Chapter 1 Mole Concept | 1 1 Mole Concept 1.1 Laws of Chemical Combinations 1.1.1 Law of Conservation of Mass (Lavoisier) Mass cannot be created or destroyed. In any physical or chemical process, the total mass of the system remain conserved. This law is not applicable to the nuclear processes, where mass and energy are inter-conversable by the Einstein’s equation, E = Dm×c2, (c is the speed of light). 1.1.2 Law of Constant Composition or Definite Proportion (Proust) The composition of a compound always remains fixed and it is independent to the source from which the compound is obtained. This law cannot be applied to the compound obtained by using different isotopes of the elements, selectively. 1.1.3 Law of Multiple Proportion (Dalton) If two elements combine to form more than one compound, then for the fixed mass of one element, the mass of other element combined will be in simple ratio. 1.1.4 Law of Reciprocal Proportion (Richter) If the two elements combine separately with a third element, the mass ratio of the first two elements combined with a fixed mass of the third element will be equal to or in simple ratio to the mass ratio of first two elements in a compound formed by their direct combination. 1.1.5 Gay Lussac’s Law of Volume Combination This law is applicable to only those chemical reactions in which at least two reaction components are gaseous. According to this law ‘the volumes of all gaseous reactants reacted and the volumes of all gaseous products formed, measured at the same pressure and temperature, bear a simple ratio. 2 | Chapter 1 Mole Concept Example 1 A 15.9 g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0 g. The two substances react, releasing carbon dioxide gas to the atmosphere. After reaction, the contents of the reaction vessel weigh 29.3 g. What is the mass of carbon dioxide given off during the reaction? Solution The total mass of reactants taken = 15.9 + 20.0 = 35.9 g. From the conservation of mass, the final mass of the contents of the vessel should also be 35.9 g. But it is only 29.3 g. The difference is due to the mass of released carbon dioxide gas. Hence, the mass of carbon dioxide gas released = 35.9 – 29.3 = 6.6 g Example 2 When a mixture of aluminium powder and iron (III) oxide is ignited, it produced molten iron and aluminium oxide. In an experiment, 5.40 g of aluminium was mixed with 18.50 g of iron (III) oxide. At the end of the reaction, the mixture contained 11.17 g of iron, 10.20 g of aluminium oxide, and an undetermined amount of unreacted iron (III) oxide. No aluminium was left. What is the mass of the iron (III) oxide left? Solution From the conservation of mass, the total mass of the reaction system should be conserved. Hence, total mass of Al and Iron (III) oxide taken = total mass of iron and aluminium oxide formed + the mass of Iron (III) oxide left or, 5.40 + 18.50 = 11.17 + 10.20 + mass of Iron (III) oxide left \ Mass of Iron (III) oxide left = 2.53 g Example 3 10 g of potassium chlorate is heated strongly by which a part of it decomposes as 2KClO3 $ 2KCl + 3O2 and the rest amount decomposes as 4KClO3 $ 3KClO4 + KCl If the mass of oxygen gas escaped out is 0.48 g, what is the total mass of the solid residue at the end of reaction? Solution As 10 g of KClO3 is decomposed, the total mass of products formed will also be 10 g. But, since 0.48 g oxygen gas is escaped out, the total mass of solid residue is 10 – 0.48 = 9.52 g Example 4 An experiment requires 43.7 g of isopropyl alcohol. Instead of measuring out the sample on a balance, a chemist dispenses the liquid into a graduated cylinder. The density of isopropyl alcohol is 0.785 g/mL. What volume of isopropyl alcohol should he use? mass, m Solution Density, d = volume, V \ V= 43.7 g m = = 55.7 mL 0.785 g/mL d Example 5 One litre of milk weighs 1.035 kg. The butter fat present in it occupies 4% of its volume. If the density of fat is 875 kg/m3, what is the density of fat free skimmed milk? Solution The total mass of milk = mass of fat + mass of fat free skimmed milk or, or, mmilk = (V.d)fat + (V.d)fat free milk 875 kg d kg m `1 L # 96 j # c m 1.035 kg = `1 L # 4 j # c 100 100 1000 L 1000 L Chapter 1 Mole Concept | 3 \ Density of fat free milk, d = 1042 kg/m3 Example 6 105 g water is saturated with ammonia gas to form a solution of density 0.9375 g/mL and containing 30% ammonia, by mass. What is the volume of solution formed? Solution The solution contains 30% ammonia and hence, from mass conservation, the rest 70% will be water. It means that 100 g of solution contains 70 g water. 70 g water results 100 g solution 100 \ 105 g water will result 105 = 150 g solution. 70 150 g m = = 160 mL Now, the volume of solution formed, V = 0.9375 g/mL d Example 7 A nugget of gold and quartz weighs 100 g. The densities of gold, quartz and the nugget are 19.3, 2.65 and 6.4 g/mL, respectively. Determine the weight of gold in the nugget. Solution Let 100 g of the nugget contains x g gold. Then, the rest (100 – x) g will be quartz. Now, as the nugget is a mixture of solids, the volume of nugget may be assumed as sum of volumes of gold and quartz present in it. Vnugget = Vgold + Vquartz or, x 100 (100 - x ) = + 19.3 6.4 2.65 \ Mass of gold present, x = 67.92 g Example 8 A clay sample contains 50% silica and 12% water. The sample is dried partially. If the partially dried clay contains 7% water, what will be the percentage of silica in it? Solution On drying, the loss will occur only in the mass of water present. The mass composition of silica and all other contents present should be constant. It means that mass % of silica mass % of silica = mass % of other contents original clay mass % of other contents partially dried clay x 50 = 100 - ( x 7) 100 - (50 12) or, percentage of silica in partially dried clay, x = 52.84% or, Example 9 The following are results of analysis of two samples of the same or two different compounds of phosphorus and chlorine. From these results, decide whether the two samples are from the same or different compounds. Also state the law, which will be obeyed by the given samples. Compound Amount P Amount Cl Compound A 1.156 g 3.971 g Compound B 1.542 g 5.297 g Solution The mass ratio of phosphorus and chlorine in compound A, mP : mCl = 1.156 : 3.971 = 0.2911 : 1.000 The mass ratio of phosphorus and chlorine in compound B, 4 | Chapter 1 Mole Concept mP : mCl = 1.542 : 5.297 = 0.2911 : 1.000 As the mass ratio is same, both the compounds are same and the samples obey the law of definite proportion. Example 10 Two oxide samples of lead were heated in the current of hydrogen and were reduced to the metallic lead. The following data were obtained (i) Weight of yellow oxide taken = 3.45 g; Loss in weight in reduction = 0.24 g (ii) Weight of brown oxide taken = 1.227 g; Loss in weight in reduction = 0.16 g. Show that the data illustrates the law of multiple proportion. Solution When the oxide of lead is reduced in the current of hydrogen, metallic lead is formed. Definitely, the loss in weight in reduction is due to removal of the oxygen present in the oxide, to combine with the hydrogen. Therefore, the composition of the yellow oxide is: oxygen = 0.24 g and lead = 3.45 – 0.24 = 3.21 g. m 3.21 13.375 The mass ratio of lead and oxygen, r1 = Pb = = mO 0.24 1.000 and the composition of the brown oxide is: oxygen = 0.16 g and lead = 1.227 – 0.16 = 1.067 g. The mass ratio of lead and oxygen, r2 = mPb 1.067 6.669 = = mO 0.16 1.000 Now, r1 : r2 = 13.375:6.669 = 2 : 1 (simple ratio) and hence, the data illustrates the law of multiple proportion. Example 11 Methane contains 75 % carbon and 25% hydrogen, by mass. Carbon dioxide contains 27.27 % carbon and 72.73% oxygen, by mass. Water contains 11.11 % hydrogen and 88.89% oxygen, by mass. Show that the data illustrates the law of reciprocal proportion. Solution Methane and carbon dioxide, both contains carbon and hence, carbon may be considered as the third element. Now, let the fixed mass of carbon = 1 g. Then, 25 1 the mass of hydrogen combined with 1 g carbon in methane = = g 75 3 and the mass of oxygen combined with 1 g carbon in carbon dioxide = 72.73 8 = g 27.27 3 Hence, the mass ratio of hydrogen and oxygen combined with the fixed mass of 1 8 1 carbon, r1 = : = 3 3 8 11.11 1 Now, the mass ratio of hydrogen and oxygen in water, r2 = = 88.89 8 As r1 and r2 are same, the data is according to the law of reciprocal proportion. Note: You may also solve the problem by taking hydrogen or oxygen as the third element. Try yourself. Example 12 2.5 mL of a gaseous hydrocarbon exactly requires 12.5 mL oxygen for complete combustion and produces 7.5 mL carbon dioxide and 10.0 mL water vapour. Chapter 1 Mole Concept | 5 All the volumes are measured at the same pressure and temperature. Show that the data illustrates Gay Lussac’s law of volume combination. Solution Vhydrocarbon : Voxygen : Vcarbon dioxide : Vwater vapour = 2.5 : 12.5 : 7.5 : 10.0 = 1 : 5 : 3 : 4 (simple ratio) Hence, the data is according to the law of volume combination. 1.2 Atoms and Molecules 1.2.1 Atom may be defined as the smallest particle of an element which does not exist free in nature but takes part directly in chemical combinations. Atom of any element is represented by the symbol of that element. For example, hydrogen atom is represented by H, sodium atom is represented by Na, etc. 1.2.2 Molecule may be defined as the smallest particle of an element or compound which exists free in nature but does not participate directly in chemical combinations. Molecule of any substance is represented by their molecular formula, which tells the exact number of atoms of same or different elements present in each molecule of that substance. For example, water molecule is represented by H2O. It tells that each water molecule contains two atoms of hydrogen and one atom of oxygen. · Atoms of inert gases exist free in nature. · The term molecule should not be used for the ionic compounds. The perfect term for them is formula unit, which represents the simple ratio of the ions present in the compound. · The smallest particles of metals are always atoms, not molecules. 1.3 Atomic Weight 1.3.1 1.3.2 The atomic weight of an element represents the number of times by which one 1 th part by weight of one atom of C12 atom of the element is heavier than 12 isotope. weight of one atom of the element Atomic weight of an element = 1 × weight of one atom of C12 isotope 12 Atomic Mass Unit (amu) unified mass (u) It represents 1 th part by weight of one atom of C12 isotope. 12 1 1u= g = 1.66 × 10–24 g NA (NA = Avogadro’s number = 6.023 × 1023) 1.3.3 1.3.4 If the atomic weight of an element is A, then the weight of one atom of the element is A amu and the weight of NA­atoms of the element is A g. When atomic weight of any element is expressed in gram, it is called g-atomic weight of that element. Weight and number of atoms of an element are related as: wt. (in gm) × NA No. of atoms = gm. at. wt 6 | 1.3.5 Chapter 1 Mole Concept Concept of g-atom One g-atom of any element contains NA atoms and hence, weighs equal to its g-atomic weight. wt. (in gm) No. of atoms No. of g-atoms = = NA gm. at. wt. 1.3.6 Chapter 1 Mole Concept | 7 The atomic weight of most of the element is fractional due to the presence of isotopes. The average atomic weight of an element may be calculated from its isotopic composition as (percentage abundance atomic weight) Average atomic weight = 100 Example 13 The atomic mass of calcium is 40. How many atoms are present in 2g calcium? wt. (in gm) 2 × NA = (6.0231023) = 3.0115 1022 gm. at. wt. 40 Solution No. of Ca-atoms = Example 14 The point given as full stop at the end of a sentence by a graphite pencil weighs 210–12 g. How many carbon atoms are present in such a point? wt. (in gm) × NA Solution No. of C-atoms = gm. at. wt. = 2 10-12 (6.0231023) = 1.004 1011 12 Example 15 What is the mass of 3.01151020 atoms of mercury? The atomic weight of mercury is 200. Solution \ wt. (in gm) × NA gm. at. wt. no. of atoms 3.0115 1020 weight = gm. at. wt. = 200 = 0.1 g NA 6.0231023 No. of atoms = Example 16 The specific gravity of the stainless steel spherical balls used in ball-bearings are 10.2. How many iron atoms are present in each ball of diameter 1 cm if the balls contain 84% iron, by mass? The atomic mass of iron is 56. Solution Volume of a ball = 4 3 4 1 × 3.14 × ( cm)3 = 0.523 cm3 pr = 3 3 2 Mass of a ball = V × d = 0.523 cm3 × 10.2 g.cm–3 = 5.33 