1
Overall rate = k [A]x [B]y
= k[A] [B]
Overall it is 2nd order reaction, so rate constant k for 2nd
order reaction is Lmol–1 S–1.
199. For
the
non-stoichiometric
reaction,
2A+B→C+D, the following kinetic data were
obtained in three separate experiments, all at
298 K.
Initial rate
Initial
Initial
concentration concentration of formation
of C
[A]
[B]
(mol L–1 s–1)
0.1M
0.1M
1.2×10–3
0.1M
0.2M
1.2×10–3
0.2M
0.1M
2.4×10–3
The rate law for the formation of C is
Objective Chemistry Volume-II
dc
= k [ A ][ B]
dt
dc
2
(c)
= k [ A ][ B]
dt
(a)
dc
2
= k [ A ] [ B]
dt
dc
(d)
= k [A ]
dt
[JEE Mains-2014]
(b)
Ans. (d) : r = k[A]x [B]y
1.210–3 = k [0.1]x [0.1]y -------(i)
1.2×10–3 = k[0.1]x [0.2]y --------(ii)
2.410–3= k[0.2]x [0.1]y---------(iii)
Dividing (ii) by (i)
1 = 2y
y=0
Dividing (iii) by (i)
2 = 2x
x=1
∴ Overall rate = k[A]1 [B]0
= k[A]
200. It has been found that for a chemical reaction
with rise in temperature by 10oC the rate
constant is
(a) nearly doubled
(b) nearly tripled
(c) increases 5 times
(d) increases 4 times.
J & K CET-(2014)
Ans. (a) : For every 10oC rise in temperature rate
constant becomes nearly doubled because rate constant
depend upon temperature.
201. Consider the reaction
2N2O5(g) → 4NO2(g) + O2(g)
The rate law for this reaction is Rate = k [N2O5]
Which of the following statements is true
regarding the above reaction?
(a) Its order is 1 and molecularity is 1.
(b) Its order is 1 and molecularity is 2.
(c) Its order is 2 and molecularity is 2.
(d) Its order is 2 and molecularity is 1.
J & K CET-(2013)
Ans. (a) : For the given reaction
2N2O5(g)→4NO2(g)+O2(g)
Rate = k[N2O5]
The reaction occurs in two steps as:
slow
Step1: N 2 O5
→ NO 2 + NO3
fast
Step2: N 2 O5 + NO3 →
3NO 2 + O 2
The slow step is umimolecualr while fast step is
bimolecular. Therefore, the above reaction is
unimolecular and its order is first because slow step is
rate determining step.
202. In a reaction, 2A + B → A2B, the reactant B
will disappear at
(a) half the rate as A will decrease
(b) the same rate as A will decrease
(c) twice the rate as A will decrease
(d) half the rate as AB will form
J & K CET-(2011)
276
YCT
Ans. (a) : For the given reaction,
2A + B →A2B
−1 d [ A ] − d [ B ] d [ A 2 B ]
=
=
rate =
2 dt
dt
dt
−d[B] −1 d[A]
=
dt
2 dt
d[B] 1 d[A]
=
dt
2 dt
Hence, the rate of disappearance of B is equal to half
the rate of A will decrease.
203. Observe the formation reaction
2A +B→ C
The rate of formation of C is
2.2 ×10-3 mol L-1 min -1What is the value of
d[A]
−
(in mol L-1min-1) ?
dt
(a) 2.2×10-3
(b) 1.1×10-3
-3
(c) 4.4×10
(d) 5.5× 10-3
JIPMER-2014
d [ C]
= 2.2 ×10−3 mol L–1min–1.
dt
For reaction , 2A+B →C
−1 d [ A ] d [ C ]
=
2 dt
dt
−d [ A ] 2d [ C ]
=
= 2 × 2.2 ×10−3
dt
dt
= 4.4 × 10−3 molL−1 min −1
Ans. (c) : Given:-
4 k [ 0.2] [ 0.4]
=
2 k [ 0.2]x [ 0.2]y
x
y
21 = 2 y
y =1
On dividing eq (iii) by a (ii)
36 k [ 0.6] [ 0.4]
=
4 k [ 0.2]x [ 0.4]1
x
9 = 3x
1
⇒ ( 3 ) = 3x
2
⇒X=2
Hence, overall rate = k[A]2 [B]
206. The time required for 100% completion of a
zero order reaction is
a
(a) ak
(b)
2k
a
2k
(c)
(d)
k
a
Karnataka-CET-2011
Ans. (c) : Half-life of a zero order reaction =
a
2k
Time taken for completion = 2t1/2 = 2 ×
a
2k
a
k
207. In the following reaction, the initial
concentrations of the reactant and initial rates
at 298 K are given
2A → C + D
Initial rate in mol L–1 s–1
[A]0, mol L–1
0.01
5.0 × 10–5
0.02
2.0 ×10–4
The value of rate constant of this reaction at
298 K is
(a) 0.01 s–1
(b) 5 × 10–3 mol L–1s–1
–2
–1
(c) 2.0 × 10 mol Ls
(d) 5 × 10–1 mol–1 Ls–1
–1
–1 –1
(e) 5.0 × 10 mol L s
Kerala-CEE-2012
Ans. (d) : For given reaction , 2A→C + D
Rate = k[Ao]x
5×10–5 = k[0.01]x -------(i)
2×10–4 = k[0.02]x -------(ii)
Dividing eqn (ii) by (i)
4 = 2X
(2)2 = 2X
X=2
Rate = k[Ao]2
From eqn (i)
5×10–5 = k[0.01]2
5 × 10−5
k=
= 5 × 10−1 mol −1 LS−1
10−4
=
204. In the first order reaction, 75% of the reactant
gets disappeared in 1.386 hr. The rate constant
of the reaction is
(a) 3.0×10-3s-1
(b) 2.8×10-4s-1
-3 -1
(c) 17.2×10 s
(d) 1.8×10-3s-1
JIPMER-2012
Ans. (b) : For 1st order reaction
t75% = t3/4 = 2t1/2 = 1.386hr = 1.386×3600 sec
2 × 0.693
t3/4 =
k
2 × 0.693
k=
= 2.77 ×10−4 sec −1 ≈ 2.8 × 10−4 sec−1
1.386 × 3600
205. For a reaction, A + B → Products, the rate of
the reaction at various concentrations are given
below
Expt. No. [A]
[B]
Rate (mol dm–3s–1)
1.
0.2
0.2
2
2.
0.2
0.4
4
3.
0.6
0.4
36
The rate law for the above reaction is
(a) r = k [A] [B]2
(b) r = k [A]3 [B]
2
2
(c) r = k [A] [B]
(d) r = k [A]2 [B]
Karnataka-CET-2012
Ans. (d) : For a reaction, A+B→Product
Rate, r = [A]x [B]y
2 = k[0.2]x [0.2]y--------(i)
4 = k[0.2]x[0.4]y--------(ii)
Objective Chemistry Volume-II
36= k[0.6]x[0.4]y-------(iii)
On dividing eqn (ii) by eqn (i)
277
YCT
208. In a reaction, 2A+B → 3C, the concentration
of A decreases from 0.5 mol L-1 to 0.3 mol L-1 in
10 minutes. The rate of production of 'C'
during this period is
(a) 0.01 mol L-1 min-1 (b) 0.04 mol L-1 min-1
(c) 0.05 mol L-1 min-1 (d) 0.03 mol L-1 min-1
(e) 0.02 mol L-1 min-1
Kerala-CEE-2014
Ans. (d) : For reaction, 2A + B →3C
Given:- d[A] =0.3–0.5 = – 0.2 mol L–1
t = 10 min
d
C
1 [ ]
1 d [A]
=−
3 dt
2 dt
d [ C]
3 ( −0.2 )
=− ×
= 0.03mol L−1 min −1 .
dt
2
10
209. Ammonium (NH4+) reacts with nitrite ion
(NO −2 ) in aqueous solution according to the
equation
NH 4+ (aq) + NO −2 (aq)
→ N 2 (g) + 2H 2O( l )
The following initial rates of reaction have been
measured
for
the
given
reactant
concentrations.
Expt.
Rate
[NH 4+ ],(M) [NO4− ],(M)
No
(M/hr)
210. Rate law for the reaction, A+B → product is
rate = k[A]2[B]. What is the rate constant; if
rate of reaction at a given temperature is 0.22
Ms–1, when [A] = 1M and [B] = 0.25 M?
(a) 3.52 M–2 s–1
(b) 0.88 M–2 s–1
–2 –1
(d) 0.05 M–2 s–1
(c) 1.136 M s
MHT CET-2014
Ans. (b) : Given, Rate = 0.22 Ms–1 ,
k=?
[A] = 1M ,
[B] = 0.25M
Rate = k [A]2 [B]
0.22
k=
= 0.88M −2S−1
2
(1) ( 0.25)
211. Which is a correct integrated rate equation ?
2.303
a
log
(a) k = −
t
a−x
−2.303
a−x
(b) k =
log
t
a
(c) – d(a–x) = kdt
(d) All are integrated rate equations
MHT CET-2009
Ans. (b) : For first order reaction, integrated rate
equation.
2.303
a
k=
log10
t
a−x
(a − x )
−
2.303
0.020
1
0.010
0.020
OR
k=
log10
0.020
0.030
2
0.015
t
a
3
0.010
0.010
0.005
212. After how many seconds will the concentration
of the reactant in a first order reaction be
Which of the following is the rate law for this
halved if the rate constant is 1.155×10–3 s–1?
reaction?
(a) 600
(b) 100
(a) Rate = k [NH +4 ][NO 2– ]4
(c) 60
(d) 10
MHT CET-2008
(b) Rate = k [NH +4 ][NO 2– ]
Ans.
(a)
:
Half
life
time
for
first
order
reaction
+
– 2
(c) Rate = k [NH 4 ][NO 2 ]
0.693
0.693
t1 =
=
(d) Rate = k [NH +4 ]2 [NO 2– ]
k
1.155 × 10−3
2
1
1
t 1 = 600sec
(e) Rate = k [NH +4 ]2 [NO 2– ]4
2
Kerala-CEE-2014
213.
The
rate of the reaction,
x
4
Ans. (c) : For given reaction, rate = k NH +4 NO −2
2NO + Cl 2 → 2NOCl is given by the rate
equation, rate = k[NO]2[Cl2]. The value of the
0.02 = k [0.01]x [0.02]y ----------(i)
x
y
rate constant can be increased by
0.03 = k[0.015] [0.02] ------(ii)
x
y
(a) increasing the temperature
0.005 =k[0.01] [0.01] -------(iii)
(b) increasing the concentration of NO
Dividing eqn (i) by eqn (iii)
(c) increasing the concentration off the Cl2
4 = 2y
⇒22 = 2y
(d) doing all of these
y=2
AIPMT -Mains 2010
Dividing eqn (ii) by eqn (i)
Ans. (a) : The value of rate constant depends on the
x
y
0.03 k [ 0.015] [ 0.02]
temperature. According to Arrhenius equation.
=
0.02 k [ 0.01]x [ 0.02]y
k = Ae − Ea / RT
On increasing the temperature of reaction the
3
1
x
= (1.5) = (1.5 )
value of rate constant k also increase.
2
214. The reaction of hydrogen and iodine
X=1
monochloride is given as:
2
Overall rate = k NH +4 NO −2
H 2( g ) + 2ICl g → 2HCl ( g ) + I 2( g )
Objective Chemistry Volume-II
278
YCT
This reaction is of first order with respect to Ans. (b) : For the given reaction
H 2( g ) and ICl ( g ) , following mechanisms were
2N2O5
→ 4NO + O
2
Given,
− d [ N 2 O5 ]
Mechanism B:
H 2( g ) + ICl ( g ) → HCl ( g ) ; slow
dt
d [ NO 2 ]
HI ( g ) + ICl ( g ) → HCl ( g ) + I 2( g ) ;fast
Which of the above mechanism(s) can be
consistent with the given information about the
reaction?
(a) A and B both
(b) Neither A nor B
(c) A only
(d) B only
(AIPMT -2007)
Ans. (d) : It is given, that, this reaction is of first order
w.r.t. H2 and ICl
Rate = k [H2] [ICl]
For mechanism A, H2(g) + 2ICl(g) →2HCl(g) +I2(g)
Rate = k[H2] [ICl]2
For mechanism B, Rate depends on slowest step which
is rate determining step.
H2(g) + ICl(g)
→ HCl( g ) ;
slow
Rate = k [H2] [ICl]
Hence, mechanism B consistent with the given
information.
215. For a reaction between A and B the order with
respect to A is 2 and order with respect to B is
3. The concentrations of both A and B are
doubled, the rate will increase by a factor of
(a) 12
(b) 16
(c) 32
(d) 10
Karnataka NEET-2013
Ans. (c) : Rate = k [A]2[B]3 = r
If concentration of A and B are doubled,
dt
d [O2 ]
dt
= k ' [ N 2 O5 ]
= k "[ N 2 O 5 ]
1
k [ N 2 O 2 ] = k "[ N 2 O 5 ]
2
k = 2k "
k" =
k
2
1
217. For the reaction N 2 O 5( g ) → 2NO 2( g ) + O 2( g ) the
2
value of rate of disappearance of N 2O5 is given
3
r ' = 32 k [ A ] [ B]
2
= k [ N 2 O5 ]
1
1
k [ N 2 O5 ] = k ' [ N 2 O 5 ]
2
4
k ' = 2k
r ' = k [ 2A ] [ 2B]
2
2
1 d [ N 2 O5 ] 1 d [ NO 2 ] d [ O 2 ]
=
=
–
2
dt
4 dt
dt
proposed.
Mechanism A:
H 2( g ) + 2ICl g → 2HCl ( g ) + I 2( g )
3
as 6.25 ×10 –3 mol L–1s –1 . The rate of formation
of NO 2 and O 2 is given respectively as:
(a) 6.25 × 10 −3 mol L–1s −1 and
6.25 × 10 −3 mol L−1s −1
(b) 1.25 × 10−2 mol L–1s −1 and
3.125 × 10 −3 mol L–1s −1
(c) 6.25 × 10 −3 mol L–1s −1 and
3.125 × 10 −3 mol L–1s −1
(d) 1.25 × 10−2 mol L–1s −1 and
6.25 × 10 −3 mol L–1s −1
AIPMT -2010
r ' = 32 r
Ans. (b) : Given:Rate increased by a factor of 32.
− d [ N 2 O5 ]
= 6.25 ×10 –3 mol L–1S–1
216. The rate of the reaction : 2N 2O 5 → 4NO 2 + O 2
dt
can be written in three ways.
1
For reaction, N2O5(g)→2NO2(g)+ O 2 (g)
–d [ N 2 O 5 ]
2
= k [N 2O5 ]
dt
−d [ N 2O5 ] 1 d [ NO 2 ]
d [O2 ]
=
=2
d [ NO 2 ]
d [O 2 ]
dt
2 dt
dt
= k ' [ N 2O5 ];
= k"[ N 2 O5 ]
d [ NO 2 ]
d [ N 2 O5 ]
dt
dt
= –2
= 1.25 ×10 –2 mol L–1S –1
The relationship between k and k' and between
dt
dt
k and k" are
d[O 2 ]
1 d [ N 2 O5 ] 6.25 × 10−3
(a) k ' = 2k, k " = k
(b) k ' = 2k, k " = k / 2
=−
=
dt
2
dt
2
(c) k ' = 2k, k " = 2k
(d) k ' = k, k " = k
-3
-1 -1
= 3.125 ×10 molL S
AIPMT-Mains-2011
Objective Chemistry Volume-II
279
YCT
Ans. (165) : Given half life period t1/2 = 70 minutes
= 70×60 sec
= 4200 sec
For first order reaction
0.693
218. For a first order reaction A → products, initial
k=
concentration of A is 0.1 M, which becomes
t1/ 2
0.001 M after 5 minutes. Rate constant for the
0.693
reaction in min−1 is
=
4200
(a) 0.2303
(b) 1.3818
= 165 ×10–6 S–1
(c) 0.9212
(d) 0.4606
NEET-17.06.2022 221. For the given first order reaction
A→B
Ans. (c) :
The half life of the reaction is 0.3010 min. The
A → products
ratio of the initial concentration of reactant to
Initial conc. Ao = 0.1 M
the concentration of reactant at time 2.0 min
Conc. After 5 min At = 0.001 M
will be equal to _______, (Nearest Integer)
t = 5 min
JEE Main-28.07.2022, Shift-I
Ao
2.303
Ans.
(100)
:
Given
that,
t
1/2 = 0.3010 min
k=
log
t
A
A
→
B
t
2.303
0.1
A
0.693 2.303
=
log
Since,
=
log 0
5
0.001
t1/ 2
2
At
k = 0.9212 min−1
A
0.693 2.303
219. The reaction between X and Y first order with or
=
log 0
0.3010
2
At
respect to X and zero order with respect to Y.
A
Initial rate
[ X]
[Y]
log 0 = 2
Experiment
-1
-1
-1
-1
At
mol L min
mol L
mol L
3.
Order of Reaction,
Molecularity
A0
I.
0.1
0.1
2×10−3
or
= 100
−3
At
II.
L
0.2
4×10
III.
0.4
0.4
M×10−3
222. For a reaction A → 2B + C the half lives are
100s and 50 s when the concentration of
IV.
0.1
0.2
2×10−3
reactant A is 0.5 and 1.0 mol L–1 respectively.
Examine the data of table and calculate ratio of
The order of the reaction is ____ (Nearest
numerical values of M and L. (Nearest Integer)
Integer)
JEE Main 29.07.2022, Shift-I
JEE Main 26.07.2022, Shift-I
Ans. (40) : The rate law expression
Ans.
(2)
:
Given,
Rate = k[X] [Y]0
Reaction A + 2B → C
Rate = k [X]
1
Using I and II, we gett1/ 2 ∝
n −1
−3
[A
0]
4 × 10
L
=
1
2 × 10−3 0.1
[100] ∝
or
L = 0.2
(0.5)n −1
Using I and III, we get1
(50) ∝ n −1
M × 10−3 0.4
(1)
=
0.1
2 × 10−3
n −1
1
or
M=8
[2] ∝
0.5
M
8
n–1
=
= 40
[2]
=
[2]
L 0.2
n–1 = 1
220. [A]
[B]
→
n =2
Reactant
Product
The order of the reaction is 2
If formation of compound [B] follows the first
order of kinetics and after 70 minutes the 223. A first order reaction is half completed in 45
min. How long does it need 99.9% of the
concentration of [A] was found to be half its
reaction to be completed?
initial concentration. Then the rate constant of
(a) 10 Hours
(b) 20 Hours
the reaction is x × 10–6 s–1. The value of x is –––
(c) 5 Hours
(d) 7.5 Hours
–––. (Nearest Integer)
Karnataka
CET-17.06.2022,
Shift-II
JEE Main 27.07.2022, Shift-II
Objective Chemistry Volume-II
280
YCT
Putting the value in the rate constant equation
0.10 − 0.075
0.0030 =
Time
Time = 8.33 sec
227. Half life periods for a reaction at initial
concentrations of 0.1M and 0.01 M are 5 and
50 minutes, respectively. The order of reaction
is
(a) 3
(b) 2
(c) 1
(d) 0
TS-EAMCET-18.07.2022, Shift-II
Ans. (b) : Given,
CH3COOC2 H5 + NaOH → CH3COONa + C2 H5OH is We know the relation –
given by the equation,
( t1 )12 a 2 n−1
Rate = k [ CH 3 COOC2 H 5 ][ NaOH ] .
=
( t 2 )12 a1
−1
If concentration is expressed in mol L , the
Where, t1, t2 are half life period and a1 and a2 are initial
unit of K is
and final concentration of 0.1 M and 0.01 M
−1 −1
−1
(a) L mol s
(b) s
respectively.
n = order of reaction.
(c) mol−2 L2s −1
(d) molL−1s −1
Karnataka CET-17.06.2022, Shift-II Given, ( t1 )1 = 5min , a1 = 0.1M
0.693
(45 min)
2.303
a
2.303
100
t=
log
=
log
k
(a − x)
k
0.1
2.303 × (45min) × 3
=
= 7.5
(0.693)
448.6 min
= 448.6 min =
60
= 7.5 hr
224. The rate of the reaction
Ans. (d) : k =
2
( t 2 ) = 50 min , a 2 = 0.01M
Ans. (a) : For a second
dx
2
= k [A]
dt
conc
2
= k [ conc]
time
mol L−1
= k mol L−1 × mol L−1
s
k = L mol−1s −1
1
n−1
1 0.01
=
10 0.1
n−1
225. Which of the following is a zero order reaction?
(a) 2HI
→ H2 + I 2
→ 2HBr
(b) H2 + Br2
∆
(c) 2N2O5
→ 4NO2 + O2
hν
→ 2HCl
(d) H2 + Cl2
TS-EAMCET-19.07.2022, Shift-II
hν
Ans. (d) : H 2 + Cl2
→ 2HCl
It is an example of zero order reaction because the
rate of the reaction is proportional to zero power of the
concentration of reactants.
226. The rate constant for a zero order reaction A
→ products is 0.0030 mol L–1s–1. How long it
will take for the initial concentration of A to
fall from 0.10 M to 0.075M?
(a) 10 s
(b) 20 s
(c) 8.33 s
(d) 1.33 s
AP-EAMCET-04.07.2022, Shift-I
Ans. (c) : For a zero order reaction,
Intialconcentration − Finalconcentration
Rate constant =
Time
Given that, Rate constant = 0.0030 mol L–1 s–1
Initial conc.[A0] = 0.10M
Final conc. [A] = 0.75 M
Time = ?
Objective Chemistry Volume-II
2
1 1
=
10 10
n–1=1
n=2
Hence, reactions are the order of 2.
228. A first order reaction goes to 90% completion
in 10minutes. The rate constant of the reaction
is.
(a) 0.2303 min–1
(b) 2.303 min–1
–1
(c) 0.02303 min
(d) 22.30 min–1
Assam-CEE-31.07.2022
Ans. (a) : For first order reaction, formula for rate
constant,
2.303
a
k=
log
t
a−x
2.303
100
k=
log
10
10
∴
k = 0.2303 min–1
229. A reaction X → Y follows 2nd order Kinetics,
doubling the concentration of X will increase
the rate of formation of Y by a factor of.
1
(a) 2
(b)
2
1
(c) 4
(d)
4
Assam-CEE-31.07.2022
Ans. (c) : For second order X → Y
Let reaction rate = r1
r1= k[A]2
……..(1)
281
YCT
When concentration of X is doubled then rate becomes
r2
∴ r2 = k [2A]2 = 4k[A]2 …….(2)
Dividing equation 2 by 1,
r1
= 4 or r2 = 4r1
r2
Hence, rate of formation of B increase by factor of 4
when concentration of A is doubled.
230. In a first order reaction, the reactant
decomposes 25% of its initial concentration at
40 minutes. What is the value of rate constant
of the reaction?
(log 3 = 0.4771, log 4 = 0.6021)
(a) 7.19×10–3 min–1
(b) 2.19×10–3 min–1
–3
–1
(d) 1.19×10–3 min–1
(c) 5.19×10 min
AP-EAPCET-12.07.2022, Shift-II
Ans. (a) : The Rate constant k for a First order reaction
is given–
[R ]
2.303
k=
log 0
t
[R]
Where,
[R0] = Initial concentration of reactant
[R]= Final concentration of reactant.
t = 40 min
2.303
100
k=
log
40
100 − 25
2.303
4 2.303
=
log =
(log 4 − log 3)
40
3
40
2.303
k=
(0.6021 − 0.4771)
40
2.303
=
× .125 = 7.19 × 10−3
40
231. The correct relation to find out the half life a
first order reaction is (k = rate constant of
reaction, [A]o = Initial concentration of
reactant)
[ A ]o
ln2
(a)
(b)
k
2k
1
1
(c)
(d)
k
k [ A ]o
k=
2.303
log 2
t 12
k=
ln 2
t 12
ln 2
k
232. In the reaction, A→products, If the
concentration of the reactant is doubled, rate of
the reaction remains unchanged. The order of
the reaction with respect to A is–
(a) 1
(b) 2
(c) 0.5
(d) 0
AP-EAMCET-08.07.2022, Shift-I
Ans. (d) : For zero order reaction,
Rate = k [A0]
When concentration [A] is doubled,
Rate = k[2A0]
Rate is unchanged, hence it is zero order reaction.
233. The unit of rate constant of a second order
reactions
(a) mol/L-s
(b) L/mol-s
(c) L2/mol2-s
(d) per second
[BITSAT – 2015]
CG PET -2009
J & K CET-(2008, 2007)
CG PET -2006
Ans. (b) : Unit of rate constant k = (mol)1–n (L) n–1s–1
Where ,
n= order of reaction
For second order reaction, n = 2
Unit of rate constant = (mol) 1–2 (L) 2–1 s–1
= L mol–1 s–1
234. The rate of reaction between two reactants A
and B decreases by a factor of 4 if the
concentration of reactant B is doubled. The
order of this reaction with respect to reactant B
is:
(a) 2
(b) –2
(c) 1
(d) –1
[BITSAT – 2017], (AIPMT -2005)
JIPMER-2007, CG PET -2007
AP-EAPCET-12.07.2022, Shift-I Ans. (b) : Rate, r = k [A]x[B]y
………(i)
If concentration of reaction B is doubled i.e. 2B then
Ans. (a) : For a first order reaction –
A→B
r
k[A]x[2B]y =
………(ii)
Rate = k[A]
4
on dividing eq. no. (ii) by eq. no. (i)
[A ]
2.303
k=
log 0
r
t
A
[ ]
x
y
4 = k [ A ] [ 2B]
Where, k = rate constant
x
y
r
A0 = initial concentration
k [ A ] [ B]
A = concentration at time t
1
= 2y
t = time
4
When, t = t1/2, then A = A0/2
−2
( 2 ) = 2y
[ A 0 ]× 2
2.303
k=
log
y = −2
t1
[A0 ]
t 12 =
2
Objective Chemistry Volume-II
282
YCT
235. Unit of the constant of zero order reaction is :
(a) time–1
(b) mol L–1 s –1
–1 –1
(c) L mol s
(d) L mol–1s–1
J & K CET - 2014, 2011
BCECE - 2004
1–n
Ans. (b) : Unit of rate constant = (mol) (L)n-1 sec-1
Where,
n= order of reaction.
For zero order reaction, (n=0)
Unit of rate constant = (mol)1−0 (L)0−1 s−1
= mol L−1 s−1
236. 75% of a first order reaction is completed in 32
min. 50% of the reaction would have been
completed in
(a) 24 min
(b) 16 min
(c) 18 min
(d) 23 min
Tripura JEE-2021
CG PET- 2015
(AP-EAMCET-1998)
Ans. (b) :
For first order reaction–
2.303
a
k=
log
t
a−x
If 75% completed in 32 min then–
2.303
100
rate constant ( k ) =
log
32
(100 − 75 )
= 0.0433
For 50% reaction completed then time
2.303
100
t=
log
0.0433
(100 − 50 )
Reason: For a zero order reaction, the rate of
reaction is independent of initial concentration.
(a) If both Assertion and Reason are correct and
Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are
incorrect.
AIIMS-26.05.2019 (Morning)
AIIMS-26.05.2018 (Evening)
Ans. (b): For zero order reaction
Rate = k [ A o ]
Rate of reaction is independent of initial concentration.
Half - life time for zero order reaction.
1 [Ao ]
t1 =
2
2 k
t 1 ∝ [Ao ]
o
2
Hence, on doubling the concentration of reactant, halflife time is also doubled.
So, both assertion and reason is correct but reason is not
correct explanation of assertion.
239. The following data are for the decomposition of
ammonium nitrite in aqueous solution.
Vol. of N2 in cc
Time (min)
6.25
10
9.00
15
11.40
20
2.303
13.65
25
t=
log 2
35.65
Infinite
0.0433
t = 16 min .
The order of reaction is
(a) zero
(b) one
237. For first order reaction, the value of slope for
(c)
two
(d)
three
[ R ]0
graph of log
→ t is –––––.
BCECE-2008, BITSAT – 2010
[R]
Ans. (b) : Let check the value of k for first order
k
2.303
reactions.
(a)
(b)
2.303
k
Vo
2.303
(i)
k=
log10
k
t
( Vo − Vt )
(c) –k
(d) −
2.303
2.303
35.65
GUJCET-2021, 2022
k=
log10
10
35.65
− 6.25)
(
Ans. (a) : We know that first order reaction is given by.
2.303
a
2.303
35.65
k=
log
k=
log10
t
a−x
10
( 29.4 )
Let initial concentration (a) = [R]0
k = 1.9 ×10−2 min −1
Final concentration (a – x) = [R]
2.303
35.65
(ii)
k=
log10
[ R ]0
k
20
( 35.65 − 11.4 )
So,
t = log
[R ]
2.303
2.303
35.65
k=
log10
k
20
∴
slope (m) =
(24.25)
2.303
k = 1.9 × 10 −2 min −1
238. Assertion: If in a zero order reaction, the
concentration of the reactant is doubled, the half- Value of k is constant then reaction follows first order
kinetics.
life period is also doubled.
Objective Chemistry Volume-II
283
YCT
240. The reaction A → B follows first order
kinetics. The time taken for 0.8 mole of A to
produce 0.6 mole of B is 1 h. What is the time
taken for conversion of 0.9 mole of A to
produce 0.675 mole of B?
(a) 1 h
(b) 0.5 h
(c) 0.25 h
(d) 2h
CG PET -2007
(AIPMT -2003)
Ans. (a) : Given :- t = 1hr when a=0.8 mol, (a–x)=0.2
mol
If a=0.9mol, (a–x) = 0.225mol then t = ?
For first order reaction
k=
2.303
a
log10
t
(a − x )
2.303
0.8
log
1
0.2
k = 2.303 2log10 2
(a)
3
2
(c) 0
1
2
4
(d)
2
(b)
Manipal-2018
JCECE – 2009
Ans. (b) : Order of reaction is calculated from the sum
of power of concentration of reactants in rate law.
Rate = k[A]3/2 [B]–1
3
1
Order of reaction = − 1 =
2
2
244. A hypothetical reaction
X2 + Y2→ 2XY follows the following
mechanism
X 2 ↽ ⇀ X + X.....fast
k=
X + Y → XY + Y….. slow
X + Y → XY ….. fast
The order of the overall reaction is
(a) 2
(b) 3/2
2.303
0.9
t=
log
(c)
1
(d) 0
2.303 × 2log10 2
0.225
NEET - 2017
1
JIPMER-2015
× log10 4
t=
2log10 2
Ans. (b) : Rate is determined from slowest step of
mechanism.
1
× 2 log10 2
t=
for the given mechanism,
2log10 2
Rate = k [X] [Y2]
……(i)
t = 1 hr
But X and Y are intermediates, then from
step −1
step − 2
241. X
→ Y
→ Z is complex reaction.
X2 ↽ ⇀ X + X
Total order of reaction is 2 and step-2 is slow
[X]2
step. What is molecularity of Step-2?
keq =
(a) 1
(b) 2
[X 2 ]
(c) 3
(d) 4
1/ 2
X = ( k eq [X 2 ])
……(ii)
GUJCET-2015, 2016
Ans. (b) : Molecularity of slowest step in complex Put the value of X on equation (i)
Rate = k. (keq[X2])1/2. [Y2]
reaction is equal to order of reaction. If step-II is
slowest step then order of reaction is equal to
= k1 [X2] 1/2 [Y2]
molecularity of reaction therefore, molecularity of step1 3
Order of reaction = 1 + =
II is 2.
2 2
242. The value of rate constant of a pseudo first
245.
Higher
order
(>3)
reactions are rate due to
order reaction:
(a)
shifting
of
equilibrium
towards reactants due
(a) Depends on the concentration of reactants
to
elastic
collisions
present in the small amount
(b) loss of active species on collision
(b) Depends on the concentration of reactants
(c) low probability of simultaneous collision of
present in excess
all reacting species
(c) Is independent of concentration of reactants
(d)
increase
in entropy as more molecules are
(d) Depends only on temperature
involved
HP CET-2018
Karnataka-CET-2021
Karnataka CET-2018
Kerala-CEE-29.08.2021
Ans. (a) : In pseudo order reaction rate does not
Ans. (c) : Molecularity and order > 3 is not possible
depends on reactants whose concentration is present in
because of Slow probability of simultaneous collision of
excess.
all the reacting species.