g Mass of iron present in a ball = 5.33 \ No. of iron atoms = = 84 = 4.48 g 100 wt. (in gm) × NA gm. at. wt. 4.48 (6.023 × 1023) = 4.82 × 1022 56 Example 17 Morphine contains 67.3% carbon, 4.6% nitrogen (by mass) and remaining are the other constituents. Calculate the relative number of carbon and nitrogen atoms in morphine. Solution Let there is 100 g morphine. Then the masses of carbon and nitrogen present are 67.3 and 4.6 g, respectively. Now, w 67.3 NA NA A C 12 No. of C-atoms 17 = = = 4.6 w No. of N-atoms 1 NA NA 14 A N 8 | Chapter 1 Mole Concept Example 18 The density of mercury is 13.6 g/mL. Calculate the diameter of an atom of mercury assuming that each atom of mercury is occupying a cube of edge length equal to the diameter of the mercury atom. (Hg = 200) Solution Let the diameter of the mercury atom as well as the edge length of cube is l cm. Now, the volume of one atom = volume of cube = (l cm)3 mass of one atom or, = (l cm)3 density 200 1.66 10-24 1/3 cm = 2.910–8 cm or, l = 13.6 Example 19 Calculate the number of neutrons present in 7 mg of C14. Solution No. of neutrons in one C14-atom = mass no. – atomic no. = 14 – 6 = 8 and no. of atoms = wt. (in gm) 7 10-3 × NA = (6.0231023) = 3.0115 gm. at. wt 14 1020 \ Total number of neutrons present = 8 × (3.0115 × 1020) = 2.409 × 1021 Example 20 Calculate the number of g-atoms in the following: (a) 3.5 g nitrogen (b) 100 mL mercury (density = 13.6 g/mL, Hg = 200) (c) 1.506 × 1022 atoms of hydrogen (d) 6.023 × 1020 molecules of ozone Solution (a) No. of g-atoms = wt. (in gm) 3.5 = = 0.25 gm. atomic wt. 14 wt. (in gm) V ×d 100 13.6 = = = 6.8 gm. atomic wt. A 200 no. of atoms 1.506 1022 (c) No. of g-atoms = = = 0.025 NA 6.0231023 no. of atoms (6.0231020 ) 3 = = 0.003 (d) No. of g-atoms = NA 6.0231023 (b) No. of g-atoms = Example 21 From 10 g calcium, 3.0115 × 1022 atoms are removed. Calculate the number of g-atoms of calcium left. (Ca = 40). Solution No. of g-atoms of Ca left = g-atoms of Ca taken – g-atoms of Ca removed wt. (in gm) no. of atoms – NA gm. atomic wt. 22 3.0115 10 10 = – = 0.20 40 6.0231023 = Example 22 In nature, lithium exists in two isotopic forms, Li6 (at. mass = 5.9456, abundance = 10%) and Li7 (at. mass = 6.9832, percentage abundance = 90%). What is the average atomic mass of lithium? Solution Average atomic mass = (percentage abundance × 100 atomic weight) Chapter 1 Mole Concept = | 9 10 5.9456 90 6.9832 = 6.8794 100 Example 23 K40 is one of the few naturally occurring radioactive isotopes of elements of low atomic number. Its percent natural abundance is 0.012%. How many K40 atoms do you ingest by drinking one cup of whole milk containing 370 mg K? Average atomic weight of K = 39. wt. (in gm) × NA Solution No. of K-atoms = gm. at. wt. = \ No. of K40-atoms = 370 10-3 (6.0231023) = 5.7141018 39 0.012 (5.7141018) = 6.8571014 100 Example 24 A sample of oxygen contains only O16 and O18 atoms in 4000:1 atomic ratio. Calculate the average neutrons per oxygen atom? Solution Average neutrons per atom = total neutrons total atoms 4000 (16 - 8) 1 (18 - 8) = 8.0005 4000 1 = 1.4 Molecular Weight 1.4.1 1.4.2 The Molecular weight of a substance represents the number of times by which 1 th part by weight of one atom of one molecule of the substance is heavier that 12 12 C isotope. weight of one molecule of the substance Molecular weight of a substance = 1 weight of one atom of C12 isotope 12 Molecular weight of any substance may be given as the sum of atomic weights of all the atoms present in each molecule. 1.4.3 If the molecular weight of a substance is M, then the weight of one molecule of the substance is M u and the weight of NA molecules of the substance is M g. When the molecular weight of any substance is expressed in gram, it is called gram molecular weight. 1.4.4 Weight and number of molecules of any substance are related as: wt. (in gm) × NA No. of molecules = gm. mol. wt. 1.4.5 Concept of g-molecule One g-molecule of any substance contains NA molecules and hence, weighs equal to its g-molecular weight. In addition to it, one g-molecule of an ideal gas occupies 22.7 L at STP (pressure = 1 bar and temperature = 273 K) or 22.4 L at 1 atm and 273 K. wt. (in gm) No. of molecules = No. of g-molecules = NA gm. mol. wt. 10 | Chapter 1 Mole Concept Volume of gas at STP Volume of gas at 1 atm and 273 K = = 22.7 L 22.4 L Note: STP means standard temperature and pressure. IUPAC and hence, NCERT has changed the value of standard pressure from 1 atm to 1 bar. Most of the books are using the earlier concept and they take the volume of an ideal gas at STP (1 atm and 273 K) equal to 22.4 L. In this book, STP is strictly used as 1 bar and 273 K and hence the corresponding volume is taken as 22.7 L. 22.4 L is used only when STP is mentioned as 1 atm and 273 K. 1.4.6 Average molecular weight of any non-reacting gaseous mixture • When percentage composition is given by volume, Mmix = (volume percent × mol. wt.) 100 (For the non-reacting gaseous mixtures, the composition by volume and mole are identical) • When percentage composition is given by weight, Weight percent 100 = M mix mol. wt. Example 25 How many molecules are present in 45 g glucose, C6H12O6? Solution Molecular weight of glucose = 6 × 12 + 12 × 1 + 6 × 16 = 180 wt. (in gm) × NA gm. mol. wt. 45 = (6.0231023) = 1.5061023 180 Now, no. of molecules = Example 26 The population of India is 1.004 billion. If you have to distribute two million cane sugar (C12H22O11) molecules to each person, how many gram of sugar should you need? Solution No. of sugar molecules needed = (1.004 × 109) × (2 × 106) = 2.008 × 1015 Now, Molecular weight of sugar = 12 × 12 + 22 × 1 + 11 × 16 = 342 no. of molecules × mol. wt. wt. of sugar needed = NA = 2.008 1015 × 342 = 1.14 × 10–6 g 6.0231023 Example 27 If molecular weight of glucose-1-phosphate is 260 and its density is 1.5 g/mL. What is the average volume occupied by each molecule of this compound? Solution Volume occupied by one molecule = mass of one molecule density 260 1.66 10-24 = 2.8810–22 mL 1.5 Example 28 The shape of Tobacco Mosaic Virus (TMV) is cylindrical, having length and diameter 3000 Å and 170 Å, respectively. The density of the virus is 0.08 g/mL. What is the molecular weight of TMV? = Solution Volume of a virus particle = pr2l = 3.14 × ( 170 10–8 cm)2 × (300010–8 cm) = 6.81 × 10–17 cm3 2 Chapter 1 Mole Concept | 11 Mass of a virus particle = V × d = (6.81 × 10–7) × 0.08 = 5.448 × 10–18 g Molecular weight of virus = (5.448 × 10–18) × (6.023 × 1023) = 3.28 × 106 Example 29 Each molecule of a substance contains 7 atoms of carbon, 14 atoms of hydrogen and 5.33×10–23 g of other elements. What is the molecular weight of substance? Solution Mass of one molecule = 712 u + 141 u + 5.3310-23 u 1.66 10-24 = 130.1 u Hence, the molecular weight of the substance = 130.1 Example 30 A mixture of 1.65 × 1021 molecules of X and 1.85 × 1021 molecules of Y weighs 0.638 g. If the molecular weight of X is 42, what is the molecular weight of Y? Solution mmix = mX + mY no. of molecules no. of molecules mol. wt. + mol. wt. = NA NA X Y or, 0.638 = or, 1.65 1021 1.85 1021 × MY 23 × 42 + 6.02310 6.0231023 MY = 170.27 Example 31 The atomic ratio of H1 to H3 in a sample of water is 1 : 8 × 10–8. Calculate the number of H3 atoms in 9.0 g of such water sample. Solution Average atomic weight of hydrogen = 1 1 3 8 10-8 1 1 8 10-8 Hence, molecular weight of water = 2 × 1 + 1 × 16 = 18 9.0 × (6.023 × 1023) = 3.0115 × 1023 Now, no. of water molecules = 18 No. of hydrogen atoms = 2 × (3.0115 × 1023) = 6.023 × 1023 8 10-8 \ No. of H3-atoms = × (6.023 × 1023) = 4.82 × 1016 1 8 10-8 Example 32 How many formula units are present in 715 mg washing soda, Na2CO3×10H2O? Solution Formula unit weight of Na2CO3×10H2O = 2 × 23 + 1 × 12 + 3 × 16 + 10 × 18 = 286 weight (in gm) × NA Now, no. of formula units = gm formula unit weight = 715 10-3 (6.0231023) = 1.506 × 1021 286 Example 33 How many hydrogen atoms are present in 9.6 g urea, NH2CONH2? Solution Molecular weight of urea = 2 × 14 + 4 × 1 + 1 × 12 + 1 × 16 = 60 No. of urea molecules = = wt. (in gm) × NA gm mol. wt. 9.6 × (6.0231023) = 9.6368 × 1022 60 12 | Chapter 1 Mole Concept As each urea molecule contain four hydrogen atoms, the total number of hydrogen atoms = 4 × (9.6368 × 1022) = 3.855 × 1023 Example 34 Calculate the number of electrons present in 1.90 g phosphate ion, PO34 - . The atomic numbers of phosphorus and oxygen are 15 and 8, respectively and their atomic masses are 31 and 16, respectively. Solution The ionic mass of phosphate ion = 1 × 31 + 4 × 16 = 95 Now, the no. of PO34 - ions = wt. (in gm) × NA gm ionic wt. 1.90 × (6.023 × 1023) = 1.2046 × 1022 95 and no. of electrons in one PO34 - ion = 1 × 15 + 4 × 8 + 3 = 50 \ Total number of electrons present = 50 × (1.2046 × 1022) = 6.023 × 1023 = Example 35 Calculate the number of g-molecules in the followings: (a) 2.45 g sulphuric acid, H2SO4 (b) 3.0115 × 1024 molecules of SO3 (c) 1.2046 × 1022 atoms of nitrogen (d) 16.8 L of oxygen gas at STP (1 atm and 0°C) Solution (a) No. of g-molecules = wt. (in gm) 2.45 = = 0.025 gm mol. wt. 98 (b) No. of g-molecules = no. of molecules 3.0115 1024 = =5 NA 6.0231023 (c) No. of g-molecules = no. of molecules (1.2046 1022 ) / 2 = = 0.01 NA 6.0231023 (d) No. of g-molecules = vol. of gas at 1 atm and 0°C (in L) 16.8 = = 0.75 22.4 22.4 Example 36 A gaseous mixture of methane and ethane contains 40% methane, by volume. What is the average molecular weight of the mixture? Solution Mmix = (volume percent × mol. wt.) 100 = 40 16 + 60 30 = 24.4 100 Example 37 A sample of ozonised oxygen contains 60% ozone, by weight. What is the average molecular weight of the sample? Solution 100 = Mmix = \ Msample = Weight percent mol. wt. 60 40 per. of O3 per. of O2 + = + = 2.5 48 32 M O3 M O2 100 = 40 2.5 Chapter 1 Mole Concept | 13 Example 38 The composition of a sample of air is N2 = 80% and O2 = 20%, by volume. What will be the composition of the air sample, by weight? Solution Mair = (vol. percent mol. wt.) 100 = 80 28 20 32 = 28.80 100 Let the weight percent of N2 in air is x. Then the percentage of O2 will be (100 – x). Now, or, Weight percent 100 = Mmix mol. wt. x 100 100 - x = + 28 28.80 32 or, x = 77.78 \ Composition of air, by weight: N2 = 77.78% and O2 = 22.22% 1.5 Mole 1.5.1 The term mole is S.I. unit to represent the quantity of any substance. One mole of any substance contains same number of particles of that kind as the number of atoms present in exactly 12 g of C-12 isotope. For example, 1 mole atom = NA atoms (it is also called 1 g-atom) 1 mole molecule = NA molecules (it is also called 1 g-molecule) 1 mole ions = NA ions (it is also called 1 g-ion) Mole can be used to represent the quantity of any substance like atom, molecule, ion, rupee, etc. As these are the molecules, which have independent existence, the term mole is mainly used for molecules and hence, mole normally appears as g-molecule. The working formula for determination of mole will be similar to the determination of g-molecule. 1.5.2 For gases, the amount may also be expressed in terms of pressure (p), volume (V) and temperature (T) in place of weight or mole (n). These physical quantities are related by ideal or perfect gas equation as pV = nRT where R = Universal gas constant = 0.082 L-atm/K-mol = 8.314 J/K-mol ≈ 2 cal/K-mol Units of Pressure and their Relation 1 atm = 76 cm of Hg = 760 mm of Hg = 760 torr (1 torr = 1mm of Hg) = 1.013 × 106 dyne/cm2 = 1.013 × 105 N/m2 or Pa = 1.013 bar (1 bar = 105 Pa) 14 | Chapter 1 Mole Concept Chapter 1 Mole Concept | 15 Units of Volume and their Relation 1 mL = 1 cm3 = 1 c.c. 1 L = 1000 mL = 1 dm3 1 m3 = 1000 L = 106 mL Units of Temperature and their Relation T = 273 + t where T = absolute temperature is Kelvin t = temperature in °C 1.5.3 1.5.4 1.5.5 w pM = V RT where, M = molecular weight of gas • The ratio of densities of two gases under identical conditions of pressure and temperature is equal to the ratio of their molar masses. Density of an ideal gas, d = Vapour density of any substance is the ratio of density of that substance to the density of hydrogen gas under identical conditions of temperature and pressure. However, in some cases, vapour density may also be given with respect to air or other gases. When vapour density is given with respect to hydrogen, it is half of the molecular weight of the substance. M -M , where The degree of dissociation of a gaseous substance is given as: a = 0 ( n -1)M M0 is the molecular weight of the original compound, M is the average molecular weight of the gaseous mixture formed after dissociation and n is the number of moles of the different gaseous substances formed from each mole of the compound dissociated. 1.5.6 Avogadro’s Hypothesis Under identical conditions of pressure and temperature, equal volumes of all the gases contain equal number of molecules. Example 39 Calculate the mass of the following: (a) 0.01 mole glucose, C6H12O6 (c) 0.8 g-molecule oxygen Solution \ (b) 2.5 mole ammonia, NH3 (d) 1.25 g-atom magnesium (Mg = 24) No. of moles, n = wt. (in gm), w gm mol. wt., M w=n×M (a) Molecular weight of C6H12O6 = 6 × 12 + 12 × 1 + 6 × 16 = 180 Now, w = n × M = 0.01 × 180 = 1.80 g (b) Molecular weight of NH3 = 1 × 14 + 3 × 1 = 17 Now, w = n × M = 2.5 × 17 = 42.5 g (c) w = n × M = 0.8 × 32 = 25.6 g (d) w = n × M = 1.25 × 24 = 30 g 16 | Chapter 1 Mole Concept Example 40 Dopamine is a neurotransmitter, a molecule that serves to transmit message in the brain. The chemical formula of dopamine is C8H11O2N. How many moles are there in 1 g of dopamine? Solution Molecular weight of dopamine = 8 × 12 + 11 × 1 + 2 × 16 + 1 × 14 = 153 Now, no. of moles, n = wt. (in gm), w 1 = = 0.00654 gm mol. wt., M 153 Example 41 The volume of one mole of water at 4°C is 18 mL. One mL of water contains 20 drops. How many molecules are present in one drop of water? 1 = 0.05 mL 20 Mass of a drop of water = V × d = 0.05 × 1 = 0.05 g Solution Volume of a drop of water = (Density of water is given as 1 g/mL, as 1 mole of water, i.e., 18 g occupy 18 mL) 0.05 Number of moles of water present in a drop = 18 0.05 \ Number of water molecules present in a drop = × (6.023 × 1023) 18 = 1.67 × 1021 Example 42 A certain nut crunch cereal is listed as containing 11 g of sugar, C12H22O11, per serving size of 60 g. How many servings (in gram) of this cereal must one eat to consume 0.0278 mole of sugar? Solution Molecular weight of sugar = 12 × 12 + 22 × 1 + 11 × 16 = 342 Hence, weight of sugar = n × M = 0.0278342 = 9.508 g 11 g of sugar is present in the serving size of 60 g 60 \ 9.508 g sugar will be present in × 9.508 = 51.86 g 11 Example 43 A given mixture consists only of pure substance X and pure substance Y. The total weight of the mixture is 3.72 g. The total number of moles is 0.06. If the weight of one mole of Y is 48 g and there is 0.02 mole of X in the mixture, what is the weight of one mole of X? Solution mMix = mX + mY = (n × M)X + (n × M)Y or, 3.72 = 0.02 × MX + (0.06 – 0.02) × 48 or, MX = 90 \ Weight of one mole of X = 90 g Example 44 0.11 g of a colourless oxide of nitrogen occupies 41 mL at 27°C and 1.5 atm. Identify the oxide. Solution The ideal gas equation is pV = nRT = \ M= w RT M 0.11 0.082 (27 273) wRT = = 44 1.5 (41 10-3 ) pV Now, let the formula of oxide is NxOy. Its molecular weight will be 14x + 16y. But from calculation it is 44. By hit and trial method, x = 2, y = 1 and hence, the formula of oxide is N2O. Chapter 1 Mole Concept | 17 Example 45 One molecule of haemoglobin will combine with four molecules of oxygen. If 1.0 g of haemoglobin combines with 1.53 mL of oxygen at body temperature (37ºC) and a pressure of 743 torr, what is the molar mass of haemoglobin? pV Solution Mole of oxygen combined with haemoglobin, n = RT 743 1.53 760 1000 = = 5.8810–5 0.082 (37 273) 4 molecules of oxygen combine with 1 molecule of haemoglobin or, 4 mole of oxygen combine with 1 mole of haemoglobin \ 5.88 × 10–5 mole of oxygen will combine with 1 (5.8810–5) 4 = 1.4710–5 mole of haemoglobin But from question, haemoglobin combined should be 1.0 g. Now, w=n×M w 1 or, molecular weight of haemoglobin, M = = = 68027.21 n 1.47 10-5 Example 46 Calculate the density of nitrogen gas at STP (1 atm and 0°C). Solution Density of any substance is independent to its amount and hence, density may be calculated by taking any amount of the matter. Let, we have 1 mole of nitrogen gas. Its weight will be 28 g and volume will be 22.4 L at STP (1 atm and 0°C). w 28 g \ Density, d = = = 1.25 g/L V 22.4 L Alternate method pM 1 28 = = 1.25 g/L Density, d = RT 0.082 273 Example 47 The density of a gaseous substance is 1.5 (C2H6 = 1). Calculate the volume occupied by 9.0 g of the gaseous substance at STP (1 atm and 0°C). d Solution From question, substance = 1.5 dC2H6 But the ratio of densities is equal to the ratio of their molecular weights and hence, Msubstance, M = 1.5 × M C2H6 = 1.5 × 30 = 45 Now, the volume occupied by the gas, V = n × 22.4 L w 9 22.4 L = 22.4 L = M 45 = 4.48 L Example 48 The density of mercury vapour is 6.92 with respect to air. If the average molar mass of air is 29 g/mol, what is the atomicity of mercury in vapour state? (Hg = 200) Solution Let the atomicity of mercury in vapour state is x. Its molecular formula may be expressed as Hgx. 18 | Chapter 1 Mole Concept From question, or, or, \ dHg x dair M Hg x M air = 6.92 = 6.92 200 x = 6.92 29 Atomicity, x 1 (as it can never be fractional) Example 49 A gaseous mixture of SO3 and SO2 contain 40% SO3, by mole. What is the percentage of SO3 in the mixture, by mass? Solution Let we have 100 mole of the mixture. Then, it will contain 40 mole SO3 and rest (100 – 40) = 60 mole SO2. Now, mass of SO3 present = n × M = 40 × 80 = 3200 g and mass of SO2 present = n × M = 60 × 64 = 3840 g \ Total mass of mixture = 3200 + 3840 = 7040 g 3200 100 = 45.45% Hence, mass percent of SO3 = 7040 Example 50 A gaseous mixture contains 4 g oxygen and 14 g nitrogen. What is the average molecular weight of the mixture? Solution or, or, \ nmix = nO2 + nN2 w w w = + M O2 M N2 M mix 4 14 4 14 = + M mix 32 28 Mmix = 28.8 Example 51 The vapour density of a gaseous mixture of N2O4 and NO2 is 40. How many moles of NO2 are present in (a) 100 mole of mixture (b) 100 g of mixture? Solution (a) Let the mole of NO2 in 100 moles of mixture is x. Then, the mole of N2O4 = 100 – x Now, mmix = mNO2 + mN2O4 or, or, \ (n × M)mix = ( n M )NO2 + ( n M )N2O4 100 × (40 × 2) = x × 46 + (100 – x) × 92 Mole of NO2, x = 26.09 (b) Let the mass of NO2 in 100 g of mixture is y g. Then, the mass of N2O4 = (100 – y) g Now, nmix = nNO2 + mN2O4 or, w w w = + M mix M NO2 M N2O4 Chapter 1 Mole Concept or, | 19 y 100 100 - y = + 46 2 40 92 mass of NO2, y = 15 g 15 = 0.326 And mole of NO2 = 46 Alternate Method \ After solving part A, part B may also be solved easily as: 100 Total moles of the mixture = = 1.25 80 Let the moles of NO2 in the mixture is z, then the moles of N2O4 in the mixture = (1.25 – z) Now, mmix = mNO2 + mN2O4 or, 100 = z 46 + (1.25 – z) 92 Hence, moles of NO2, z = 0.326 Note: Any problem of non-reacting gaseous mixture may be solved by applying mass and mole conservations. Example 52 The molecular weight of a sample of PCl5 is found to be 180. What is the degree of dissociation of PCl5 into PCl3 and Cl2? Solution Degree of dissociation, a = Here, and \ M0 - M ( n -1)M M0 = M PCl = 1 × 31 + 5 × 35.5 = 208.5 5 M = 180 (from question) n = 2 (for the reaction: PCl5 PCl3 + Cl2) M -M 208.5 -180 a = 0 = = 0.1139 ( n -1)M (2 -1) 180 Example 53 The density of a sample of SO3 is 6.2 g/L at 27°C and 2.05 atm. What is the degree of dissociation of SO3 into SO2 and O2? PM Solution Density, d = RT dRT 6.2 0.082 300 \ M= = = 74.4 P 2.05 Now, M0 = M SO3 = 1 × 32 + 3 × 16 = 80 And, \ n = 1.5 (for the reaction: 2SO3 2SO2 + O2) M -M 80 -74.4 = = 0.15 a= 0 ( n -1)M (1.5 -1) 74.4 1.6 Percentage Composition of Compounds 1.6.1 1.6.2 The composition of any compound represents the relative amount of all the constituent elements, by weight. The percentage composition of an element in a compound can be determined by using the following formula, if the composition (by atoms) of the compound is known. 20 | Chapter 1 Mole Concept Z A 100% M Where, Z = Number of atoms of that element in each molecule A = atomic weight of the element M = Molecular weight of the compound. 1.6.3 The composition of an element can also be determined by some experimental informations. For example, the percentage of carbon and hydrogen in an organic compound can be determined by burning a known weight of the compound completely and then weighing the carbon dioxide and water formed. Let w1 g of the organic compound gives w2 g of CO2 and w3 g H2O, on complete combustion, then w 12 Percentage of carbon = x 2 x 100% w1 44 w 2 x 3 x 100% Percentage of hydrogen = w1 18 Percentage of an element = Similarly, the percentage of other elements can also be determined. The basic concept behind such determination is that the source of that element should be only the original compound. The reaction or series of reactions are performed in such a way that in the final product, that element comes only from the original compound. Example 54 Calculate the percentage composition of glucose, C6H12O6. Solution Molecular weight of C6H12O6 = 612 + 121 + 616 = 180 \ and 6 12 100 = 40% 180 12 1 Percentage of hydrogen = 100 = 6.67% 180 6 16 100 = 53.33% percentage of oxygen = 180 Percentage of carbon = Example 55 Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The weight percentage of carbon in cortisone is 69.98%. What is the molecular weight of cortisone? Z A Solution Percentage of an element = 100% M or, \ 69.98 = 21 12 100 M M = 360.1 Example 56 A polystyrene of formula Br3C6H2(C8H8)n was prepared by heating styrene with tribromobenzyl peroxide in the absence of air. It was found to contain 10.46% bromine by weight. Find the value of n. (Br = 80) Z A Solution Percentage of an element = 100% M 3 80 or, 10.46 = 100 314 104n \ n = 19.04 19 Chapter 1 Mole Concept | 21 Example 57 A sample of copper sulphate pentahydrate, CuSO4×5H2O, contains 3.782 g of Cu. How many grams of oxygen are in this sample? (Cu = 63.5) Solution Each formula unit of CuSO4×5H2O contains 1 atom of copper and 9 atoms of oxygen. The relative contributions of these elements are 63.5 unit copper and 9 × 16 = 144 units oxygen. 63.5 g of copper is present with 144 g oxygen 144 3.782 = 8.577 g oxygen \ 3.782 g copper will be present with 63.5 Example 58 An organic compound contains carbon, hydrogen, nitrogen and oxygen only. 0.135 g of this compound on combustion produced 0.198 g of CO2 and 0.108 g H2O while the same amount gave 16.8 mL of nitrogen at 0°C and 76 cm Hg of pressure. Calculate the percentage of oxygen in the compound. wt. of CO2 12 Solution Percentage of carbon = 100% wt. of org. comp. 44 = 12 0.198 100% = 40% 44 0.135 Percentage of hydrogen = wt. of H2O 2 × wt. of org. comp. × 100% 18 = 2 0.108 × × 100% = 8.89% 18 0.135 wt. of N 2 × 100% wt. of org. comp. 16.8 28 22400 × 100% = 15.56% = 0.135 Percentage of nitrogen = \ Percentage of oxygen = 100 – (40 + 8.89 + 15.56) = 35.55% Example 59 Penicillin-V was treated chemically to convert sulphur to barium sulphate. An 8.19 mg sample as penicillin gave 5.46 mg of barium sulphate. What is the percentage of sulphur in penicillin-V, by weight? If there is one S-atom in the molecule, what is the molecular weight? (Ba = 137) Solution Molecular weight of BaSO4 = 137 + 32 + 416 = 233 233 g BaSO4 contains 32 g sulphur 32 \ 5.46 mg will contain 5.46 = 0.75 mg sulphur 233 As the source of sulphur in BaSO4 is only the organic compound, the percentage of sulphur = 0.75 × 100 = 9.16% 8.19 Z A 100% M 1 32 or, 9.16 = 100 M \ Molecular weight of compound, M = 349.34 Now, Percentage of an element = 22 | Chapter 1 Mole Concept Example 60 There is 4 g mixture of NaCl and NaBr. If Na is 30% of the total mass of the mixture, calculate the moles of each in the mixture. (Na = 23, Cl = 35.5, Br = 80) Solution Let the mixture contains x mole NaCl and y mole NaBr. Then, the total mole of sodium in the mixture will be (x + y). Now, total mass of the mixture, m = mNaCl + mNaBr or, 4 = x × 58.5 + y × 103 … (1) and the mass of sodium present = 30% of total mass 30 × 4 = (x + y) × 23 … (2) or, 100 From Eq. (1) and (2), mole of NaCl = x = 0.0309 and mole of NaBr = y = 0.0213 Example 61 The composition of a sample of Wurztite is Fe0.93O1.00. What percentage of the iron is present in the form of Fe(III)? Solution The stoichiometric oxides of oxygen are only FeO and Fe2O3. Wurztite is a non-stoichiometric oxide of iron which may be assumed as a mixture of FeO and Fe2O3 in x : y mole ratio. Fe0.93O1.00 = x FeO + y Fe2O3 From Eq. (1) and (2), the values of x and y may be determined by atomic conservations. On conserving the iron atoms, 0.93 = x + 2y … (1) On conserving the oxygen atoms, 1.00 = x + 3y … (2) x = 0.79, y = 0.07 2y × 100 0.93 2 0.07 × 100 = 15.05% = 0.93 \ Percentage of iron as Fe(III) = 1.7 Molecular and Empirical Formula of Compounds 1.7.1 Molecular Formula Molecular formula of any compound represents the exact number of atoms of different elements present in each molecules of the compound. 