Hence, for pseudo first order reaction rate depends on
Note:- If a chemical reaction contains one or two
the concentration of present in the small amount.
reactants, then probability of collision between them to
243. What is the order of a reaction which has a rate produce the product is very high because the orientation
expression
of the reactants also suits well till the reaction contains
reactants up to two.
rate = k[A]3/2[B]-1?
Objective Chemistry Volume-II
284
YCT
246. If the rate constant for a first order reaction is
k, the time (t) required for the completion of
99% of the reaction is given by
4.606
2.303
(E)
(a) t =
(b) t =
k
k
0.693
6.909
© t=
(d) t =
k
k
Choose from the options given below, the
Karnataka-CET-2021
correct one regarding order of reaction is
Kerala-CEE-29.08.2021
(a) (B) zero order (C) and (E) first order
Ans. (a) : For 90% completion,
(b)
(A) and (B) zero order (E) first order
(a – x) = 1%
st
(c)
(B) and (D) zero order (E) first order
According to 1 order reaction,
(d) (A) and (B) zero order (C) and (E) first order
2.303
a
[JEE Main 2021, 25 July Shift-I]
t=
log10
k
(a − x)
Ans. (b)
(A) Rate doesn’t depends on initial concentration of
2.303
100
t99%
=
log10
reactants then it follows zero order reaction
k
1
(B) For zero order reaction half-life time is directly
4.606
proportional to the initial concentration.
t=
k
a
t1 =
247. A first-order reaction has k = 5.48 × 10–14 sec–1.
2
2k
Determine its two-third life (t2/3).
(C) For first order reaction concentration varies
(a) 2.005 × 1013 sec
(b) 5.002 × 1014 sec
exponential with time.
(c) 2.005 × 1014 sec
(d) 5.002 × 1013 sec
A t = A 0 e− kt
Tripura JEE-2022 (D) This graph is not possible for any order of the
Ans. (a): Given that,
reactions.
−14
−1
(E)
For
first order reaction, rate is directly proportional
k = 5.48×10 sec
to the concentration of reactants
We know that,
Rate = k [ A ]
[A0 ]
2.303
tn =
log
249. For a reaction of order n, the unit of the rate
k
[A ]
constant is
(a) mol1–n L1–n s
(b) mol1–n L2n s–1
Let,
[A0 ] = a
1–n n–1 –1
(c) mol L s
(d) mol1–n L1–n s–1
2
a
[JEE Main 2021, 27 July Shift-I]
[A] = a − a =
3
3
Ans. (c) : unit of rate constant = (mol L−1)1−n s−1
Therefore,
= (mol)1−n Ln−1 s−1
2.303
a
250. For a first order reaction, the ratio of the time
t2/3 =
log
= 2.005 × 1013 sec
5.48 × 10−14
a
for 75% completion of a reaction to the time
for 50% completion is ................
3
(Integer answer)
248. For the following graphs
[JEE Main 2021, 31 Aug Shift-I]
Ans. (2) : For first order reaction
At
t 75% , a = 100%,
a − x = 25%
(A)
(B)
t 75% =
t 75% =
At
(C)
100 2.303
2.303
log10
× 2× log 2
=
25
k
k
a = 100% a − x = 50%
2.303
100
log10
k
50
2.303
t 50% =
log10 2
k
t 50% = 2× t 75%
= 1:2
t 50% =
(D)
∴
Objective Chemistry Volume-II
t 50%
2.303
a
log10
k
a
−
( x)
285
YCT
251. The first order rate constant for the
decomposition of CaCO3 at 700 K is 6.36 × 10–3
s–1 and activation energy is 209 kJ mol–1. Its
rate constant (in s–1) at 600 K is x × 10–6. The
value of x is ......... .
(Nearest integer)
[Given, R = 8.31 JK–1 mol–1, log 6.36 × 10–3 = –
2.19, 10–4.79= 1.62×10–5]
[JEE Main 2021, 27 Aug Shift-II]
Ans. (16) : According to Arrhenius theory,
K = Ae–Ea/RT
Ea
In K = InA –
RT
At temperature T1,
Ea
In K1 = In A –
……..(i)
RT1
At temperature T2,
Ea
In K2 = In A –
…….. (ii)
RT2
Subtracting Equation (ii) and (i)
Ea 1
1
In K1 − InK 2 =
−
R T2 T1
K
Ea 1
1
In 1 =
−
K2
R T2 T1
At,
K
Ea 1
1
2.303log 1 =
−
K2
R T2 T1
K
Ea 1
1
−
log 1 =
K 2 2.303 T2 T1
T1 = 700K
K1 = 6.36×10–2 s–1
Ea = 209 kJ mol–1
R = 8.314 Jk–1 mol–1
T2 = 600K
K2 = x × 10–6 s–1
Objective Chemistry Volume-II
Ans. (d): A x × By = r
For entry (1) and (2) we get,
2 × 10−2
0.02 0.02
×
=
0.02 0.04
2 × 10−2
x
x
y
y
1 1
× =1
1 2
From entry (2) and (3) we get,
2 × 10−2
0.02 0.04
×
=
0.04 0.04
8 ×10−2
x
x
y
y
1
1 1
× =
4
2 1
So, the rate of reaction –
From entry (rate) (1),
r = k(A) '(B) '
k=
r
2 × 10−2
=
(A) '(B) ' (0.02)(.02)
2 × 10−2
1
100
× 104 = × 102 =
= 50
4
2
2
254. What will be the overall order of a reaction for
which that rate expression is given as
Rate = k[ A]1 / 2 [B]3 / 2
(a) second order
(b) first order
(c) zero order
(d) third order
TS EAMCET 04.08.2021, Shift-I
Ans. (a) : Rate = k[A]1/2 [B]3/2
1 3
Hence, the overall order of reaction = +
2 2
4
= =2
2
The order of reaction is second order reaction.
255. Gaseous cyclobutene isomerizes to butadiene in
a first order process which has 'k' value of
3.3×10–4 s–1 at 153°C. The time in minutes it
takes for the isomerization to proceed 40% to
completion at this temperature is______.
(Rounded off to the nearest integer)
JEE Main-24.02.2021, Shift-I
=
1
6.36×10−3
209
1
=
−
−6
x ×10
2.303×8.314 600 700
100
209
=
2.303×8.314 420000
x = 16
252. Higher order (>3) reactions are rare due to
(a) Shifting of equilibrium towards reactants due
to elastic collisions
(b) Loss of active species on collision
(c) Low probability of simultaneous collision of
all reacting species
(d) Increase in entropy as more molecules are
involved
Kerala-CEE-29.08.2021
Ans. (c) : For higher order (>3) reaction to occur, 3 or
more molecules (having energy equal to or greater than
activation energy) must simultaneously collide with
proper orientation. The probability for such collision is
very low. Hence, reaction are rare.
log
253. For a reaction A + B → P, the following data
are provided
Initial rate
Entry
[A] in M
[B] in M
(M/s)
1
0.02
0.02
2×10–2
2
0.02
0.04
2×10–2
3
0.04
0.04
8×10–2
The rate constant for this reaction in standard
unit is
(a) 5
(b) 1.2
(c) 2.4 × 10–4
(d) 50
TS-EAMCET (Engg.), 06.08.2021
286
YCT
Ans. : (26 min) The isomers of cyclobutene is :
| → CH2=CH–CH=CH2
Cyclobutene
Given data,
kt = ln
[A o ]
[ A ]t
k = 3.3×10–4 sec–1
for first order kinetics–
2.303
a
k=
log
t
a–x
2.303
100
or
t=
log
–4
3.3 × 10
60
t = 1547.956 sec
t = 25.799 min
t = 26 min
256. In a 1st order reaction, reactant concentration
C varies with time t as:
1
(a)
increases linearly with t
C
(b) log C decreases linearly with t
1
(c) C decreases with
t
1
(d) log C decreases with
t
[BITSAT – 2021]
Ans. (b) :
For first order reaction,
[C]=[Co]e–Kt
Taking log on both side
log [C] = log [Co] – kt
For straight line graph, y = mx + c
When y = log [C], m = slope = –k, x = t
If graph is plot in log c v/s t. slope decrease linearly
257. Which statement among the following is
incorrect?
(a) Unit of rate of disappearance is M s–1
(b) Unit of rate of reaction is M s–1
(c) Unit of rate constant k depends upon order of
reaction
(d) Unit of rate constant k for a first order
reaction is M s–1
AP EAPCET 20.08.2021 Shift-I
Ans. (d) : Unit of rate constant for first order reaction is
sec–1 so, the option d is correct.
258. For zero order reaction, a plot of t1/2 versus
[A]0 will be ––
(a) A straight line passing through the origin and
slope = k
(b) A horizontal line (parallel to x – axis)
Objective Chemistry Volume-II
(c) A straight line with slope –k
(d) A straight line passing through origin and
1
slope =
2k
AP EAPCET 20.08.2021 Shift-I
Ans. (d) : For zero order reactionConcentration
k=
time
Since, half-life time
[A]0
Concentration =
2
Time = t1/2
So, t 1 2 =
[A]0
2k
i.e,
straight line passing through origin and slope = 1/2 k
259. If the rate constant for a first order reaction is
2.303 × 10–3 s–1 find the time required to reduce
4 g of the reactant to 0.2g.
(a) 1.30 hours
(b) 21.60 hours
(c) 0.36 hours
(d) 2.60 hours
AP EAPCET 19-08-2021 Shift-I
Ans. (c) : Given that:k = 2.303 × 10 –3 s –1
t=?
a = 4g
a – x = 0.2g
From the equation of first order kinetics2.303
a
k=
log
t
a–x
2.303
a
or
t=
log
k
a–x
2.303
4
or t =
log
–3
–1
2.303 × 10 (s )
0.2
t = 1000 (s) × log 20
t = 1000 (s) × 1.3010
t = 1301.05
1301.0
t=
hours
3600
t = 0.36 hours
260. The rate constant for a first order reaction is
60 s–1. How much second will it take to reduce
the initial concentration of the reactant to its
1
th value?
16
(a) 2.3 × 10–2
(b) 9.5 × 10–2
–2
(c) 4.6 × 10
(d) 6.9 × 10–2
GUJCET-2021
287
YCT
Ans. (c) : Given that :k = 60 sec–1
a – x = 1/16
a=1
For first order reaction 2.303
a
t=
log
k
a−x
2.303
1
t=
log
60
1/16
2.303
× 1.204 = 0.04621
t=
60
∴ t = 4.6 × 10–2 sec.
261. x2(g)+y2(g)→2xy(g) the
We get,
2.303
B
log
t
A
B
kt = 2.303 log
A
kt = 2.303 [log B – log A]
k=
So the decomposition of A and the group show the
data formation of B.
1
indicates that rate = k [ x 2 ][y 2 ] 2 .The Since time increase reduce the A.
263. It is true that
molecularity and the order of reaction,
(a) a second order reaction is always a multistep
respectively, are
reaction
1
3
(b)
a
zero order reaction is a multistep reaction
(a) 1 and
(b) 2 and
(c) a first order reaction is always a single step
2
2
reaction
3
2
(d)
a
zero order reaction is a single step reaction
(c) 2 and 1
(d)
and
[JEE Main 2020, 3 Sep Shift-I]
2
3
TS EAMCET 10.08.2021, Shift-II Ans. (b) : Zero order reactions are multi step reaction
because it is complex reactions.
Ans. (b) :
264. If 75% of a first order reaction was completed
x2 (g) + y2 (g) → 2xy(g)
in 90 minutes, 60% of the same reaction would
1/
be completed in approximately (in minutes)
Rate = k[ x2 ][ y2 ] 2
.......... .
(Take : log2 = 0.30; log2.5 = 0.40)
1 3
order of reaction =1+ =
[JEE Main 2020, 4 Sep Shift-I]
2 2
Ans. (60) : Given:- t 75% = 90 Minute
Molecularity = No. of moleof x2 + No. of moleof y2
For 1st order reaction
=1+1
2× 0.693
t 75% = 2× t 1 =
=2
2
k
262. If the decomposition reaction A(g) → B(g) follows
0.693
×
2
first order kinetics, then the graph of rate of
k=
90
formation of B, denoted by R, against time 't'
For 60% completion, (a–x) = 40
will be –––––––
2.303
a
t=
log10
k
(a − x )
experimental
(a)
(b)
(c)
(d)
2.303
100
= log10
0.693× 2
40
90
2.303×90
=
log10 (10) − log10 (4)
0.693× 2
t = 60 min
=
265. A flask contains a mixture of compounds A and
B. Both compounds decompose by first-order
kinetics. The half-life for A and B are 300 s and
180 s, respectively. If the concentrations A and
AP-EAPCET 25.08.2021, Shift-II
B are equal initially, the time required for the
concentration of A to be four times that of B (in
Ans. (c) : The first order reaction –
s) is (Use In 2 = 0.693)
2.303
a
(a) 120
(b) 180
k=
log
t
a−x
(c) 300
(d) 900
Decomposition of A to formation of B,
[JEE Main 2020, 5 Sep Shift-I]
Objective Chemistry Volume-II
288
YCT
( ) = 300 s
Half-life time of B = ( t ) = 180 s
Ans. (d) : Half-life time of A = t 1
1
2 A
2 B
Initially, concentration of A = concentration of B = x
Final concentration of A = 4× final concentration of B
Let
Final concentration of A = 4y
Final concentration of B = y
For
First order reaction,
2.303
a
t=
log10
k
(a − x )
kA =
0.693 −1
s
300
kB =
0.693 −1
s
180
Ans. (d) : Given, t60% = 50 min
Then a = 100%
(a – x) = 40%
for first order reaction,
2.303
a
t=
log10
k
(a − x)
2.303
100
t60% =
log10
= 50 min
k
40
2.303
10
k=
log10
……(i)
50
4
For t93.6%
(a – x) = 6.4%
2.303
100
t93.6% =
log10
k
6.4
Put the value of k from equation (i)
x
2.303×300
log10
4y
0.693
t
=
x
t
2.303×180
log10
y
0.693
x = 8y
t93.6% =
2.303
100
log10
2.303
10
64
log10
50
4
3
50
10
log10
10
4
log10
4
2.303×300
8y
t=
log10 = 900sec
t93.6% = 150 min
4y
0.693
268. Which of the following graph has intercept
266. Consider the following reactions
equal to zero?
A→P1; B→P2; C →P3; D→P4,
[ R ]0
The order of the above reactions are a, b, c and
(a) log [ R ] → t
(b) log
→t
d, respectively. The following graph is obtained
[ R]
when log[rate] vs log[conc.] are plotted:
1
(c) log K →
(d) [R] → t
T
GUJCET-2020
=
[ R ]o
[R ]
vs t for first
k
2.303
with zero
Ans. (b) : When graph plotted for log
order reaction, slope is equal to
is equal to zero.
Among the following, the correct sequence for intercept
st
For
1
order
reaction,
the order of the reactions is
(a) D>A>B>C
(b) A>B>C>D
[ R ]o
kt = 2.303 log
(c) C>A>B>D
(d) D>B>A>C
[R ]
[JEE Main 2020, 5 Sep Shift-I]
[ R ]o
kt
order
log
=
Ans. (d) : Rate = k [conc.]
[ R ] 2.303
Taking ‘log’ of above equation on both side
For straight line graph, y=mx + c
log [ rate] = log k + order log [conc.]
Here intercept, c = 0
For straight line graph, y = mx + C
269. The results given in the below table were
obtained during kinetic studies of the following
Slope = m = order
reaction:
Therefore, greater the value of slope higher the order of
2A + B → C + D
reaction
Experiment
[A]/mol [B]/mol
Initial
D>B>A>C
–1
–1
rate/mol
L
L
267. The time required for 60% completion of a
L–1 min–1
first order reaction is 50 min.
I.
0.1
0.1
6.00×10–3
The time required for 93.6% completion of the
II.
0.1
0.2
2.40×10–2
same reaction will be
III.
0.2
0.1
1.20×10–2
(a) 100 min
(b) 83.8 min
(c) 50 min
(d) 150 min
IV.
X
0.2
7.20×10–2
Karnataka-CET-2020
V.
0.3
Y
2.88×10–1
Objective Chemistry Volume-II
289
YCT
X and Y in the given table are respectively
(a) 0.4, 0.4
(b) 0.4, 0.3
(c) 0.3, 0.4
(d) 0.3, 0.3
[JEE Main 2020, 2 Sep Shift-II]
Ans. (c) : Rate = k [ A ] [ B]
m
n
6×10−3 = k [0.1] [0.1] ____(i)
m
n
2.4×10−2 = k [0.1] [0.2] ____(ii)
m
n
1.2×10−2 = k [0.2] [ 0.1] ____(iii)
On dividing equation (ii) by equation (i)
L1 = 2 n
m
n
n=2
On dividing equation (iii) by equation (i)
2 = 2m
m =1
Rate
= k [ A ][ B]
From equation (i)
2
k=
(a) 0.02231 min–1
(c) 0.06231 min–1
(e) 0.1231 min–1
Kerala-CEE-2019
Ans. (a) : Given, at time (t) = 0, a = 25ml
at t = 40, (a – n) = 9.6
So, the first order reaction
2.303
a
k=
–––– (1)
log
t
(a − x )
Putting vale in eqn (1) we get
2.303
25
k=
log
40
9.6
2.303
k=
× 0.4156
40
k = 0.02231 min–1
272. Which among the following reaction is an
example of a zero order reaction?
(a)
−3
6×10
(0.1)×(0.1)
2
−2
k = 6mol L min
2
(b) 0.04231 min–1
(d) 0.08231 min–1
C12 H 22O 11 (aq) + H 2O(l)
→ C6 H12 O6 (aq)
+ C6 H12O6 ( aq )
Pt
(b) 2NH3 ( g )
→ N 2 ( g ) + 3H 2
−1
(c) 2H 2O 2 ( l )
→ 2H 2 O ( l ) + O2 ( g )
7.20×10−2 = k [ X ][ 0.2]
2
7.2×10−2
= 0.3mol L−1
6× 0.04
2
2.88×10−1 = k [0.3][ Y ]
(d) H 2 ( g ) + I 2 ( g )
→ 2HI ( g )
[X] =
2.88×10−1
= 0.16
[Y] =
6× 0.3
[ Y = 0.4]
2
MHT CET-02.05.2019, SHIFT-III
Pt
Ans. (b) : 2NH 3 (g )
→ N 2 (g ) + 3H 2 (g )
Rate = k NH 3
°
Rate = k
Decomposition of ammonia on platinum surface follows
zero order of reaction.
270. For a 2nd order reaction, rate constant k has Rate of these reaction directly proportional to zeroth
unit
power of the reactant concentration.
(a) s–1
(b) mol–1 s–1
273.
A first order reaction has a rate constant of
(c) L mol−1
(d) L mol−1 s–1
2.303 ×10 –3 s –1 . The time required for 40 g of
Assam CEE-2020
this reactant to reduce to 10 g will be
Ans. (d) : General rule for unit of rate constant may be
[Given that log10 2 = 0.3010]
given as −
(a) 230.3 s
(b) 301 s
Unit of rate constant :(c) 2000 s
(d) 602 s
n −1
1
−1
(Odisha NEET-2019)
=
× time
unit of concentration
Ans. (d) : Given:– k = 2.303×10−3 S−1
n −1
n −1
1
liter
−1
−1
a = 4g
(a − x ) = 10g
=
× sec = mole × sec
mole
/
liter
For first order reaction,
For second order reaction n = 2
2.303
a
2−1
t=
log10
liter
−1
k
a
−
x)
(
=
× sec
mole
40
2.303
So, unit of rate constant = mol−1 liter sec−1
=
log10
−3
10
2.303×10
271. For a first order reaction, A(g) → B(g) AT 35ºC,
t = 602 sec
the volume of "A" left in the reaction vessel at
various times are given below. [Given data: log 274. Rate of a first-order reaction after 10 minutes
(5/4) = 0.0969]
and 20 minutes are 0.04 mol lit–1 s–1 and 0.03
t / minutes 0
10
20
30
40
mol lit–1 s–1 respectively. Calculate the half-life
v / mL
25
20
15.7 12.5 9.6
of the reaction.
Objective Chemistry Volume-II
290
YCT
(a) 32.4 min
(c) 48.6 min
(b) 24.3 min
(d) 97.2 min
Tripura JEE-2019
Ans. (b) : Given:– t1 = 10 min, r1 = 0.03mol L−1 S−1
t 2 = 20 min,
For 1 order reaction,
Rate = k [ A ]
r2 = 0.03mol L−1 S−1
st
In (2) reaction the value of rate constant is higher than
(1) reaction. Hence, reaction (1) is slower than reaction
(2).
276. The given plots represent the variation of the
concentration of a reaction R with time for two
different reactions (i) and (ii). The respective
orders of the reactions are
r1 = 0.04 = k [ A ]10 ........(i)
r2 = 0.03 = k [ A ]20 .......(ii)
On dividing equation (i) by equation (ii)
[ A ]10 4
=
[ A ]20 3
When time is taken from 10min to 20min
2.303
a
k=
log10
t
(a − x )
k=
(a) 1, 1
(c) 0, 1
[A]
2.303
log10 10
10
[ A ]20
4
= 0.2303log10
3
Half-life time of first order reaction,
0.693
t1 =
2
k
0.693
=
4
0.2303log10
3
≈ 24.3min
275. Consider the following two first order reactions
occurring at 298 K with same initial
concentration of A:
(1) A→ B; rate constant, k = 0.693 min–1
(2) A→ C; half-life, t1/2 = 0.693 min
Choose the correct option.
(a) Reaction (1) is faster than reaction(2).
(b) Reaction (1) is slower than reaction (2).
(c) Both reactions proceed at the same rate.
(d) Since two different products are formed, rates
cannot be compared.
WB-JEE-2019
Ans. (b) : For first order reaction,
Rate = k [ A ]
If [A] is constant then rate depend on value of rate
constant (k).
For (1) reaction, A → B
k = 0.693min −1
For (2) reaction, A → C
t 1 = 0.693
2
k=
0.693
t1
2
0.693
k=
= 1 min −1
0.693
Objective Chemistry Volume-II
(b) 0, 2
(d) 1, 0
[JEE Main 2019, 9 April Shift-I]
Ans. (d) : For first order reaction
R = R 0 e−Kt
Taking ‘ln’ on both side
ln [ R ] = ln [ R 0 ]− kt
For straight line graph, y = mx + C
Where y = ln [ R ],
x = t,
m = −k
C = ln [ R 0 ]
For zero order reaction
kt = [ R 0 ]−[ R ]
[ R ] = [ R 0 ]− Kt
For straight line graph y = mx + C
Where y = [ R ],
x = t,
m = −k,
C = [R 0 ]
277. The following results were obtained during
kinetic studies of the reaction:
2A + B → Products
Experiment
[A]
[B]
Initial rate
(mol
(mol L– of reaction
1
L–1)
)
(in mol L–1
min–1)
I.
0.10
0.20
6.93×10–3
II.
0.10
0.25
6.93×10–3
III.
0.20
0.30
1.386×10–2
291
YCT
The time (in minutes) required to consume half On dividing (iv) and (i), we get,
of A is
2 = 2m
(a) 5
(b) 10
m =1
(c) 100
(d) 1
Put
the
value
of m on equation (iii)
[JEE Main 2019, 9 Jan Shift-I]
1
+
n
=3
m
n
Ans. (b) : Rate = K [ A ] [ B]
2
Rate
= k [ A ][ B]
m
n
−3
6.93×10 = k [0.1] [0.2 ] _____(i)
Order of reaction = 2
m
n
6.93×10−3 = k [ 0.1] [0.25] _____(ii)
279. The reaction, 2X→B is a zeroth order reaction.
If the initial concentration of X is 0.2 M, the
m
n
−2
1.386×10 = k [0.2 ] [0.3] ____(iii)
half-life is 6 h. When the initial concentration
On dividing eq. (ii) from eq. (i)
of X is 0.5 M, the time required to reach its
n
0
n
final concentration of 0.2 M will be
5
5 5
1 =
⇒ =
(a) 7.2 h
(b) 18.0 h
4 4
4
(c) 12.0 h
(d) 9.0 h
⇒n=0
[JEE Main 2019, 9 Jan Shift-II]
On dividing eq. (iii) and eq. (i)
Ans. (b) : Given:- t 1 = 6h, [ A o ] = 0.2M
0
2
m 6
2 = (2)
1
4
A o = 0.5M A1t = 0.2 t = ?
1
m
( 2 ) = ( 2)
For zero order reaction
⇒m =1
[A ]
t1 = o
1
0
2
2k
Rate
= k [ A ] [ B]
0.2
1
Rate
= k [A]
k=
= mol L−1 S−1
2× 6 60
6.93×10−3
kt = [ A o ]− A1t
k=
= 69.3×10−3 min −1
0.1
0.5 − 0.2 = 0.3
Time required to consume half of A is
0.3
0.693
0.693
= 18h
t=
t1 =
=
1
2
k
69.3×10−3
= 10 min
60
278. For the following reaction,
280. Decomposition of X exhibits a rate constant of
2A+B → Products
0.05 µg/year. How many years are required for
When concentration of both (A and B) becomes
the decomposition of 5 µg of X into 2.5µg?
double, then rate of reaction increases from
(a) 20
(b) 25
0.3 mol L–1 s–1 to 2.4 mol L–1 s–1.
(c) 40
(d) 50
When concentration of only A is double, the
[JEE Main 2019, 12 Jan Shift-I]
rate of reaction increases from
Ans. (d) : It follows zero order reaction by seeing the
0.3 mol L–1 s–1 to 0.6 mol L–1 s–1.
unit
of rate constant
Which of the following true?
Half-life
time for zero order reaction
(a) The whole reaction is of 4th order
a
(b) The order of reaction w.r.t. B is one
t1 =
(c) The order of reaction w.r.t. B is 2
2
2k
(d) The order of reaction w.r.t. A is 2
5
=
[JEE Main 2019, 9 Jan Shift-II]
2× 0.05
m
n
Ans. (c) : Let Rate k [ A ] [ B] = −0.3,mole−1s−1 ___(i)
t 1 = 50 year
2
If concentration of both the reactant is doubled
281. A bacterial infection in an internal wound
then
grown as N'(t) = N0 exp (t), where the time t is
m
n
'
−1 −1
r = k [ 2A ] [ 2B] = 2.4mole s ____(ii)
in hours. A dose of antibiotic, taken orally,
On dividing equation (ii) from equation (i), we get,
needs 1 hour to reach the wound. Once it
reaches there, the bacterial population goes
B = 2 m.2 n = 2 m+n
down as
3
(m+ n )
⇒ m + n = 3 ___(iii)
( 2) = 2
N
dN
= –5N 2 . What will be the plot of 0 vs t
When only concentration of A is doubled then
dt
N
m
n
r ' = 0.6 mole−1 s−1 = k [ 2A ] [ B] ____(iv)
after 1 hour?
Objective Chemistry Volume-II
292
YCT
282. Formula for half-life of a zero order reaction is
C
C
(b) o
(a) o
(b)
(a)
k
2k
2Co
2Co
(c)
(d)
k
2k
JIPMER-2019
Ans. (b) : Half -life time zero order reaction,
(c)
(d)
C
[Ao] = Co , then [At] = o
2
[At] = [Ao] – kt
kt = [Ao] – [At]
[JEE Main 2019, 12 April Shift-I]
Expression for zero order rate constant
Ans. (a) : Given, Bacterial injection growth rate for 1hr
[A ] − [A t ]
k= o
N '( t ) = N o e t
t1
2
dN
2
After 1hr growth rate,
= −5N
C
C
dt
k.t1/2 = Co – o = o
2
2
For
0 to 1hr
Co
t1/2 =
N ' = N oet
2k
283.
For
a
first
order reaction, a plot of log (a – x)
At
t = 1,
N ' = eN o
against time is a straight line with a negative
slope equal to
dN
After 1hr,
= −5N 2
−k
dt
(a)
(b) −2.303k
2.303
dN
2
= −5N
2.303
Ea
N2
(c)
(d) −
k
2.303R
Integrating both the sides
AIIMS-25
May
2019 (Evening)
N dN
t
Ans.
(a)
:
For
first
order
reaction,
=
−
5
dt
∫N oe N 2
∫1
a
N
kt = 2.303 log
−1
(a − x )
= −5( t −1)
N No e
= 2.303 log ( a ) − log ( a − x )
−1 1
kt
− = −5( t −1)
= log ( a ) – log ( a − x )
Noe N
2.303
kt
No 1
log (a – x) = log (a) –
− = 5N o ( t −1)
2.303
N e
For straight line graph
No
1
y = mx + C
= 5N o ( t −1) +
N
e
A plot of log (a – x) v/s t slope is on
1
k
No
m=–
= 5N o t + − 5N o
e
2.303
N
284. Find out time period of Ist order reaction when
For straight line equation, y = mx + C
2
rd of the reaction completes. The value of
No
Where, y =
, x=t
3
N
rate constant is 4.3 × 10–4.
(a) 0.0025 × 103 sec
(b) 0.25 × 103 sec
1
3
Slope = m = 5N o
C = − N o
(c) 0.025 × 10 sec
(d) 2.5 × 103 sec
e
AIIMS-25 May 2019 (Morning)
2 1
Ct = 1 − =
Ans. (d): Given:- Co = 1
3 3
For 1st order reaction
2.303
C
t=
log o
k
Ct
Objective Chemistry Volume-II
293
YCT
=
Ans. (b) : 2A→A2
By formula :-
2.303
1
log
4.3 × 10 –4
(1/ 3)
2.303 × 104
log 3
=
4.3
= 2.5 × 103 sec
285. The decomposition of NH3 on Pt surface is a
zero order reaction. If the value of rate
constant is 2 × 10–4 mol L–1 S–1.
The rate of appearance of N2 and H2 are
respectively:
N2
H2
−4
−1 −1
−4
(a) 1 ×10 mol L s ,
3 × 10 mol L−1s −1
r1 c1
=
r2 c 2
n
r c
=
3r 27c
1 1
=
3 27
n
n
3n
1 1
=
3 3
1 = 3n
1 ×10 −4 mol L−1s −1
(b) 3 × 10−4 mol L−1s −1 ,
1
n=
(c) 2 × 10 −4 mol L−1s −1 ,
6 × 10−4 mol L−1s −1
3
288. In a reaction A → B, if the concentration of
(d) 3 × 10−4 mol L−1s −1 ,
3 × 10 −4 mol L−1s −1
reactant is increased by 9 times then rate of
AIIMS-26.05.2019 (Morning)
reaction increases 3 times. What is the order of
Ans. (c): For reaction,
reaction?
(a) 2
(b) 3
2NH3
→ N 2 + 3H 2
–4
–1 –1
1
1
Rate = k = 2 × 10 mol L S
(c)
(d)
2
3
1 d[NH 3 ] d[N 2 ]
Rate of reaction = −
=
GUJCET-2019
2
dt
dt
Ans. (c) : Rate = r = k [A]n
______(1)
1 d[H 2 ]
If concentration of reaction increased by 9 times then
=
=k
3 dt
Rate, r1 = k [9A]n = 3r
______(2)
Hence, the rate of appearance of N2 and H2 are On dividing eq. (2) by eq. (1), we get,
n
respectively–
3r k [9A ]
d[N 2 ]
=
= 9n
−4
−1 −1
n
∴
= k = 2 × 10 mol L S and
r
k
A
[ ]
dt
n
3
=
(9)
d[H 2 ]
= 3k
3 = (32)n
dt
3 = (3)2n
−4
= 3 × 2 × 10
2n = 1
−4
−1 −1
1
= 6 × 10 mol L S
n=
2
286. The rate expression of a reaction is
3/2
–1
289.
For
a
first order reaction A→P, the
Rate= k [A] [B] the order of the reaction
temperature
(T) dependent rate constant (k)
will be of
was found to follow the equation log k = –
(a) Zero order
(b) Half order
1
(c) 2nd order
(d) 3rd order
+ 6.0 . The pre-exponential factor A
(2000)
T
Assam CEE-2019
and the activation energy Ea, respectively, are
3
dx
−1
2
(a) 1.0 ×106 s −1 and 9.2 kJ mol −1
Ans. (b) : The rate expression is
= k [ A ] [ B]
at
(b) 6.0s −1 and 16.6 kJ mol−1
order wrt A is = 3/2
(c) 1.0 ×106 s −1 and16.6 kJ mol −1
Order wrt B is = −1
3
1
(d) 1.0 ×106 s −1 and 38.3kJ mol−1
overall order is = − 1 =
JEE Main-2019
2
2
287. In a reaction 2A→ A2, the rate is increased by Ans. (d) : Given a first order reaction A → P.
three times if the concentration of A is
Ea
∴
log k = log A −
…..(1)
increased by 27 times. The order of the
2.303RT
reaction is
1
(a) 3
(b) 1/3
Also given log k = 6.0 – (2000)
…(2)
T
(c) 1/2
(d) 3/2
Comparing equations (1) and (2) we get
Assam CEE-2019
Objective Chemistry Volume-II
294
YCT
(a)
(b)
(c)
(d)
(e)
at high temperature
at high partial pressure of HI
at low partial pressure of HI
at high partial pressure of H2
at high partial pressure of I2
Kerala-CEE-2018
Ans. (b) : The given reaction,
△
2HI
→ H 2 (g ) + I 2 (g )
follows zero order of the reaction at high partial
pressure of HI.
294. 99% of a first order reaction was completed in
32 min. When will 99.9% of the reaction
complete?
1
(a) 50 min
(b) 46 min
Rate= k[A][B] 2
(c) 49 min
(d) 48 min
Manipal-2018
1
∴order = 1 +
Ans.
(d)
:
t
=
32
min,
then
(a
–
x)
=
1%
2
99%
At t99.9% = ?
then (a – x) = 0.1%
3
=
for first order reaction,
2
2.303
a
Order of reaction = 1.5
k=
log10
t 99%
(a − x )
291. At 518°C, the rate of decomposition of a sample
of gaseous acetaldehyde, initially at a pressure
100
2.303
of 363 Torr, was 1.00 Torr s–1 when 5% had = 32 log10 1
–1
reacted and 0.5 Torr s when 33% had
2.303× 2
reacted. The order of the reaction is:
=
32
(a) 2
(b) 3
At t99.9%
(c) 1
(d) 0
100
[JEE Main 2018]
2.303
t99.9% =
log10
0.1
2.303× 2
Ans. (a) : When 5% is reacted
32
a = 100%, ( a − x ) = 95%, rate, r = 1Torr.s−1
3
When 33% is reacted
= ×32
2
a = 100%, ( a − x ) = 66%, rate, r ' = 0.5Torr s−1
t 99.9% = 48min
m
Rate = k (a − x )
295. The correct difference between first and second
m
order reactions is that
r 95
1
=
=
(a) the rate of a first-order reaction does not
r ' 66.7
0.5
depend on reactant concentrations; the rate of
2 = (1.42)m
a second-order reaction does depend on
reactant concentrations
⇒m = 2
(b) the half-life of a first-order reaction does not
292. What will be the correct unit of rate constant K
depend on [ A ]0 ; the half-life of a secondfor a reaction whose order is three?
log A = 6
A = 106
Ea
And
= 2000
2.303R
Ea = 2000 × 2.303 × 8.314
Ea = 38.29 kJ mol–1
290. If the rate equation for reaction is, rate
=k[A][B]1/2, the order of reaction is
(a) 0.5
(b) 1
(c) 1.5
(d) 2
COMEDK-2019
Ans. (c) : Given the rate of reaction-
(a) mole–1 lit sec–1
(c) sec–1
(b) mole2 lit sec–1
(d) mole–2 lit2 sec–1
J & K CET-(2018)
Ans. (d): For third order reaction
Rate = k[A]3
1−n
Unit of rate constant (mol)
n−1
−1
(lit ) (sec)
Where n = order of reaction
For third order reaction, n = 3
1−3
3−1
−1
Unit of rate constant = (mol) (lit ) (sec)
= mole–2 lit2 sec–1
293. Under what condition the order of the reaction,
2HI → H2(g) + I2(g), is zero
Objective Chemistry Volume-II
order reaction does depend on [ A ]0
(c) a first-order reaction can be catalysed; a
second-order reaction cannot be catalysed
(d) the rate of a first-order reaction does depend
on reactant concentrations; the rate of a
second-order reaction does not depend on
reactant concentrations.
(NEET 2018)
Ans. (b) : The half-life of first order reaction does not
depend on [Ao]
0.693
t1 =
2
k
t 1 α[A o ]
295
2
YCT
Ans. (b) : If unit of rate constant(K) is equal to unit of
rate then order of reaction is zero order reaction.
Rate = k [A]0
Rate = k
299. For a first order reaction A → B, the reaction
a
rate at reactant concentration of 0.01 M is
t1 α
found to be 2.0×10-5 mol L-1 s-1. The half life
2
[ Ao ]
period of reaction is:
−1
Or
t 1 α [ Ao ]
(a) 30 s
(b) 220 s
2
(c) 300 s
(d) 347 s
296. Graph plotted between log t1/2 versus log
HP CET-2018
concentration (a) is a straight line. What
Ans.
(d)
:
Given:–
conclusion can you draw from the graph?
[A] = 0.01M, Rate = 2×10−5 mol L−1s−1
For first order reaction
Rate = k [A]
The half-life of a second order reaction does depends on
[Ao]
1
t1 =
2
[A o ]k
2×10−5
= 2×10−3 s−1
0.01
Half life period of first order reaction,
0.693
t1 =
2
k
0.693
t1 =
2
2×10−3
t 1 = 346.5
k=
n = order of reaction
1
a
0.693
=
K
1
xxa
(a) n = 2, t1/ 2 =
(b) n = 1, t1/ 2 =
(c) n = 1, t1/ 2
(d) None of these
2
CG PET-2018
Ans. (c) : For the given graph, we can say that half-life
time is independent of initial concentration of reactants.
This is possible when order of reaction is first (n=1)
Half-life time for first order reaction is
0.693
t1/ 2 =
k
297. Half-life period for a first order reaction is 10
minutes. How much time is required to change
the concentration of the reactants 0.08 M to
0.01 M?
(a) 20 min
(b) 30 min
(c) 40 min
(d) 50 min
CG PET-2018
Ans. (b) : Given :- t1/2 = 10 min
For first order reaction
0.693
t1/ 2 =
k
0.693
k=
min −1
10
When, a = 0.08 M and (a–x) = 0.01 M
2.303
0.08
t=
log
0.693 10 0.01
10
t =33.23×3log102
t = 30 min
298. The units for the rate constant and the rate of
reaction are same for a reaction. What will be
the order of the reaction?
(a) Second
(b) Zero
(c) First
(d) Third
GUJCET-2018
Objective Chemistry Volume-II
t 1 ≈ 347 s
2
300. Which option is valid for zero order reaction?
3
4
(b) t 1 = t 1
(a) t 1 = t 1
2
2
2 4
2 4
(c) t1/2 = 2t1/4
(d) t1/4 = (t1/4)2
[AIIMS-26.05.2018 (M)]
Ans. (c): For zero order reaction.
1A
t1 =
2
2 k
1A
t1 =
4
4 k
t1
4
2
= =2
t1
2
4
t1 = 2 t1
2
4
301. The ratio of the half-life time (t1/2), to the three
quarter life time (t3/4) for a reaction that is
second order.
(a) depends directly on concentration of reactant
(b) is independent of concentration of reactants
(c) depends inversely on the concentration of
reactant
(d) depends-directly
to
the
square
of
concentration of reactants.
AMU-2018
Ans. (b) : For second order reaction,
1
t1/ 2 =
ka
For quarter life time (t3/4)
a
[Ao] = a
[At] =
4
296
YCT
kt3/4 =
(a) –1, 1, 3/2
(c) 1, 3/2, –1
1
1
4 1 3
−
= − =
[A t ] [Ao ] a a a
3
ka
1
t1/ 2 ka 1
=
=
3 3
t 3/ 4
ka
∴ Ratio of t1/2 to t3/4 for a reaction that is second order
is independent of concentration of reactant.
302. Hydrolysis of ester is a first order reaction. For
this reaction, the correct statement is
(a) the rate depends only on concentration of
water
(b) the amount of ester present in the reaction
mixture is very high
(c) the amount of water present in the reaction
mixture is very high
(d) the rate is proportional to the square root of
concentration of both
Assam CEE-2018
Ans. (c) : The amount of water present in the reaction
mixture is very high.
• Hydrolysis is a most important reaction of esters.
Acidic hydrolysis of an esters gives a carboxylic acid
and alcohol.
CH 3COOC2 H 5 + H 2 O → CH 3COOH + C 2 H 5OH
t3/4 =
303. He time taken for 10% completion of a first
order reaction is 20 min. Then for 19%
completion, the reaction will take ?
(a) 40 min
(b) 60 min
(c) 30 min
(d) 50 min
BCECE-2018
Ans. (a) : Given :- t10% = 20min , t19% = ?
For first order reaction,
2.303
a
t=
log10
k
(a − x )
When the reaction is 10% completed
2.303
100
log10
k
90
2.303
10
k=
log10
20
9
2.303
100
t19% =
log10
2.303
10
81
log10
20
9
t19% = 40 min .
304. What is order with respect to A, B, C,
respectively
t=
[ A ] [ B] [ C ]
rate(M / sec.)
0.1 0.02
8.08 × 10−3
0.1 0.2 0.02
2.01×10−3
0.1 1.8 0.18
6.03 × 10−3
0.2
6.464 ×10−2
0.2
0.1 0.08
Objective Chemistry Volume-II
(b) –1, 1, 1/2
(d) 1, –1, 3/2
VITEEE-2018
Ans. (d) : From the rate law equation –
r = k[A]x[B]y[C]z
…..(1)
From data first –
8.08 × 10–3 = k[0.2]x [0.1]y [0.02]z .….(2)
From data second –
2.01 × 10–3 = k[0.1]x [0.2]y [0.02]z ….(3)
By the dividing of equation (1) and (2) – we get
x –y = 2
…..(4)
Similarly, from the given data of (2) and (3) –
1
y+z =
…..(5)
2
And from the given data of (1) and (4) –
3
z=
…..(6)
2
From equation (4), (5) and (6) – we get
3
x = 1, y = −1, z =
2
305. What will be the order of the reaction for
hydrolysis of methylacetate with NaOH by
using the data provided?
Time (min)
: 0
5
10
15
Vol. of acid (mL) : 10.12 6.23 1.40 0.981
(a) 1
(b) 2
(c) 0
(d) 3
J & K CET-(2017)
Ans. (b) : Hydrolysis of methylacetate with NaOH
follow second order kinetics.
306. A first order reaction is 50% completed in 1.26
× 1014 s. How much time would it takes for
100% completion?
(a) 1.26 × 1015 s
(b) 2.52 × 1014 s
28
(c) 2.52 × 10 s
(d) Infinite
JIPMER-2017
Ans. (d) : Given,
t 50% = 1.26 × 1014 s, t100% = ?
For first order reaction, t100% is completed after infinite
time.
a
kt = 2.303 log10
(a − x)
kt = 2.303 log10
100
(0)
kt = 2.303 log10 (∞)
t=∞
307. Which among the following reactions is an
example of pseudo first order reaction?
(a) Inversion of cane sugar
(b) Decomposition of H2O2
(c) Conversion of cyclopropane to propane
(d) Decomposition of N2O5
MHT CET-2017
297
YCT
Ans. (a) : When concentration of any reactant present in
excess then it doesn't participate in rate law. This
reactions are called pseudo-order reactions. Inversion of
cane sugar follows pseudo first order reaction.
C12 H 22 O11 + H 2 O
→ C6 H12 O6 + C 6 H12 O6
(glucose )
(iv)
(iii)
(fructose)
Concentration of H2O is present in excess. Hence, [H2O]
(b) (i), (ii), (iv)
(d) (i), (iii), (ii)
308. A first order reaction has a specific reaction
TS EAMCET-2017
rate of 10 –2 sec –1 . How much time will it take Ans. (b) : For zero order reaction –
for 20 g of the reactant to reduce to 5 g?
[A]0
(a) 138.6 sec
(b) 346.5 sec
k=
(c) 693.0 sec
(d) 238.6 sec
t
(NEET-2017) And half life time
Ans. (a) : Given:– k = 10−2 sec−1 , a = 20g
t = t1
2
a
−
x
=
5g,
t
=
?
(
)
[A]0
For first order reaction,
So, t 1 =
2k
2
2.303
a
t=
log10
Hence the option (b) is correct representation of a zero
k
(a − x )
order reaction.
20
2.303
HCl
311. R – COOR' + H 2O
→ R – COOH + R'OH.
= −2 log10
5
10
What type of reaction is this?
= 138.6sec
(a) Pseudounimolecular (b) Third order
309. Which graph represents the zero order
(c) Second order
(d) Unimolecular
reaction [A(g) →B(g)]
COMEDK-2017
Ans. (a) : Given that reaction is HCl
R-COOR'+H2O
→ R – COOH + R 'OH
(b)
(a)
The reaction which follows different order of a reaction
with respect of different reactant is known as pseudounimolecular reaction.
312. For a second order reaction (2A → Product),
1
vs t is represented as
A
[
]
(c)
(d)
does not appear in the rate law expression.
(a) (i), (ii), (iii)
(c) (ii), (iii), (iv)
(a)
1
[A]
(b)
1
[A]
(c)
1
[A]
(d)
1
[A]
UPTU/UPSEE-2017
Ans. (a) : For zero order reaction
d [ B]
Rate =
dt
Rate ∝ t
d [ B]
dt
d [ B]
∝t
AMU-2017
∝ t gives straight line graph.
Ans. (a) : For second order reaction,
dt
1
1
310. Which of the following are the correct
kt =
−
A
A
[ ] [ o]
representations of a zero order reaction, where
A represents the reactant ?
1
1
=
+ kt
[A ] [Ao ]
(ii)
(i)
For straight line graph, y = mx + c
1
Where y =
, x = t, slope = m = k,
A
[ ]
Hence,
Objective Chemistry Volume-II
298
YCT
Intercept = c =
1
[Ao ]
It can be observed that the slope of the straight line is
equal to the value of the rate constant.
313. In the reaction, A → Products, when the
concentration of A was reduced from 2.4×10-2
M to 1.2×10-2 M, the rate decreased 8 times at
the same temperature. The order of the
reaction is
(a) 0
(b) 1
(c) 2
(d) 3
(e) 0.5
Kerala-CEE-2016
Ans. (d) : Rate = k [A]x
If [A] = 2.4 × 10–2 then rate = r
r = k [2.4 × 10–2]x
….(i)
r
IF [A] = 1. 2 × 10–2 then rate =
8
r
= k [1.2 × 10–2]x
…..(ii)
8
on dividing eqn (i) by eqn (ii)
r
= (2) x
r
8
8 = (2)x
=x=3
314. The rate of
first-order reaction is
0.04mol L–1s –1 at 10 seconds and 0.03 mol L–1s –1
at 20 seconds after initiation of the reaction.
The half-life period of the reaction is
(a) 44.1 s
(b) 54.1 s
(c) 24.1 s
(d) 34.1 s
(NEET-I 2016)
Ans. (c) : Given:– r1 = 0.01 mol–1 S–1 , t1 = 10 s
r2 = 0.03 mol L–1 S–1 , t2 = 20 s
For first order reaction,
Rate = k[A]
r1 = 0.04 = k[A]10 .........(i)
r2 = 0.03 = k[A]20 ..........(ii)
On dividing (i) by (ii)
[A]10 4
=
[A]20 3
2.303
a
k=
log10
t
a
−
( x)
When time is taken from 10s to 20s
2.303
[A]10
k=
log10
10
[A]20
Objective Chemistry Volume-II
4
k = 0.303 log10
3
Half-life time of first order reaction
0.693
t1 =
2
k
0.693
=
4
0.2303log10
3
t 1 = 24.15
2
315. The decomposition of phosphine (PH3) on
tungsten at low pressure is a first-order
reaction. It is because the
(a) rate is proportional to the surface coverage
(b) rate is inversely proportional to the surface
coverage
(c) rate is independent of the surface coverage
(d) rate of decomposition is very slow.
(NEET-II 2016)
Ans. (a) : The decomposition of phosphine (PH 3 ) on
tungsten at low pressure is a first order reaction because
surface coverage is proportional to partial pressure of
PH3. Therefore rate of reaction is proportional to the
surface coverage.
316. For the reaction, 2SO2+O2 (excess) → 2SO3 the
order of reaction with respect to O2 is
(a) two
(b) three
(c) zero
(d) one
UPTU/UPSEE-2016
Ans. (c): For the reaction
2SO 2 + O 2 → 2SO3
(excess)
Order of reaction w.r.t O2 is zero because when
concentration of any reactant is taken in excess then it
does not participate in rate law.
317. Which of the following is correct for a first
order reaction?
(b) t1/ 2 ∝ a 2
(a) t1/ 2 ∝ a o
1
(c) t1/ 2 ∝ a
(d) t1/ 2 ∝
a
UPTU/UPSEE-2016
Ans. (a): Half-life time for first order reaction,
0.693
t1 =
2
k
t 1 ∝ ao
2
Half-life period of first order reaction is independent of
initial concentration of reactants.
318. Decomposition of H2O2 follows a first order
reaction. In 50 min, the concentration of H2O2
decreases from 0.5 to 0.125 M in one such
decomposition. When the concentration of
H2O2 reaches 0.05 M, the rate of formation of
O2 will be
(a) 6.93×10–4 mol min–1 (b) 2.66 Lmin–1 at STP
(c) 1.34×10–2 mol min–1 (d) 6.93×10–2 mol min–1
[JEE Main 2016]
299
YCT
Ans. (a) : Decomposition of H2O2
2H 2 O 2
→ 2H 2 O + O 2
m
When concentration is reduced from 0.5M to 0.125M
1
Which is reduced to th times then time t 3 is 50min
4
4
For first order reaction half-life time is independent of
concentration of reactants than
t3
t 1 = 4 = 25 min
2
2
0.693 0.693
k=
=
min −1
t1
25
( )
2
Rate
= k[A] =
d[H 2O2 ]
dt
0.693
=
× 0.05
25
= 1.386 ×10−3 mol L−1 min −1
1 d [ H 2O 2 ] d [ O 2 ]
=
2
dt
dt
d [O 2 ] 1.386×10−3
=
= 6.93×10−4 mol L min −1
dt
2
319. The order of the reaction for the oxidation of
ferrous sulphate into ferric sulphate by
potassium chlorate in the presence of sulphuric
acid is
(a) 2
(b) 0
(c) 1
(d) 3
J & K CET-(2016)
Ans. (a) : Oxidation of ferrous sulphate into sulphate by
potassium chlorate in presence of sulphuric acid follows
second order of the reaction.
320. For the reaction, 2A + B → C + D, on the basis
of the following data, find the order of reaction.
Exp. [A] mol L–1 [B] mol L–1 Initial rate
No.
(mol L–1 min–1)
0.1
0.1
6 × 10–3
1.
0.3
0.2
7.2 × 10–2
2.
0.3
0.4
2.88 × 10–1
3.
0.4
0.1
2.40 × 10–2
4.
(a) 1
(b) 2
(c) 3
(d) 4
JCECE-2016
Ans. (c) : Kinetic studies of the reaction,
2A + B → C + D
Rate = k[A]x [B]y
Considering the experiment I and IV suppose rate law
of the reaction is
R ( I)
k[A]1m × [B]1n
=
n
R ( IV ) k[A]mIV × [B]IV
1 1
=
⇒ m =1
4 4
Now, considering the experiments II and III
R ( II ) k[A]mII × [B]nII
=
R ( III ) k[A]mIII × [B]nIII
7.2 × 10−2
k[0.3]m × [0.2]n
=
2.88 ×10−1 k[0.3]m × [0.4]n
m
1 1
=
⇒n=2
4 2
∴ So rate law of the reaction is
Rate = k[A]1 [B]2
Overall order = 1+2 = 3
321. Look at the graph,
−
6.0 × 10−3
k[0.1]m × [0.1]n
=
2.40 × 10−2 k[0.4]m × [0.1]n
Objective Chemistry Volume-II
Choose the correct equation from the following
which best suited to the above graph
(a) [At]= [A]0 – kt
(b) [At]= [A]0 + kt
(c) [At]= [A0]e – kt
(d) [At]= kt2 + [A]0
JIPMER-2016
Ans. (b) : By seeing the graph, we conclude that this
graph is for zero order reaction.
For zero order reaction
[At]= [A]0 + kt
322. Collision theory is used to explain how
chemical species undergo a reaction. Using this
theory and the kinetic molecular model, which
of the following does not influence the rate of a
chemical reaction?
(a) The temperature of the system.
(b) The geometry or orientation of the collision.
(c) The velocity of the reactants at the point of
collision.
(d) All of the above influence the rate.
[AIIMS-2016]
Ans. (d): Rate of a chemical reaction depends on total
number of collision. Fraction of collision with proper
orientation and with sufficient energy. Total number of
collision further depends on average speed of
molecules, area and number of molecules. Rate of
chemical reaction also depends on temperature.
According to collision theory,
8KT
.πσ 2AB N A .ρe − Ea / RT
πµ
323. 50 % of a first order reaction is complete in 23
minutes. Calculate the time required to
complete 90% of the reaction.
(a) 70.4 minutes
(b) 76.4 minutes
(c) 38.7 minutes
(d) 35.2 minutes
AMU-2016
Rate
300
=
YCT
Ans. (b) : given :- t1/2 = 23 min , t90% = ?
Half life time for first order,
0.693
t1/2 =
k
0.693
k=
23
If 90% of reaction is completed then a = 100
(a–x) =10
2.303
100
t90% =
log
0.693
10
23
t90% = 76.4 min.
324. The first order gaseous decomposition of N2O4
into NO2 has a k value of 4.5 × 103 s–1 at 1oC
and an energy of activation of 58 kJ mole–1. At
what temperature would k be 1.00 × 104 s–1?
(a) 274 K
(b) 283 K
(c) 273 K
(d) 293 K
AMU-2016
3 –1
o
Ans. (b) : Given :- k1 = 4.5 × 10 s , T1 = 1 C = 274 K
Ea = 58×103 J mol–1
k2 = 1×104 S–1
According to Arrhenius Equation,
log10
log
Ea 1 1
k2
=
−
k1 2.303R T1 T2
104
58 ×103 1
1
=
−
3
4.5 ×10
2.303 × 8.314 274 T2
T2 = 281.69 K ≈ 283K
325. In the reaction,
P + Q
→?R+S
The time taken for 75% reaction of P is twice
the time taken for 50% reaction of P. The
concentration of Q varies with reaction time as
shown in the figure. The overall order of the
reaction is
326. Total order to reaction X + Y → XY is 3. The
order of reaction with respect to X1 is 2. State
the differential rate equation for the reaction.
d[x]
2
= k [ X ][ Y ]
(a) dt
d[x]
0
3
(b) = k [X] [Y]
dt
d[x]
2
(c) = k [X] [Y]
dt
d[x]
3
o
= k [X] [Y]
(d) dt
GUJCET-2016
Ans. (c) : Overall order of reaction is calculated from
power of concentration terms in rate low.
−d[x]
= k[X]2 [Y]1
dt
Order = 3
327. In a zero order reaction, when dx/dt is plotted
against (a–x) _____ is obtained.
(a) a line passing through the origin
(b) a curve
(c) a parabola
(d) a line parallel to x-axis
SRMJEEE – 2016
Ans. (d) : As we know in a zero order reaction, rate is
independent of concentration. Hence, plot of rate (a – x)
be a horizontal line
328. For a reaction X → Y, the graph of the product
concentration (x) versus time (t) came out to be
a straight line passing through the origin.
–d[X]
(a) 2
(b) 3
Hence, the graph of
and time would be
dt
(c) 0
(d) 1
(a) Straight line with a negative slope and and
[BITSAT – 2016]
intercept on y-axis
Ans. (d) : Overall Rate = k [P]x [Q]y
(b)
Straight
line with a positive slope and an
If
t75% = 2×t50%
st
intercept
on y-axis
Then order of reaction is first. So P follows 1 order
(c) A straight line parallel to x-axis
kinetics.
(d) A hyperbola.
From the given graph for Q
COMEDK-2016
[Q] = [Qo] – kt
Ans. (c) : For a zero order reactionkt = [Qo] – [Q]
This is integrated rate law for zero order reaction. So
x
k=
follow zero order kinetics.
t
Overall rate = k[P][Q]o
Thus, the graph would be a straight line passing through
= k [P]
origin. So the given information is for zero order
Order of reaction = 1+ 0= 1
reaction, and rate of reaction is constant.
Objective Chemistry Volume-II
301
YCT
332. The half-life period of a first order reactions
d[x]
having rate constant k = 0.231 × 10-10 s–1 will be
vs time will be a
dt
(a) 32 × 1010 s
(b) 2 × 1010 s
10
straight line parallel to x-axis.
(d) 2 × 10-10 s
(c) 3 × 10 s
-12
(e) 3 × 10 s
Kerala-CEE-2015
Ans. (c) : Given,
k = 0.231 × 10–10 s–1
Half life period for first order reaction,
0.693
329. Which concentration plot is linear for a first t1/2 =
k
order reaction?
0.693
(a) [A] versus time
=
0.231× 10−10
(b) ln [A] versus time
t
=
3 × 1010 s
(c) log [A] versus 1/time
1/2
(d) square root of [A] versus time
333. For the reaction X → Y, the concentrations of
AMU-2015
'X' are 1.2 M, 0.6 M, 0.3 M and 0.15 M at 0, 1,
2 and 3 hours respectively. The order of the
Ans. (b) : For first order reaction,
reaction is
[A ]
(a) zero
(b) half
kt = ln o = ln [ A 0 ] − ln [ A ]
[A]
(c) one
(d) two
(e) three
ln [A] = ln [Ao] – kt
Kerala-CEE-2015
For linear graph , y = mx + c
Hence, graph is plot for ln [A]v/s t with negative slope.
Ans. (c) : After every 1hr, concentration of reactant is
reduced to half of the initial concentration. This shows
half-life time is independent of initial concentration of
i.e., half- life time is always same in every
concentration. So, It follows first order kinetics.
0.643
330. In a first order reaction with time the
t1/2 =
concentration of the reactant decreases
k
(a) linearly
(b) exponentially
t1/2 = ∝ (a)°
(c) no change
(d) None of these
334. For the reaction O3(g) + O(g) → 2O2(g), if the
[BITSAT – 2015]
rate law expression is, rate = k[O3][O], the
Ans. (b) : For first order reaction.
molecularity and order of the reaction
respectively are
[At]=[Ao]e–Kt
(a) 2 and 2
(b) 2 and 1.33
Concentration of reactants decreases exponentially with
(c) 2 and 1
(d) 1 and 2
time for first order reaction.
MHT CET-2015
331. In a first order reaction, the concentration of
the reactant is reduced to 12.5% in 1hr. When Ans. (a) : For the given reaction,
was it half completed?
O3 (g ) + O (g )
→ 2O 2 (g )
(a) 3 hr
(b) 20 min
Rate = k[O3] [O]
(c) 30 min
(d) 15 min
Order of the reaction → Sum of the power of the
Karnataka-CET-2015
reactant concentration.
Ans. (b) : Given:
Order of reaction = 1 + 1 = 2
t12.5% = 1hr = 60min, t12.5% = ?
Molecularity of reaction → In the elementary reaction
100% → 12.5%
taking part by the reacting species.
Molecularity of reaction = 2
ln [ A 0 ]
= kt
335. In a first order reaction, the concentration of
[A0 ]
the reactant, decreases from 0.8 M to 0.4 M in
15 min. The time taken for the concetration to
ln [100]
= k(1)
change from 0.1 M to 0.025 M is
[12.5]
(a) 30 min
(b) 15 min
k = ln8
(c) 7.5 min
(d) 160 min
k = 2.079
UPTU/UPSEE-2015
5k = ln8
Ans.
(a)
:
t
=
15min
1
k = 2.079
Therefore, plot rate vs time i.e. –
2
2.693
t1/2 =
= 0.333hr × 60
2.079
t1/ 2 = 20 min
Objective Chemistry Volume-II
Half-life time of first order reaction,
0.693
t1 =
2
k
302
YCT
0.693
min−1
15
When a = 0.1M, (a–x) = 0.025M
According to first order reaction,
2.303
a
t=
log10
k
(a − x )
Similarly,
r1
k[A]x [B]y
=
r3 k[A]x [2B]y
2.303×15
0.1
=
× log10
0.693
0.025
t = 30 min
336. Consider the following reaction
H2(g)+I2(g) → 2HI(g) and
Rate – k[H2][I2]
Which one of the following statements is
correct?
(a) The reaction must occur in a single step
(b) This is a second order reaction overall
(c) Raising the temperature will cause the value
of k to decrease
(d) Raising the temperature lowers the activation
energy for the reaction
SCRA-2015
Ans. (b): Given reaction
H2(g)+I2(g) → 2HI(g)
Rate = k [H2] [I2]
A chemical reaction in which the rate of reaction is
proportional to the concentration of each of two reacting
molecules. This reaction is a second order reaction
overall.
337. In a chemical reaction A + 2B → products,
when concentration of A is doubled, rate of the
reaction increases 4 times and when
concentration of B alone is doubled rate
continues to be the same. The order of the
reaction is
(a) 1
(b) 2
(c) 3
(d) 4
COMEDK-2015
Ans. (b) : Given the chemical reactionA+2B
→ products
Let the order of reaction w.r.t A is x and w.r.t B is y.
r1 = k[A]x [B]y
If A is doubled
r2 = k[2A]x [B]y = 4
It B is doubled
r3 = k[A]x [2B]y = 1
1 1
or = ⇒ y = 0
2 2
Hence, the rate law equation is –
Rate = k[A]2 [B]0
∴ Order of reaction is 2 (second order)
338. The dissociation of HI molecule, as shown
below occurs at a temperature of 639K. The
rate constant k = 3.02×10-5 M-1 s-1
2HI(g) → H 2 (g) + I 2 (g)
What is the reaction order ?
(a) 0
(b) 1
(c) 2
(d) 3
SCRA-2012
Ans. (c) : The given rate constant,
k = 3.02 × 10–5 M–1 sec–1
k=
r1
k[A]x [B]y
=
r2 k[2A]x [B]y
1 1
=
4 2
2
x
1 1
=
2 2
x=2
x
Objective Chemistry Volume-II
1 1
=
1 2
y
y
0
1
1
HI → (H 2 ) + I 2
2
2
Rate = k[Concentration]n
k=
Rate
M / sec
=
[Concentration]n
Mn
Order, k = M/sec
Ist order, k = sec–1
IInd order, k =
M / sec
= M −1 sec −1
M2
So, the reaction order is 2.
339. For a first order reaction, the ratio of times to
complete 99.9% and half of the reaction is:
(a) 8
(b) 9
(c) 10
(d) 12
MPPET - 2012
2.303
a
Ans. (c) : t =
log
k
a-x
For first order reaction Let initial concentration (a) = 100
For 99.9% complete reaction
2.303
100
t1 =
log
k
100 − 99.9
2.303
t1 =
log103 ..............(i)
k
2.303
100
t2 =
log
for 50% complete reaction
k
100 − 50
2.303
t2 =
log2 ....................(ii)
k
from equation (i) and (ii)
t1 log 103
=
= 9.96 ≈ 10
t2
log 2
303
YCT
(c) a straight line passing through origin
340. The reaction, 2NO(g) + O2 (g) → 2NO2(g), is
second order with respect to NO and first order
(d) a curve.
with respect to O2. If the volume of reactant is
SRMJEEE – 2009
suddenly reduced to half value, the rate of Ans. (b) : In a first order reaction the expression of rate
reaction would be:
constant is :
(a) one-fourth of original value
2.303
a
(b) one-eighth of original value
k=
log
t
a
−
x
(c) eight times of original value
(d) four times of original value
or
kt = 2.303 log a − log ( a − x )
MPPET - 2012
kt
or
= log a – log (a – x)
Ans. (c) :
2.303
2NO(g) + O2(g)→ 2NO2(g)
−kt
R1 = k [NO]2 [O2]1.............(i)
log (a – x) =
+ log a
....(i)
2
1
2.303
R2 = k [2NO] [2O2] .......... (ii)
Comparing the equation from y = mx+c
2
k [ NO 2 ] [ O 2 ]
R1
The plot of against log (a – x) gives a straight line with
=
−k
R 2 k [ 2NO ]2 [ 2O 2 ]
slope equal to
2.303
1 1
1
R1
=
=
2
R 2 ( 2) ( 2) 8
⇒ R2 = 8R1
On making the volume half, concentration doubles
and rate of reaction becomes eight times.