1.7.2 Empirical Formula Empirical formula of any compound represents the simplest atomic ratio of the different elements present in the compound. Compound Ethane Ethanol Acetic Acid Glucose Molecular formula C2H6 C2H6O C2H4O2 C6H12O­6 Empirical formula CH3 C2H6O CH2O CH2O · Two different compounds can have the same empirical formula. · Two different compounds can have the same molecular formula. Chapter 1 Mole Concept | 23 · Molecular formula = (Empirical formula)n molecular formula weight where, n = empirical formula weight · For most of the ionic compounds, the formulae represented are their empirical or simplest formulae. Example 62 An organic compound contains carbon = 40%, hydrogen = 6.67% and rest oxygen. The molecular weight of the compound is 60. Determine the empirical and molecular formula of the compound. Solution Element Percentage composition Atomic weight % composition at. wt. Simplest ratio C 40 12 40 = 3.33 12 3.33 =1 3.33 H 6.67 1 6.67 = 6.67 1 6.67 =2 3.33 O 100 – (40 + 6.67) = 53.33 16 53.33 = 3.33 16 3.33 =1 3.33 Hence, the empirical formula of compound = C1H2O1 = CH2O Now, n= mol. formula wt. 60 = =2 emp. formula wt. 12 2 1 16 \ Molecular formula = (CH2O)2 = C2H4O2 Alternate method (without using tabular form) As the molecular formula represents the exact number of atoms of each element present in each molecule, mole of atoms in one mole of the compound will represent its atomic contribution in the molecular formula. 40 60 100 =2 Mole of C-atoms in 1 mole of compound = 12 6.67 60 100 = 4 Mole of H-atoms in 1 mole of compound = 1 53.33 60 100 = 2 Mole of O-atoms in 1 mole of compound = 16 \ The molecular formula of the compound = C2H4O2 and its empirical formula = CH2O Example 63 At room temperature, pyridine is a colourless liquid with an extremely unpleasant odour. Pyridine occurs naturally in tobacco and hemlock. It contains only the elements C, H and N and is 75.92% C and 6.37% H, by weight. At 383 K and a pressure of 630 mm of Hg, the density of gaseous pyridine has been measured as 2.12 g/L. Calculate the molecular formula of the pyridine. 24 | Chapter 1 Mole Concept Solution From ideal gas equation, d = pM RT \ Molecular weight of pyridine, M = 2.12 0.082 383 dRT = = 80.32 630 p 760 75.92 80.32 Now, mole of C-atoms in 1 mole of pyridine = 100 = 5.08 5 12 6.37 80.32 = 5.11 5 Mole of H-atoms in 1 mole of pyridine = 100 1 100 - (75.92 6.37) 80.32 100 And mole of N-atoms in 1 mole of pyridine = 14 = 1.02 1 \ Molecular formula of pyridine = C5H­5N Example 64 An oxide of iron contains 70% iron, by weight. Determine the simplest formula of the oxide. Solution Element Percentage composition Atomic weight Fe 70 56 O 100 – 70 = 30 16 % composition at. wt. 70 = 1.25 56 1.25 = 1 × 2 =2 1.25 30 = 1.875 16 1.875 = 1.5 × 2 = 3 1.25 Simplest ratio Hence, the simplest formula of the oxide = Fe2O3 Alternate method (without using tabular form) Empirical formula simply represents the atomic ratio of the atoms of each elements present, which may be determined directly. w w 70 30 : =2:3 NFe : NO = N A : N A = A Fe A O 56 16 Hence, the simplest formula of the oxide = Fe2O3 Example 65 A hydrocarbon contains 10.5 g of carbon per g of hydrogen. One litre vapours of hydrocarbon at 127°C and 1 atm pressure weighs 2.8 g. Find molecular formula of hydrocarbon. Solution From ideal gas equation, pV = nRT 2.8 × 0.082 × (127 + 273) M \ Molecular weight of hydrocarbon = 91.84 Now, from question, the hydrocarbon contains 10.5 g carbon per g of hydrogen. It means that there are 10.5 g carbon and 1 g hydrogen in each 11.5 g hydrocarbon. 1×1= Chapter 1 Mole Concept | 25 Hence, 10.5 91.84 Mole of C-atoms in 1 mole of hydrocarbon = 11.5 = 6.98 7 12 1 91.84 11.5 = 7.98 8 And mole of H-atoms in 1 mole of hydrocarbon = 1 \ Molecular formula of hydrocarbon = C7H8 Example 66 A hydrated salt loses 36.4% water on heating and resulting anhydrous salt has the following composition: Cu = 39.6%, S = 20.2%, O = 40.2%. Calculate the simplest formula of anhydrous and hydrated salt. (Cu = 63.5) Solution The empirical formula of anhydrous salt may be determined directly as NCu : NS : NO = 39.6 20.2 40.2 × NA : × NA : × NA 1 : 1 : 4 63.5 32 16 Hence, the simplest formula of anhydrous salt = CuSO4 Now, let the simplest formula of hydrated salt is CuSO4×nH2O From question, mass of water present in it = 36.4% of mass of salt or, 18n = 36.4 × (63.5 + 32 + 4 × 16 + n × 18) 100 or, n = 5.07 5 Hence, the simplest formula of hydrated salt = CuSO4×5H2O Example 67 Thiophene is a liquid compound of the elements C, H and S. A sample of thiophene weighing 7.96 mg was burned in oxygen, giving 16.65 mg CO2. Another sample was subjected to a series of reactions that transformed all the sulphur in the compound to barium sulphate. If 4.31 mg of thiophene gave 11.96 mg of barium sulphate, what is the empirical formula of thiophene? Its molecular weight is 84. What is its molecular formula? (Ba = 137) 12 16.65 Solution Percentage of carbon present in thiophene = 100 = 57.05% 44 7.96 32 11.96 × × 100 = 38.11% Percentage of sulphur present in thiophene = 233 4.31 and percentage of hydrogen = 100 – (57.05 + 38.11) = 4.84% 57.05 84 \ Mole of C-atoms in 1 mole of thiophene = 100 = 3.99 4 12 38.11 84 Mole of S-atoms in 1 mole of thiophene = 100 =1 32 4.84 84 = 4.06 4 and mole of H-atoms in 1 mole of thiophene = 100 1 \ Molecular formula of thiophene = C4H4S. It will also be the empirical formula. 26 | Chapter 1 Mole Concept Example 68 Carbohydrates are compounds containing carbon, hydrogen and oxygen only having the atomic ratio of H : O = 2 : 1. When heated in the absence of air, these compounds decompose to form carbon and water. (a) If 310 g of the carbohydrate leave a residue of 124 g of C, what is the empirical formula of the carbohydrate? (b) If 0.0833 mole of the carbohydrate contains 1.00 g hydrogen, what is its molecular formula? Solution (a) Let the empirical formula of carbohydrate is CxH2yOy. Its empirical formula weight is 12x + 18y. (12x + 18y) g carbohydrate should leave a residue of 12x g carbon \ 310 g carbohydrate should contain 12 x × 310 g carbon 12 x 18 y 12 x × 310 = 124 12 x 18 y or, x:y=1:1 \ Empirical formula of carbohydrate = CH2O But from question, (b) Now, let the molecular formula of carbohydrate = (CH2O)n 1 mole of carbohydrate contains 2n × 1 = 2n g hydrogen 2n \ 0.0833 mole carbohydrate will contain × 0.0833 g hydrogen 1 But from question, 2n 0.0833 = 1 or, n=6 \ Molecular formula of carbohydrate = C6H12O6 Chapter 1 Mole Concept | 27 Exercise 1 (Subjective Questions) 1. When chlorine gas is bubbled into a solution of sodium bromide, the sodium bromide reacts to give bromine, a red-brown liquid, and sodium chloride (ordinary table salt). A solution was made by dissolving 20.6 g of sodium bromide in 100.0 g of water. After passing chlorine through the solution, investigators analyzed the mixture. It contained 16.0 g of bromine and 11.7 g of sodium chloride. How many grams of chlorine reacted? 2. The mass of a sample of gold (specific gravity = 19.3) is 1.93 mg. It may be converted into a transparent sheet of area 14.5 cm2, by hammering. What will be the average thickness of the sheet? 3. Sodium bicarbonate, NaHCO3, can be purified by dissolving it in hot water (60ºC), filtering to remove the insoluble impurities, cooling to 0ºC to precipitate solid NaHCO3, and then filtering to remove the solid, leaving soluble impurities in solution. Any NaHCO3 that remains in the solution is not recovered. The solubility of NaHCO3 in hot water at 60ºC is 164 g/L. Its solubility in cold water at 0ºC is 69 g/L. What is the percentage yield of NaHCO3 when it is purified by this method? 4. A porous catalyst has an internal surface area of 800 m2 per cm3 of bulk material. Fifty percent of the bulk volume consists of the pores (holes), while the other 50% of the volume is made of solid substance. Assume that the pores are cylindrical tubules of uniform diameter, d, and length, l, and that the measured surface area is the total area of the curved surfaces of the tubules. What is the diameter of the pores? 5. There is available 10 tons of a coal sample containing 2.5% sulphur. Two coal samples containing 0.8% and 1.1% sulphur are also available. How many tons of each of the later two samples should be mixed with the original 10 tons to give 20 tons sample containing 1.7% sulphur? 6. Antarctica, almost completely cover in ice, has an area of 1.51013 m2 with an average height of 2250 m. With out the ice, the height would be only 450 m. Estimate the mass of this ice. The density of ice is 0.917 g/cm3. 7. Cobalt (II) sulphate heptahydrate has pink-colored crystals. When heated carefully, it produces cobalt (II) sulphate monohydrate, which has red crystals. What are the formulas of these hydrates? If 3.548 g of the heptahydrate yields 2.184 g of the monohydrate, how many gram of the anhydrous cobalt (II) sulphate could be obtained? 8. LSD is a complex compound whose mass is made up of 74.27% carbon, 7.79% hydrogen, 12.99% nitrogen and 4.95% oxygen. What percent of the atoms in LSD are carbon atoms? 9. Element X reacts with oxygen to produce a compound of formula X2O5. In an experiment, it is found that 2.0769 g of pure X produce 3.7076 g of pure X2O5. Using the known atomic weight of oxygen, calculate the atomic weight of X. 28 | Chapter 1 Mole Concept 10. On electrolysis of an aqueous solution of cupric ion, the following reaction takes place at the cathode: Cu2+ + 2e $ Cu. 1.93 105 Coulombs is needed to deposit one gram atomic weight of copper. Given the clue that an electron carries 1.6 × 10–19 Coulombs, calculate the number of atoms in one gram atomic weight of copper. 11. While making silver ornaments, a certain amount of copper is added in the silver to obtain the desired properties. A silver ornament contains 1.0 1010 copper atoms per cubic centimeter. What relative masses of silver and copper wire mixed to make the ornament? Density of pure silver is 10.5 g/mL. (Cu = 63.5, Ag = 108) 12. The density of a metal of atomic weight 55 is 7.50 g/mL. Assuming that the atoms are spherical, calculate its atomic radius. 13. Does 1 g of all the element contains nucleons equal to the Avogadro’s number? 14. Hydrogen sulphide is a gas with the odour of rotten eggs. The gas can sometimes be detected in automobile exhaust. It is a compound of hydrogen and sulphur in the atomic ratio 2 : 1. A sample of hydrogen sulphide contains 0.587 g H and 9.330 g S. What is the atomic mass of S relative to H? 15. An oxide of nitrogen has molecular weight 92. Find out the total number of electrons in one mole of the oxide. 16. If the volume occupied in the crystal by one formula unit of NaCl is 4.7 × 10-23 ml, calculate the volume of a crystal weighing 1.17 g. 17. Assume that a polyethylene chain is truly linear. If a polymer chain had a molecular weight of 1 × 106, what will be the length of one polyethylene molecule? A carboncarbon single bond length is 154 pm. 18. 0.315 g of a substance when introduced into a Hofmann’s tube generated 128.5 mL of vapour at 30°C, the level of Hg inside being 430 mm higher than outside the tube, and the barometer reading was 758 mm, calculate the molecular weight of the substance. 19. 0.607 g of a silver salt of a tribasic organic acid was quantitatively reduced to 0.370 g of pure silver. Calculate the molecular weight of the acid. (Ag = 108) 20. 0.532 g of the chloroplatinate of a diacidic base left 0.195 g of residue of Pt on ignition. Calculate molecular weight of the base. (Pt = 195) 21. Avogadro noted that two volumes of hydrogen combine with one volume of oxygen to form two volumes of water vapour. He also gave the density of water vapour relative to that of air as 0.625 and hydrogen as 0.0732 (in units in which air had unit density). Use this information and Avogadro’s principle to compute the molar masses of water vapour and oxygen relative to that of hydrogen. 22. A sample of ammonia contains only H1 and H2 isotopes of hydrogen in 4 : 1 ratio and N14 and N15 isotopes of nitrogen in 3 : 1 ratio. How many neutrons are present in 1.785 mg ammonia? 23. At 343 K and 755 mm pressure, 0.246 g of N2O4 gave 120.5 mL vapour. Calculate the apparent molecular weight and degree of dissociation of the gas. 24. A gaseous mixture contains 40% H2 and 60% He, by volume. Calculate the total Chapter 1 Mole Concept | 29 number of moles of gases present in 10 g of such mixture. 25. Atoms of elements A, B and C combine to form a compound in the atomic ratio of 1 : 6 : 2. Atomic masses of A, B and C are 64, 4 and 16 u, respectively. Calculate the maximum mass of the compound formed from 1.28 g of A, 3.0 × 1023 atoms of B and 0.04 mole atom of C. 26. A drug marijuana owes its activity to tetrahydrocannabinol, which contains 70 percent as many carbon atoms as hydrogen atoms and 15 times as many hydrogen atoms as oxygen atoms. The number of moles in a gram of tetrahydrocannabinol is 0.00318. Determine its molecular formula. 27. When acetylene is passed through red hot metal tubes, the molecules trimerised to form benzene. The molecular weight of the gaseous mixture, when acetylene is passed through the tube, is 60. Calculate the degree of trimerisation of acetylene. 28. Recent controversial efforts to generate energy via “cold fusion” of deuterium atoms have centered on the remarkable ability of palladium metal to absorb as much as 900 times its own volume in hydrogen or deuterium gas at 1 atm and 273 K. Calculate the ratio of deuterium atoms to Pd atoms in a piece of fully saturated Pd metal. The density of Pd is 12.02 g/mL. (Pd = 106.4) 29. One mole of mixture of N2, NO2 and N2O4 has a mean molar mass of 55.4 g. On heating to a temperature, at which all the N2O4 may be dissociated into NO2, the mean molar mass tends to a lower value of 39.6 g. What is the mole ratio of N2, NO2 and N2O4 in the original mixture? 30. A solution contains 0.18 g per mL of a substance X, whose molecular weight is approximately 68000. It is found that 0.27 mL of oxygen at 760 mm and 310 K will combine with the amount of X contained in 1 mL of the solution. How many molecules of oxygen will combine with one molecule of X? 31. A mixture of 0.5 mole of CO and 0.5 mole of CO2 is taken in a vessel and allowed to effuse out through a pinhole into another vessel which has vacuum. If a total of A moles has effused out in time, t, show that M1A + M2 (1 - A) = 36, where M1 and M2 are mean molar masses of the mixture that has effused out and the mixture still remaining in the vessel, respectively. 32. The chemical formula of chelating agent versene is C2H4N2(C2H2O2Na)4. If each mole of this compound could bind 1 mole of Ca2+, what would be the rating of pure versene, expressed as mg CaCO3 bound per gram of chelating agent? Here the Ca2+ is expressed in terms of the amount of CaCO3 it could form. 33. A mixture of hydrogen and methane has a density 0.25 times the density of oxygen at the same pressure and temperature. Find the percentage of methane in the mixture, by weight. 34. The average molar mass of the vapour above solid NH4Cl is nearly 26.74 g/mol. What is the composition, by weight, of this vapour? 35. Hemoglobin is the oxygen carrying molecule of red blood cells, consisting of a protein and a non-protein substance, heme. A sample of heme weighing 35.2 mg contains 3.19 mg of iron. If a heme molecule contains one atom of iron, what is the molecular weight of heme? 30 | Chapter 1 Mole Concept 36. A purified pepsin isolated from a bovine preparation was subjected to an amino acid analysis. The amino acid present in the smallest amount was lysine, C6H14N2O2 and the amount of lysine was found to be 0.43 g per 100 g protein. What is the minimum molecular weight of the protein? 37. A polymeric substance, tetrafluoroethylene, can be represented by the formula (C2F4)x , where x is a large number. The material was prepared by polymerising C2F4 in the presence of a sulphur bearing catalyst that serves as a nucleus upon which the polymer grew. The final product was found to contain 0.012% S. What is the value of x if each polymeric molecule contains one sulphur atom? Assume that the catalyst contributes a negligible amount to the total mass of the polymer. (S = 32, F = 19) 38. A typical analysis of pyrex glass showed 12.9% B2O3, 2.2% Al2O3, 3.8% N2O, 0.4% K2O and the balance SiO2. What is the ratio of silicon to boron atoms in the glass? (B = 10.8, Si = 28, Al = 27, K = 39) 39. Three different brands of liquid chlorine are available in the market for the use in purifying water of swimming pools. All are sold at the same rate of Rs. 10 per litre and all are water solutions. Brand A contains 10% hypochlorite (ClO) (w/V), brand B contains 7% of available chlorine (w/v) and brand C contains 14% sodium hypochlorite (NaClO) (w/V). Which of the three is most economical to buy? 40. A green pigment called hemovanadin can be isolated from the sea squirt a marine chordate resembling tadpole. The hemovanadin molecule has a molecular weight of 240000 daltons, and is 0.51% vanadium by weight. How many vanadium atoms are present per hemovanadin molecule? (V = 51, 1 dalton = 1 g/mol) 41. Nitrogen content in a sample of urea is 42.5%. What is the percentage purity of urea in the urea sample? 42. Determine the empirical formula of clay having the following percentage composition: Al2O3 = 39.77; SiO2 = 46.33; H2O = 13.9. (Al = 27, Si = 28) 43. An unknown oxide of manganese is reacted with carbon to form manganese metal and CO2. Exactly 31.6 g of the oxide, MnxOy , yields 13.2 g of CO2. Find the simplest formula of the oxide. (Mn = 55) 44. A complex compound of cobalt has the following composition: Co = 22.58%, H = 5.79%, N = 32.2%, O = 12.26% and Cl = 27.17%. When this compound is heated, it loses ammonia to the extent of 32.63% of its mass. How many molecules of ammonia are present in the complex compound? Derive the empirical formula for the compound. (Co = 58.9) 45. The formula of a compound should be represented as [PdCxHyNz] (ClO4)2. Analysis showed that the compound contains 30.15% carbon and 5.06% hydrogen. When converted to the corresponding thiocyanate, [PdCxHyNz](SCN)2, the analysis was 40.46% carbon and 5.94% hydrogen. Calculate the values of x, y and z. (Pd = 106.4) 46. The following experiment was performed in order to determine the molecular formula of a gaseous compound known to contain B and H. An evacuated gas bulb of volume 952 mL and mass 73.2684 g was filled with the gaseous borane at 273 K and a pressure of exactly 1 atm. The mass of the gas filled bulb was 75.5398 g. This sample of the compound was then burnt in excess oxygen, which completely converted it to water, and the solid white boron oxide, B2O3. The mass of the B2O3 was found to be Chapter 1 Mole Concept | 31 5.9316 g. What is the correct molecular formula of the borane? (B = 10.8) 47. Cupric ammonium sulphate was found to contain 27.03% water of crystallisation. Upon strongly heating it gave cupric oxide corresponding to 19.89% of starting mass. Find the empirical formula of cupric ammonium sulphate. (Cu = 63.5) 48. A hydrated calcium phosphate contained 7.14% water of crystallisation. The molar mass of this phosphate is 252.2. The percentage is 15.18% Ca, 24.6% P, 2.4% H and 57.11% O. What is the molecular formula of the phosphate? 49. There are 20 amino acids that are building blocks of proteins. The amino acid histidine is known to contain only carbon, hydrogen, oxygen and nitrogen. The composition by weight of histidine is 46.45% C; 5.85% H; 27.08% N. What is the empirical formula of histidine? 50. A 2.5 g sample of uranium was heated in the air. The resulting oxide weighed 2.949 g. Determine the empirical formula of the oxide. (U = 238) 32 | Chapter 1 Mole Concept Exercise 2 (Objective Questions) Section I (Only One Correct Option) 1. Two elements X (atomic mass 16) and Y (atomic mass 14) combine to form compounds A, B and C. The ratio of different masses of Y which combines with a fixed mass of X in A, B and C is 1 : 3 : 5. If 32 parts by mass of X combines with 84 parts by mass of Y in B, then in C, 16 parts by mass of X will combine with ......... parts by mass of Y. (A) 14 (B) 42 (C) 70 (D) 84 2. Law of definite proportion does not apply to nitrogen oxide because (A) atomic weight of nitrogen is not constant (B) molecular weight of nitrogen is variable (C) equivalent weight of nitrogen is variable (D) atomic weight of oxygen is variable 3. In a textile mill, a double-effect evaporator system concentrates weak liquor containing 4% (by weight) caustic soda to produce a lye containing 25% solids (by weight). Calculate the weight of the water evaporate per 100 kg feed in the evaporator. (A) 125.0 g (B) 50.0 kg (C) 84.0 kg (D) 16.0 kg 4. Zinc ore (zinc sulphide) is treated with sulphuric acid, leaving a solution with some undissolved bits of material and releasing hydrogen sulphide gas. If 10.8 g of zinc ore is treated with 50.0 mL of sulphuric acid (density 1.153 g/mL), 65.1 g of solution and undissolved material remains. In addition, hydrogen sulphide (density 1.393 g/L) is evolved. What is the volume (in litres) of this gas? (A) 4.3 (B) 3.35 (C) 4.67 (D) 2.40 5. A sample of an ethanol–water solution has a volume of 54.2 cm3 and a mass of 49.6 g. What is the percentage of ethanol (by mass) in the solution? (Assume that there is no change in volume when the pure compounds are mixed.) The density of ethanol is 0.80 g/cm3 and that of water is 1.00 g/cm3. (A) 18.4% (B) 37.1% (C) 33.95% (D) 31.2% 6. Zinc metal reacts with yellow crystals of sulphur in a fiery reaction to produce a white powder of zinc sulphide. A chemist determines that 65.4 g of zinc reacts with 32.1 g of sulphur. How many grams of zinc sulphide could be produced from 20.0 g of zinc metal? (A) 29.8 g (B) 9.8 g (C) 97.5 g Chapter 1 Mole Concept | 33 (D) 31.8 g 7. A student gently drops an object weighing 15.8 g into an open vessel that is full of ethanol, so that a volume of ethanol spills out equal to the volume of the object. The experimenter now finds that the vessel and its contents weigh 10.5 g more than the vessel full of ethanol only. The density of ethanol is 0.789 g/cm3. What is the density of the object? (A) 6.717 g/cm3 (B) 4.182 g/cm3 3 (D) 2.352 g/cm3 (C) 1.563 g/cm 8. A person needs 2.0 mg of riboflavin (vitamin B2) per day on an average. How many gram of butter should be taken by the person per day if it is the only source of riboflavin? Butter contains 5.5 microgram riboflavin per gram. (A) 363.6 g (B) 2.75 mg (C) 11 g (D) 19.8 g 9. A sample of clay contains 40% silica and 15% water. The sample is partially dried by which it loses 5 g water. If the percentage of water in the partially dried clay is 8, calculate the percentage of silica in the partially dried clay. (A) 21.33% (B) 43.29% (C) 75% (D) 50% 10. The density of quartz mineral was determined by adding a weighed piece to a graduated cylinder containing 51.2 mL water. After the quartz was submersed, the water level was 65.7 mL. The quartz piece weighed 38.4 g. What was the density of quartz? (A) 1.71 g/mL (B) 1.33 g/mL (C) 2.65 g/mL (D) 1.65 g/mL 11. Some bottles of colourless liquids were being labelled when the technicians accidentally mixed them up and lost track of their contents. A 15.0 mL sample withdrawn from one bottle weighed 22.3 g. The technicians knew that the liquid was either acetone, benzene, chloroform, or carbon tetrachloride (which have densities of 0.792 g/cm3, 0.899 g/cm3, 1.489 g/cm3, and 1.595 g/cm3, respectively). What was the identity of the liquid? (A) Carbon tetrachloride (B) Acetone (C) Chloroform (D) Benzene 12. An ammonium sulphate solution of concentration 0.05 kilo mol reacts with calcium hydroxide. How many litres of a solution (specific gravity 0.92) containing 20.5% by weight of ammonia can be prepared using this reaction? (A) 12.0 L (B) 9.0 L (C) 18.0 L (D) 4.5 L 13. Atomic mass of an element is (A) the actual mass of the one atom of the element (B) the relative mass of an atom of the element (C) the average relative mass of different atoms of the element (D) much different from the mass number of the element 34 | Chapter 1 Mole Concept 14. The use of C12 scale has superseded the older scale of atomic mass based on O16 isotope, one important advantage of the former being (A) the atomic masses on C12 scale became whole numbers (B) C12 is more abundant in the earth’s crust than O16 (C) the difference between physical and chemical atomic masses got narrowed down significantly (D) C12 is situated midway between metals