341. For the chemical reaction A → B it is found
that the rate of the reaction doubles when the
concentration of A is increased four times. The
344. Which one of the following is a second order
order in A for this reaction is
reaction?
(a) Two
(b) One
(a) 2NO + O2 → 2NO2
(c) Zero
(d) Half
(b)
Dissociation of HI on the surface of gold
MPPET- 2009
n
(c)
Dissociation
of H2O2 in solution
Ans. (d) : Rate of reaction (r1) = k [A] ----- (i)
n
(d)
K
S
O
+
2KI
→ 2K2SO4 + I2
2
2
8
(r2) = 2r1 = k (4A) ----(ii)
SRMJEEE – 2009
Divide equation (2) by (1) we get –
Ans. (d) : (a) The reaction
n
r2 k(4A)
=
2NO(g) + O2(g)
→ 2NO2(g)
r1
k(A) n
is second order in NO and first order in O2
r2
(b) Dissociation of HI on the surface of gold is the zero
= 4n
order reaction.
r1
(c) Dissociation of H2O2 in solution is the first order
2r1
reaction.
= 4n
(∴ 2r1 = r2)
r1
(b) K2S2O8 + 2KI
→ 2K2SO4 + I2
21 = 22n
The given reaction follows the second order kinetics.
∴
2n =1
345. Identify the formula which is applicable to the
n =1/2
conversion of 20% of the initial concentration
So, the order of reaction is half .
of the reactant to the product in a first order
reaction. (Rate constant = k)
342. The order of photochemical reaction between
H2 and Cl2 is
2.303
100
(a) t20% =
log
(a) Zero
(b) First
5
20
(c) Second
(d) Third
2.303
100
SCRA-2010
(b) t20% =
log
Sunlight
20
k
Ans. (a) : H 2 + Cl 2 → 2HCl
2.303
5
The order of all photochemical reaction is zero because
(c) t20% =
log
it does not depend upon the concentration of reactants.
k
4
343. In a first order reaction, the plot of log(a – x)
2.303
k
against times 't' gives
(d) t20% =
log
100
80
(a) a straight line with positive slope
SRMJEEE – 2012
(b) a straight line with negative slope
Objective Chemistry Volume-II
304
YCT
Ans. (c) : For the first order Kinetics –
2.303
a
k=
log
t
a−x
Given,
Rate constant = k
a × 20 a
x=
=
100
5
2.303
a
∴ k=
log
a
t 20%
a−
5
2.303
a
k=
log
4a
t 20%
5
2.303
5
or t20% =
log
k
4
346. The reaction 2NO(g) + O2 (g) → 2NO2(g) is of
first order. If volume of reaction vessel is
reduced to 1/3, the rate of reaction would be
(a) 1/3 times
(b) 2/3 times
(c) 3 times
(d) 6 times
Ans. (c) : Given reaction 2NO(g) + O2(g) → 2NO2(g)
1
When the volume of vessel is reduced to then
3
concentration of reactant becomes three times. The rate
of reaction for first order is proportional to
concentration.
Rate of reaction(R1 )
k[NO 2 ]
=
Rateof reaction(R 2 ) k × 3[NO 2 ]
R2 = 3R1
347. The half-life of two samples are 0.1 and 0.8 s.
Their respective concentration are 400 and 50
respectively. The order of the reaction is
(a) 0
(b) 2
(c) 1
(d) 4
VITEEE-2014
Ans. (b) : Given that,
( t1/ 2 )1 = 0.1 sec, ( t1/ 2 ) = 0.8 sec
a1 = 400,
We know that
( t1/ 2 )1
( t1/ 2 )2
a2 = 50
n −1
a
= 2
a1
Where n = order of the reaction
( n −1)
0.1 50
=
0.8 400
Taking log on both sides −
0.1
50
log
= ( n − 1) log
0.8
400
1
1
log = ( n − 1) log
8
8
n −1 = 1
⇒n=2
Objective Chemistry Volume-II
348. Acid hydrolysis of ester is first order reaction
and
rate
constant
is
given
by
2.303
V − V0
k=
log ∞
where, V0,Vt and V∞ are
t
V∞ − Vt
the volume of standard NaOH required to
neutralize acid present at a given time, if ester
is 50% neutralized then
(b) V∞ = ( Vt − V0 )
(a) V∞ = Vt
(c) V∞ = 2Vt − V0
(d) V∞ = 2Vt + V0
VITEEE-2011
Ans. (c) : Given the hydrolysis of ester –
+
H
RCOOR '+ H 2O
→ RCOOH + R 'OH
At t = 0
At t= t
At t = ∞
At
or
x
0
0
x–α
α
α
x–x
x
x
t = 0, V0 = Volume of NaOH due to H+
Vt = x + V0
V∞ = α + V0
x
If ester is 50% hydrolyzed then α =
2
x
Vt = + V0
2
x = 2Vt − V0
∴
V∞ = 2Vt − 2V0 + V0
or
V∞ = 2Vt − V0
349. If a plot of log10C versus t gives a straight line
for a given reaction, then the reaction is
(a) zero order
(b) first order
(c) second order
(d) third order
VITEEE- 2008
Ans. (b) : Any first order reaction follows the equation
−k
log [ A ] =
t + log [ A ]o
2.303
∵ it resembles equation of straight line
y= mx + C
y = log [A] i.e. log10 C
−k
m=
if x = t & C = log [ A ]o
2.303
Hence, the plot is for a 1st order reaction.
350. The order of the reaction
1
N 2O5 → N 2O 4 (g) + O 2 (g)is
2
(a) 3
(b) 2
(c) 1
(d) 0
VITEEE- 2006
Ans. (c) : It is a first order reaction because rate of
reaction ∝ [N2O5]
305
YCT
351. Which one of the following plots is correct for a
first order reaction?
(b)
(a)
(c)
(d)
AP-EAMCET (Engg.) 2013
Ans. (c) :
For first order reaction
2.303
a
k=
log
t
a−x
kt
= log a − log ( a − x )
2.303
−k
or log ( a − x ) =
t + log a
2.303
It is similar to straight line equation where intercept is
−k
equal to log a and slope equal to :
2.303
−k
2.303
352. For a first order reaction at 27°C, the ratio of
time required for 75% completion to 25%
completion of reaction?
(a) 3.0
(b) 2.303
(c) 4.8
(d) 0.47
AP EAMCET (Engg.)-2009
Ans. (c): For first order reaction.
2.303
a
k=
log10
t
a
–
( x)
slope =
2.303
a
log10
k
(a – x )
For t75%,a = 100%,
a – x = 25%
For t25%,a = 100%,
a – x = 75%
2.303
100
log10
t 75%
k
25 = log10 ( 4 )
=
t 25% 2.303 log 100
4
log10
10
k
75
3
2log10 ( 2 )
=
2log10 ( 2 ) – log10 ( 3)
= 4.8
353. Which one of the following statements is
correct for the reaction?
CH3COOC2H5(aq)
+
NaOH(aq)
→
CH3COONa(aq) + C2H5OH(aq)
t=
Objective Chemistry Volume-II
(a)
(b)
(c)
(d)
Order is two but molecularity is one
Order is one but molecularity is two
Order is one but molecularity is one
Order is two but molecularity is two
AP-EAMCET (Engg.)-2012
Ans. (d) : For the given reaction,
CH3COOC2 H5 ( aq ) + NaOH ( aq )
→ CH 3COONa ( aq ) +
C2 H5 OH ( aq )
∴ Molecularity is two because two molecules are taking
part.
And rate of the reaction is
r = k [ CH 3COOC2 H5 ][ NaOH ]
∴ order of reaction = 1+1=2 (second order)
354. In a first order reaction, the concentration of
the reactant decrease from 0.6 M to 0.3 M in 15
min. The time taken for the concentration to
change from 0.1 M to 0.025 M in minutes is
(a) 1.2
(b) 12
(c) 30
(d) 3
AP-EAMCET (Engg.) -2014
Ans. (c) : For first order2.303
a
∵k =
log
t
a−x
Where A0 = a
A = a−x
2.303
A
k=
log 0
t
A
2.303
0.6
=
log
15
0.3
2.303
=
log 2 _______ ( i )
15
2.303
A
Put the value of k in t =
log 0
k
A
A0
2.303 15
×
t=
log
2.303 log 2
A
15
0.1
t=
log
log 2
0.025
15
t=
log 4
log 2
15
2 log 2
=
log 2
= 2×15
t = 30 min
355. Which one of the following is an example for
pseudo first order reaction?
(a) Alkali hydrolysis of an ester
(b) Acid hydrolysis of sucrose
(c) Decomposition of PCl5
(d) Dissociation of HI
COMEDK-2011
Ans. (b) : Pseudo first order reaction are second order
chemical reaction that are found to behave as first order
reactions, eg. of pseudo first order reaction is acid
hydrolysis of sucrose.
306
YCT
356. The time required for 100% completion of zero
order reaction is
a
a
(a)
(b)
k
2k
2a
k
(c)
(d)
k
a
COMEDK-2012
Ans. (a) : Rate constant for zero order reaction is
1
k = [ a – (a – x) ]
t
Where,
a = initial concentration
(a – x) = final concentration
As the reaction is 100% completed (a–x)=0
a
a
Hence, k = or t =
t
k
357. Decomposition of NH3 on the surface of
platinum has a rate constant of 2.5×10–4 mol
dm–3 s–1 at 350 K. The order of reaction is
(a) 0
(b) 1
(c) 3
(d) 2
COMEDK-2012
Ans. (a) : Decomposition of NH3 on the surface of
platinum has a rate constant of 2.5×10–4 mol dm–3 s–1.
The order of reaction is zero and unit of zero order
−1
reaction is mol L sec–1
358. aP + bQ → Products, when [P] is doubled
keeping [Q] constant rate increases 2 times,
when [P] is constant and [Q] is doubled, rate
increases four times. The overall order is
(a) 1
(b) 2
(c) 3
(d) 2.5
COMEDK-2014
Ans. (c) : Given the reaction isaP + bQ
→ Products
Ans. (c) : The properties of molecularity is that
molecularity of a reaction can never be zero and can
never be fractional. Order, rate constant and half- life of
a reaction may be in fraction.
360. For a first order reaction, the initial
concentration of a reactant is 0.05 M. After 45
minutes it is decreased by 0.015 M. Calculate
half reaction time. (t1/2)
(a) 87.42 min.
(b) 25.90 min.
(c) 78.72 min.
(d) 77.20 min.
GUJCET-2011
Ans. (a): Given that,
initial concentration (Ao) = 0.05M
Final concentration (A) = 0.05 – 0.015 = 0.035
t = 45 minutes:
For first order reaction, half life (t1/2) is
0.693
t1/ 2 =
...(1)
k
2.303
A
∵k=
log o
t
A
2.303
0.05
∴k=
log
45
0.035
2.303
k=
× 0.154
45
k = 0.00792
Putting the value of k in equation (1) we get 0.693
t1/ 2 =
0.00792
t1/2 = 87.5 min
361. A reaction is of the first order relative to A and
is of second order relative to B. What will be
the effect on rate if the concentrations of A and
B are doubled?
(a) Velocity remains constant (b) 4 times
(c) 2 times
(d) 8 times
GUJCET-2007
Ans. (d) : The first order reaction and second order
reaction which is related to A and B respectively can be
written as–
A + 2B → Product
The rate expression will be–
r = k[P]a [Q]b ................(i)
When [P] is doubled keeping [Q]
constant rate increase 2 times.
2r = k[2P]a[Q]b.................(ii)
2
When [P] is constant and [Q] is doubled rate increases
r1 = [ A ] [ B]
.....(1)
four times then,
If
concentration
of
A
and
B
is
doubled
then
the rate
4r = k[P]a [2Q]b...............(iii)
constant rate will be –
On dividing equation (ii) by (i) , we get
2
r2 = [ 2A ] [ 2B]
2 = 2a
2
a=1
r2 = 8[ A ] [ B] ( from equation (1) )
On dividing equation (iii) by (i) we get
r2 = 8r1
4=2b
362. For Fourth Order reaction, what is the unit of k?
22 =2b
−3
(a) ( mole / litre )
b=2
∴ Overall order = a + b = 1 + 2 = 3
−3
(b) ( mole / litre ) second
359. For a chemical reaction ….. can never be a
+3
fractional number:
(c) ( mol / litre ) second –1
(a) Order
(b) Half-life
–3
(d) ( mole / litre ) second –1
(c) Molecularity
(d) Rate constant
GUJCET-2008
AP-EAMCET (Medical), 2003
Objective Chemistry Volume-II
307
YCT
Ans. (d) : The unit of rate constant is:
k = ( mol ) ( lit ) sec
Given- n = 4 (for fourth order reactions)
1− n
n −1
∴ k= ( mol )
−1
( lit ) sec−1
−3
3
∴ k= ( mol ) ( lit ) sec−1
1− 4
4 −1
−3
mol
−1
k=
sec
lit
363. Assertion: The order of a reaction can have
fractional value.
Reason: The order of a reaction cannot be written
from balanced equation of a reaction.
(a) If both Assertion and Reason are correct and
the Reason is the correct explanation of
Assertion.
(b) If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
(c) If Assertion is correct but Reason is incorrect.
(d) If both the Assertion and Reason are
incorrect.
[AIIMS-2008]
Ans. (b): Order of reaction can be positive. Negative
and fractional cannot be calculated from balanced
chemical equation of a reaction. Order of reaction can
be calculated from rate low.
Hence, both assertion and reason is correct but reason
doesn't correctly explain the assertion.
364. A reaction which is of first order w.r.t. reactant
A, has a rate constant 6 min-1. If we start with
[A] = 0.5 mol L-1, when would [A] reach the
value of 0.05 moL-1
(a) 0.384 min
(b) 0.15 min
(c) 3 min
(d) 3.84 min
[AIIMS-2013]
Ans. (a): For first order reaction,
a
kt = 2.303log
(a − x )
2.303
0.5
log
6
0.05
t = 0.384 min
365. T50 of a first order reaction is 10 min. Starting
with 10 mol L-1, rate after 20 min is
(a) 0.0693 mol L-1 min-1
(b) 0.0693 × 2.5 mol L-1 min-1
(c) 0.0693 × 5 mol L-2 min-1
(d) 0.0693 × 10 mol L-1 min-1
[AIIMS-2008]
Ans. (b): Given,
t50% = 10 min, [Ao] = 10 mol L-1
t = 20 min
[A] = ?
Half life time for first order reaction
0.693 0.693
k=
=
= 0.0693 min–1
t
10
∵ Concentration at starting (i.e. [A]o) = 10 mol L−1
t=
Objective Chemistry Volume-II
∴ Concentration after 20 min (two half-lives) = 2.5 M
dx
∵ Rate = = k [ A ]
dt
Therefore, Rate = 0.0693 × 2.5 mol L-1 min-1
366. The first order rate constant for a certain
reaction increases from 1.667 × 10-6 s-1 at 727°C
to 1.667 × 10-4 s-1 at 1571°C.
The rate constant at 1150°C, assuming
constancy of activation energy over the given
temperature range is (Given : log 19.9 = 1.299)
(a) 3.911 × 10-5 s-1
(b) 1.139 × 10-5 s-1
-5 -1
(c) 3.318 × 10 s
(d) 1.193 × 10-5 s-1
[AIIMS-2009]
Ans. (c): Given,
k1 = 1.667x10-6s–1,
T1 =727+273=1000K
k2 = 1.667x10−4 s-1 ,
T2 = 1571+273 = 1844K
k3 = ?
T3 = 1150+273 = 1423K
k
Ea 1
1
log10 2 =
−
k1 2.303R 1000 1844
Ea
1.667 ×10−4
844
=
−6
1.667 ×10
2.303R 1844 × 1000
E a 2 × 2.303 × 1844 × 1000
=
R
844
1
k 3 2 × 2.303 × 1844 ×1000 1
log10
=
−
k1
844 × 2.303
1000 1423
log10
= 4369.67 ×
422
= 1.296
1423 ×1000
k3
= 1.29
1.677 ×10−6
k3
= anti log(1.29) = 19.9
1.677 ×10−6
k 3 = 19.9 × 1.667 × 10 −6
log10
k 3 = 3.318 ×10 −5 s −1
367. For a first order reaction, to obtain a positive
slope, we need to plot (where [A] is the
concentration of reactant A)
(a) – log10 [A] v/s t
(b) – loge[A] v/s t
(c) log10 [A] v/s log t
(d) [A] v/s t
[AIIMS-2008]
Ans. (b): According to first order reaction,
[A ]
kt = log e o = log e [ A o ] − log e [ A ]
[A]
− log e [ A ] = kt − log e [ A o ]
For straight line graph, y = mx + c
Where y = –loge[A] and x = t
So, we need to plot –loge [A] v/s t plot.
368. For the second order reaction,
A+B→Products
When a moles of A reacts with b moles of B, the
rate equation is given by
308
YCT
1
b (a − x)
ln
(a − b ) a (b − x)
When a > b, the rate expression becomes that
of
(a) first order
(b) zero order
(c) unchanged, second order (d) third order
AMU-2012
Ans. (a) : When concentration of one reaction is in
excess then it doesn’t depend on its concentration.
Therefore second order reaction becomes first order
when a > b.
369. For the reaction , 2A +B → C + D the order of
reaction is
(a) one with respect to [B]
(b) two with respect to [A]
(c) three
(d) can’t be predicted.
AMU – 2009
Ans. (d) : Given the reaction.
2A + B → C + D
Hence, the order of reaction is an experimentally
determined value and cannot be predicted from the
reaction.
370. The plot between concentration versus time for
a zero order reaction is represented by
k 2t =
372. The plot concentration vs time or rate vs time
for different order of reaction are given below:
The plot which is not possible to describe an
order of reaction is
(a) I
(b) II
(c) III
(d) IV
AMU–2006
Ans. (b) : The plot of rate vs. concentration describes
the order of the reaction, not the plot of rate vs. time .
373. The rate constant of a reaction is 1.0×107 L2
mol–2 sec–1. The order of this reaction will be,
(a) zero
(b) first
(c) second
(d) third
AMU–2006
th
Ans. (d) : For an n order reaction, unit of rate constant
is (moles)1–n(lit)n–1 (time)–1 . The unit of the given
reaction is L2 mol–2 sec–1.
Hence, the order of the given reaction is 3 .
374. For the first order reaction A → products, the
rate constant is 60×10-4 sec-1. What
(a)
(b)
concentration of A would gives rate of 2.4 × 10-3
mol L-1sec-1?
(a) 0.4 mol L-1
(b) 2.5 mol L-1
-5
-1
(c) 1.44×10 mol L
(d) 1.44×10-2 mol L-1
(c)
(d)
AMU-2004
AMU – 2009
Ans. (a) : k = 60×10–4sec–1
Ans. (d) : The plot of concentration of the reactant
–3
–1
–1
versus time for a zero order reaction gives a straight line Rate of reaction =2.4×10 mol L sec
For first order reaction,
with k-slope as mentioned below in given figure.
Rate of reaction = k[A]
2.4 × 10−3
= 0.4 mol L–1
60 ×10−4
375. For a reaction A → B, it is found that rate of
the reaction doubles when the concentration of
A is increased four times. The order of the
reaction is
(a) two
(b) half
(c) one
(d) zero
Assam CEE-2014
371. The rate constant of a reaction is 3.00×103 L
mol–1 sec–1 . The order of this reaction will be
Ans. (b) :
(a) zero
(b) first
Rate (r1) = k[A]n (where, n is order of reaction ) …..(i)
(c) second
(d) third
2r1 = k[4A]n ……(ii)
AMU – 2007 On solving equation (i) and (ii) :n
Ans. (c) : The rate constant of reaction is
k [A ]
r1
3.00×103 L mol–1 sec–1
=
2r1 k [ 4A ]n
For an nth order reaction, unit of rate constant is
(moles)1–n(lit)n–1 (time)–1 . The unit of the given reaction
n
1 1
is L mol–1 sec–1.
=
2 4
Hence, the order of the given reaction is 2 .
[A]=
Objective Chemistry Volume-II
309
YCT
Thus, 8 mol change to 4 mole in = 4 min
And 4 mol change to 2 mole in = 2 min
Thus, total time = 6 min.
379. For the reaction H2(g) + Br2(g) ⇌ 2HBr(g),
1
The rate law is rate = k[H2][Br2]1/2
n=
Which of the following statement is true about
2
this
reaction?
376. The first order rate constant for the
(a) The reaction is of second order
decomposition of N2O5 is 6.2 × 10–4s–1. The half(b) Molecularity of the reaction is 3/2
life in seconds is nearly
(c) The unit of k is s–1
(a) 1117.7
(b) 111.7
(d) Molecularity of the reaction is 2
(c) 223
(d) 160
[BITSAT – 2006]
BCECE-2009
–4 –1
Ans.
(d)
:
Molecularty
is
calculated
by
stoichiometry of
Ans. (a) : Given :- k = 6.2x10 s , t1/2 = ?
reactants
and
order
of
reaction
is
calculated
from rate
For first order reaction,
law.
0.693
For the given reaction,
We know that t1/ 2 =
k
H 2 ( g ) + Br2 ( g ) ⇌ 2HBr ( g )
0.693
1
t1/ 2 =
6.2 × 10−4
Rate = k [ H 2 ][ Br2 ] 2
t1/ 2 = 1117.7sec
1 3
Order = 1 + =
377. 3A → B + C
2 2
It would be zero order reaction when
Molecularity = 1 + 1 = 2
(a) The rate of reaction is proportional to square 380. The reaction 2N O ⇌ 2N O + O is
2 5
2 4
2
of concentration of A
(a) Bimolecular and second order
(b) The rate of reaction remains the same at any
(b) Unimolecular and first order
concentration of A
(c) Bimolecular and first order
(c) The rate remains unchanged at any
(d)
Bimolecular and zero order
concentration of B and C
[BITSAT – 2005]
(d) The rate of reaction doubles if concentration
of B is increased to double
Ans. (c) : For given reaction, 2N 2O5 ⇌ 2N 2O 4 + O 2
BCECE-2012
Rate = k[N2O5]
Ans. (b) : For zero order reaction,
Order = 1
Molecularity = 2
dx
or Rate = k[A]o
Hence,
it
is bimolecular and first order reaction.
dt
381. For a reaction A → B, the rate increases by a
Rate = k
factor of 2.25 when the concentration of A is
Rate of reaction remains the same at any concentration
increased by 1.5. What is the order of the
of A.
reaction?
378. For the zeroth order reaction, sets I and II are
(a) 3
(b) 0
given, hence x is –
(c) 2
(d) 1
[BITSAT – 2008]
Ans. (c) : Given,
r2
A
= 2.25 and 2 = 1.5
r1
A1
Rate = k [A]n
1 1
=
2 2
1 = 2n
2n
r2 [ A 2 ]
n
=
⇒ 2.25 = (1.5 )
r1 [ A1 ]n
n=2
Hence it is second order reaction.
(a) 2 min
(b) 4 min
382. Two substances R and S decompose in solution
(c) 6 min
(d) 8 min
independently, both following first order
BCECE-2013
kinetics. The rate constant of R is twice that of
Ans. (c) : By set I :S. In an experiment, the solution initially
Half-life is 2 min.
contained 0.5 millimoles of R and 0.25 of S. The
In set II, number of moles have been doubled thus halfmolarities of R and S will be equal Just at the
life is also doubled i.e. now it is 4 min.
end of time equal to
n
Objective Chemistry Volume-II
310
YCT
(a)
(b)
(c)
(d)
∵1 − n = 1
twice the half life of R
twice the half life of S
the half life of S
the half life of R
[BITSAT – 2014]
Ans. (a) : Given :- Rate constant of R (kR) = 2×Rate
constant of S (kS)
Moles of R = 0.5 m mol
Moles of S=0.25 m mol
Molarity of R = Molarity of S
For 1st order reaction, t1/2 is independent of initial
concentration of reactants.
0.693
k
0.693 0.693
=
( t1/ 2 )R =
kR
2k S
t1/ 2 =
( t1/ 2 )S =
0.693
kS
1
2 ( t1/ 2 )S
Hence, half life time of S is twice the half life of R.
383. The thermal decomposition of a compound is of
first order. If a sample of the compound
decomposes 50% in 120 min, what time will is
take to undergo 90% decomposition?
(a) Nearly 400 min
(b) Nearly 45 min
(c) Nearly 480 min
(d) Nearly 240 min
CG PET- 2011
Ans. (a) : Given,
t50% = 120 min
t90% = ?
For first order reaction,
0.693
t50% =
k
0.693
k=
min −1
120
2.303
a
For
t 90% =
log10
k
(a − x )
( t1/ 2 )R =
2.303
100
log
0.693 10 10
120
= 400 min
384. If the initial concentration of the reactant is
doubled, the time for half reaction is also
doubled. The order of the reaction is
(a) zero
(b) one
(c) two
(d) three
CG PET- 2012
Ans. (a) : Generally half life time
t1/2 ∝ a(1–n)
if initial concentration is doubled then half life is also
doubled, therefore,
t1/2 ∝ a
=
Objective Chemistry Volume-II
n=0
Hence order of reaction is zero. Half-life time for zero
order reaction is
a
t1/2=
2Lk
385. A first order reaction completed its 20% in 200
minute. How much time it will take to complete
its 80%?
(a) 400 min
(b) 800 min
(c) 1400 min
(d) 1000 min
CG PET- 2012
Ans. (c) : For the first order reaction,
2.303
a
k=
log
t
(a − x)
For 20% completion of the reaction,
2.303
100
k=
log
200
(100 − 20)
2.303
k=
log1.25
200
2.303× 0.0969
k=
200
For 80% completion of the reaction,
2.303
100
k=
log
t
(100 − 80)
2.303
k=
log 5
t
2.303× 0.699
k=
t
Putting the value of K, we get,
2.303× 0.0969 2.303× 0.699
⇒
=
200
t
200× 0.699
t=
0.0969
t = 1442 min ≈ 1400 min
386. 2A → B + C
It would be a zero order reaction when
(a) the rate of reaction is proportional to square
of concentration A
(b) the rate of reaction remains same at any
concentration of A
(c) the rate remains unchanged at any
concentration B and C
(d) the rate of reaction doubles if concentration of
B is increased to double
CG PET-2007
Ans. (b) : For zero order reactionRate = k[A]0
Rate = k
So, rate doesn’t depends on the concentration of A.
Hence, the rate remains unchanged at any concentration
of A.
311
YCT
(c) Inversely proportional to concentration
387. The reaction 2NO 2 ( g ) + O 2 ( g ) ↽ ⇀ 2NO 2 ( g ) is
(d) Inversely proportional to the square of the
of first order. If volume of reaction vessel is
concentration
reduced to 1/3, the rate of reaction would be
GUJCET-2014
1
2
Ans.
(b)
:
Half-life
period
of
first
order
reaction is
(a) times
(b) times
3
3
independent of initial concentration of reactants
(c) 3 times
(d) 6 times
0.693
t1 =
2
CG PET-2005
k
391. Higher order (>3) reactions are rare due to
Ans. (c) : For 1st order reaction,
(a) low probability of simultaneous collision of
mole
Rate = k[A]
⇒r=k
all the reacting species
v
(b) increase in entropy and activation energy as
more molecules are involved
1
v
If volume is reduced to
i.e.
then
(c)
shifting
of equilibrium towards reactants due
3
3
to elastic collisions
(d) loss of active species on collision
mole
mole
r' = k
=
3k
JEE Main-2015
v
v
Ans. (a) : Higher order reactions are rare due to low
3
probability of simultaneous collision with proper
r' = 3k [A]
orientation of reacting species.
r' = 3r
392. H2 gas is adsorbed on the metal surface like
Rate of reaction increases by a factor of 3.
tungsten. This follows ........ order reaction.
(a) third
(b) second
388. Unit of first order rate constant is
−1 −1 −1
(c)
zero
(d) first
(a) mo1 Ls
(b) mol L s
[AIEEE-2002]
(d) s −1 , min −1 etc.
(c) mol −1s −1
Ans. (c) : Adsorption of H2 gas on metal surface like
CG PET-2006 tungsten follows zero order reaction.
393. The following mechanism has been proposed
Ans. (d) : Unit of rate constant = (mol L–1)1–n s–1 or
for the reaction of NO with Br2 to form NOBr
(mol L)1–n min–1
NO(g)+Br2(g) ↽ ⇀ NOBr2(g)
Where n= order of reaction
st
For 1 order reaction, n = 1
NOBr2(g)+ NO(g)
→ 2NOBr(g)
Unit of rate constant = (mol–1)1–1 s–1
If
the
second
step
is
the
rate determining step,
Unit of k = s–1 or min–1
the order of the reaction with respect to NO(g)
is
389. The expression for velocity constant for the
(a) 1
(b) 0
second order reaction is
(c)
3
(d) 2
2.303
a
1
x
log10
(b) k =
(a) k =
[AIEEE-2006]
t
a−x
t a [a − x ]
Ans. (d) : Order of reaction calculated from power of
concentration of reactants in rate determining step on
1
x2
(c) k = 2
(d) None of the above
2
slowest step.
t a [a − x ]
For
NOBr2 (g) + NO(g) → 2NOBr(g) (Slow )
CG PET-2006
Rate
= k [ NOBr2 ][ NO ] ____(i)
Ans. (b) : Integrated rate expression for second order
NOBr2 Is an intermediate so rate doesn’t depends on it.
reaction is ,
Therefore,
1
1
kt =
−
[ NOBr2 ]
[A t ] [Ao ]
k eq =
NO ][ Br2 ]
[
Let
[At] = (a-x)
[Ao] = a
kt =
1
1
x
− =
(a − x ) a a (a − x )
[ NOBr2 ] = k eq [ NO ][ Br2 ] _____(ii)
Putting the value of [ NOBr2 ] on equation (i)
k=
x
t.a ( a − x )
Rate
= k1 [ NO ] [ Br2 ]
2
Order of reaction with respect to NO is 2.
390. The half life period for a first order reaction 394. t can be taken as the time taken for the
1/4
is….
concentration of a reactant to drop to 3/4 of its
(a) Proportional to concentration
initial value. If the rate constant for a first
order reaction is K, the t1/4 can be written as
(b) Independent of concentration
Objective Chemistry Volume-II
312
YCT
(a) 0.75/k
(c) 0.29/k
(b) 0.69/k
397. Certain reactions follow the relation between
concentrations of the reactant vs time as
(d) 0.10/k
[AIEEE 2005]
Ans. (c) : For first order reaction
2.303
a
t=
log10
k
a
−
( x)
3
a = a (a – x) = a
4
2.303
a
2.303
4
=
t1 =
log10
log10
4
3
k
k
3
a
4
0.29
t1 =
4
k
395. The following data was obtained for chemical
reaction given below at 975 K.
2NO(g)+2H2(g) → N2(g)+2H2O(g)
[NO]
[H2]
[Rate]
mol L–1
mol L–1
mol L–1
A.
8×10–5
8×10–5
7×10–9
–5
–5
B.
24×10
8×10
2.1×10–8
–5
–5
C.
24×10
32×10
8.4×10–8
The order of the reaction with respect to NO is
......... [Integer answer]
JEE Main-2021
Ans. (1) :
For reaction,
2NO(g) + 2H 2 (g) → N 2 (g) + 2H 2 O(g)
If
Rate = k [ NO ] [ H 2 ]
x
y
7 ×10−9 = k 8×10−5 8×10−5 ___(i)
x
y
2.1×10−8 = k 24×10−5 8×10−5 ____(ii)
x
398. For the reaction R → P a graph of [R] against
time is found to be a straight line with negative
slope. What is the order of reaction?