and non-metals in the periodic table 15. One atomic mass unit in kilogram is (A) 1/NA (C) 1/1000 NA (B) 12/NA (D) 1000/NA 16. An element, X, have three isotopes X20, X21 and X22. The percentage abundance of X20 is 90% and its average atomic mass of the element is 20.18. The percentage abundance of X21 should be (A) 2% (B) 8% (C) 10% (D) 0% 17. The number of atoms present in 0.5 g-atoms of nitrogen is same as the atoms in (A) 12 g of C (B) 32 g of S (C) 8 g of oxygen (D) 24 g of Mg 18. A graph is plotted for an element, by putting its weight on X-axis and the corresponding number of number of atoms on Y-axis. Determine the atomic weight of the element for which the graph is plotted. (A) infinite (B) 40 (C) 0.025 (D) 20 no. of atoms 22 0 ) 1 .5 × –1 (1 tan = θ wt. (in gm) 19. The O18/O16 ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. The average mass of an atom of oxygen in these meteorites is ............. that of a terrestrial oxygen atom? (A) equal to (B) greater than (C) less than (D) None of these 20. If isotopic distribution of C12 and C14 is 98.0% and 2.0% respectively, then the number of C14 atoms in 12 g of carbon is (B) 1.20 × 1022 (A) 1.032 × 1022 (D) 6.02 × 1023 (C) 5.88 × 1023 21. At one time there was a chemical atomic weight scale based on the assignment of the value 16.0000 to naturally occurring oxygen. What would have been the atomic weight, on such a table, of silver, if current information had been available? The atomic weights of oxygen and silver on the present table are 15.9994 and 107.868. (A) 107.908 (B) 107.864 (C) 107.868 (D) 107.872 22. In what atomic ratio O16 and O17 atoms should be taken in a sample such that the average atomic mass becomes 16.72? (A) 7 : 18 (B) 18 : 7 (C) 19 : 8 (D) 8 : 19 Chapter 1 Mole Concept | 35 23. Two isotopes of an element Q are Q97 (23.4% abundance) and Q94 (76.6% abundance). Q97 is 8.082 times heavier than C12 and Q94 is 7.833 times heavier than C12. What is the average atomic weight of the element Q? (A) 94.702 (B) 78.913 (C) 96.298 (D) 94.695 24. The element silicon makes up 25.7% of the earth’s crust by weight, and is the second most abundant element, with oxygen being the first. Three isotopes of silicon occur in nature: Si28 (92.21%), which has an atomic mass of 27.97693 u; Si29 (4.70%), with an atomic mass of 28.97649 u; and Si30 (3.09%), with an atomic mass of 29.97379 u. What is the atomic weight of silicon? (A) 28.0856 (B) 28.1088 (C) 28.8342 (D) 29.0012 25. The oxide of a metal contains 30% oxygen by weight. If the atomic ratio of metal and oxygen is 2 : 3, determine the atomic weight of metal. (A) 12 (B) 56 (C) 27 (D) 52 26. The mercury content of a stream was believed to be above the minimum considered safe limit (1 part per billion by weight). An analysis indicated that the concentration was 1.68 parts per billion. How many Hg atoms was present in 15 L of the water, the density of which is 0.998 g/mL? (Hg = 200) ? (A) 7.57 × 1013 (B) 7.57 × 1019 (C) 7.57 × 1016 (D) 5.37 × 1016 27. There are 2.6191022 atoms in 1.0 g of sodium. Assume that sodium atoms are spheres of radius 0.186 nm and that they are lined up side by side. How many miles, in length, is the line of sodium atoms? (A) 9.74 × 1012 (B) 1.56 × 1010 9 (C) 1.44 × 10 (D) 6.089 × 109 28. Ammonia is a gas with a characteristic pungent odour. It is sold as a water solution for use in household cleaning. The gas is a compound of nitrogen and hydrogen in the atomic ratio 1: 3. A sample of ammonia contains 7.933 g N and 1.712 g H. What is the atomic mass of N relative to H? (A) 14 (B) 27.80 (C) 13.90 (D) 14.10 29. The waste of nuclear power plant contains C12 and C14 in the ratio of 4:1 by moles. What is the molecular weight of CH4 gas produced from this disposed waste? Given abundance of C12 and C14 are 98% and 2%, respectively. (A) 15.998 (B) 16.0053 (C) 16 (D) 16.4 30. Pheromones are a special type of compound secreted by the females of many insect species to attract the males for mating. One pheromone has the molecular formula C19H38O. Normally, the amount of this pheromone secreted by a female insect is about 1 × 10–12g. How many molecules are in this quantity? (A) 1 × 1012 (B) 3.55 × 1015 (C) 2.135 × 109 (D) 6.023 × 1023 36 | 31. Chapter 1 Mole Concept Molecular weight of dry air is (A) less than moist air (C) equal to moist air (B) greater than moist air (D) may be greater or less than moist air 32. At room temperature, the molar volume of hydrogen fluoride gas has a mass of about 50 g. The formula weight of hydrogen fluoride is 20. Gaseous hydrogen fluoride at room temperature is therefore, probably a mixture of (A) H2 and F2 (B) HF and H2F2 (D) H2F2 and H3F3 (C) HF and H2.5F2.5 33. The density of a DNA sample is 1.1 g/mL and its molar mass determined by cryoscopic method was found to be 6 × 108 g/mol. What is the volume occupied by one DNA molecule? (A) 5.45 × 108 mL (B) 1.83 × 10–9 mL –16 (D) 1.09 × 10–13 mL (C) 9.06 × 10 mL 34. The density of a plant virus is 1.66 g/cm3. If each virus particles are spheres of diameter 6Å, the molecular weight of virus should be (A) 36 p (B) 72 p (D) 18 p (C) 5.976 × 10–23 p 35. Total number of valence electrons present in 6.4 g peroxides ion (O22–) is (A) 0.2 NA (B) 3.2 NA (C) 3.6 NA (D) 2.8 NA 36. A gaseous mixture contains 70% N2 and 30% unknown gas, by volume. If the average molecular weight of gaseous mixture is 37.60, the molecular weight of unknown gas is (A) 42.2 (B) 60 (C) 40 (D) 50 37. A gaseous mixture contains CO2 and SO3. The average molecular weight of the mixture may be (A) 40 (B) 80 (C) 70 (D) 90 38. The number of F– ions in 4.2 g AlF3 is (A) 0.05 (B) 9.03 × 1022 22 (C) 3.01 × 10 (D) 0.15 1 If the atomic mass were given by as th part and molecular mass as 1 th part by 6 12 mass of one atom of C12 isotope, what would be the molecular mass of water? Atomic masses of hydrogen and oxygen on new scale are, 1 and 16, respectively. (A) 18 (B) 9 (C) 36 (D) unpredictable 39. 40. Each molecule of a tear gas, Lewisite contains 2 hydrogen atoms, 1.78 × 10–22 g chlorine, 2 carbon atoms and 1.25 × 10–22 g of an unknown metal. Its molecular weight will be (A) 195 (B) 208.5 (C) 280 (D) 188.3 Chapter 1 Mole Concept | 37 41. 20 molecules of SO3 will weigh as much as …......... molecules of oxygen. (A) 100 (B) 50 (C) 15 (D) 8 42. If the mass of neutron is doubled and that of proton is halved, the molecular weight of H2O containing H1 and O16 atoms only, will (A) increase by about 25% (B) decrease by about 25% (C) increase by about 17% (D) increase by about 14% 43. Which of the following contains largest number of oxygen atoms? (A) 2 g atomic oxygen (B) 2 g oxygen molecules (C) 2 g ozone molecules (D) all have same number 44. The number of hydrogen atoms in 0.9 g glucose, C6H12O6, is same as (B) 0.17 g ammonia, NH3 (A) 0.048 g hydrazine, N2H4 (C) 0.30 g ethane, C2H6 (D) 0.03 g hydrogen, H2 45. Volume occupied by 6 g of a mixture of nitrogen and oxygen is 4.48 L at 1 atm and 0°C. The percentage by mass of nitrogen in the mixture is (A) 50% (B) 46.67% (C) 53.33% (D) 28% 46. The average density of the universe as a whole is estimated as 3 × 10–29 g per mL. If we assume that the entire mass is only H atoms, what is the average volume of space that contains one H atom? (A) 110 L (B) 9 × 10–6 L (C) 55 L (D) 220 L 47. The number of molecules present in 1 mL of an ideal gas at 273 K and 1 atm is called .......... and its value is ................. . (B) Loschmidt number, 2.7 × 1019 (A) Avogadro’s number, 6.023 × 1023 (C) Rydberg number, 1.09 × 107 (D) Universal gas constant, 0.082 unit 48. Mass of one proton is 1.675 × 10-24 g. How many moles of proton are present in its 1 kg? 1 10 (A) (B) 104 1.675 6.023 1.675 6.023 1 1 (C) (D) 1024 1027 6.75 1.675 Assuming that 1, 3, 5-hexatriene has only pure double bonds and pure single bonds, how many grams of it contain one mole of double bonds? (A) 13.3 g (B) 26.7 g (C) 40 g (D) 80 g 49. 50. 2 moles of H atoms at 1 atm and 0°C occupies a volume of (A) 11.2 L (B) 44.8 L (C) 2 L (D) 22.4 L 51. Which of the following will occupy greater volume under the similar conditions of pressure and temperature? (A) 6 g oxygen (B) 0.98 g hydrogen (C) 5.25 g nitrogen (D) 1.32 g helium 38 | 52. Chapter 1 Mole Concept The degree of dissociation, a, (dt is the theoretical density of original compound and d0 is the observed density of the sample of compound) be calculated by using the d - d0 . This formula is applicable at formula: a= t ( n -1)d0 (A) constant volume and constant temperature (B) constant volume and variable temperature (C) constant pressure and constant temperature (D) all of the above conditions 53. 112.0 mL of NO2 at 1 atm and 0°C was liquefied, the density of the liquid being 1.15 g.mL–1. Calculate the volume and the number of molecules in the liquid NO­2. (A) 0.10 mL and 3.01 × 1022 (B) 0.20 mL and 3.01 × 1021 (C) 0.20 mL and 6.02 × 1023 (D) 0.40 mL and 6.02 × 1021 54. In an experiment, it is found that 2.0769 g of pure X produces 3.6769 g of pure X2O5. The number of moles of X is (A) 0.04 (B) 0.06 (C) 0.40 (D) 0.02 55. The volume occupied by 20 g water at 1.2 atm and 4°C is about (A) 20 mL (C) 20 0.082 4 L 18 1.2 (B) 20 0.082 227 L 18 1.2 (D) 20 L 56. If common-salt (molecular weight = 58.5) costs Rs. 7.00 per kg and sugar (molecular weight = 342) costs Rs. 14.00 per kg, what would be the cost of 1 mole of salt and sugar? (A) Both will have the same cost (B) The cost of sugar will be half the cost of salt (C) The cost of sugar will be more than that of salt (D) The cost of sugar will be twice the cost of salt 57. Number of g-atoms of oxygen present in 0.8 mole of tartaric acid C4H6O6 would be (A) 3.6 (B) 1.8 (C) 2.4 (D) 4.8 58. Equal masses of oxygen, hydrogen and methane are taken in identical conditions. What is the ratio of the volumes of the gases under identical conditions? (A) 16 : 1 : 8 (B) 1 : 16 : 2 (C) 1 : 16 : 8 (D) 2 : 16 : 1 59. While resting, the average 70 kg human male consumes 14 L of oxygen per hour at 25ºC and 100 kPa. How many moles of oxygen are consumed by the 70 kg man while resting for 1 hour? (A) 0.57 (B) 57292.5 (C) 0.58 (D) 58037.32 Chapter 1 Mole Concept | 39 60. 3.0 g of a solid compound produced 672 mL of a gas at STP (1 atm and 0°C), on strong heating. After the reaction the weight of the solid residue was 1.68 g. What is the molecular weight of the gas obtained? (A) 19.84 (B) 25.25 (C) 44 (D) 39.6 61. When a sample of hydrogen fluoride is cooled to 303K, most of the molecules undergo dimerisation. If the vapour density of such a sample is 19, what percent of hydrogen fluoride molecules are in dimer form? (F = 19) (A) 94.7 (B) 89.9 (C) 97.3 (D) 5.3 62. 2.3 g of a mixture of NO2 and N2O4 has a pressure of 0.82 atm, at temperature T K in a container of volume V litres such that the ratio, T : V is 300 : 1 in magnitude. What is the degree of dissociation of N2O4? (A) 0.17 (B) 0.33 (C) 0.67 (D) 0.70 63. 1 kg of each substance is taken. Which of them will have largest mass of nitrogen? (B) NH3NO3 (A) KNO3 (D) HNO3 (C) (NH4)2HPO4 64. A 10 g sample of ore containing 2.8 g of HgS. What is the percentage of mercury in the ore? (Hg = 200, S = 32) (A) 24.1 (B) 29.01 (C) 28 (D) 39 65. The mineral haematite is Fe2O3. Haematite ore contains unwanted material called gangue in addition to Fe2O3. If 5 kg of ore contains 2.78 kg of Fe, what percentage of ore is gangue? (Fe = 56) (A) 55.6% (B) 44.4% (C) 20.6% (D) 79.4% 66. The hydrated salt Na2SO4×nH2O undergoes 55×9% loss in weight on heating and becomes anhydrous. The value of n will be (A) 5 (B) 3 (C) 7 (D) 10 67. The commonly used pain reliever, aspirin, has the molecular formula C9H8O4. If a sample of aspirin contains 0.968 g of carbon, what is the mass of hydrogen in the sample? (A) 0.717 g (B) 0.0717 g (C) 8.000 g (D) 0.645 g 68. For CuSO4×5H2O, which is the correct mole relationship? (A) 9 × mole of Cu = mole of O (B) 5 × mole of Cu = mole of O (C) 9 × mole of Cu = mole of O2 (D) mole of Cu = 5 × mole of O 40 | Chapter 1 Mole Concept 69. Nitrogen (N), phosphorus (P) and potassium (K) are the main nutrients in plant fertilizers. According to an industry convention, the numbers on the label refer to the mass percent of N, P2O5 and K2O, in that order. What is N : P : K ratio of a 30 : 10 : 10 fertilizer in terms of moles of atoms of each element, expressed as x : y : 1.0 ? (N = 14, P = 31, K = 39) (A) 10 : 0.66 : 1.0 (B) 20 : 0.66 : 1.0 (C) 8.4 : 1.3 : 1.0 (D) 16.8 : 1.3 : 1.0 70. Chloroflurocarbons such as CCl3F (M = 137.5) and CCl2F2 (M = 121) have been linked to ozone depletion in Antartica. As of 2004, these gases were found in 275 and 605 parts per trillion (1012), by volume. What are the concentrations of these gases under conditions typical of Antartic stratosphere (200 K and 0.05 atm)? (A) [CCl3F] = 8.38 × 10–13 mol L–1, [CCl2F2] = 1.84 × 10–12 mol L–1 (B) [CCl3F] = 2.00 × 10–12 mol L–1, [CCl2F2] = 5.00 × 10–12 mol L–1 (C) [CCl3F] = 2.00 × 10–9 mol L–1, [CCl2F2] = 5.00 × 10–9 mol L–1 (D) [CCl3F] = 6.56 × 10–7 mol L–1, [CCl2F2] = 1.64 × 10–8 mol L–1 71. Caffeine has a molecular mass of 194. If it contains 28.9% by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is (A) 4 (B) 6 (C) 2 (D) 3 72. A certain mixture of MnO and MnO2 contains 66.67 mol % of MnO. What is the approximate weight percent of Mn in it? (Mn = 55) (A) 66.67 (B) 33.33 (C) 72 (D) 28 73. A compound contains 36% carbon by weight. If each molecule contains two carbon atoms, the number of moles of compound in its 10 g is (A) 66.67 (B) 0.15 (C) 0.30 (D) 0.60 74. What will be percentage concentration of a solution that is obtained by mixing 300g of 25% solution of NaCl with 150 g of 40% solution of NaCl? All percentage are mass percentages. (A) 25% (B) 30% (C) 35% (D) 40% 75. In the Dumas method, 0.2 g of an organic nitrogenous compound gave 27 mL of N2 (volume reduced to 1 atm and 0°C). What is the percentage of nitrogen by weight in the compound? (A) 16.9 (B) 19.6 (C) 33.1 (D) 13.3 76. 