(a) Second order
(b) Third order
(c) First order
(d) Zero order
J & K CET-(2010)
Ans. (d) : For zero order reactions
kt = [ R o ]−[ R ]
[ R ] = [ R o ]− kt
y
8.4×10−8 = k 24×10−5 32×10−5 _____(iii)
On dividing equation (ii) from equation (i)
x
What is the expected order for such reactions?
(a) 0
(b) 1
(c) 2
(d) Infinity
J & K CET-(2012)
Ans. (b) : For first order reaction
A t = A o e−kt
Concentration of reactants varies exponentially with the
time
y
For straight line graph, y = mx + C
Where y = [ R ], x = t,
C = [R o ]
Slope = m = −k
In
[ R ] V S time ( t )
⇒
x =1
Negative slope is obtained.
Hence order of the reaction w.r.t NO is 1
399. The rate constant for a first order reaction is
396. Consider following two reactions,
6.909 min–1. Therefore, the time required in
minutes for the participation of 75% of the
d [ A]
0
A
→ Product; –
= k1 [ A]
initial reactant is
dt
2
2
d [B ]
(a) log 2
(b) log 4
B
→ Product; –
= k 2 [B]
3
3
dt
3
3
k1 and k2 are expressed in terms of molarity
(c) log 2
(d) log 4
(mol L–1) and time (s–1) as
2
2
(a) s–1, Ms–1L–1
(b) Ms–1, Ms–1
J & K CET-(2009)
(c) s–1, M–1 s–1
(d) Ms–1, s–1
Ans. (a) : Given:[AIEEE 2008]
k = 6.909 min −1 ,
t 75% = ?
Ans. (d) : k1 is rate constant for zero order reaction unit
Let
a = 100%,
(a − x ) = 25%
of rate constant for zero order reaction is
For
first
order
reaction,
mol L−1 s−1 or Ms−1
2.303
a
t=
log10
k2 is rate constant for first order reaction unit of rate
–1
k
a
−
x)
(
constant k2 = s
3 = (3)
x
Objective Chemistry Volume-II
313
YCT
2.303
100
log10
6.909
25
2
t 75% = log 2
3
400. Inversion of cane-sugar in dilute acid is
(a) bimolecular reaction
(b) pseudo uni-molecular reaction
(c) uni-molecular reaction
(d) tri-molecular reaction
J & K CET-(2007)
Ans. (b) : Inversion of cane-sugar in dilute acid follows
pseudo uni-molecular reaction.
t 75% =
r = ( 2) r
y
1 = ( 2)
y=0
When concentration of C is doubled
y
r '" = k [ A ] [ B] [ 2C ] = 4r
x
y
z
4r = ( 2) r
z
z=2
Hence, Rate = K [ A ][C ]
Order of reaction = 1 + 2 = 3
403. The rate constant of a reaction is found to be
H+
3×10–3 mol L–1 min–1. The order of reaction is
C12 H 22 O11 + H 2O
→ C6 H12 O 6 + C6 H12 O 6
(a) zero
(b) 1
401. For the first order reaction half life is 14 sec,
(c) 2
(d) 1.5
the time required for the initial concentration
J & K CET-(2006)
to reduce to 1/8 of its value is
Ans. (a) : By seeing the value of rate constant we said
(a) (14)3 sec
(b) 28 sec
that the order of reaction is zero.
1− n
1
(c) 42 sec
(d) (14)2 sec
mol
mol
−1
−1
k=
J & K CET-(2007)
min =
min
liter
liter
Ans. (c) : Given:n=0
t 1 = 14sec
404. The half life period of a reaction is independent
2
of initial pressure. Predict the order of the
For first order reaction,
reaction.
0.693 0.693
(a) 0
(b) 1
k=
=
t1
14
(c)
2
(d) 3
2
J & K CET-(2001)
2.303
a
t=
log10
Ans. (b) : In initial concentration of reactants doesn't
k
(a − x )
depends on half-life period, then reaction follows first
order kinetics.
2.303
1
=
log10
0.693
1
0.693
t1 =
2
k
8
14
t 1 ∝ (a)º
2.303×14
2
=
×3log 2
405. The concentration of a reactant X decreases
0.693
from 0.1 M to 0.005 M in 40 min. If the
= 42sec
reaction follows first order kinetics, the rate of
402. A reaction involving A, B and C as reactants is
the reaction when the concentration of X is 0.01
found to obey the rate law, rate = K [A]x[B]y
M will be
[C]z. When the concentrations of A, B and C
(a) 1.73×10-4 M min-1 (b) 3.47×10-4 M min-1
are doubled separately, the rate is also found to
(c) 3.47×10-5 M min-1 (d) 7.53×10-4 M min-1
increase two, zero and four times respectively.
JCECE - 2009
The overall order of the reaction is
Ans. (d) : Given: [Ao] = 0.1M, [At] = 0.005M t = 40
(a) 1
(b) 2
min
(c) 3
(d) 4
For first order reaction.
J & K CET-(2008)
2.303
[A ]
k=
log10 o
x
y
z
Ans. (c) : Rate = k [ A ] [ B] [C ] = x
t
[A t ]
When the concentration of A is doubled
2.303
0.1
=
log10
x
y
z
40
r ' = k [ 2A ] [ B] [ C] = 2r
0.005
–2
–1
k = 7.4 × 10 M min
x
2r = ( 2 ) r
Rate of reaction when concentration of x is 0.01 M will
be
x =1
Rate
= k [x]
When the concentration of B doubled,
= 7.4 × 10–2 × 0.01
x
y
z
r" = k [ A ] [ 2B] [ C ] = r
= 7.4 × 10–4 M min–1
Objective Chemistry Volume-II
314
2
YCT
406. The rate constant of a first order reaction at
2.303
100
t=
log
27o C is 10–3 min–1. The temperature coefficient
0.00578
100 − 90
of this reaction is 2. What is the rate constant
2.303
t=
log10 = 398.4 ≈ 398.8 min
(in min–1) at 17oC for this reaction?
0.00578
-3
-4
(a) 10
(b) 5×10
410. For a zero order reaction
(c) 2×10-3
(d) 10-2
(a) t1/2∝ R0
(b) t1/2∝ 1/R0
JCECE - 2009
2
(d) t1/2∝ 1/ R 02
(c) t1/2∝ R 0
Ans. (b):
JIPMER-2012
Rateconstant ( at 27º C )
Ans.
(a)
:
For
a
zero
order
reaction
Temperature coefficient =
Rateconstant ( at 17º C )
[R] = −kt + [C]
−3
When
t = 0, [R] = [R0]
10
2=
∴
[C] = [R0]
k t ºc
∴
[R] = − kt + [R0]
ktºc = 5 × 10–4 min–1
[R ]
407. The order of a reaction with rate equal to When, t = t1/2, [R] = 0
2
-1/2
KC3/2
A CB is
[R 0 ]
∴
− [ R 0 ] = − kt1/ 2
1
(a) 1
(b) −
2
2
[R ]
3
t1/ 2 = 0
∴
(d) 2
(c) −
2k
2
t1/ 2 ∝ [R 0 ]
JCECE - 2010
411. For a reaction, A+B →Product, the rate is
Ans. (a) : Given, r = kCA3/2CB–1/2
given by,
3 1
r = k[A]1/2[B]2
∴
Order of reaction = + −
2 2
What is the order of the reaction?
3 −1
(a) 0.5
(b) 2
=
=1
(c) 2.5
(d) 1
2
JIPMER-2010
408. Order of reaction is decided by
Ans. (c): Order of reaction is calculated from Sum of
(a) temperature
power of concentration of reactants in rate law.
(b) mechanism of reaction
r = k [A]1/2 [B]2
(c) molecularity
1 5
(d) pressure
Order of reaction = 2 + = = 2.5
2 2
JCECE - 2011
Ans. (b) : Order of reaction is decided by the 412. For a reaction, the half–life is independent of
initial concentration. What is the order of that
mechanism of the reaction. It depends upon the slowest
reaction?
step of reaction (which is also called the rate
(a) Zero
(b) One
determining step of the reaction).
(c)
Two
(d) Three
409. The thermal decomposition of a molecule
JIPMER-2004
shows first order kinetics. The molecule
decomposes 50% in 120 min. How much time it Ans. (b) : For first order reaction, the half-life is
independent of initial concentration of reactants.
will take to decompose 90%?
(a) 300 min
(b) 360 min
0.693
t1 =
(c) 398.8 min
(d) 400 min
2
k
JCECE - 2012
t 1 ∝ [A ]0
º
2
2.303
a
Ans. (c) : t =
log
413. If 50% of the reactant is converted into a
k
a−x
product in a first order reaction in 25 min, how
(I) When a = 100, x = 50 and t = 120 min
much of it would react in 100 min?
2.303
100
(a) 93.75%
(b) 87.5%
k =
log
120
100 − 50
(c) 75%
(d) 100%
2.303
Karnataka-CET-2013
k=
log 2
Ans. (a) : Given,
120
t1/2 = 25 min
2.303 × 0.3010
k=
= 0.00578min−1
0.693
120
t1/2 =
(II) When a = 100, x = 90, t = ?
K
Objective Chemistry Volume-II
315
YCT
0.693
2.303
10
=
log10
…….(i)
25
20
4
Now, for first order
t84% = ?
(a – x) = 16
2.303
a
2.303
100
t=
log
t84% =
log10
K
(a − x)
k
16
a
t × k 100 × 0.693
Put the value of k from equation (i)
log
=
=
(a − x) 2.303 25 × 2.303
2
2.303
10
a
t84% =
log10
4
log
= 4log 2 = log 2
2.303
10
4
log10
(a − x)
20
4
Taking anti log both side
=
20
×
2
a
= 24 = 16
= 40 min
(a − x)
416. Time required for 100 percent completion of a
If intial conc = 100
zero order reaction is
100
= 16
a
2k
100 − x
(a)
(b)
x = 93.75%
2k
a
414. At 300 K, a gaseous reaction A → B + C was
a
(c)
(d) ak
found to follow first order kinetic. Starting
k
with pure A, the total pressure at the end of 20
Karnataka-CET, 2010
min was 100 mm of Hg. The total pressure
after the completion of the reaction is 180 mm Ans. (c) : For zero order reactions,
of Hg. The partial pressure of A (in mm of Hg)
[A] = –kt + [A]0
is
where, [A]0 = initial concentration = a
(a) 90
(b) 180
[A] = remaining concentration = a – a = 0
(c) 80
(d) 100
On putting values of [A]0 and [A], we get
Karnataka-CET-2012
a
Ans. (c) : For the given reaction
t=
k
A
→ B+C
At , t = 0
Po
0 0
417. For a chemical reaction A → B the rate of the
At, t = 20 min
Po − P
P P
reaction is 2 × 10–3 mol dm–3 s–1, when the
At, t = ∞
0
Po Po
initial concentration is 0.05 mol dm–3. The rate
PT =100 mm of Hg
of the same reaction is 1.6 × 10–2 mol dm–3 s–1
PT' ' = 2 =180mm of Hg
when the initial concentration is 0.1 mol dm–3.
PT = (Po – P) + 2P = Po + P = 100
…..(1)
The order of the reaction is
PT = 2Po = 180
(a) 2
(b) 0
Put the value of Po in equation (i)
(c)
3
(d)
1
100 = 90 – P
Karnataka-CET,
2009
P = 10 mm of Hg
Ans. (c) : Given that,
Partial pressure of A = Po – P
= 90 – 10 = 80 mm of Hg
A = 0.05 mol dm–3
415. A first order reaction is 60% complete 20 Let the rate equation for the reaction be –
minutes. How long will the reaction take to be A → B
84% complete?
Rate = k (A)n
(a) 68 mm
(b) 40 mm
(Rate)1 = k (0.05)n = 2 × 10–3 mol dm–3 s–1 … (i)
(c) 76 mm
(d) 54 mm
n
–2
–3 –1
… (ii)
Karnataka-CET-2012 (Rate)2 = k (0.1) = 1.6 × 10 mol dm s
Dividing
the
Eq.
(ii)
by
Eq.
(i).
Ans. (b) : Given that,
t60% = 20min
(Rate) 2
k(0.1)n
1.6 × 10−2
=
=
For 60% completion,
(a – x) = 40%
(Rate)1 k(0.05) n
2 × 10−3
According to 1st order reaction,
(2)n = 8 or (2)n = 23
2.303
a
k=
log10
∴
n=3
t 60%
(a − x)
Order of reaction = 3
100
2.303
418.
For the decomposition of a compound AB at
k=
log10
20
(40)
600K, the following data were obtained
k=
Objective Chemistry Volume-II
316
YCT
[AB] mol dm3
Rate of decomposition of
AB in mol dm–3 s–1
0.20
2.75 × 10–8
0.40
11.0 × 10–8
0.60
24.75 × 10–8
The order for the decomposition of AB is
(a) 1.5
(b) 0
(c) 1
(d) 2
Karnataka-CET, 2009
Ans. (d) : AB → Product
Rate = k [AB]n
(Rate)1 = k [0.20]n = 2.75 × 10–8
… (i)
(Rate)2 = k [0.40]n = 11.0 × 10–8
… (ii)
Dividing Eq. (ii) by Eq. (i),
(Rate) 2 k(0.40) n 11.0 ×10−8
=
=
(Rate)1 k(0.20) n 2.75 ×10−8
2n = 4
or
2n = 22
Hence, n = 2
419. Which of the following is a second order
reaction?
(a) H 2 + Br2
→ 2HBr
(b) NH 4 NO3
→ N 2 + 3H 2 O
sunlight
(c) H 2 + Cl2 →
2HCl
→
(d) CH 3 COOCH 3 + NaOH
CH 3 COONa + H 2 O
Karnataka-CET-2007
Ans. (d) : The reaction is said to be of second order if
its reaction rate is determined by the variation of two
concentration terms of reactants.
CH 3COOH 3 + NaOH
→ CH 3COONa + H 2 O
is an example of second order reaction.
420. For a zero order reaction, the plot of
concentration of reactant vs time is (intercept
refers to concentration axis)
(a) linear with + ve slope and zero intercept
(b) linear with – ve slope and zero intercept
(c) linear with – ve slope and non-zero intercept
(d) liner with + ve slope and non-zero intercept
(e) A curve asymptotic to concentration axis
Kerala-CEE-2008
Ans. (c): For zero order reaction,
[A] =[Ao] – kt
For straight line graph,
y = mx + c
y = [A],
x = t,
c = [Ao],
m = –k
The plot of concentration of reactant v/s time is linear
with negative slope and non-zero intercept.
421. At 500K, the half-life period of a gaseous
reaction at an initial pressure of 80 kPa is 350s.
When the pressure is 40 kPa, the half-life
period is 175s. The order of the reaction is
(a) zero
(b) one
(c) two
(d) three
(e) half
Kerala-CEE-2007
Ans. (a) : To find the order of reaction, the value of rate
constant is same for different conception.
Half - life time of zero-order reaction,
a
t1 =
2
2k
k=
a
2t 1
2
If pressure is used in place of concentration then,
p
k=
2t 1
2
When p = 80kPa
t1/2 = 350s
80 ×10
= 114.28 mol l–1 s–1
2 × 350
When p = 40kPa
t1/2 = 175s
3
k=
40 ×103
= 114.28 mol l–1 s–1
2 × 175
Hence, order of reaction is zero.
k=
422. The reaction 2A + B + C
→ D + E is found
to be first order in A, second in B and zero
order in C. What is the effect on the rate of
increasing concentration of A, B and C two
times?
(a) 72 times
(b) 8 times
(c) 24 times
(d) 36 times
(e) none of the these
Kerala-CEE-2006
Ans. (b) : For the given reaction, 2A + B + C → D + E
Rate = k [A] [B]2 = r
…(i)
If the concentration of A, B and C are doubled then.
r = k[2A] [2B]2
= 2 × 4. k [A] [B]2
= 8 k [A] [B]2
…(ii)
On dividing equation (ii) and equation (i)
r'
=8
r
Rate increased by a factor of 8.
423. The following homogeneous gaseous reactions
were experimentally found to be second order
overall.
1. 2NO
→ N2 + O2
2. 3O 2
→ 2O 3
3. N 2O 3
→ NO + NO 2
4. H 2 + I 2
→ 2HI
Objective Chemistry Volume-II
317
YCT
Which of these are most likely to be elementary
reactions that occur in one step?
(a) 3 only
(b) 1 and 3
(c) 1 and 4
(d) 3 and 4
(e) 1, 2 and 3
Kerala-CEE-2006
Ans. (c) : For elementary reaction,, molecularity of
reaction is equal to the order of reaction.
(1) 2NO → N2 + O2
Rate = k [NO]2
Order molecularity = 2
Hence, it is an elementary reaction.
(2) H2 + I2 → 2HI
Rate = k [H2] [I2]
Order of reaction = 2
Molecularity = 2
Hence, it is also an elementary reaction.
424. Consider the following statements in respect of
zero order reaction.
I. The rate of the reaction is independent of
reactant concentration.
II. The rate of the reaction is independent of
temperature.
III. The rate constant of the reaction is
independent of temperature.
IV. The rate constant of the reaction is
independent of reactant concentration.
Choose the correct statement/s
(a) I only
(b) I and II only
(c) III and IV only
(d) I and III only
(e) I and IV only
Kerala-CEE-2011
Ans. (e) : For zero order reaction
Rate = k [A]°
Rate = k
So, the rate of reaction is independent of reactant
concentration.
Rate constant of reaction depend on the temperature.
According to Arrhenius equation, on increasing
temperature rate constant also increase.
k = A e–Ea/RT
So, state is also depends on temperature for zero order
reaction.
Rate constant of reaction is always independent of
concentration of reactants.
Hence, only option I and IV are correct.
425. The rate of the reaction A → products, at the
initial concentration of 3.24 × 10-2 M is nine
times its rate at another initial concentration of
1.2 × 10–3 M. The order of the reaction is
1
3
(a)
(b)
2
4
3
2
(c)
(d)
2
3
1
(e)
3
Kerala-CEE-2011
Objective Chemistry Volume-II
Ans. (d) : Given that,
r1 = 9 r2,[A1] = 3. 24 × 10–2 M
[A2] = 1. 2 × 10–3 M
n
r1 = k [A1]
….(i)
r2 = k [A2]n
….(ii)
On dividing equation (i) by equation (ii)
r1 A1
=
r2 A 2
n
n
3.24 × 10−2
9=
−3
1.2 × 10
n
9 = (27)
(3)2 = = [(3)3]n
2
n=
3
426. The rate constant of a first order reaction is
doubled when the temperature is increased
from 20oC to 25oC. How many times the rate
constant will increase if the temperature is
raised from 20oC to 40oC?
(a) 4
(b) 8
(c) 16
(d) 32
(e) 64
Kerala-CEE-2012
Ans. (c) : For every 5ºC rise in temperature the rate
constant becomes double. For 20ºC rise (5×4) rise in
temperature rate constant (K) = 24= 16.
427. The reaction, A + B → products is first order
with respect to A and second order with respect
to B. When 1.0 mole each of A and B were
taken in one litre flask, the initial rate of the
reaction is 1.0 × 10-2 mol L-1s-1. The rate of the
reaction when 50% of the reactants have been
converted into products is
(a) 1.00 × 10-3 mol L-1 s-1
(b) 0.05 × 10-2 mol L-1 s-1
(c) 1.25 × 10-3 mol L-1 s-1
(d) 4.00 × 10-2 mol L-1 s-1
(e) 2.00 × 10-3 mol L-1 s-1
Kerala-CEE-2013
Ans. (c) : the reaction, A + B → Product
Rate = k [A] [B]2
A
+
B → Product
At, t = 0
1M
1M
0M
At, t =t
0.5M
0.5M
0.5M
At, t = 0
1.0 × 10–2 = k [1] [1]2
k = 10–2 L2 mol–2 S–1
At, t = t
Rate = 10–2 [0.5] [0.5]2
Rate = 1.25 × 10–3 mol L–1 S–1
428. A first order reaction is given as A → products.
Its integrated equation is
2.303
a−x
1
a
(a) k =
log
(b) k = log
t
a
t
a−x
318
YCT
log 2
log 2
1
a−x
(a)
(b)
(d) −k = log
k
k 0.5
t
a
MHT CET-2011
ln 2
0.693
(c)
(d)
Ans. (c) : According to first order reaction
k
0.5k
–kt
(AIPMT -2007)
(a – x) = Ae
Ans. (c) : Given,
Integrated equation of first order reaction
a
[ Ao ] = 0.5M, t 1 = ?
Kt = ln
2
−
a
x
(
)
Half-life time of first order reaction,
2.303
a
1 [A ]
k=
log10
t 1 = ln o
t
(a − x )
2
k [A L ]
429. The first order integrated rate equation is
1 0.5
t 1 = ln
x
2
0.25
k
(a) k =
t
ln 2
t1 =
2.303
a
2
k
log
(b) k = −
In first order reaction, in half life period concentration
t
a−x
reduced to half.
1
a
(c) k = 1n
433. A reaction is 50% complete in 2 hours and
t a−x
75% complete in 4 hours. The order of reaction
is
1
x
(d) k =
(a) 1
(b) 2
t a (a − x)
(c) 3
(d) 0
MHT CET-2010
(Karnataka NEET-2013)
Ans. (c) : Integrated rate equation of first order Ans. (a) : For first order reaction, t is two times of
75%
reaction,
t 50%
1
a
k = ln
t 75% = 2× t 50%
t (a − x )
t 50% is 2 hrs So, t 75% is 4 hrs i.e. twice of t 50% . Hence it
430. For which order reaction, the unit of rate follows first order kinetics.
constant is time–1 ?
434. Which one of the following statements for the
(a) Zero order
(b) First order
order of a reaction is incorrect?
(c) Second order
(d) third order
(a) Order can be determined only experimentally.
MHT CET-2009
(b) Order is not influenced by stoichiometric
1−n
N−1
coefficient of the reactants.
Ans. (b) : Unit of rate constant = (mol) (L) S−1
(c) Order of a reaction is sum of power to the
For first order reaction
concentration terms of reactants to express
n=1
the rate of reaction.
−1
−1
(d)
Order
of reaction is always whole number.
Unit of rate constant = s or time
(AIPMT -2011)
431. For a reaction, A+2B → C, rate is given by
Ans. (d) : Order of reaction is positive, negative or
d [ C]
+
= k [ A ][ B ] , hence, the order of the having fractional values. Hence, the statement that order
of reaction is always whole number is wrong.
dt
435. Which of the following does not depend upon
reaction is
the concentration of reactants?
(a) 3
(b) 2
(a) Zero order reaction
(c) 1
(d) 0
(b)
First order reaction
MHT CET-2008
(c) Second order reaction
Ans. (b) : Order of the reaction is calculated from the
(d) Third order reaction
rate law. sum of powers of concentration of reactants. in
UP CPMT-2014
rate law tells the order of reaction.
Ans. (a) : In case of zero order reaction rate of reaction
d [C]
does not depend upon the concentration of reactants.
Rate =
= k [ A ][ B]
dt
A → Product
Order of reaction = 1 + 1= 2
Rate = k[A]n
432. In a first-order reaction, A → B, if k is rate When,
n=0
constant and initial concentration of the
rate = k
reactant A is 0.5 M, then the half-life is
(c) k =
2.303
a
log
t
a−x
Objective Chemistry Volume-II
319
YCT
436. It times for the completion of 75% of a reaction
1 1
−
= kt
is 40 min, then 50% of the reaction was
C Co
completed in
[Co − C]
(a) 16 min
(b) 25 min
= kCo t
(c) 18 min
(d) 20 min
C
UP CPMT-2014
[Co − C]
So, the plot of
v/s time t will be linear with
Ans. (d) Given, t 75% = 40 minutes
C
slope kCo (according to y = mx + c).
t =?
50%
Solution- k =
k=
a
2.303
log
t 75%
(a − x)
100
2.303
log
40
(100 − 75 )
2.303
100
log
40
25
k = 0.057 × log 4
k=
[Graph for second order reaction]
438. Units of specific reaction rate for second order
reaction is
k = 0.057 × 0.602
∵ log 4 = 0.602
(a) s–1
(b) mol L–1s–1
−1
2
–2 –1
k = 0.035min
(c) L mol s
(d) L mol–1s–1
UP CPMT-2011
Time required for 50% completion of reaction
Ans.
(d)
:
For
second
order
reaction,
a
2.303
t 50% =
log
dx
k
= k(conc.)2
(a − x )
dt
2.303
100
conc.
1
t 50% =
log
k=
×
0.035
100 − 50
time (conc.) 2
2.303
= conc.–1 time–1
t 50% =
log 2
0.035
= L mol–1s–1
2.303 × .3010
439. The rate constant of a reaction is 2.5×10–2
t 50% =
∵log 2 = .3010
Min–1. The order of the reaction is
0.035
(a) 0
(b) 1
t 50% = 19.8 ≈ 20 min
(c)
2
(d) 3
[C − C]
437. For a certain reaction, a plot of 0
against
UP CPMT-2010
C
th
Ans.
(b)
:
For
n
order
reaction,
the
units of rate
time t, yields a straight line. C0 and C are
constant
concentrations of reactants at t = 0 and t = t
1– n
respectively. The order of reaction is
mol
=
min –1
(a) Zero
(b) 1
L
(c) 2
(d) 3
Given, the unit of rate constant = min–1
UP CPMT-2013 On comparing,
Ans. (c) : A → products
1–n=0
dC A
– n = –1
2
For second order reaction −
= k[C A ]
∴
n=1
dt
∴ The order of the reaction is 1.
On integrating,
1
440. Order of reaction can be
= kt + I (where, I = integration constant)
(a) zero
C
(b) fraction
at t = 0, C = Co
(c) whole number
1
∴I=
(d) integer, fraction, zero
Co
UP CPMT-2007
by putting this value of I in formula
Ans. (d) : The total number of molecules or atoms
1
1
whose concentration determine the rate of reaction as
= kt +
order of reaction.
C
Co
Order of reaction may be zero, negative, positive or in
1 1
fraction and greater than three while molecularity of
−
= kt
C Co
reaction cannot be zero, negative or fractional.
Objective Chemistry Volume-II
320
YCT
441. The following data are for the decomposition of
ammonium nitrite in aqueous solution
Vol. of N2 in cc
Time (min)
6.25
10
9.00
15
11.40
20
13.65
25
35.65
Infinity
The order of reaction is
(a) zero
(b) one
(c) two
(d) three
UP CPMT-2006
Ans. (b) : For first order reaction
2.303
a
k=
log
t
a−x
a ∝ V∞ (= 35.05) and x ∝ Vt
Constant values of k calculated for different times
shows first order reaction.
442. For the reaction system,
2NO (g) + O2 (g) → 2NO2 (g)
Volume is suddenly reduced to half its value by
increasing the pressure on it. If the reaction is
of first order with respect to O2 and second
order with respect to NO; the rate for reaction
will
(a) Diminish to one – fourth f its initial value
(b) Diminish to one-eighth of its initial value
(c) Increase to eight times of its initial value
(d) Increase to four times of its initial value
UPTU/UPSEE-2007
For first order of reaction
2.303
a
t=
log10
k
(a − x )
2.303
1
log10
1
5.48×10−14
3
2.303
=
log10 (3)
5.48×10−14
t 2 = 2×1013 sec
t2 =
( )
3
3
444. Decomposition of H2O2 is a first order reaction.
Initially, solution of H2O2 having half-life time
15 min is 16 volume. When solution becomes 1
volume.
(a) 4 min
(b) 15 min
(c) 30 min
(d) 60 min
UPTU/UPSEE-2012
Ans. (d) : Given:–
t 1 = 15 min, initial volume = Vo = 16
2
Final Volume = 1 =Vt , t = ?
Half-life time of first order reaction,
0.693
k=
t1
2
0.693
k=
min−1
15
2.303
a
k=
log10
t
(a − x )
2.303
16
log
0.693 10 1
15
t = 60 min
445. Consider the following reaction,
–
–
(CH3)3 CCl + OH → (CH3)3 C – OH + Cl If
–
OH does not take part in slow step, the
molecularity and order of the reaction are
respectively,
(a) 2, 2
(b) 2, 1
(c) 1, 2
(d) 1, 1
2
UPTU/UPSEE-2010
r ' = 8 k [O 2 ][ NO ]
Ans. (b) : For the reaction.
r ' = 8x
(CH 3 )3 CCl + OH− → (CH 3 )3 C − OH + Cl−
The rate of reaction will be increased to eight times of
its initial value.
Molecularity = 1 + 1 = 2
443. What is the two third life of a first order
Rate = k (CH 3 )3 CCl
-14 -1
reaction having K = 5.48 ×10 s ?
Order of reaction = 1
(a) 1×1013 s
(b) 2 × 1013 s
Thus in the given reaction, molecularity is 2 and order
is 1.
(c) 8 × 1013 s
(d) 5 × 1014 s
UPTU/UPSEE-2012 446. A reaction proceeds by first order, 75% of this
reaction was completed in 32 min. The time
Ans. (b) : Given,
required for 50% completion is
k = 5.48×10−14 s−1
(a) 8 min
(b) 16 min
t=
Ans. (c) : Rate = k [O 2 ][ NO ] = x
On decreasing the volume by half then
M
Concentration =
V
2
M M
O2
NO 2
Rate = r ' = k
V V
2 2
2
r ' = k [ 2O 2 ][ 2NO ]
2
After
2
1
rd of completion, (a − x ) =
3
3
Objective Chemistry Volume-II
(c) 20 min
321
(d) 24 min
UPTU/UPSEE-2008
YCT
449. Which one of the following is wrong about
Ans. (b) : Given,
molecularity of a reaction?
t 75% = 32 min t 50% = ?
(a) It may be whole number of fractional
For first order reaction, the relation between t75% and
(b)
It is calculated from reaction mechanism
t50%.
(c)
It is the number of molecules of the reactants
t 75% = 2× t 50%
taking part in a single step chemical reaction.
t 75%
(d)
It
is always equal to the order of elementary
t 50% =
reaction
2
32
WB-JEE-2012
=
Ans. (a) : Molecularity is the number of molecules of
2
the reactants taking part in an elementary reaction. It is
= 16min
447. The following graph shows how t1/2 (half-life) of calculated from reaction mechanism. It is always in
a reactant R changes with the initial reactant whole number not in fractional.
concentration a0. The order of the reaction will 450. Acid catalyzed hydrolysis of ethyl acetate
follows a pseudo-first order kinetics with
be
respects to ester. If the reaction is carried out
with large excess of ester, the order with
respect to ester will be
(a) 1.5
(b) 0
(c) 0.5
(d) 1
(a) 0
(b) 1
WB-JEE-2013
(c) 2
(d) 3
Ans.
(b)
:
Acid
catalyzed
hydrolysis
of
ethyl acetate
WB-JEE-2009
follows a pseudo-first order kinetic with respects to
1
Ans. (c) : If graph is plotted for t v/s
a straight line ester. If the reaction is carried out with large excess of
ao
ester, the order with respect to ester will be zero
because the concentration of ester doesn’t participate in
slope is obtained in second order of reaction.
rate law.
Half-life time for second order reaction
451. The rate of a certain reaction is given by, rate
1
1
kt = −
= k [H+]n. The rate increases 100 times when
at ao
the pH changes from 3 to 1. The order (n) of
1
1
the reaction is
kt 1 =
−
ao ao
2
(a) 2
(b) 0
2
(c) 1
(d) 1.5
1
WB-JEE-2014
t1 =
2
ka o
+ n
Ans. (c) : Rate = k H
1
t 1 ∝ is a property of second order reaction.
pH = − log H +
2
a
448. In the hydrolysis of an organic chloride in If
pH = 3 = − log H1+
presence of large excess of water,
RCl +H2O→ROH + HCL
H1+ = 10−3
(a) molecularity and order of reaction both are 2
If
pH = 1 = − log H +2
(b) molecularity is 2 but order of reaction is 1
(c) molecularity is 1 but order of reaction is 2
H +2 = 10−1
(d) molecularity is 1 and order of reaction is also
n
1
k H +2
r2
WB-JEE-2010
= 100 =
n
r1
k H1+
Ans. (b) : For reaction,
RCl + H 2O → ROH + HCl
n
10−1
Molecularity is the number of reactants taking part in a
100 = −3
reaction.