0.2 g of an organic compound containing, C, H and O, on combustion yielded 0.147 g CO2­and 0.12 g water. The percentage of oxygen in it is (A) 73.29% (B) 78.45% (C) 83.23% (D) 89.50% Chapter 1 Mole Concept | 41 77. Haemoglobin of a blood corpuscle contains 0.33% iron. The molecular weight of haemoglobin was found to be 67000 g. What is the number of iron atoms present in each molecule of haemoglobin? (Atomic weight of Iron = 56) (A) 2 (B) 3 (C) 4 (D) 5 78. The empirical formula of an organic compound containing carbon and hydrogen is CH2. The mass of one litre of this organic gas is exactly equal to that of one litre of nitrogen, under identical conditions. Therefore, the molecular formula of the organic gas is (A) C2H4 (B) C3H6 (C) C6H12 (D) C4H8 79. 60 g of an organic compound has 24 g carbon and rest hydrogen and oxygen. The empirical formula of the compound is (A) C2H8O2 (B) C2H4O (D) CH2O (C) CH4O 80. A hydride of silicon prepared by the reaction of Mg2Si with acid exerted a pressure of 306 torr of 299 K, in a bulb of capacity 57 mL. If the mass of the hydride was 0.0861 g, what is its molecular formula? (Si = 28) (A) SiH4 (B) Si2H6 (C) Si3H8 (D) Si3H6 81. A compound have carbon, hydrogen, and oxygen in 3 : 3 : 1 atomic ratio. If the number of moles in 1 g of the compound is 6.06 × 10-3, the molecular formula of the compound will be (A) C3H3O (B) C6H6O2 (C) C9H9O3 (D) C12H12O4 82. On analysis, a certain compound was found to contain iodine and oxygen in the ratio of 254 g of iodine and 80 g of oxygen. The atomic mass of iodine is 127 and that of oxygen is 16. Which of the following is the formula of compound? (A) IO (B) I2O (C) I5O3 (D) I2O5 83. Iron forms two oxides. For the same amount of iron, amount of oxygen combined in the first oxide is two-third of the amount of oxygen combined in the second oxide, the ratio of valencies of iron in first and second oxide is (A) 1 : 1 (B) 2 : 3 (C) 3 : 2 (D) 2 : 5 84. A compound contains equal masses of the elements A, B and C. If the atomic weights of A, B and C are 20, 40 and 60 respectively, the empirical formula of the compound is (A) A3B2C (B) AB2C3 (C) ABC (D) A6B3C2 85. A compound contains elements X and Y in 1 : 4 mass ratio. If the atomic masses of X and Y are in 1 : 2 ratio, the empirical formula of compound should be (A) XY2 (B) X2Y (C) XY4 (D) X4Y 42 | Chapter 1 Mole Concept A semiconductor, YBa2Cu3O7, is prepared by a reaction involving Y2O3, BaO2 and Cu2O. The mole ratio in which these compounds should combine, is (A) 1 : 2 : 4 (B) 1 : 2 : 3 (C) 1 : 4 : 6 (D) 1 : 3 : 4 87. Assume that the atomic weight of oxygen is 7. A sample of 11 g of an oxide of uranium contains 10 g of uranium. Which of the following formula for the oxide is compatible with the data? (A) Uranium oxide is UO and the atomic weight of U is 70 (B) Uranium oxide is U3O8 and the atomic weight of U is 240 (C) Uranium oxide is UO2 and the atomic weight of U is 105 (D) Uranium oxide is U2O3 and the atomic weight of U is 105 88. A sample of protein was analysed for metal content and analysis revealed that it contained magnesium and titanium in equal amounts, by weight. If these are the only metallic species present in the protein and it contains 0.016% metal by weight, the minimum possible molar mass of the protein is (Mg = 24, Ti = 48) (A) 600000 (B) 150000 (C) 300000 (D) 1200000 89. A series of experiments were carried out with a fixed mass of a metal and variable mass of chlorine. The result is represented by the adjacent graph. The empirical formula of compound formed in each case is found to be MCl2. The atomic weight of the metal is (Cl = 35.5) (A) 40 (B) 47.33 (C) 118.33 (D) 106.5 90. Mass of compound formed (gm) 10 86. 6 Mass of chlorine (gm) In the blood of an infant, there are about 1.3 × 1012 red blood cells, which contain a total of about 0.15 g iron. On the average, how many iron atoms are present in each red blood cell of the infant? (A) 8.8 × 1023 (B) 4.7 × 1013 (C) 1.2 × 109 (D) 3.0 × 108 Section II (One or More than One Correct Option) 91. Which of the following pairs do not contain equal number of atoms? (A) 11.2 mL of N2 (at 1 atm and 0°C) and 0.015 g of NO (B) 22.4 L of N2O and 22.4 L of NO under identical conditions (C) 1 millimole of HCl and 0.5 millimole of H2S (D) 1 mole of H2O2 and 1 mole of N2O4 Chapter 1 Mole Concept | 43 92. For which of the following reactions, the degree of dissociation cannot be calculated dt - d0 , where dt and d0 are the theoretical and applying the formula: a = ( n -1)d0 experimental densities, respectively. (A) PCl5 PCl3 + Cl2 (B) 2NH3 N2 + 3H2 (C) 2HI H2 + I2 (D) 2KClO3 2KCl + 3O2 93. 11.2 L of gas at STP (1 atm and 0°C) weighs 14.0 g. The gas may be (A) N2O (B) NO2 (C) N2 (D) CO 94. 31 mol of 14 ions contains 7 N (A) 7 NA electrons (C) 7 NA neutrons (B) 7 NA protons (D) 14 NA protons 95. The composition of universe is approximately 90% hydrogen and 10% helium, by weight. The composition represents that (A) there are 18 hydrogen atoms in the universe per atom of helium. (B) there are 9 hydrogen atoms in the universe per atom of helium. (C) there is 4.5 g hydrogen in the universe per gram of helium. (D) the average molar mass of universe is 2.20 g per mole. 96. The formula weight of hydrogen fluoride is 20. The vapour density of a sample of hydrogen fluoride gas is measured by an experiment as 20. It may represent that (A) some molecules of hydrogen fluoride are dissociated (B) some molecules of hydrogen fluoride are in dimer form (C) all hydrogen fluoride molecules are in dimer form (D) some hydrogen fluoride molecules are in trimer form 97. Which of the following statement(s) is/are correct for water? (A) H and O are in 2 : 1 atomic ratio (B) H and O are in 2 : 1 weight ratio (C) H and O are in 1 : 8 weight ratio (D) H and O are in 2 : 1 volume ratio 98. The atomic weight of a diatomic gaseous element is 19. Which of the following statement(s) is/are correct regarding the element? (A) weight of one atom is 19 u (B) weight of NA molecules is 38 g (C) volume of NA atoms is 22.7 L at STP (D) weight of NA atoms is 19 g 99. Three isotopes of an element have mass numbers M, (M + 1) and (M + 2). If the mean mass number is (M + 0.5), then which of the following ratio(s) may be accepted for M, (M + 1) and (M + 2) in the order (A) 1 : 1 : 1 (B) 4 : 1 : 1 (C) 9 : 6 : 1 (D) 2 : 1 : 1 100. The mass of 0.01 mole of o-nitrophenol is not nearly equal to (A) 0.0434 moles of oxygen (B) 2.616 × 1022 molecules of oxygen (C) 1.744 × 1022 molecules of ozone (D) 1.308 × 1022 molecules of methane 44 | Chapter 1 Mole Concept 101. Which of the following statement(s) is/are correct about the Avogadro’s number? (A) It is the number of atoms contained in one mole of atoms of any monoatomic element (B) It is the number of electrons required to deposit one mole of atoms of any metallic element from a solution of the metal salt (C) It is the number of grams of any element which contains 6.023 × 1023 atoms of that element (D) It is the number of particles (atoms, molecules or ions) required to make one gram of the substance under consideration 102. The non-stoichiometric compound, titanium monoxide, has a continuous range of composition from Ti0.75O to TiO0.69. Which of the following is/are the correct statement regarding the compound? (Ti = 48) (A) The maximum percentage by weight of oxygen in the compound is 30.8 (B) The minimum percentage by weight of titanium in the compound is 69.2 (C) The minimum percentage by weight of oxygen in the compound is 18.7 (D) The minimum percentage by weight of titanium in the compound is 82.3 103. 1 g-atom of nitrogen represents (A) 6.023 × 1023 N2 molecules (B) 14 g of nitrogen (C) 11.2 L of nitrogen gas at 1 atm and 273 K (D) 1 g of nitrogen 104. Which of the following statement(s) is/are correct? (A) One g-atom of each element contains the same number of atoms (B) One mole of each substance contains the same number of molecules (C) One g-molecule of each covalent compound contains the same number of molecules (D) One g-ion of each ionic compound contains the same number of ions 105. An oxide of nitrogen has 30.43% nitrogen (At. wt. of N = 14) and its one molecule weight 1.527 × 10–22 g. Which of the following statement regarding the oxide is (are) true? (B) Its empirical formula is NO2 (A) Its empirical formula is N2O (C) Its molecular formula is N2O4 (D) Its molecular formula is N4O2 Section III (Comprehension Based Questions having Only One Correct Option) Comprehension I (For Q. Nos. 106-108) A sample of bright blue mineral was weighed in air, then weighed again while suspended in water. An object is buoyed up by the mass of the fluid displaced by the object. In air, the mineral weighed 7.35 g; in water, it weighed 5.40 g. The densities of air and water are 1.205 g/dm3 and 0.9982 g/cm3, respectively. 106. What is the density of the mineral? (A) 0.4255 g/cm3 (C) 3.429 g/cm3 (B) 3.759 g/cm3 (D) 0.2068 g/cm3 Chapter 1 Mole Concept | 45 107. What is the volume of the mineral taken? (B) 9.429 cm3 (A) 1.956 cm3 (C) 2.221 cm3 (D) 6.823 cm3 108. What would be the observed mass of a piece of mineral of volume 50 cm3, when suspended in water? (A) 187.95 g (B) 49.91 g (C) 138.04 g (D) 187.88 g Comprehension II (For Q. Nos. 109-111) The first concept of atomic weight was given by Dalton. He defined that the absolute mass of an atom cannot be determined but we may compare the masses of atoms of different elements, perfectly, by knowing the chemical formula and percentage composition, by mass, of the compound formed by the elements concerned. The chemical or molecular formula can be determined with the help of Avogadro’s hypothesis that is, under the similar conditions of pressure and temperature, equal volume of all the gases have equal number of molecules. Dalton defined the atomic weight of an element as the number of times by which one atom of the element is heavier than one atom of hydrogen. In order to determine the atomic weight of nitrogen, the following data are observed by experiments, for a compound containing only nitrogen and hydrogen atoms: Data I : The compound contains 88% nitrogen and 12% hydrogen, by mass. Data II : 10 mL of this gaseous compound exactly gives 10 mL nitrogen and 20 mL hydrogen, on complete decomposition. (All volumes are at the same temperature and pressure) 109. What is the molecular formula of the compound if nitrogen and hydrogen, both are diatomic? (A) NH3 (B) N­2H­4 (C) N4H2 (D) N2H2 110. What is the atomic weight of nitrogen on this hydrogen scale? (A) 14 (B) 14.67 (C) 14.33 (D) 13.67 111. What would be the molecular formula of the compound if nitrogen were triatomic and hydrogen were diatomic? (B) N3H4 (A) NH3 (C) N­4H (D) N3H2 ­ 3 Comprehension III (For Q. Nos. 112-114) Vitamin C (M = 176) is a compound of C, H and O found in many natural sources, especially citrus fruits. When a 1.0 g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: Mass of CO2 absorber after combustion = 85.35 g Mass of CO2 absorber before combustion = 83.85 g Mass of H2­O absorber after combustion = 37.96 g Mass of H2O absorber before combustion = 37.55 g 112. What is the percentage of carbon, by weight in vitamin C? (A) 66.67% (B) 40.9% (C) 20% (D) 60% 46 | Chapter 1 Mole Concept 113. What is the percentage of hydrogen, by weight in vitamin C? (A) 4.55% (B) 41% (C) 20.5% (D) 9.11% 114. What is the empirical formula of vitamin C? (B) C3H4O3 (A) CH­2O (C) C6H8O6 (D) CHO Comprehension IV (For Q. Nos. 115-117) Figure shows a scheme, for concentrating a dilute solution of NaOH. Water Vapour Feed stream 25000 kg/hr (2% NaOH, 80% water 18% NaCl, by weight) Concentrated solution (50% NaOH, 50% water by weight) Slurry (90% NaCl, 5% NaOH, 5% water, by weight) 115. How much water evaporated per hour? (A) 5000 kg (C) 19500 kg (B) 500 kg (D) 20000 kg 116. How much concentrated solution obtained per hour? (A) 5000 kg (B) 500 kg (C) 19500 kg (D) 20000 kg 117. How much slurry obtained per hour? (A) 5000 kg (C) 19500 kg (B) 500 kg (D) 20000 kg Comprehension V (For Q. Nos. 118-120) You have a piece of gold jewellery weighing 9.35 g. Its volume is 0.654 cm3. Assume that the metal is an alloy (mixture) of gold and silver, which have densities of 19.3 g/cm3 and 10.5 g/cm3, respectively. Also assume that there is no change in volume when the pure metals are mixed. 