10
n
Molecularity = 2
2
10 = 10−1+3
Rate = k [ RCl ]
10 2 = 102n
Rate of reaction does not depend on [H O] because H O
2
2
2 = 2n
n=1
is present in excess.
Order of reaction = 1
Objective Chemistry Volume-II
322
YCT
4.
Arrhenius Equation
Ans. (c) : Endothermic reactions are chemical reaction
in which reactants absorb heat energy from the
surrounding to form product. Higher the activation
energy means potential energy of product is greater than
reactants.
454. According to Arrhenius equation, the rate
constant (k) is related to temperature (T) as
452. The temperature dependence of rate constant
(K) of a chemical reaction is written in terms of
Arrhenius equation, k = Ae-E*/RT. Activation
energy (E*) of the reaction can be calculated by
ploting.
E 1 1
k
(a) ln 2 = a −
1
1
k1 R T1 T 2
(b) log k vs
(a) log k vs
T
log T
E 1 1
k
1
(b) ln 2 = − a −
(c) k vs T
(d) k vs
k
R T1 T 2
1
log T
JCECE - 2008, UPTU/UPSEE-2007
E 1 1
k
(c) ln 2 = a +
(AIPMT-2003)
k1 R T1 T 2
Ans. (a) : According to Arrhenius equation,
E 1 1
k
Ea
(d) ln 2 = − a +
log10k = log10A −
k1
R T1 T 2
2.303RT
A straight line graph with negative slope is obtained
AMU-2011, VITEEE- 2007
1
Ans. (a) : According to Arrhenius equationwhen graph is plotted between log k Vs .
T
k = Ae − Ea / RT
−E a
Taking log both sides
Thus, slope =
2.303R
E 1
ln k = ln A − a ×
453. An endothermic reaction with high activation
R T
energy for the forward reaction is given by the
E
1
diagram.
∴ ln k1 = ln A − a ×
....(i)
R T1
Ea 1
×
....(ii)
R T2
Subtracting equation (i) from (ii)
ln k 2 = ln A −
ln k 2 − ln k1 =
Ea
R
1 1
−
T1 T2
Ea 1 1
−
k1 R T1 T2
455. Rate of a reaction can be expressed by
Arrhenius equation k = A ⋅ e − Ea / RT . In this
equation, Ea represents
(a) The energy above which not all the colliding
molecules will react
(b) The energy below which colliding molecules
will not react
(c) The total energy of the reacting molecules at
temperature T
(d) The fraction of molecules with energy greater
than the activation energy of the reaction
AP - EAMCET (MEDICAL) - 2009
[AIEEE 2006]
Ans. (b) : Rate of reaction can be expressed by
Arrhenius equation–
ln
k2
=
−
Ea
k = Ae RT
Where, k = Rate constant
A = Frequency factor
Ea = Activation energy
T = Absolute temperature
COMEDK-2019, [AIIMS-2005]
Objective Chemistry Volume-II
323
YCT
Ea represents the energy of activation which implies it is
the energy below which colliding molecules will not
react. Arrhenius equation gives the dependence of the
rate constant k of a chemical reaction.
456. How enzymes increases the rate of reactions?
(a) By lowering activation energy
(b) By increasing activation energy
(c) By changing equilibrium constant
(d) By forming enzyme substrate complex
COMEDK-2016, (AIPMT-2000)
Ans. (a) : Enzymes are used as a catalyst for the
reaction. Presence of enzyme increases the rate of
reaction by lowering the activation energy of the
reactant.
457. The rate constant of a reaction at temperature
200K is 10 times less than the rate constant at
400K. What is the activation energy Ea of the
reaction? (R = Gas constant)
(a) 1842.4 R
(b) 921.2 R
(c) 460.6 R
(d) 230.3 R
AP-EAMCET (Medical), 2003
VITEEE- 2011
Ans. (b) : Given that,
k1 = k, T1 = 200K, k2 = 10 × k, T2 = 400K, Ea = ?
R = Gas constant
∆H = (E a )F − ( E a )R
(E a )F > (E a )R or (E a )F < (E a )R
For exothermic reaction, Ea for reverse reaction is
higher than Ea for forward reaction. For endothermic
reaction, Ea for reverse reaction is lesser than Ea for
forward reaction.
460. In respect of the equation K = Ae–Ea/RT in
chemical kinetics, which one of the following
statements is correct?
(a) k is equilibrium constant
(b) A is adsorption factor
(c) Ea is energy of activation
(d) R is Rydberg constant
[AIEEE-2003], JCECE - 2012
Ans. (c) : According to Arrhenius equation,
k = A e−Ea / RT
Where k is rate constant of reaction,
A is pre-exponential frequency factor
Ea is actuation energy
R is universal gas constant
T is temperature in Kelvin scale
461.
In
a zero-order reaction, for every 10 ºC rise of
E a T2 – T1
k
Now, log 2 =
temperature,
the rate is doubled. If the
k1 2.303R T2 T1
temperature is increased from 10 ºC to 100 ºC,
the rate of the reaction will become
E a 400 – 200
10k
log
=
(a) 256 times
(b) 512 times
k
2.303R 400× 200
(c) 64 times
(d) 128 times
E 200
(AIPMT
-2012), JCECE - 2005
2.303 log1010 = a
or
R 400× 200
Ans. (b) : Given:or
E a = 921.2R
Rateconstant at ( t + 10 ) °C
µ=2=
458. The effect of a catalyst in chemical reaction is
Rate constant at t o C
to change the
∆T
r2
(a) activation energy
= ( µ ) 10
(b) heat of reaction
r1
(c) final products
∆T = 100 − 10 = 90o C
(d) concentration of products
90
r2
9
AMU–2003, 2002
= ( 2 ) 10 = ( 2 )
r
Ans. (a) : Catalysis decrease the activation energy
1
required in a chemical reaction. It permits more and
r2
= 512
more molecules to take part in the chemical reaction,
r1
leading to an increase in the rate of reaction because
molecules will have energy equal or greater than the 462. If the activation energy of a reaction is 80.9 kJ
lowered threshold energy.
mol–1 , the fraction of molecules at 700 K,
459. The activation energy for a simple chemical
having enough to react to form products is e–x .
reaction A → B is Ea in forward direction. The
The value of x is _____. (Rounded off to the
activation energy for reverse reaction
nearest integer) [Use R = 8.31J K–1 mol–1]
(a) is always less than Ea
JEE Main 26.02.2021, Shift-II
(b) can be less than or more than Ea
Ans. : Given that,
(c) is always double of Ea
Ea=80.9 KJ/mole.
(d) is negative of Ea
[BITSAT-2010], (AIPMT -2003) T=700K –1
–1
Ans. (b) : Activation energy for reverse reaction may be R=8.31 JK mol
of molecules able to cross energy barrier =
less than or more than activation energy of forward Fraction
e–Ea/RT =e–x
reaction.
Objective Chemistry Volume-II
324
YCT
Ea
T – 500
⇒ 1 = 10,000 2
RT
500T2
3
80.9 × 10
20(T2 − 500)
x=
1=
8.31× 700
T2
x = 13.91
T2 = 20T2 – 500 × 20
x ≈ 14
19T2 = 10,000
463. For the reaction, aA + bB → cC + dD, the plot T2 = 526.3 K
1
464. The decomposition of formic acid on gold
is given below :
of log K vs
surface follows first order kinetics. If the rate
T
constant at 300 K is 1.0×10-3 s-1 and the
activation energy Ea = 11.488 kJ mol-1, the rate
constant at 200 K is _________ ×10-5 s-1.
(Round off to the Nearest Integer).
(Given : R = 8.314 J mol-1 K-1)
JEE Main-16.03.2021, Shift-I
x=
(
)
Ans. 10 × 10 −5 s −1 :
The temperature at which the rate constant of
the reaction is 10–4s–1 is _____ K.
[Rounded off to the nearest integer]
[Given : The rate constant of the reaction is
10–5 s–1 at 500 K]
Given that,
T2 = 300K, T1 = 200K
k 2 = 1.0 × 10 −3 ,k1 = ? and E a = 11.488KJmol−1
∴ Now, from the equation of activation energy–
JEE Main 25-02-2021, Shift-I
Ans. : Given that,
k1 = 10–5, k2 = 10-4, T1 = 500K, T2 = ?
Slope = –10,000 K
Rate constant (K) = 10–4s–1
⇒
From Arrhenius equation,
– Ea
K = Ae RT
lnK = lnA –
Ea
RT
log10 K = log10 A –
Ea
2.303RT
y = mx + c
From integrated form of Arrhenius equation
⇒ log10
Ea 1 1
k2
=
–
k1 2.303R T1 T2
1
10 –4
1
⇒ log10 –5 = 10,000
–
10
500 T2
Objective Chemistry Volume-II
log
k2
Ea 1 1
=
−
k1 2.303R T1 T2
log
1.0 × 10−3 s −1 11.488 × 1000 1
1
=
−
k1
2.303 × 8.314 200 300
log
10−3
3− 2
= 600 ×
k1
600
log
10−3
=1
k1
10 =
10−3
k1
⇒
k1 = 10–4
So,
x × 10–5 = 10–4
⇒
x = 10
[10×10–5 sec–1]
465. The rate constant of a reaction increases by five
times on increase in temperature from 27°C to
52°C. The value of activation energy in kJ mol–
1
is
. (Rounded-off to the nearest integer)
[R = 8.314 J K–1 mol–1]
JEE Main-25.02.2021, Shift-II
Ans. :
Given that,
T1 = 27°C, k1= ?, k2= 5K
T1= 27+273
= 300K
T2 = 52°C
= 52+273
= 325K
From integrated form of Arrhenius equation,
log10
325
Ea 1 1
k2
=
−
k1 2.303R T1 T2
YCT
⇒ log10
Ans. (b) : According to Arrhenius equation,
E
ln k = ln A − a
RT
1
ln k Vs
graph.
T
E
Slope = – a = –5 ×10 –3 k
R
Ea= 5×103 × 8.3 Jk–1 mol–1
Ea = 41.5 kJ mol–2
469. If for a hypothetical reaction, Ea = 0 at 273K,
then find the ratio of the rate constants at 383
K.
(a) 10
(b) 1
(c) 0
(d) 100
AP EAPCET 25.08.2021, Shift-II
Ans. (a) : For hypothetical reaction given
Ea = 0 and T = 273K
5K
1
Ea
1
=
−
K 2.303 × 8.314 300 325
⇒ log10 5 =
E a 325 − 300
19.147 300 × 325
Ea
25
×
19.15 300 × 325
⇒ E a = 19.15 × 12 × 325 × 0.699
⇒ log10 5 =
E a = 52, 204.815 Joule / mol
E a = 52kJ mol−1
Hence, the energy of activation of the reaction is ≈ 52
kJ mol–1.
466. An exothermic reaction X→Y has an activation
energy 30 kJ mol–1. If energy change ∆E during
the reaction is –20 kJ mol–1, then the activation
energy for the reverse reaction in kJ is .........
[JEE Main 26.02.2021, Shift-I] So,
Ans. (50) : For exothermic reaction, ∆E = –20kJ mol −1
f
Ea for backward reaction, (E a ) b = ?
∆E = ( E a )f – ( E a )b
b
f
= 50kJ / mol .
467. For a reaction A → B , enthalpy of reaction is –
4.2 kJ mol–1 and enthalpy of activation is 9.6 kJ
mol–1. The correct potential energy profile for
the reaction is shown in option.
(b)
(c)
(d)
RT
700K
A
→ Product;
500K
A
→ Product
Catalyst
It was found that the Ea is decreased by 30
kJ/mol in the presence of a catalyst. If the rate
remains unchanged, the activation energy for
catalysed reaction is (Assume pre exponential
factor is same):
(a) 198 kJ/mol
(b) 105 kJ/mol
(c) 75 kJ/mol
(d) 135 kJ/mol
[JEE Main-09.01.2020, Shift-I]
Ans. (c) : Given:- T1 = 700K
T2 = 500K
E a 2 = E a1 – 30 …….(i)
Ans. (c) : Given that,
∆H = – 4.2 kJ/mol,
(Ea)f = 9.6 kJ/mol from the
given data we said the reaction is exothermic i.e., ∆H =
–ve.
So, (Ea)f < (Ea)b for exothermic reaction. Hence only
option (c) is correct.
1
468. The slope of Arrhenius plot lnk vs of first
T
3
order reaction is –5 ×10 K. The value of Ea of
the reaction is Given : R = 8.314 JK –1mo –1
−1
(b) 41.5 kJ mol
(d) 166 kJ mol −1
(NEET-2021)
Objective Chemistry Volume-II
Ea
k = 1. e R ×383
k=1
470. For following reactions:
(NEET-2021)
(a) −83 kJ mol
(c) 83.0 kJ mol −1
−
−0
= ( E a ) – ∆E = 30 – ( –20 )
−1
0
−
k = A. e
(a)
Ea
RT
−
k = A. e R×273
k=A=1
If T = 383K
Ea for forward reaction, (E a ) = 30 kJ mol–1
( Ea )
k = A. e
326
k1 = k2 = k
A1 = A2 = A
According to Arrhenius equation,
In absence of catalyst
k = Ae − a1 1
In presence of catalyst
E / RT
……. (ii)
k = Ae − a 2 2
…….(iii)
On comparing equation (ii) and (iii)
E
–E / RT
–E
/ RT
/ RT
Ae a1 1 = Ae a 2 2
Taking log on both sides and substituting the
value of E a 2 from eqn. (i)
E a1
E a − 30
= 1
700R
500R
E a 1 E a1 – 30
=
7
5
YCT
1
R
(d) 2R
[JEE Main-05.09.2020, Shift-II]
E a 2 = Activation energy of reaction in presence
Ans. (d) : According to Arrhenius equation,
of catalyst
E × 103
lnk = lnA – a
E a 2 = E a1 – 30 = 75 kJ/mol.
⇒
RT
471. A sample of milk splits after 60 min. at 300 K
From straight line graph y = mx + c,
and after 40 min. at 400 K when the population
103
of lactobacillus acidophilus in it doubles. The
y = logk, x =
,
C = log A
T
activation energy (in kJ/mol) for this process is
closest to (Given, R = 8.3 J mol–1 K–1
E
And, slope = m = – a
2
–3
R
In =0.4, e = 4.0)
3
−E a
−10
⇒
slope =
= tan θ =
= −2
(a) 39.8
(b) 19.9
R
5
(c) 3.98
(d) 7.96
⇒
Ea = 2R kJ mol–1.
[JEE Main-09.01.2020, Shift-II]
473. The rate of a reaction decreased by 3.555 times
Ans. (c) : Given, T1 = 300K,
T2 = 400K
when the temperature was changed from 40°C
Time (t1) = 60 min ,
t2 = 40min
to 30°C. The activation energy (in kJ mol–1) of
We know that,
the reaction is .........
Rate of reaction ∝ Rate constant
(Take R = 8.314 J mol–1K–1, In 3.555 = 1.268)
[concentration ]
[JEE Main-06.09.2020, Shift-II]
And, Rate of reaction =
time
Ans. (100.00) : Given,
1
r1
Rate of reaction ∝
= 3.555,
Ea = ?
time
r2
1
T2 = 30oC = 303K ,
T1 = 40oC = 313K
Also, Rate constant ∝
time
r2 k 2
1
=
=
k 2 t1
r1 k1 3.555
∴
=
k1 t 2
According to Arrhenius Equation,
According to Arrhenius equation,
k −E a 1 1
ln 1 =
k2
Ea 1 1
–
R T1 T2
log10 =
k2
–
k1 2.303R T1 T2
−E a 1
1
ln (3.555) =
−
t1
Ea
1
1
8.314 313 303
log10 =
300 − 400
t
2.303
×
8.3
2
⇒
Ea = 100kJ/mol.
2.303 × 8.3 × 300 × 400
60
474.
The
rate of a certain biochemical reaction at
Ea =
× log10
physiological
temperature (T) occurs 106 times
100
40
faster with enzyme than without. The change in
∴ Ea = 3.98 kJ/mol
the activation energy upon adding enzyme is:
472. The rate constant (k) of a reaction is measured
(a) +6 RT
(b) – 6(2.303) RT
at different temperatures (T), and the data are
(c) +6(2.303) RT
(d) –6 RT
plotted in the given figure. The activation
[JEE Main 2020, 8 Jan Shift-I]
energy of the reaction in kJ mol–1 is (R is gas
constant)
Ans. (b) : Given,
r2
= 106
r1
r2 = Rate of reaction in presence of enzyme
r1 = Rate of reaction in absence of enzyme.
T = Temperature
Ea = ?
According to Arrhenius equation,
E a1 = Activation energy of reaction in absence of
catalyst
E a1 = 105 kJ/mol
2
R
(c) R
(a)
ln
Objective Chemistry Volume-II
327
(b)
(
E a1 – E a 2
k2
=
k1
RT
)
YCT
∵
k 2 r2
= = 106
k1 r1
∴
2.303 log10 (106) =
(E
a1
− Ea
2
Ans. (d) : The temperature dependence of the rate of a
chemical reaction can be explained by Arrhenius
equation,
)
k = A.e Ea / RT
RT
But change in activation energy,
∆E a = E a 2 − E a1 = –2.303 × 6RT .
(
⇒
)
475. The rate constant of a reaction is given by
k = PZe – Ea / RT under standard notation. In
order to speed up the reaction, which of the
following factors has to be decreased?
(a) Z
(b) Both Z and T
(d) T
(c) Ea
Karnataka-CET-2020
Ans. (c) : The rate constant of a reaction is given by
k = PZe – Ea / RT
In order to speed up the reaction, the value of Ea
(activation energy) has to be decreased.
476. An endothermic reaction A → B has an
activation energy of 13 kJ mol–1 and the
enthalpy change for the reaction is 2 kJ mol–1.
The activation energy of the reaction B → A is
(a) 15 kJ mol–1
(b) 11 kJ mol–1
–1
(c) 2 kJ mol
(d) –15 kJ mol–1
–1
(e) 26 kJ mol
Kerala-CEE-2020
Ans. (b) : For an endothermic reaction, the activation
energy for the backward reaction is equal to the
difference of the activation energy of the forward
reaction and the enthalpy change of the reaction.
For an endothermic reaction, the activation
Energy,
Eab = Eaf – ∆H
i.e.
Eaf = Eab + ∆H
For B → A Eab = 13 – 2
= 11 kJ mol–1
So, option (b) is correct.
477. The activation energy of a reaction is zero. The
rate constant of this reaction :
(a) increases with an increase of temperature
(b) decreases with an increase of temperature
(c) decreases with decreases of temperature
(d) is independent of temperature
Manipal-2020
Ans. (d) : From Arrhenius equation,
k = Ae – Ea / RT
If Ea = 0
k = A e −0
k=A
Hence, rate constant of reaction is independent
temperature.
478. Find the correct equation among the following.
E
E
(a) ln k − ln A = a
(b) k = a
RT
RT
Ea
Ea
(c) ln k + ln A =
(d)
= ln A − ln k
RT
RT
AP EAMCET (Engg.) 17.09.2020 Shift-I
Objective Chemistry Volume-II
lnk = lnA −
Ea
RT
Ea
= lnA − lnk
RT
479. The rate of reaction doubles, when the
temperature is changed from 300 K to 310 K.
Activation energy of the change is........
(R = 8.314 J K–1mol-1, log 2 = 0.301)
(b) 48.6 kJ mol–1
(a) 53.6 kJ mol–1
(c) 58.5 kJ mol–1
(d) 60.5 kJ mol–1
AP EAMCET (Engg.) 18.09.2020, Shift-I
Ans. (a) : From Archenius equation,
r2 k 2
=
=2
r1 k1
⇒
Now, 2.303 log
log
k 2 E a T2 − T1
=
k1 R T1T2
k 310
E a 310 − 300
=
k 300 2.303R 300 × 310
Ea
10
−3
2.303 × 8.314 × 10 300 × 310
–1
Ea = 53.6 kJ mol
480. For the reaction of H2 with I2, the rate constant
is 2.5×10–4dm3 mol–1s–1 at 327°C and 1.0dm3
mol–1s–1 at 527°C. The activation energy for the
reaction, in
kJ mol–1 is (R=8.314 JK–1 mol–1)
(a) 59
(b) 72
(c) 150
(d) 166
[JEE Main-2019, 9 April Shift-II]
⇒ log 2 =
Ans. (d) : Given:– k1 = 2.5×10−4 dm 3 mol−1 s−1
k 2 = 1.0dm3 mol−1 s−1 , T1 = 327°C = 600K
T2 = 527°C = 800K,
Ea = ?
According to Arrhenius equation,
k
E a 1
1
log10 2 =
−
k1 2.303R T1 T2
1
1
Ea
1
=
⇒
log10
−
2.5×10−4 2.303×8.314 600 800
2.303×8.314× 600×800×3.6
⇒
Ea =
200
−1
⇒
E a ≈ 166 kJ mol .
481. Consider the given plots for a reaction obeying
Arrhenius equation (0°C<T<300°C) : (K and Ea
are rate constant and activation energy
respectively)
328
YCT
Choose the correct option.
(a) Both I and II are wrong
(b) Both I and II are correct
(c) I is wrong but II is right
(d) I is right but II is wrong
[JEE Main-2019, 10 Jan Shift-I]
Ans. (b) : According to Arrhenius equation,
k = A e−Ea / RT
In k vs Ea graph, rate constant (K) decreases on
increasing the value of activation energy (Ea).
So, k exponentially decreases on increasing Ea.
In k vs T graph, on increasing temperature the value of
rate constant also increases exponentially
483. Which is a wrong statement?
(a) Rate constant k = Arrhenius constant A : if Ea
=0
(b) e –Ea / RT gives the fraction of reactant
molecules that are activated at the given
temperature
1
plot is a straight line
(c) ln k vs
T
(d) presence of catalyst will not alter the value of
Ea
Karnataka-CET-2019
Ans. (d) : A catalyst may increase or decrease the rate
of reaction by decrease or increase the value of
activation energy (Ea).
Hence, presence of catalyst will after the value of Ea.
484. The activation energy of a reaction is zero. Its
rate constant at 280 K is 1.6×10–6s–1, the rate
constant at 300 K is
(a) 3.2×10–6s–1
(b) zero
(c) 1.6×10–6s–1
(d) 1.6×10–5s–1
MHT CET-02.05.2019, SHIFT-II
Ans. (c) : Given,
Ea = 0, k1 = 1.6 × 10–6 s–1,
k2 = ?, T1 = 280K
, T2 = 300K
If Ea = 0 ,
k=A
A is constant for the reaction, then
k1 = k2 = 1.6 × 10–6 sec–1
485. For a reaction, activation energy Ea = 0 and
the rate constant at 200 K is 1.6 ×106 s –1 . The
rate constant at 400 K will be
[Given that gas constant R = 8.314 JK –1mol –1 ]
(a) 3.2 × 10 4 s −1
(b) 1.6 × 106 s −1
3 −1
(c) 1.6 × 10 s
(d) 3.2 × 106 s −1
(Odisha NEET-2019)
Ans. (b) : Given:Ea = 0 k1= 1.6 × 106 s–1
T1 = 200K,
T2 = 400K, A1 = A2 = A
k2 = ?
If Ea = 0,
k=A
then k1 = k2 = 1.6 × 106 s–1
486. The rate constant is doubled when temperature
increases from 27°C to 37°C. Activation energy
in kJ is
(a) 34
(b) 54
(c) 100
(d) 50
[VITEEE-2019]
Ans. (b) : Given that,
T1 = 27°C = 273 + 27 = 300K
T2 = 37°C = 37 + 273 = 310K
k2
=2
k1
Ea = ?
482. If a reaction follows the Arrhenius equation,
the plot ln K vs 1/(RT) gives straight line with a
gradient (–y) unit.
The energy required to activate the reactant is
y
unit
(b) –y unit
(a)
R
(c) yR unit
(d) y unit
[JEE Main-2019, 11 Jan Shift-I]
Ans. (d) : According to Arrhenius equation,
k = Ae − Ea / RT
E
ln k = ln A − a ………(i)
RT
1
Since plot of lnk vs gives straight line with a
RT
gradient (-y) unit. Therefore, on comparing equation (i)
with y= mx+c, we have
Slope = m = −E a = −y (given)
∴ y = Ea
∴
Thus, Activation energy = y unit .
Objective Chemistry Volume-II
329
log
Ea 1 1
k2
=
−
k1 2.303R T1 T2
YCT
log 2 =
298 × 333
R log e 2.1
35
298 × 333
2.1
(d)
R log e
35
1.5
1
Ea
1
−
2.303 × 8.314 300 310
(c)
300 × 310
E a = 0.3010 × 2.303 × 8.314 ×
10
[BITSAT-2018]
Ea= 53598.59 J mol–1
Ans. (b) : Given that,
Ea = 54 kJ
−3
−2
487. What is the activation energy (kJ/mol) for a k1 = 1.5×10 , k 2 = 2.1×10
reaction if its rate constant doubles when the T1 = 25 + 273 = 298K
T2 = 60 + 273 = 333K
temperature is raised from 300 K to 400 K? (R
Ea = ?
= 8.314 J mol–1K–1)
(a) 68.8
(b) 6.88
According to Arrhenius equation,
(c) 34.4
(d) 3.44
k
E a 1
1
AIIMS-26 May 2019 (Morning)
log10 2 =
−
k1 2.303R T1 T2
k
Ans. (b): Given:– 2 = 2
k1=? , k2= 2
2.1×10−2 E a 1
1
k1
=
log e
−
−3
1.5
×
10
R
298
333
T1 = 300K,
T2 = 400K, E a = ?
According to Arrhenius equation,
2.1×10−2 E a
35
log e
=
−3
k 2
1.5×10
R 333× 298
E a 1
1
log10 =
−
21
R × 298×333
k1 2.303R T1 T2
Ea =
log e
1.5
35
1
2
Ea
1
log10 =
−
490. The following equation is the Arrhenius
1 2.303×8.314 300 400
equation, k = Ae- Ea /RT , where Ea is the
8.314× 0.3×300× 400× 2.303
Ea =
minimum energy molecules must possess in
100
order to react to form a product, k is the rate
E a = 6.8kJ mol−1
constant, A is the frequency factor, R is the gas
constant and T is the Kelvin temperature.
488. Plots showing the variation of the rate constant
(k) with temperature (T) are given below. The
Under normal circumstances, the Arrhenius
plot that follows. Arrhenius equation is
plot is obtained by plotting
(a) logarithm of the inverse of rate constant 1/k,
versus the inverse temperature 1/T
(a)
(b) k
(b) logarithm of the rate constant k, versus the
temperature T
(c) logarithm of the rate constant k, versus the
inverse temperature 1/T
(c)
(d)
(d) logarithm of the inverse of rate constant 1/k,
versus the temperature T.
[BITSAT – 2019]
J & K CET-(2018)
Ans. (a) : According to Arrhenius equation,
Ans.
(c):
From
the
Arrhenius
equation,
k = A e−Ea / RT
K = Ae – Ea / RT
log k = log A
Taking log10 on both sides, we have
K varies exponentially with temperature (T) when graph
Ea
is plot between K vs T
log10 k = log10 A –
2.303RT
For straight line graph, y = mx + c
1
Ea
y = log10k, x = , m = −
,c = log10 A
T
2.303R
Hence, logarithm of rate constant K versus the inverse
489. The rate constant of a reaction is 1.5 ×10–3 at
1
25oC and 2.1 ×10–2 at 60oC. The activation temperature graph is plotted.
T
energy is
−2
491. For certain reaction, the values of A and Ea in
35
2.1×10
R log e
(a)
Arrhenius equation are 4 × 1013 s–1 and 98.6
−2
333
1.5 × 10
kJ/mol. If the reaction is 1st order, at what
298 × 333
21
temperature will its half-life period be 10
(b)
R log e
35
1.5
minutes?
Objective Chemistry Volume-II
330
YCT
(a) 325.60 K
(c) 300 K
(b) 311.35 K
Ans. (a) : Given,
(d) 510.05 K
2576
….(i)
JCECE - 2018 ln k = – T + 12.1
Ans. (b) : Given,
On comparing equation (i) with Arrhenius equation
A = 4 × 1013 s–1, Ea = 98.6 kJ/mol,
E
ln k = ln A – a
t1/2 = 10 min
RT
For 1st order reaction,
Ea
= 2576
0.693 0.693
k=
=
R
t1/ 2
10
Ea = 2576 R = 2576×8.314
= 0.0693 min–1
⇒
Ea = 21.416 kJ
= 0.001155 s–1
At T = 27oC = 300 K
As we know, Arrhenius equation is;
2576
+ 12.1
ln k = −
Ea
300
log k = log A –
2.303RT
ln k = 3.51
Ea
⇒
k = 0.335 × 102 min–1.
= log A − log k
494. Which among the following equation
2.303RT
13
represents Arrhenius equation?
= log(4 × 10 ) – log(0.001155)
(a) k = AeEa/RT
(b) k = A.eRT/Ea
= 13.602 + 2.937 = 16.539
A
A
98600
(d) k = RT / E
(c) k = E / RT
T=
a
a
e
e
2.303 × 8.314 × 16.539
MHT CET-2017
T = 311.36 K.
Ans.
(c)
:
From
Arrhenius
equation,
492. For gaseous reaction, the following data is
k = Ae – Ea / RT
given
A
A→ B , K1 = 1015e−2000/T
k = Ea / RT
e
C→ D , K2 = 1014e−1000/T
Where,
The temperature at which K1 = K2 is
k = rate of constant
(a) 2000K
(b) 868.8K
T = Temperature
(c) 434.2K
(d) 1000K
Ea = Activation energy
CG PET-2017
A = A is called frequency
Ans. (c) : Given that,
495. For a reaction A → B; ∆H = 20 kJ mol–1 the
k1= k2 and
activation energy of the forward reaction is 85
A → B, k1 = 1015.e −2000 / T
kJ/mol. The activation energy of the backward
reaction will be
C → D, k 2 = 1014 ,e −1000 / T
(a) 105 kJ/mol
(b) 65 kJ/mol
1015
(E −E ) / T
(c) 45 kJ/mol
(d) 75 kJ/mol
Thus, 14 = e a1 a 2
10
[VITEEE-2017]
or,
10 =e ( 2000−1000) / T 10 = e1000 / T
Ans. (b) : Given reaction A → B
∆H = 20 kJ mol–1
1000
1000
or, ln 10 =
or, log10 10 =
Energy of activation for forward reaction
T
( 2.303) T
(Ea) = 85 kJ/mol and ∆H= 20 KJ/mol
1000
∴ Energy of activation for backward reaction
or,
T=
= 434.2 K
2.303
(Eb) = Ea – ∆H
493. What will be the value of activation energy
= 85 – 20
(Ea in kJ) and rate constant (K in min-1) for the
(Eb) = 65 KJ/ mole.
given equation at 27oC? The equation ln K = 496. For an endothermic reaction where ∆H
–2576/T + 12.1
represent the enthalpy of reaction in kJ/mol,
(a) Ea = 21.416 kJ and k = 0.335 × 102 min-1
the minimum value for energy of activation
(b) Ea = 11.46 kJ and k = 0.335 × 102 min-1
(for forward reaction) will be
(a) less than ∆H
(b) zero
(c) Ea = 20.23 kJ and k = 0.43 × 102 min-1
(c) more than ∆H
(d) equal to ∆H
(d) Ea = 21.416 kJ and k = 1.44 × 102 min-1
[BITSAT – 2017]
J & K CET-(2017)
Objective Chemistry Volume-II
331
YCT
499. The slope of the graph drawn between lnk and
1
as per Arrhenius equation gives the value
T
(R = gas constant, Ea = Activation energy)
R
E
(a)
(b) a
Ea
R
Ans. (c) : For an endothermic reaction,
H is positive
∆H = ( E a )f − (E a )b
−R
Ea
TP-EAMCET – 2016
Ans. (c) : According to Arrhenius equationk = Ae− Ea / RT
Taking log both sides, we get
Ea
log k = log A −
2.303RT
Ea
or lnk = ln A −
RT
(c)
For endothermic reaction ∆H is positive. Minimum
value for energy of actuation for forward reaction will
be more than H.
1
497. A graph is ploted between log K virsus for
T
calculation of activation energy (Ea). The
correct plot is
(a)
−Ea
R
(d)
(b)
Ea
and intercept 'lnA'
R
[BITSAT – 2017] 500. The formation of H2O2 in the upper
atmosphere follows the mechanism :
Ans. (b) : According to Arrhenius equation,
H 2 O + O
→ 2OH
→ H 2O 2
k = A e−Ea / RT
−1
∆H = 72 kJ mol ,Ea = 77 kJ mol −1 . Then, Ea for
Taking logarithm on both side,
the backward reaction will be – (per mol)
Ea
log k = log A −
(a) –149 kJ
(b) + 149 kJ
2.303× RT
(c) –5 kJ
(d) + 5 kJ
For straight line graph y = mx + c
BCECE-2016
The value of log k should increase uniformly with T or
Ans. (d) : ∆H = (E a )forward − (E a )backward
1
decrease with .
∆H = 72kJ mol−1 , (E a )forward=77kJ mol−1
T
1
(E a )backward = (E a )forward − ∆H
498. A plot of ln K against
(abscissa) is expected
T
= (77 − 72) kJmol–1
to be a straight line with intercept on ordinate
axis equal to
= 5kJ mol−1
∆S°
∆S°
501. A hydrogenation reaction is carried out at 500
(a)
(b)
2.303R
R
K. If the same reaction is carried out in the
presence of a catalyst at the same rate, the
∆S°
(c) −
(d) R × ∆S°
temperature required is 400 K. If the catalyst
R
lowers the activation barrier by 40 kJ mol the
[VITEEE-2016]
activation energy of the reaction will be
Ans. (b) : We know that (a) 100 kJ mol-1
(b) 200 kJ mol-1
-1
∆G° = –RT ln k = ∆H° – T∆S°
(c) 300 kJ mol
(d) 175 kJ mol-1
–∆G° = RT ln k = T∆S° – ∆H°
CG PET-2016
∆H° ∆S°
Ans.
(b)
:
Given
that,
⇒ ln k = −
+
RT
R
T1 = 500K, T2 = 400K
Thus, a plot of log k versus 1/T (abscissa) will be
(E a )1 = (E a )2 + 40
straight line.
According to Arrhenius equation,
Comparing with y = mx + c
In absence of catalyst
o
∆S
− E a / RT1
∴ y intercept is
.
k = A e 1 ......(i)
R
(c)
(d)
Objective Chemistry Volume-II
The slope of the graph is −
332
YCT
The factor by which rate of catalysed reaction
is
increased, is
k = A e 2 ........(ii)
(a) 21
(b) 2100
Rate constant is same in both the cases because catalyst
(c) 2000
(d) 1200
only change the actuation energy from equation (i) and
[VITEEE-2015]
equation (ii)
Ans. (c) : Given that,
− E a / RT1
− E a / RT2
Ae 1
−Ae 2
T = 300K
Taking log on both side,
Ea (uncatalysed) = 76 kJ/mole
E a1
Ea2
Ea (catalysed) = 57 kJ/mole
=
For Arrhenius equation,
500R 400R
k = Ae − Ea / RT
E a 2 + 40 E a 2
=
Taking log both side we get5
4
−1
Ea
E a 2 = 160 kJ mol
∴
log k = log A −
2.303RT
Ea1 = E a 2 + 40 = 200 kJ mol−1
E a(uncatalysed)
log k1 = log A −
..….(i)
502. The addition of a catalyst during a chemical So,
2.303RT
reaction alters which of the following
E a (uncatalysed)
quantities?
log k 2 = log A −
…..(ii)
(a) Enthalpy
(b) Activation energy
2.303RT
(c) Entropy
(d) Internal energy
Subtract (ii) from equation (i) we get(NEET-I 2016)
k
1
E a (uncat.) − E a (cat.)
log 2 =
Ans. (b) : The addition of a catalyst during a chemical
k1 2.303RT
reaction may increase or decrease the value of
k
1
activation energy. A catalyst is a substance which alters
log 2 =
( 76000 − 57000 )
the reaction but itself remains unchanged in amount
k1 2.303 × 8.314 × 300
chemical composition at the end of reaction.
k
19000
503. Which graph shows zero activation energy for
log 2 =
k1 2.303 × 8.314 × 300
reaction ?
In presence of catalyst
− E a / RT
k2
190
=
k1 6.9 × 8.314
On taking antilogk2
= 2000
k1
log
(a)
(b)
506. Energy of activation of forward reaction for an
endothermic process 90 kJ. If enthalpy change
(d)
(c)
for the reaction is 50 kJ then activation energy
for backward reaction will be
(a) 40 kJ
(b) 140 kJ
(c)
90
kJ
(d) 50kJ
UPTU/UPSEE-2016
J & K CET-(2015)
Ans. (a): Minimum energy required for the reacting
modules to undergo reaction is known as actuation Ans. (a) : Given, Ea of forward reaction, (Ea)f = 90 kJ
energy Hence, option (a) is with zero activation energy.
Change in enthalpy = 50 kJ = ∆H
Ea of backward reaction, (Ea)b = ?
504. The unit of activation energy is
(a) sec–1
(b) JK–1 mol–1
For endothermic reaction, ∆H = +ve
(c) J mol–1
(d) K–1 mol–1
⇒
∆H = ( E a ) f – ( E a ) b
SRMJEEE – 2015
( E a )b = 90 – 50 = 40kJ .
Ans. (c) : Activation energy is defined as the minimum ⇒
amount of extra energy required by a reacting molecule 507. The activation energy of a reaction can be
to get converted into product. The activation energy of
determined from the slope of which of the
reaction is measured in Jmol–1 or kJmol–1
following graphs?
505. The following data were obtained for a given
1
T
1
(a) ln k vs
(b)
vs
reaction at 300 K.
T
ln
k
T
Reaction
Energy of activation
ln k
(kJ mol–1)
(c) ln k vs T
(d)
vs T
(i) uncatalysed
76
T
(ii) catalysed
57
(NEET-2015, Cancelled)
Objective Chemistry Volume-II
333
YCT
Ans. (a) : According to Arrhenius equation,
E
ln k = ln A – a
RT
For straight line graph, y = mx + c
k2
0 310 − 300
=
= 0
4.2 × 105 R 310 × 300
k2
log
=0
4.2 × 105
1
k2
y = ln k ,
c = ln A,
x=
or
= 100 = 1
T
4.2 × 105
E
or
k2 = 4.2 × 105 sec–1
m=− a
R
Since the activation energy is zero the rate constant is
1
On plotting the graph between ln k Vs , the activation independent of the change in temperature.
T
510. The excess energy which a molecule must
energy of a reaction can be determined from the slope.
possess to become active is known as
(a) kinetic energy
(b) threshold energy
508. For a reversible reaction A ↽ ⇀ B, which one
of the following statements is wrong from the
(c) potential energy
(d) activation energy.
given energy profile diagram?
SRMJEEE – 2011
Ans. (d) The excess energy which a molecule must
possess to become active is known as activation energy.
The molecules with energy less than activation energy
will not give fruitful collisions. Such molecules will not
form products.
511. Consider the following reaction,
log
(a) Activation energy of forward reaction is
greater than backward reaction
(b) The forward reaction is endothermic
The reaction is of first order in each diagram,
(c) The threshold energy is less than that of
with
an equilibrium constant of 104. For the
activation energy
conversion of chair form to boat form e–Ea/RT =
(d) The energy of activation of forward reaction
4.35 × 10-8 m at 298 K with pre-exponential
is equal to the sum of heat of reaction and the
factor of 1012 s-1. Apparent rate constant
energy of activation of backward reaction
(=kA/kB) at 298 K is
AP EAMCET-2008
(a) 4.35 × 104 s-1
(b) 4.35 × 108 s-1
Ans. (c) :
(c) 4.35 × 10–8 s-1
(d) 4.35 × 1012 s-1
For a reversible reaction A ↽ ⇀ B
[VITEEE-2013]
• Activation energy of forward reaction is greater
Ans.
(b)
:
Given
that,
than backward reaction.
k = 104 (equilibrium constant)
Ef > Eb
Hence, statement is true.
T = 298 k
• Product level is higher than reactant level. Forward
A = 1012 sec–1 (pre – exponential factor)
reaction is endothermic. Statement is true.
e − Ea / RT = 4.35 × 10–8 m
• The threshold energy is greater than that of
so,
kB = A e − Ea / RT
activation energy. Statement is false.
= 1012 × 4.35 × 10–8
• The energy of activation of forward reaction is
equal to the sum of heat of reaction and the energy
= 4.35 × 104 sec–1
of activation of backward reaction. Statement is
k
true.
Also equilibrium constant k = A =104
kB
509. For a reaction, Ea = 0 and k = 4.2 × 105 sec–1 at
4
300 K, the value of k at 310 K will be
∴ kA=kB×10
(a) 4.2 × 105 sec–1
(b) 8.4 × 105 sec–1
k A = 104 × 4.35 ×10 4 = 4.35 ×108 sec −1
(d) unpredictable
(c) 7.4 × 105 sec–1
SRMJEEE – 2010 512. Which one of the following is true for an
exothermic reaction A ↽ ⇀ B, if Ef and Eb are
Ans. (a) : Given that,
Ea = 0, k2 = ?, T2 = 310K
the activation energies of forward and
k1 = 4.2 × 105sec–1, T1 = 300K
backward reactions respectively?
Now, from the Arrhenius equation –
(a) E f > E b
(b) E f = E b
k 2 E a T2 − T1
(c) E f = −E b
(d) E f < E b
log
=
k1
R T1 T2
AP-EAMCET- (Engg.)-2011
Objective Chemistry Volume-II
334
YCT
515. The rate constant (k1) of one of the reaction is
found to be double that of the rate constant (k2)
of another reaction. The relationship between
Eb
the corresponding activation energies of the two
Enthalpy of reaction is related to activation energy for
reaction Ea1 and Ea2 will be
forward reaction (Ef) to backward reaction (Eb) is
∆r H = Ef − Eb
(a) E a1 < E a 2
(b) E a1 > E a 2
For exothermic reaction
(c) E a1 = E a 2
(d) E a1 = 2E a 2
∆ rH < 0
AMU-2014
⇒
Eb > Ef
Ans.
(a)
:
According
to
Arrhenius
equation,
513. An endothermic reaction. A → B has an
k = A e−Ea / RT
activation energy as x kJ/mol. If the energy
change of the reaction is y kJ, the activation Taking logarithm on both side,
energy of the reverse reaction is
E
(a) –x
(b) x–y
log k = log A − a
RT
(c) x+y
(d) y–x
At constant temperature,
COMEDK-2014
k
1
Ans. (b): For endothermic reaction.
log10 2 =
(E − E a2 )
k1 2.303RT a1
When, k1 = 2k 2
Ans. (d) : Given for the exothermic reaction–
E
A↽ f ⇀B
k2
< 1, then ( E a1 − E a 2 ) < 1
k1
E a1 < E a 2
516. The activation energy for a reaction at the
temperature T K was found to be 2.303 RT J
mol–1. The ratio of the rate constant to
Arrhenius factor is–
(a) 10–1
(b) 10–2
–3
(c) 2 × 10
(d) 2 × 10–2
∆E= Ea(f) – Ea(b)
BCECE-2014
y= x–Ea(b)
Ans. (a) : Given that, temperature = TK
Ea(b) = x–y
E a = 2.303 RT J
514. Assertion: According to transition state theory for
the formation of an activated complex, one of the
k
=?
vibrational degree of freedom is converted into a
A
translational degree of freedom.
According to Arrhenius equation,
Reason: Energy of the activated complex is
k = A e−Ea / RT
higher than the energy of reactant molecules.
2.303RT
(a) If both Assertion and Reason are correct and
−
k
the Reason is a correct explanation of the
= e RT
A
Assertion.
k
(b) If both Assertion and Reason are correct but
2.303 log10 = −2.303
Reason is not a correct explanation of the
A
Assertion.
k
(c) If the Assertion is correct but Reason is
log10 = −1
A
incorrect.
(d) If both the Assertion and Reason are
k
= 10−1
incorrect.
A
(e) If the Assertion is incorrect but the Reason is 517. A reactant (A) forms two products
correct.
k1
A
→ B, Activation energy Ea 1
[AIIMS-2006]
Ans. (b): The activated complex has higher energy than
reactants. It is a special molecule which decomposes
such that one vibrational degree of freedom is converted
into translational degree of freedom along the reaction
coordinate.
Hence, both assertion and reason are correct but reason
is not a correct explanation of the assertion.
Objective Chemistry Volume-II
335
k2
A
→ C, Activation energy Ea 2
If Ea 2 = 2 Ea 1 , then K1 and K2 are related as
(a) k1 = 2k 2 e
(c) k 2 = k1e
E a 2 / RT
E a 2 / RT
(b) k1 = k 2 e
E a 1 / RT
E / RT
(d) k1 = Ak 2 e a1
[AIEEE-2011]
YCT
Ans. (b) : According to Arrhenius equation,
k1
A
→ B, E a1
For reaction, k 1 = A e−Ea / RT
For reaction,
K2
k
→ C, E a 2
− E a 2 / RT
k 20 = A e
...........(ii)
E a 2 = 2E a1 .........(iii)
On dividing equation (i) by (ii)
−(E +E
)
a1
a2
k1
= e RT
k2
Substituting the value of Ea from equation (III)
k1
=e
k2
=e
−
(E a + E a 2 )
RT
E a1 / RT
k1 = k 2 e
E a1 / RT
518. The rate of a reaction double when its
temperature changes from 300 K to 310 K. The
activation energy of such a reaction will be
(R=8.314 JK–1 mol–1 and log 2= 0.301)
(a) 53.6 kJ mol–1
(b) 48.6 kJ mol–1
–1
(c) 58.5 kJ mol
(d) 60.5 kJ mol–1
[JEE Main-2013]
Ans. (a) : Given:– T1 = 300K, T2 = 310K
r2
= 2,
r1
E a = ?, R = 8.314JK−1mol−1
r2 k 2
= =2
r1 k1
According to Arrhenius equation,
k
E a 1
1
log10 2 =
−
k1 2.303R T1 T2
1
Ea 1
2.303log10 ( 2) =
−
8.314 300 310
8.314× 2.303× 0.301×300×310
Ea =
10
= 53598.59 J mol−1
∴
= 53.6 KJ mol−1
520. The reaction A + B → C + D + 40 kJ has an
activation energy of 18 kJ. Then the activation
energy for the reaction C + D → A + B is
(a) 58 kJ
(b) –40 kJ
(c) –18 kJ
(d) 22 kJ
J & K CET-(2011)
Ans. (a) : For the reaction, A + B → C + D + 40kJ
∆H = – 40 kJ
(Ea)f = 18 kJ
For C + D → A + B, Ea = (Ea)b = ?
∆H = (Ea)f – (Ea)b
(Ea)b = 18 – (– 40) = 58kJ.
521. In a reversible reaction, the enthalpy change
and the activation energy in the forward
direction are respectively –x kJ mol–1 and y kJ
mol–1. Therefore, the energy of activation in the
backward direction in kJ mol–1 is
(a) y – x
(b) (x + y)
(c) (x – y)
(d) – x – y
J & K CET-(2009)
Ans. (a) : Given:∆H = – x kJ/mol (Ea)f = y kJ/mol
(Ea)b = ?
Here, the given enthalpy change is negative. So, the
reaction is exothermic.
For exothermic reaction,
∆H = (Ea)b –(Ea)f
(Ea)b = (Ea)f + ∆H = y + (–x)
(Ea)b = (y – x) kJ/mol.
522. Activation energy of a chemical reaction can be
determined by
(a) determining the rate constants at standard
temperature
(b) determining the rate constants at two
temperatures
(c) determining probability of collision
(d) using catalyst
JCECE - 2013
Ans. (b) : In Arrhenius equation, by knowing k1 and k2
at T1 and T2, activation energy (Ea) can be determined.
523. The half-life period for first order reaction
having activation energy 39.3 kcal mol–1 at
300ºC and frequency constant 1.11 × 1011 s–1
will be
(a) 1 h
(b) 1.68 h
(c) 1.28 h
(d) 1.11 h
JCECE - 2014
Ans. (b) : Given,
A = 1.11 × 1011s–1; T = 573 K,
Ea = 39.3 × 103 cal mol–1; R = 1.987 cal;
∵
k = Ae- Ea /RT
519. The rate of reactions exhibiting negative
activation energy.
(a) decreases with increasing temperature
(b) increases with increasing temperature
(c) does not depend on temperature
(d) depends on the height of the potential barrier.
J & K CET-(2012)
∴
Ans. (a) : According to Arrhenius equation,
k = A e-Ea /RT
or
when Ea = –ve;
If activation energy has negative value, then on
increasing temperature, rate of reaction decreases.
Objective Chemistry Volume-II
336
Ea
2.303RT
log10 k = log10 (1.11 × 1011)
39.3 × 103
−
2.303 × 1.987 × 573
log10 k = log10 A −
YCT
k = 1.14 × 10–4s–1
k
Ea 1 1
log10 2 =
0.693
0.693
–
But half-life ( t1/ 2 ) =
=
k1 2.303R T1 T2
−4
k
1.14 × 10
2.303R × 300 × 270 ×
1
= 6074s = 1.68 h.
Ea = −
log10
30
10
524. According to the Arrhenius equation, a straight
Ea = 2.303 × 2.7 R kJ/mol–1
line is to be obtained by plotting the logarithm
of the rate constant of a chemical reaction (log 527. The activation energy of a reaction at a given
k) against
temperature is found to be 2.303 RTJ mol–1.
(a) T
(b) log T
The ratio of rate constant to the Arrhenius
factor is
1
1
(c)
(d) log
(a) 0.01
(b) 0.1
T
T
(c) 0.02
(d) 0.001
JCECE - 2014
Karnataka-CET-2011
Ans. (c) : According to the Arrhenius equation, a
E
/
RT
−
straight line is obtained, when logK is plotted against Ans. (b) : k = Ae a
1/T.
k
= e − Ea / RT
1
A
525. A plot of
vs ln k for a reaction gives the
T
k −Ea
4
slope –1 × 10 K. The energy of activation for
In =
–1
–1
A RT
the reaction is (Given, R = 8.314 K mol )
–1
–1
(a) 8314 J mol
(b) 1.202 kJ mol
k −Ea
2.303log =
(c) 12.02 J mol–1
(d) 83.14 kJ mol–1
A RT
Karnataka-CET-2014
−E a
k
Ans. (d) : According to Arrhenius equation.
log =
A 2.303 RT
E
ln k = ln A − a
k −2.303RT
RT
log =
1
A 2.303 RT
A plot of
Vs ln k gives
T
k
log = –1
Ea
A
slope = −
= −1× 104 k
R
k
= 0.1
E a = 1× 10 4 × 8.314
A
= 83.14 k J mol −1
528. The activation energies of two reactions are E1
and E2 (E1>E2). If the temperature of the
526. A given sample of milk turns sour at room
system in increased from T1 to T2, the rate
temperature (27oC) in 5. In a refrigerator at
constant of the reaction changes from k1 to k'1
–3oC, it can be stored 10 times longer. The
in the first reaction and k2 to k'2 in the second
energy of activation for the souring of milk is
reaction. Predict which of the following
(a) 2.303 × 5 R kJ·mol–1
–1
expression is correct?
(b) 2.303 × 3R kJ·mol
–1
k'
k'
k'
k'
(c) 2.303 × 2.7 R kJ·mol
(a) 1 = 2
(b) 1 > 2
–1
(d) 2.303 × 10 R kJ·mol
k1 k 2
k1 k 2
Karnataka-CET-2012
k'
k'
k'
k'
(c) 1 < 2
(d) 1 = 2 = 1
Ans. (c) : Given that:k1 k 2
k1 k 2
t1 = 5 hrs
t2 = 10t1 = 50 hrs
o
o
k'
k'
T2 = 27 C = 300 k
T2 = – 3 C = 270 k
(e) 1 = 2 = 0
Ea = ?
k1 k 2
Rate ∝ k
Kerala-CEE-2009
concentration ]
[
Ans.
(b)
:
Rate =
Given,
time
E1 > E2
1
Rate ∝
From Arrhenius equation,
time
or
1
time
k 2 t1 1
= =
k1 t 2 10
According to Arrhenius equation
k∝
Objective Chemistry Volume-II
ln
ln
337
k1' E1 1 1
=
−
k1 R T1 T2
k '2
k2
=
E2 1 1
−
R T1 T2
–––––––– (i)
–––––––– (ii)
YCT
ln
k '1
k1
> ln
Ans. (a) : Given:T1 = 20o C = 293 K,
T2 = 35o C = 308K
r2 k 2
=
=2
r1 k1
From Arrhenius equation,
k2
k2
k '2
k1 k 2
529. For the two gaseous reactions, following data
k
Ea 1 1
are given
log10 2 =
–
A → B; k1 = 1, 1010 e –20,000/ T
k1 2.303R T1 T2
A →D; k2 = 1, 1012 e –24,606/ T
Ea
1
1
log102 =
The temperature at which k1 becomes equal to
293 − 308
2.303
×
8.314
k2 is
0.301× 2.303 × 8.314 × 293 × 308
(a) 400 K
(b) 1000 K
Ea =
(c) 800 K
(d) 1500 K
15
(a) 500 K
Ea = 34.7 kJ mol–1
Kerala-CEE-2008
532. Activation energy ( Ea ) and rate constants
10 –20000/T
Ans. (b):
k1 = 10 e
12 –24606/T
k2 = 10 e
( k 1 and k 2 ) of a chemical reaction at two
If k1 = k2
different temperatures ( T1 and T2 ) are related
Then,
10
–20000/T
by
10 = e
−
20000
24606
E 1 1
k
+
T
(a) ln 2 = − a −
= e T
= 102.e 4606 / T = 102
k1
R T1 T2
Taking log both side,
E 1 1
4606
k
(b) ln 2 = − a −
= log102
k1
R T2 T1
2.303T
4606
2 log 10×T=
E 1 1
k
(c) ln 2 = − a +
2.303
k1
R T2 T1
4606
T=
2.303 × 2
E 1 1
k
(d) ln 2 = a −
4606
k1 R T1 T2
=
= 1000 K
4.606
(AIPMT Mains 2012)
530. The rate constants k1 and k2 for two different
Ans.
(b,
d)
:
According
to
Arrhenius
equation,
reactions are
1016 × e-2000/T and 1015 × e-1000/T ,
⇒
k '1
>
respectively. The temperature at which k1= k2 is
1000
(a) 2000 K
(b)
K
2.303
2000
(c) 1000 K
(d)
K
2.303
(AIPMT-2008)
16
−2000 / T
Ans. (b) : k1 = 10 × e
,
k = A e – Ea / RT
Taking log on both side
E
ln k1 = ln A − a −(i)
RT1
Ea
RT2
Subtracting equation (i) from (ii)
ln k2 = ln A –
k 2 = 1015 × e −1000 / t
E
E
ln k2 – ln k1 = – a – – a
If k1 = k2
RT
2
RT1
1016 × e −2000 / T = 1015 e –1000 / T
E 1 1
1000
k
ln 2 = a −
ln 10 =
k1 R T1 T2
T
1000
T=
K
k
E 1 1
2.303
or ln 2 = − a −
k1
R T2 T1
531. What is the activation energy for a reaction if
its rate doubles when the temperature is raised 533. In Arrhenius plot, intercepts is equal to
from 20 ºC to 35ºC? ( R = 8.314 J mol –1 K –1 )
−E a
(a)
(b) ln A
(a) 34.7 kJ mol−1
(b) 15.7 kJ mol−1
R
(c) ln K
(d) log10 a
(c) 342 kJ mol−1
(d) 269 kJ mol −1
UP CPMT-2006
(NEET-2013)
Objective Chemistry Volume-II
338
YCT
Ans. (b) : Arrhenius equation is
−E a
ln k =
+ ln A
RT
Hence intercept of Y-axis = ln A
534. The minimum energy required for the reacting
molecules to undergo reaction is:
(a) Potential energy
(b) Kinetic energy
(c) thermal energy
(d) activation energy
UPTU/UPSEE-2005
Ans. (d) : Activation energy is the minimum energy
required for the reacting molecules to undergo reaction.
Activation energy is the energy required to initiate a
chemical reaction.
535. For a chemical reaction at 27oC, the activation
energy is 600 R. The ration of the rate
constants at 327oC to that of at 27oC will be
(a) 2
(b) 40
(c) e
(d) e2
WB-JEE-2013
Ans. (c) : Given,
Ea = 600 R,
T2 = 327oC = 600 K
T1 = 27oC = 300K,
k2
=?
k1
From Arrhenius equation
Ans. (c) : Reaction M∆H = –ve, and no intermediate is formed So its rate is
faster and exothermic.
Reaction N∆H = less negation,
Due to formation of inter mediate, rate is slower.
5.
Half-life Time of Chemical Reaction
537. For the decomposition of azomethane.
CH2N2CH3(g)→ CH3CH3(g) + N2(g) a first
order reaction, the variation in partial pressure
with time at 600 K is given as
The half life of the reaction is_____ × 10−5s.
[Nearest integer]
JEE Main 25.07.2022, Shift-II
Ans. (2) : For first order reaction,
1
P
k=
ln 0
t1/ 2 P
P
ln 0 = kt1/ 2
P
k E 1 1
log 2 = a −
ln 2
0.693
k1 R T1 T2
∴ t1/ 2 =
=
= 2 × 10−5
k
3.465 × 104
k 600R 1
1
log 2 =
538. The half life for the decomposition of gaseous
300 − 600
k
R
1
compound A is 240 s when the gaseous
pressure was 500 Torr initially. When the
600R
=
=1
pressure was 250 Torr, the half life was found
600R
to be 4.0 min. The order of the reaction is
k
……… (Nearest integer)
= 2 =e
JEE Main 25.07.2022, Shift-I
k1
Ans.
(1)
:
Given Data,
536. The correct statement regarding the following
Half life of gaseous compound A at pressure 500 torr
energy diagrams is
t 1 = 240 s
2
(a) Reaction M is faster and less exothermic than
reaction N
(b) Reaction M is slower and less exothermic
than reaction N
(c) Reaction M is faster and more exothermic
than reaction N
(d) Reaction M is slower and more exothermic
than reaction N
WB-JEE-2014
Objective Chemistry Volume-II
Half life of gaseous compound A t 1 at 250 torr
2
pressure = 4.0 min
As t1/2 is independent of initial pressure.
Hence order is 1.
539. At 30oC, the half life for the decomposition of
AB2 is 200 s and is independent of the intial
concentration of AB2. The time required for
80% of the AB2 to decompose is (Given : log2 =
0.30; log3 = 0.48)
(a) 200s
(b) 323s
(c) 467s
(d) 532s
JEE Main-26.07.2022, Shift-II
339
YCT
Ans. (a) : Half life period is independent of initial
concentration of section for first order reaction, second
order reaction, zero order reaction and third order
reaction. Here are 1st order reaction.
t1/2 = 200sec
A0
2.303log 2 2.303
=
k=
log
200
t
0.2A 0
Where A0 = initial concentration
log 2 1
= log 5
200 t
7
t = × 200 = 466.67s = 467s
3
540. The time required for completion of 93.75% of
a first order reaction is x minutes. The half life
of it (in minutes) is
(a) x/8
(b) x/2
(c) x/4
(d) x/3
AP-EAMCET-04.07.2022, Shift-II
Ans. (c) : Let a = 100 , x = 93.75
2.303
a
So,
t=
log
k
a−x
2.303
100
t=
log
k
100 − 93.75
2.303
100
t=
log
k
6.25
2.303
4
t=
log (2)
k
2.303log 2
t = 4×
k
2.303log 2
t = 4×
k
0.693
t = 4×
k
0.693
We know that, t 1 =
2
k
So,
t = 4× t 1
2
Here, t = x (given)
x
∴
= t1
2
4
541. The rate equation for a reaction A → B is r = k
[A]0. If the initial concentration of the reactant
is a mol dm–3, the half-life period of the
reaction is
a
k
(a)
(b)
2k
a
a
2a
(c)
(d)
k
k
JIPMER-2011
Karnataka-CET, 2009
Ans. (a) : A → B
∵
r = k[A]0
or
r=k
Objective Chemistry Volume-II
According to the rate equation given,
This is a zero order reaction.
a
∴
t1/ 2 =
2k
542. Half-lives of a first order and a zero order
reactions are same. Then, the ratio of the initial
rates of first order reaction to that of the zero
1
(a)
(b) 2 × 0.693
0.693
2
(c) 0.693
(d)
0.693
(e) 6.93
[AIIMS-2014], Kerala-CEE-2010
Ans. (b) : Given,
(t1/2)0 = (t1/2)1 = t1/2
a
Half-life time for zero order (t1/ 2 ) = 0
2k 0
0.693
Half-life time for 1st order (t1/ 2 ) =
k1
(rate)1 k1[a 0 ] (a 0 )0.693/ t1/ 2
=
=
(rate)0
k0
a 0 / 2t1/ 2
(rate)1
= 0.693 × 2
(rate)0
543. The rate of a first order reaction is
1.5 ×10 –2 mol L–1 min –1 at 0.5 M concentration
of the reactant. The half-life of the reaction is
(a) 0.383 min
(b) 23.1 min
(c) 8.73 min
(d) 7.53 min
UPTU/UPSEE-2018, (AIPMT -2004)
Ans. (b) : Given:–
Rate= 1.5×10–2 mol L–1 min–1, [A] = 0.5 M
For 1st order reaction (n) = 1
Rate = k[A]n = k[A]1
Rate
k=
[A]
1.5 ×10−2
= 3 × 10−2 min −1
0.5
0.693 0.693
t1/ 2 =
=
k
3 ×10−2
t1/2 = 23.1 min
544. Which condition among the following holds
true at the stage of half - completion for the
reaction A ⇌ B
(a) ∆G° = 0
(b) ∆G° > 0
(c) ∆G° < 0
(d) ∆G° > 0
AP-EAMCET 25-08-2021 Shift - I
Ans. (a) : For half completion of reaction–
A⇌B
Now, ∆G° = –RTlnk
At the stage of half completion of reaction [A] = [B]
therefore, K = 1
∆G° = –RTln(1)
∆G° = 0
340
k=
YCT
545.
A and B decompose via first order kinetics with
half-lives 54.0 min and 18.0 min respectively.
Starting from an equimolar non reactive mixture
of A and B, the time taken for the concentration
of A to become 16 times that of B is ---------- min.
(Round off to the Nearest Integer).
JEE Main 16.03.2021, Shift-II
Ans. (108) : Given data(t1/2)A = 54 min
(t1/2)B = 18 min
and
16[A] = [B]
Equation for first order kinetics is2.303
k=
log A (for A)
t
2.303 × 0.3010 2.303
______(1)
or
=
log A
t
( t1/ 2 )A
Similarly, for B 2.303 × 0.3010 2.303
______(2)
=
log B
t
( t1/ 2 )B
Substracting equation (2) from (1) and putting the
respective value we get
(a) 4.50
(c) 6.75
(b) 2.25
(d) 3.45
AP EAPCET-6 Sep. 2021, Shift-II
Ans. (c) : Given rate equation Ro= k[Ao]2 [Bo]
If concentration of A is increased to 1.5 times
the original and B is increased by 3 times then,
New rate of reaction= k [1.5 Ao]2 [3Bo]
= k (1.5)2 [Ao]2 3×[Bo]
= 6.75k [Ao]2 [Bo]
= 6.75 Ro
548. The half–life period of a first order reactor is 60
minutes. If initial amount of the reactant taken
is 50 g. The amount left after 4 hrs will be
(a) 6.25 g
(b) 12.5 g
(c) 3.125 g
(d) 1.25 g
AP EAPCET-6 Sep. 2021, Shift-II
Ans. (c): Given, t1/2 = 60 min
For first order reaction,
T = n × t1/ 2
..... (1)
We know that, n=
1
N=No×
2
0.3010 1 1
A
− 1 = log
18 3 t
B
0.3010 2 1
A
or
−
× = log
18
3 t
16A
0.3010 2 1
−
× = × 4 × ( −0.3010 )
18
3 t
or
t = 108 minutes
546. If the rate constant for a first order reaction is
k, then find the time required for completion of
80% of the reaction.