118. What is the percentage of gold (by mass) in the alloy? (A) 58.24% (B) 41.76% (C) 74.08% (D) 43.12% 119. What is the percentage of gold (by volume) in the alloy? (A) 56.88% (B) 41.76% (C) 58.24% (D) 43.12% Chapter 1 Mole Concept | 47 120. The relative amount of gold in an alloy is measured in karats. Pure gold is 24 karats; an alloy of 50% gold is 12 karats. State the proportion of gold in the jewelry in karats. (A) 10.02 (B) 13.98 (C) 17.78 (D) 10.35 Comprehension VI (For Q. Nos. 121-123) A granulated sample of aircraft alloy (Al, Mg, Cu) weighing 8.72 g was first treated with alkali to dissolve the aluminium, then with very dilute HCl to dissolve the magnesium, leaving residue of copper. The residue after alkali-boiling weighed 2.10 g, and the acid insoluble residue from this weighed 0.69 g. 121. What is the percentage of aluminium in the alloy? (A) 75.9% (B) 16.2% (C) 7.9% (D) 24.1% 122. How many atoms of magnesium are present per gram of alloy? (B) 3.55 × 1022 (A) 4.06 × 1021 22 (C) 9.75 × 10 (D) 6.75 × 1020 123. The density of copper is 8.94 g/cm3. What volume of solid residue will be left per 100 g of alloy? (B) 70.74 cm3 (A) 0.885 cm3 3 (C) 1.81 cm (D) 8.49 cm3 Comprehension VII (For Q. Nos. 124-126) A sample of hydrogen fluoride gas (only HF molecules) is collected in a vessel and left for some time. Then, a constant molar mass of the sample is experimentally determined as 34 g/mol. Assume that this abnormal molar mass is due to dimerisation as well as trimerisation of some HF molecules (no molecules in any other polymeric forms) and the mole ratio of monomeric and trimeric form of hydrogen fluoride molecules present is 4 : 1. 124. What percentage of hydrogen fluoride molecules is dimerised? (A) 50 (B) 58.8 (C) 76.47 (D) 17.65 125. What percentage of hydrogen fluoride molecules is trimerised? (A) 40 (B) 58.8 (C) 76.47 (D) 17.65 126. What percent of total molecules present in the final sample are H2F2 molecules? (A) 10 (B) 40 (C) 50 (D) 58.8 Comprehension VIII (For Q. Nos. 127-129) The vapour density of a gaseous mixture containing only Ar and N2O4 gases, is 40. When the mixture is left for some time, the vapour density decreased and finally becomes 37.5. It happened due to dissociation of some N2O4 into NO2. (Ar = 40) 127. What is the degree of dissociation of N2O4? (A) 0.086 (B) 0.133 (C) 0.067 (D) 0.0625 48 | Chapter 1 Mole Concept 128. What is the initial mole ratio of Ar and N2O4? (A) 1 : 1 (B) 3 : 10 (C) 1 : 3 (D) 1 : 5 129. What is the final mole ratio of Ar, N2O4 and NO2? (A) 1 : 1 : 1 (B) 1 : 3 : 11 (C) 45 : 137 : 26 (D) 4 : 13 : 3 Comprehension IX (For Q. Nos. 130-132) A compound which contains one atom of X and two atoms of Y for each three atoms of Z is made by mixing 5.0 g of X, 1.15 × 1023 atoms of Y and 0.03 g-atoms of Z. It is observed that only 4.40 g of the compound results. The atomic weights of X and Z are 60 and 80, respectively. 130. What is the atomic weight of Y? (A) 70 (C) 65 (B) 75 (D) 90 131. How many gram of Y is left uncombined? (A) 13.37 g (B) 11.97 g (C) 1.40 g (D) 12.67 g 132. How many moles of atoms of X are left uncombined? (A) 0.073 (B) 0.063 (C) 0.053 (D) 0.01 Comprehension X (For Q. Nos. 133-135) When the Baeyer’s process is used for recovering aluminium from siliceous ores, some aluminium is always lost because of the formation of an unworkable mud having the following average formula: 3Na2O×3Al2O3×5SiO2×5H2O. Since aluminium and sodium ions are always in excess in the solution from which this precipitate is formed, the precipitation of the silicon in the mud is complete. A certain ore contains 13% (by weight) kaolin, Al2O3×2SiO2×2H2O and 87% gibbsite, Al2O3×3H2O. (Al = 27, Si = 28) 133. What percent of the total aluminium in this ore is recoverable in the Baeyer’s process? (A) 80 (B) 90 (C) 85 (D) 75 134. What is the percentage of silica present in the ore, by weight? (A) 2.82 (B) 3.02 (C) 0.465 (D) 6.05 135. How many mole of Al2O3 are present per mole of ore? (A) 1.000 (B) 0.083 (C) 0.222 (D) 0.242 Chapter 1 Mole Concept | 49 Section IV (Assertion-Reason Based Questions) The following questions consist of two statements. Mark (A)If both Statements are CORRECT, and the Statement-II is the CORRECT explanation of Statement-I. (B)If both Statements are CORRECT, but the Statement-II is not the CORRECT explanation of Statement-I. (C)If Statement-I is CORRECT but Statement-II is INCORRECT. (D)If Statement-I is INCORRECT but Statement-II is CORRECT. 136. Statement-I : The composition of any chemical compound is independent from the method from which it is produced. Statement-II : In any chemical reaction, the total mass of the reaction mixture always remains conserved. 137. Statement-I : One atom of an element weighs equal to its atomic mass, expressed in amu. Statement-II : One amu represents 1 th part by mass of one atom of C12 isotope. 12 138. Statement-I : The molecular mass of any substance is the sum of atomic masses of all the atoms present in each molecule of the substance. Statement-II : The atomic as well as molecular masses are defined on the same carbon scale. 139. Statement-I : Equal masses of all the gases contain equal volume, measured under identical conditions of pressure and temperature. Statement-II : At constant pressure and temperature, the volume occupied by any gas is directly proportional to its number of moles. 140. Statement-I : The number of atoms in 2 g hydrogen and 28 g nitrogen is same. Statement-II : The number of atoms in one mole of any substance is always same. 141. Statement-I : The number of atoms in a given mass of dioxygen (oxygen) and trioxygen (ozone) gases is same. Statement-II : The number of atoms depends on atomic mass, not on molecular mass. 142. Statement-I : Compounds having the same empirical formula have the same chemical composition. Statement-II : The chemical composition of different compounds may be same only when the molecular formula of the compounds is same. 143. Statement-I : All the members of any class of hydrocarbon (alkane, alkene, alkyne or arene) have the same empirical formula. Statement-II : Compounds having the same general formula may have different empirical formula. 144. Statement-I : The density of moist air is less than that of dry air. Statement-II : Moist air contains relatively lighter water molecules along with nitrogen and oxygen molecules. 50 | Chapter 1 Mole Concept 145. Statement-I : Vapour density of sulphur vapour relative to oxygen is 2 because sulphur atom is twice as heavy as that of oxygen atom. Statement-II : Vapour density depends upon the molecular state of the substance in vapour state. Section V (Match the Column may have One or More than One Match) 146. Match the following Column I (A) Number of ideal gas molecules in (P) 1 mL at 273 K and 1 atm. (B) Number of ideal gas molecules in (Q) 22.4 L at 273 K and 1 atm. (C) Number of atoms present in exactly (R) 12 g of C12 – isotopes (D) Number of atoms present in exactly (S) 1 g of C12 – isotope (T) 147. Match the following Column I (A) Gay- Lussac law (P) (B) 16 g of methane (Q) (R) (C) 92.91 g phosphorus (P4) (D) 32 g of oxygen (S) (T) Column II Loschmidt number 2.689 × 1019 Avogadro’s number 6.023 × 1023 5.019 × 1022 Column II 6.02 × 1023 molecules 4.517 × 1023 molecules 6.02 × 1024 electrons Gases react in simple ratio of their volumes. 1.807 × 1024 atoms 148. Match the following Column I Atomic masses Isotope I Isotope II (A) Z–1 Z+2 (B) Z+1 Z+3 (C) Z 3Z (D) Z–1 Z+1 Average Z Z+2 2Z Z (P) (Q) (R) (S) 149. Match the following Column I (A) 18 g glucose (P) (B) 1 mole water (Q) (C) Dalton’s atomic theory (R) (D) Dulong and Petit’s law (S) (T) Column II Percentage composition of the heavier isotope 33.33% by mole 50% by mole % by mass depends on Z 75% by mass Column II Atomic mass 7.22 × 1023 hydrogen atoms 6.02 × 1023 oxygen atoms The atom is indestructible 3.61 × 1023 oxygen atoms Chapter 1 Mole Concept | 51 150. Match the following Column I (P) (A) 0.875 mole O2 gas (Q) (B) 1.00 mole N2 gas (R) (C) 2.00 mole NaNO3­ (D) 0.4375 mole K2SO4­(S) (T) Column II 28 g 22.4 L at 0oC and 1 atm 1.2046 × 1024 atoms of nitrogen 1.0540 × 1024 atoms of oxygen 76.125 g 151. Match the following Column I (P) (A) 9.54 g SO2 (Q) (B) 26.56 g NH3 (C) 25 g methane (R) (D) 1 atom of lead (Pb = 208) (S) (T) Column II 3.76 × 1024 hydrogen atoms 9.41 × 1023 molecules 0.1491 mol 2.823 × 1024 hydrogen atoms 345 × 10–24 g 152. Match the following Column I (A) 8 g methane (B) 1 u (C) 88 g CO2 (D) 122.5 g of KClO3 (P) (Q) (R) (S) (T) Column II 1 ` 12 j th mass of C-12 24.08 × 1023 oxygen atoms 18.06 × 1023 oxygen atoms 3.01 × 1023 molecules 12.04 × 1024 carbon atoms 153. Match the following Column I (A) 0.635 g copper (P) (B) Na2CO3×10H2O (Q) (R) (C) K2CO3 (S) (D) Mass of one molecule of CO2 (T) Column II 7.308 × 10–23 g 8.69 % of carbon 34.79% of oxygen 6.02 × 1021 atoms 4.19% of carbon 154. Match the following Column I (A) 0.254 g copper (P) (B) 1 g He gas at 273 K and 1 atm (Q) (R) (C) 222 g CaCl2 (S) (D) 1 mole of Ca(OH)­2 (T) Column II 5.6 litres 2.4 × 1021 atoms 12.04 × 1023 atoms 1.506 × 1023 atoms 30.1 × 1023 atoms 155. Match the following Column I (A) 3 mole of Co(NH3)4SO4 (B) 1 mole FeKCo(NO2)6 (C) 1.5 mole [Fe(H2O)5SCN]SO3 (D) 0.75 mole K2Cu(SCN)4 Column II (P) 3 mole of S atoms (Q) 1 mole Fe (R) 12 mole O atoms (S) 6 mole N atoms (T) 1.5 mole K atoms 52 | Chapter 1 Mole Concept Exercise 3 (Previous Years’ IIT JEE Questions) 1. An evacuated glass vessel weighs 50.0 g when empty, 148.0 g when filled with a liquid of density 0.98 g/mL and 50.5 g when filled with an ideal gas at 760 mm Hg at 300 K. Determine the molar mass of the gas. (1998, 3 M) 2. A plant virus is found to consist of uniform cylindrical particles of 150 Å in diameter and 5000 Å long. The specific volume of the virus is 0.75 mL/g. If the virus is considered to be a single particle, find its molecular weight. (1999, 3 M) 3. At 100°C and 1 atm, if the density of liquid water is 1.0 g/cm–3 and that of water vapour is 0.0006 g/cm–3, then the volume occupied by water molecules in 1 L of steam at that temperature is (2000 (s), 3 M) (A) 6 mL (B) 60 mL (C) 0.6 mL (D) 0.06 mL 4. How many mole of electron weigh one kilogram? 1 (B) (A) 6.023 × 1023 1031 9.108 1 1 (C) (D) 1054 108 9.108 9.108 6.023 (2002(s), 3 M) 5. Which has maximum number of atoms? (A) 24 g of C (B) 56 g of Fe (C) 27 g of Al (D) 108 g of Ag (2003(s), 3 M) 6. Given that the abundances of isotopes 54Fe, 56Fe and respectively. The atomic mass of Fe is (A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05 7. Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on surface of area 10–12 m2 can be expressed in scientific notation as (2010, 3 M) y × 10x. The value of x is 57Fe are 5%, 90% and 5%, (2009, 3 M) Answers Exercise 1 1. 7.1 g 4. 7. 2.5 mm 5. 6.7 ton, 3.3 ton . . CoSO4 7H2O, CoSO4 H2O, 1.957 g 9. 50.9 11. 1 × 10–13 g Cu per g Ag 2. 690 Å 3. 57.93% 6. 2.48 1019 g 8. 40.8% 10. 6.03 × 1023 12. 1.427 Å 14. 31.79 15. 2.77 × 17. 11 mm 18. 141.16 1025 13. Yes 16. 0.566 mL 19. 210