3.2
1.6
(b)
(a)
k
k
4.8
0.8
(c)
(d)
k
k
AP EAPCET 24.08.2021 Shift-II
Ans. (b): For first order reaction
2.303
a
K=
log
t
a−n
Where, a = initial concentration
a – n = reduced concentration
At 80 % of reaction completion –
2.303
100
So,
t=
log
k
20
2.303
t=
log 5
k
1.6
t=
.
k
547. For a reaction. The initial rate is given as R0
=k[A0]2[B0]. By what factor. the initial rate of
reaction will increase if the concentration of A
taken is increased to 1.5 times the original and
that of B is tripled?
Objective Chemistry Volume-II
4
1
n
4
1
1
N=50× =50×
16
2
=3.125 gram,
549. For
an
elementary
reaction
X(g)
X ( g ) −−−→ Y( g ) + Z ( g ), the t1/2 is 10 minutes. In
what period of time would the concentration of
X be reduced to 10% of its original
concentration?
(a) 20 minutes
(b) 33.2 minutes
(c) 15 minutes
(d) 25.2
AP EAPCET 20.08.2021 Shift-II
Ans. (b): given, t1/2 = 10 min
0.693
k=
= 0.0693min −1
10
When a =100%, (a – x) = 10%
For first order reaction,
2.303
a
t=
log10
k
a
−
( x)
2.303
100
log10
0.0693
10
= 33.2 min
t10% =
t10%
550. Sucrose hydrolyses in acid solution into glucose
and fructose following first order rate law with
a half-life of 3.33 h at 25°C. After 9 h, the
fraction of sucrose remaining is f.
341
The value of log10 1 is ....... ×10–2
f
(Rounded off to the nearest integer).
[Assume, In 10 = 2.303, In 2 = 0.693]
[JEE Main 2021, 24 Feb Shift-II]
YCT
Ans. (81): Sucrose → glucose + fructose
Half − life time for first orders reaction
0.693 0.693 −1
k=
=
hr
t1/ 2
3.33
After 9 hr let a = 1, (a − x ) = f
2.303
a
k=
log10
t
a
−
( x)
Ans. (a) : Given,
t1/ 2 = 36 min , a = 10 gm,
For 1st order reaction,
0.693 0.693
k=
=
min −1
t1/ 2
36
kt = 2.303 log10
t =2 hrs,
(a − x) = ?
a
(a − x )
0.693 × 2 × 60
10
kt
0.693 × 9
1
= log10
log10 =
=
= 81× 10−2
36
×
2.303
a
( − x)
f 2.303 3.33 × 2.303
551. A reaction has a half-life of 1 min. The time log10 10 = 1
(a − x)
required for 99.9% completion of the reaction
is .......... min (Round off to the nearest integer). ( a − x ) = 1g
[Use : In 2 = 0.69, In 10 = 2.3]
554. Time taken for 12.8 g of a radioactive
[JEE Main 2021, 18 March Shift-II]
substance to decay to 0.4 g is: (half is 138s)
Ans. (10) : Given, t1/2= 1min, t99.9%=?
(a) 720s
(b) 690s
For 1st order reaction,
(c) 245s
(d) 69s
[AIIMS-27 May, 2018 (E)]
0.693 0.693
−1
k=
=
= 0.693min
Ans. (b): Given, t1/2 = 138s
t1/ 2
1
a = 12.8g
2.303
a
(a – x) = 0.4g
]t=
log10
k
(a − x)
Half - life time for 1st order reaction,
0.693 0.693 −1
2.303
100
k=
=
S
t 99.9% =
log10
t1/ 2
138
0.693
0.1
t 99.9% = 10 min
k=
2.303
a
log10
t
a
−
( x)
552. For the first order reaction, A→2B, 1 mole of
reactant A gives 0.2 moles of B after 100
2.303 × 138
12.8
minutes. The half-life of the reaction is ........ t = 0.693 log10 0.4
min. (Round off to the nearest integer). [Use : t = 690 sec.
In 2 = 0.69, In 10 = 2.3; properties of
555. For first order reaction as time duration goes
x
y
from 10 min to 30 min, rate of reaction
logarithms : In x = y in x; In = In x – In y]
decreases from 0.4 Ms-1 to 0.04 Ms-1. What is
y
the half life of the reaction?
(Round off to the nearest integer)
(a) 8 min
(b) 4 min
[JEE Main 2021, 27 July Shift-II]
(c) 6 min
(d) 2 min
Ans. (693): For reaction, A
→ 2B
[AIIMS-27 May, 2018 (M)]
At t = 0
1 mole
0
Ans. (c): For first order reaction.
At t = 100 min
1–x
(2x)
d[A]
Rate =
= k [ A t ] = k [ A o ] e− kt1
0.9 mol
0.2 mol
dt
For 1st order reaction,
0.4 = k [ A o ] e−10k –––––––(i)
2.303
a
k=
log10
0.04 = k [ A 0 ] e −30k –––––––(ii)
t
(a − x )
On dividing (i) and (ii)
2.303
1
−3
−1
30 −10 )k
=
×
k=
log10
1.05
10
min
10 = e(
100
0.9
ln10
0.693
0.693
k=
t1/ 2 =
=
=
693min
20
k
1.05 ×10−3
0.693 0.693 × 20
=
= 6 min
553. If half-life of a substance is 36 minutes. Find t1/ 2 =
k
ln10
amount left after 2 hrs? Initial amount is 10
556. If a first order reaction is 80% complete in 60
gm.
minutes. Find the t1/2 of reaction?
(a) 1 g
(b) 2 g
(a) 30 min
(b) 42 min
(c) 3 g
(d) 4 g
(c) 25.85 min
(d) 14.28 min
JIPMER-2019
[AIIMS-26 May, 2018(E)]
Objective Chemistry Volume-II
342
YCT
Ans. (c): Given, t80% = 60 min, t1/2=?
For first order reaction
2.303
a
k=
log10
t
(a − x )
For first order reaction
P
1
1
80
1
k = ln 0 =
ln
ln2
=
t P0 − P 20 80 − 40 20
Again, t1/ 2 =
2.303
100 2.303
k=
log10
= 60 log10 (5)
60
20
0.693
Half life time, t1/ 2 =
k
0.693 × 60
t1/ 2 =
= 25.85min
2.303log10 (5)
Hence,
t1/2
t3/4
t2/3
Data insufficient to predict
[BITSAT – 2018]
Ans. (a) : For reaction A →B
At the point of intersection the concentration of both A
and B are equal i.e. 50%. Hence, inter-section
represents half - life time (t1/2).
558. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(a) is halfed
(b) is doubled
(c) is tripled
(d) remains unchanged.
(NEET-2018)
Ans. (b) : Half–life period of zero order reaction is
directly proportional to initial concentration of
reactions. So, half-life will double if the initial
concentration of the reactant is double.
[A ]
t 1/ 2 = o
2k
559. The half life of the first order reaction
CH3CHO (g) → CH4 (g) + CO (g)
If initial pressure of CH3CHO (g) is 80 mm Hg
and the total pressure at the end of 20 minutes
is 120mm Hg
(a) 80 min
(b) 120 min
(c) 20 min
(d) 40 min
VITEEE-2017
Ans. (c) : CH 3CHO(g) → CH 4 (g) + CO(g)
When t = 0 P0
0
0
When t = t P0 – P
P
P
∴ Total pressure (Pt)= P0 –P + P + P = 120 mm Hg
P0 + P = 120 mm Hg;
P = 120 – 80
P = 40 mm Hg
Objective Chemistry Volume-II
ln 2
× 20 = 20 min.
ln 2
560. 20% of a first order reaction was found to be
completed at 10 am. at 11.30 am. on the same
day, 20% of the reaction was found to be
remaining. The half life period in minutes of
the reactions is
(a) 90
(b) 45
(c) 60
(d) 30
AP EAMCET-2017
Ans. (b): Given, for first order reaction
Initial concentration = completed at 10 am. to
11.30 am. of 20%
= 100 – 20
= 80
Final concentration = 100 – 80 = 20
T (time) = 11.30 am – 10 am.
= 1.5 hr = 1.5 × 60
= 90 min
2.303
80
k=
log
90
20
2.303
k=
log 4
90
k = 0.015 min
0.693 0.693
t1/ 2 =
=
k
0.015
or t1/ 2 = 46.2 min ≈ 45min
t1/ 2 =
557. The accompanying figure depicts a change in
concentration of species A and B for the
reaction A → B, as a function of time. The point
of inter section of the two curves represents
(a)
(b)
(c)
(d)
ln 2
k
561. The half-life period for a radioactive substance
is 15 minutes. How many grams of this
radioactive substance is decayed from 50 gram
of substance after one hour?
(a) 37.5
(b) 25
(c) 43.75
(d) 46.875
GUJCET-2017
Ans. (d) : Given, t1/ 2 = 15min, a = 50 gm, t=1hr = 60
min
0.693 0.693
k=
=
min −1
t1/ 2
15
k=
2.303
a
log10
t
(a − x )
2.303
50
log10
0.693
a
( − x)
15
(a – x) = 3.125
Decayed substance = 50 – 3.125 = 46.875 gm
60 =
343
YCT
562. The two-third life (t2/3) of a first order reaction
in which k = 5.48 × 10–14 per sec, is :
2.303
2.303
(a)
log 3
(b)
log 2
−14
5.48 × 10
5.48 × 10−14
2.303
1
2.303
2
(c)
log
(d)
log
−14
−14
5.48 × 10
3
5.48 × 10
3
Manipal-2017
Ans. (a) : Given,
1
k = 5.48 × 10–14 s–1, (a–x) =
3
For first order reaction
2.303
a
k=
log10
t
(a – x)
2.303
1
2.303
t 23 =
log10
=
log10 3
5.48 ×10 –14
( 13 ) 5.48 × 10 –14
563. The rate constant for a first order reaction is
7.0×10–4 s–1. If initial concentration of reactant
is 0.080 M, what is the half life of reaction?
(a) 990 s
(b) 79.2 s
(c) 12375 s
(d) 10.10×10–4 s
MHT CET-2017
Ans. (a) : Given,
k = 7.0 ×10–4 s–1 , [A0] = 0.08 M
Half life time for 1st order reaction
0.693
0.693
t 12 =
=
k
7.0 ×10−4
= 990 sec
564. In a reaction A → Products, when start is made
from 8.0 × 10–2 M of A, half-life is found to be
120 minute. For the initial concentration 4.0 ×
10–2 M, the half-life of the reaction becomes 240
minute. The order of the reaction is:
(a) zero
(b) one
(c) two
(d) 0.5
VITEEE-2016
Ans. (c) : From the half life reaction-
Ans. (b) : Given,
k = 4 × 10−3 mol L−1 min −1 ,
[ A o ] = 2 ×10−2 mol / L
It follows zero order kinetics by seeing the unit of rate
constant.
Half - life time for zero order reaction,
[ A ] 2 ×10−2 = 2.5min = 150sec
t1/ 2 = o =
2k
2 × 4 × 10−3
566. The half-life period of a first order reaction is
60 min. What percentage will be left over after
240 min?
(a) 6.25%
(b) 4.25%
(c) 5%
(d) 6%
Karnataka-CET-2016
Ans. (a) : Given,
t1/ 2 = 60 min,
% left over at 240 min = ?
Half – life time of first order reaction is constant for
every concentration of reactants.
Let take % initial concentration = 100%
t1/ 2
t1/ 2
t1/ 2
∴ 100%
→ 50%
→ 25%
→12.5%
60 min
60 min
60 min
t1/ 2
→ 6.25%
60min
(a − x )
a – x = 6.25%
567. The rate constant and half-life of a first order
reaction are related to each other as
0.693
(a) t1/ 2 =
(b) t1/2 = 0.693k
k
1
(c) k = 0.693 t1/2
(d) kt1/ 2 =
0.693
MHT CET-2016
Ans. (a) : Half-life time of first order reaction
0.693
t 12 =
k
568. A first order reaction is one-fifth completed in
40 minutes. The time required for its 100%
completion is
n −1
( t1/ 2 )1 a 2
(a) 100 minutes
(b) 200 minutes
=
( t1/ 2 )2 a1
(c) 350 minutes
(d) Infinity
n
−
1
SCRA-2015
120 4 × 10−2
=
Ans. (d): Given that, t = 40 min
240 8 ×10−2
For first order reaction
n −1
1 1
2.303
a
=
k=
log
2 2
t
a−x
1=n–1
2.303
1
k=
log
n=2
40
1− 1
Hence, the order of reaction is two.
5
565. The rate constant for a reaction at the initial
2.303
5
k=
log = 0.00557
concentration of 2 × 10–2 mol L–1 is 4 × 10 –3
40
4
mol L–1 min–1. The half-life period for this The time required for100% completion is –
reaction in seconds is
2.303
100
(a) 300
(b) 150
∴
t=
log
0.00557
100 − 100
(c) 180
(d) 240
J & K CET-(2016) or
t=∞
Objective Chemistry Volume-II
344
YCT
569. The time required for a first order reaction to
complete 90% is 't'. What is the time required
to complete 99% of the same reaction?
(a) 2t
(b) 3t
(c) t
(d) 4t
AP-EAMCET (Engg.) 2015
Ans. (a) : From the first order reaction–
2.303
a
t=
log
K
a−x
Let t be the time required for 90% Completion and
t1 be the time required for 99% completion
2.303
100
So, t =
log
k
100 – 90
2.303
t=
log 10
.....(i)
k
2.303
100
For t1 =
log
k
100 – 99
2.303
t1 =
log 100
.....(ii)
k
From equation (i) and (ii)
2.303
log 10
t1
= k
⇒ t1 = 2t
t 2.303 log 100
k
570. Half-life period of a first order reaction is 10
min. Starting with initial concentration 12 M,
the rate after 20 min is
(a) 0.0693 M min–1
(b) 0.693 × 3 M min–1
–1
(c) 0.0693 × 3 M min (d) 0.0693 × 4 M min–1
Karnataka-CET-2015
Ans. (c) : Given, t1/2 = 10 min
Half life time for 1st order reaction,
0.693 0.693
k=
=
= 0.0693min −1
t1/ 2
10
If [Ao] = 12 M, Rate after 20 min =?
[At] = 3 M
Rate = k [At] = 0.0693×3 M min–1
571. Half-life period of a first order reaction, A→
product is 6.93 h. What is the value of rate
constant?
(a) 1.596 h–1
(b) 0.1 h–1
–1
(c) 4.802 h
(d) 10 h–1
MHT CET-2015
Ans. (b) : Given, t½ = 6.93 h
k=?
Half-life period of first order reaction
0.693
k=
t 12
(a) t1/2 = [A0] 2k
(b) t1/2 =
A
(c) t1/ 2 = 0
2k
(d) t1/ 2 =
0.693
k
2 A 0
k
MHT CET-2015
Ans. (c) : The relationship between rate constant and
half-life period of zero order reaction is given as–
[A ]
t1 = 0
2
2k
Where, Ao is the initial concentration and k is the rate
constant for zero order reaction.
573. When initial concentration of a reactant is
doubled in a reaction, its half-life period is not
affected. The order of the reaction is
(a) second
(b) more than zero but less than first
(c) zero
(d) first
(NEET-2015, Cancelled)
Ans. (d) : Half-life period of first order reaction is
independent of initial concentration of reaction.
Hence, on doubling the initial concentration of a
reactant half–life period is not affected.
0.693
t 1/ 2 =
k
574. The half life period of second order reaction is
proportional to
(a) a1
(b) a0
–2
(c) a
(d) a–1
SRMJEEE – 2011
Ans. (d) :
For the second order kinetics –
k2 =
1 1
1
−
t ( a − x ) a
When, x =
a
, t=t1 ,
2
2
Hence,
1 1
1
k2 =
−
a a
t 1
2 a −
2
1 2 1
k2 =
−
t 1 a a
2
or
or
1 1
t1 = ×
2
k2 a
t 1 ∝ a–1
2
575. The half life period of a first order chemical
reaction is 6.93 minutes. The time required for
0.693
the completion of 99% of the chemical reaction
=
= 0.1 h −1
6.93
will be (log 2 = 0.301)
572. The relationship between rate constant and
(a) 23.03 minutes
(b) 46.06 minutes
half-life period of zero order reaction is given
(c) 460.6 minutes
(d) 230.03 minutes
by
[AIEEE 2009]
Objective Chemistry Volume-II
345
YCT
Ans. (b) : Given, that –
t1/2 = 6.93 minutes
0.693
t1/ 2 =
K
0.693
k=
= 0.1 min–1
6.93
For first order reaction2.303
a
k=
log
t1/ 2
a−x
2.303
100
0.1 =
log
t
1
2.303 × 2log10
0.1 =
t
2.303 × 2
t=
0.1
t = 46.06 minutes.
576. For a first order reaction at 27°C, the ratio of
time required for 75% completion to 25%
completion of reaction is
(a) 3.0
(b) 2.303
(c) 4.8
(d) 0.477
VITEEE- 2009
Ans. (c) : For a first order reaction,
2.303
a
t=
log10
λ
a−x
Let initial amount of reactant is 100.
100
log
t1
100 − 75 [∵λ remains constant]
=
t 2 log 100
100 − 25
100
log
25 = log 4
=
100 log 4 / 3
log
75
log 4
=
log 4 − log 3
2 × 0.3010
=
2 × 0.3010 − 0.4771
0.6020
=
= 4.81
0.1249
577. In a first order reaction, the concentration of
the reactant decrease from 0.6 M to 0.3 M in 15
minutes. The time taken for the concentration
to change from 0.1 M to 0.025 M in minutes is
(a) 1.2
(b) 12
(c) 30
(d) 3
AP-EAMCET (Engg.) - 2014
Ans. (c) : Concentration of reactant decreases from 0.6
M to 0.3 M at 15 min.
So,
t1/2 = 15 min.
From first order reaction
0.693.
t1/2 =
K
Objective Chemistry Volume-II
0.693
= 0.0462 min–1
15
Again the concentration changes from 0.1 M to 0.025
M.
2.303
0.1
∴
t=
log
0.0462
0.025
2.303
t=
log 4
0.0462
t = 49.84 × 2 × 0.301
t = 30 minutes.
578. The standard potentials, at 25°C, for the halfreactions are given against them below :
Zn2+ + 2e– → Zn
E° = –0.762 V
Mg2+ + 2e– → Mg
E° = –2.37 V
When Zn dust is added to a solution of MgCl2,
(a) magnesium is precipitated
(b) zinc dissolves in the solution
(c) zinc chloride is formed
(d) no reaction will takes place
AP - EAMCET(MEDICAL) - 2009
Ans. (d) : The standard reduction potential of Zn is
higher than that of Mg. So, it is placed after Mg in the
electrochemical series. Therefore, it can not replace Mg
from MgCl2.
Zn + MgCl2→ ZnCl2+Mg
∴ E ocell = E oZn / Zn +2 + E omg +2 / mg = + 0.762 + ( –2.37 )
k=
= –1.605V
∆G = –nFE cell
Value of ∆G is positive for negative value of Ecell
therefore, no reaction takes place.
579. The fraction of element disintegrated after 4
half-lifes in percentage is
(a) 75%
(b) 87.5%
(c) 93.75%
(d) 92.5%
AP - EAMCET (Medical) - 2007
Ans. (c) : Given data, n = 4
n
1
N = N0
2
Where, N = Final amount
N0 = Initial amount
n = number of half-life
Now,
4
1
1
N = 100 = 100 × = 6.25%
2
16
∴ Fraction of element disintegrated = N0 – N
= 100 – 6.25
= 93.75%
580. If 0.4 Curie be the activity of 1 gram of a
radioactive sample whose atomic mass is 226,
then what is the half life period of the sample ?
(1 Curie = 3.7 × 1010 disintegrations/sec)
(a) 1.2 × 1011 sec
(b) 1.8 × 1011 sec
10
(c) 1.2 × 10 sec
(d) 1.8 × 1010 sec
SCRA - 2009
∴
346
YCT
Ans. (a) : We know that,
0.693 W
Rate of decay =
× × 6.022 × 1023
t1/ 2
M
0.4 × 3.7 ×1010 =
0.693 1
×
× 6.022 × 1023
t1/ 2
226
0.693 × 6.022 × 1023
226 × 0.4 × 3.7 ×1010
4.17 ×1023−10
=
334.48
= 0.01246 ×1013
t1/ 2 =
t1/ 2
t1/ 2
t1/ 2 = 1.246 × 1011 sec.
581. The 6.25% of radioactive substance is left after
480 min. The half life period is
(a) 60 min
(b) 90 min
(c) 150 min
(d) 120 min
COMEDK-2012
Ans. (d) : For the first order reaction2.303
a
k=
log
t
(a – x)
2.303
100
log
480
6.25
k = 5.78 × 10 –3 min –1
(∴log16=1.204)
0.693
0.693
∴t1/ 2 =
=
= 119.89
k
5.78 × 10 –3
t1/2 ≈ 120 minutes
582. A reaction is 50% complete in 2 hours and
75% complete in 4 hours. The order of reaction
is
(a) 1
(b) 2
(c) 3
(d) 0
Karnatake NEET-2013
Ans. (a) : As t75%= 2×t50%,
∴ The order of the reaction is one, A is first order
reaction.
583. Give relation between half reaction time (t1/2)
and initial concentration of reactant for (n –1)
order reaction.
k=
(a) t1/2 ∝ [R]0
(b) t1/2 ∝ [ R ]0
(c) t1/2 ∝ [ R ]0
(d) t1/2 ∝ [ R ]0
n +1
2− n
n −2
GUJCET-2011
Ans. (b): Relation between t1/2 and initial concentration
of reactant for (n – 1) order reaction.
t1/ 2 ∝ [R]02−n
584. For a first order reaction, the initial
concentration of a reactant is 0.05 M. After 45
minutes it is decreased by 0.015 M. Calculate
half reaction time. (t1/2)
(a) 87.42 min.
(b) 25.90 min.
(c) 78.72 min.
(d) 77.20 min.
GUJCET-2011
Objective Chemistry Volume-II
Ans. (a): Given that, initial concentration (N0) = 0.05M
Final concentration (N) = 0.05 – 0.015 = 0.035
t = 45 minutes:
For first order reaction, half life (t1/2) is
0.693
t1/ 2 =
...(1)
k
2.303
N
log o
∵k=
t
N
2.303
0.05
∴k=
log
45
0.035
2.303
k=
× 0.154
45
k = 0.00792
Putting the value of k in equation (1) we get 0.693
t1/ 2 =
0.00792
t1/2 = 87.5 min
585. What is the formula to find the value of t1 2 for
a Zero Order Reaction?
0.693
[R° ]
(b)
(a)
K
2K
2K
K
(c)
(d)
[R° ]
[R° ]
GUJCET-2008
Ans. (b) : The expression for zero order reaction is–
[ R ] —– (1)
k=
t
where –
k = Rate constant
[R] = Final concentration
t = time
[ R ] (where, R = Initial concentration)
∵ [R ] = °
0
2
and t = t1 2
∴ From equation (1), we get
[R ]
k= °
2t1 2
or t1 2 =
[R ° ]
2k
586. For nth order reaction, the half-life period, t1/2 is
proportional to initial concentration as–
1
(a) n −1
(b) a n +1
a
1
(d) n
(c) a n −1
a
BCECE-2010
Ans. (a) : Half – life time for nth order reaction,
1
t1/ 2 ∝ n −1
a
587. In a first- order reaction A→B, if k is rate
constant and initial concentration of reactant A
is 0.5 M then the half- life is
347
YCT
ln 2
k
log 2
(c)
2k
0.693
0.5
log 2
(d)
k 0.5
CG PET -2007
Ans. (a) : Half-life time for first order reaction
2.303
a
1
a
k=
log10
= ln
t
(a − x) t (a − x)
ln 2
1
0.5
t1/2 = ln
⇒ t1/ 2 =
k 0.5
k
2
588. The time for half-life period of a certain
reaction, A→ products is 1h. When the initial
concentration of the reactant 'A' is 2.0 mol L–1,
how much time does it take for its
concentration to come from 0.50 to 0.25 mol L–
1
, if it is a zero order reaction?
(a) 4 h
(b) 0.5 h
(c) 0.25 h
(d) 1 h
[AIEEE-2010]
Ans. (c) : Given : t1/2 = 1hr, [Ao] = 2 mol L–1
For zero order reaction,
a
a
t1/ 2 =
⇒ k=
2k
2t1/ 2
2
k=
= 1mol L−1hr −1
2 ×1
When [Ao] = 0.5 mol L–1, [At] =0.25mol L–1, t1/2 =?
a
0.5
t1/2 =
=
= 0.25hr
2k 2 × 1
589. A first order reaction has a rate constant K =
3.01 × 10–3/s. How long it will take in
decompose half of the reaction?
(a) 2.303 s
(b) 23.03 s
(c) 230.3 s
(d) 2303 s
J & K CET-(2012)
Ans. (c) : Given, k = 3.01× 10−3 S−1 , t1/ 2 = ?
For first order reaction,
0.693
t1/ 2 =
= 230.3sec
3.01× 10−3
590. The rate constant of a second order reaction,
2A → products, is 10–4 L mol–1 min–1. The
initial concentration of the reactant is 10–2 mol
L–1. What is its half-life (in min)?
(a) 10
(b) 1000
(c) 100
(d) 106
J & K CET-(2006)
Ans. (d) : Given,
k = 10 −4 L mol −1 min −1 , [A o ] = 10 −2 mol L−1
By seeing the unit of rate constant we said that it
follows second order kinetics.
Half - life time for 2nd order reaction.
1
1
t1/ 2 =
= −4
= 106 min
k [ A o ] 10 × 10−2
(a)
(b)
Objective Chemistry Volume-II
591. If ‘a’ and “t1/2” are initial concentration of
reactant and half-life of a zero order reaction
respectively, which of the following is correct?
1
(a) t1/ 2 ∝
(b) t1/ 2 ∝ a
a
1
(c) t1/ 2 ∝ 2
(d) t1/ 2 ∝ a 2
a
J & K CET-(2006)
Ans. (b) :
a
Let [ A o ] = a
t = t1/ 2
[At ] = 2
Half life time for zero order reaction
kt = [Ao] – [At]
a a
kt1/ 2 = a − =
2 2
a
t1/ 2 =
2k
∴ t1/ 2 ∝ a
592. If the 75% of a first order reaction is complete
in 8 min, then time taken to decompose 50% of
its initial amount is :
(a) 2 min
(b) 4 min
(c) 12 min
(d) 1 min
JCECE - 2006
Ans. (b) : Given,
t 75% = 8min, t 50% = ?
For first order reaction,
t75% = 2 × t 50%
8
= 4 min
2
593. Half-life of a substance is 6 min. If its initial
amount is 32g, then amount present after 18
min is:
(a) 4 g
(b) 8 g
(c) 16 g
(d) 2 g
JCECE - 2006
Ans. (a) : Given,
t1/2 = 6 min,
a = 32 g, t = 18 min, a – x = ?
Half-life time of first order reaction independent of
initial concentration of reactant.
t 50% =
t1/ 2
t1/ 2
t1/ 2
32g
→16g
→ 8g
→ 4g
6min
6 min
6min
(a − x )
t = 6 + 6 + 6 = 18 min
(a – x) = 4g
594. For a first order reaction A
→ B, the reaction
rate at reactant concentration of 0.01 M is
found to be 2.0 × 10-5 mol L-1s-1. The half-life
period of the reaction is
(a) 220s
(b) 30s
(c) 300s
(d) 347s
JIPMER-2007
348
YCT
Ans. (b) : Given,
Ans. (d) :
Rate = k [Ao]
k = 5.48 × 10–14 s–1, (a–x) =
−5
2 ×10
= 2 × 10−3 s −1
0.01
Half – life time of first order reaction,
0.693
0.693
t1 =
=
= 347sec
k
2 × 10−3
2
595. The half-life of a first order reaction having
rate constant k = 1.7 × 10–5 s–1 is:
(a) 12.1 h
(b) 9.7 h
(c) 11.3 h
(d) 1.8 h
JIPMER-2006
Ans. (c) : Given,
k = 1.7 × 10 −5 s −1 , t1/ 2 = ?
Half – life time for 1st order reaction,
0.693
0.693
=
= 11.3h
t1/ 2 =
k
1.7 × 10−5
∆
596. A(g)
→ P(g) + Q (g) + R(g), follows first
order kinetic with a half-life of 69.3 s at 500oC.
Starting from the gas 'A' enclosed in a
container at 500oC and at a pressure of 0.4 atm,
the total pressure of the system after 230 s will
be
(a) 1.15 atm
(b) 1.32 atm
(c) 1.22 atm
(d) 1.12 atm
Karnataka-CET-2014
Ans. (d) : For reaction,
k=
∆
A(g)
→ P(g) + Q(g) + R(g)
t=0
P0
P0
2
t = 230s P0 – P
For first order reaction
0.693 0.693
k=
=
= 10−2 s –1
t 12
69.3
t 1 2 = 69.3s
2.303
a
k=
log10
t
(a – x)
log10
a
10 –2 × 230
=
=1
(a – x)
2.303
0
0
0
P0
2
P
P0
2
P
P0
2
P
1
3
For first order reaction
2.303
a
k=
log10
t
(a – x)
1
2.303
2.303
log10 =
log10 (3)
–14
1
5.48 ×10
5.48
× 10 –14
3
t 2 3 = 2.01× 1013 s
t 23 =
598. For a first order reaction, the half-lisfe period
is
(a) dependent on the square of the initial
concentration
(b) dependent on first power of initial
concentration
(c) dependent on the square root of initial
concentration
(d) independent on initial concentration
MHT CET-2008
Ans. (d) : The half-life period of first order reaction is
independent of initial concentration of reactants
a
Let a = a
(a – x) =
2
2.303
a
2.303
a
k=
log10
=
log10
a
t 12
(a – x)
t 12
( 2)
2.303
log10 (2)
k
0.693
t 12 =
k
599. The half-life of a substance in a certain
enzyme-catalysed reaction is 138 s. The time
required for the concentration of the substance
to fall from 1.28 mg L-1 to 0.04 mg L-1 is
t 12 =
(a) 414 s
(c) 690 s
(b) 552 s
(d) 276 s
(AIPMT Mains 2011)
Ans. (c) : Given:– t1/2 = 138 s
Half life time of first order reaction
0.693 0.693 −1
k=
=
s
t1/ 2
138
a
= 10
(a – x)
2.303
a
t=
log 10
0.4
k
(a − x)
= 10
(a – x)
2.303 × 138
1.28
t=
log 10
(P0 – P) = (a–x) = 0.04 atm
0.693
0.04
P = 0.4 – 0.04 = 0.36 atm
t = 5×138= 690 second
PT = P0 + 3P = 0.04 + 3× 0.36 = 1.12 atm
600.
Half-life
period of a first order reaction is 1386
597. Find the two third life (t2/3) of a first order
–14
seconds.
The specific rate constant of the
reaction in which k = 5.48 10 per second
13
13
reaction
is
(a) 201 10 s
(b) 2.01 10 s
(a) 0.5 × 10 −2 s −1
(b) 0.5 × 10 −3 s −1
(c) 201 1020 s
(d) 0.201 1010 s
−2 −1
(e) none of these
(c) 5.0 × 10 s
(d) 5.0 × 10 −3 s −1
Kerala-CEE-2006
(AIPMT -2009)
Objective Chemistry Volume-II
349
YCT
603. The reaction
2A + B 2
→ 2AB
is an
elementary reaction.
For a certain quantity of reactants, if the
volume of the reaction vessel is reduced by a
factor of 3, the rate of the reaction increases by
a factor of ________. (Round off to the Nearest
Integer).
JEE Main 17.03.2021, Shift-II
Ans.
(27)
:
The
given
reaction
is:
6. Factor Affecting the Rate of
2A
+
B
→
2AB
2
Reaction
The rate expression can be written as for the above
601. A reaction is
