Chemistry Paper-1
Chapter-1
Max. Marks: 07
Marks with Options: 10
Introduction
• Solution:
 Homogeneous mixture of two or more components
 Mixture obtained when one substance is completely
dissolved in the other
 Examples:
Salt dissolved in water
Ethanol in acetone
Hard water
Etc.
• Components of Solution:
• Solvent
• Solute
– Solvent:
Medium of the solution
OR Larger component of a Solution
– Solute:
Substance being dissolved
OR Smaller component of a Solution
Note:
The two components of solution must be non-reacting.
Example:
When sugar is dissolved in water, water is solvent and sugar is
solute.
•
•
•
•
•
•
•
All components in one phase
All particles are molecules or ions
Uniformly distributed particles
Particles cannot be seen in any microscope
Particles cannot be filtered out by any filter paper
Particles do not reflect or diffract the light
May be solid, liquid or gas
• Concentration
Amount of solute dissolved in given amount of
solution or solvent.
• Dilute solution
 Contains lesser amount of solute
 i.e. Low concentration of solute
 Such as dilute HCl
• Concentrated solution
 Contains larger amount of solute
 i.e. High concentration of solute
 Such as Concentrated H2SO4
• Ideal Solution:
– Solution which follows the Raoult’s at all temperatures and
concentrations.
– Other solutions are called Real Solutions
– In practice no solution is Ideal Solutions
– Some solutions behave ideally at very dilute concentrations
and moderate temperatures
– Examples: Benzene + toluene, n-hexane + n-heptane
• Properties of Ideal solution:
•
•
•
•
No heat of dilution
Vsolution = V1 + V2 + … + Vn
Follows the Law of Gases
Follows the Raoult’s law
• Amount of solute dissolved in given amount of solution
or solvent
• Different units of Concentration are used in different
areas of Science
• A solution whose concentration is exactly known is called
as Standard Solution
• Temperature units involving “volume” are dependent on
temperature
• Temperature units not involving “volume” but “mass” or
weights only are independent of temperature
Normality:
Molarity
Molality
Mole fraction
Percent
• Definition:
No. of gram equivalents per liter of solution
• How to Calculate:
normality
gram equivalents 1000
ml of solution
OR
normality
grams of solute 1000
equivalent wt. ml of solution
• 1 N Solution:
1 gram equivalent dissolved in 1 L solution
• Gram Equivalent:
Equivalent wt. expressesed in grams.
gram equivalent
grams of solute
equivalent wt.
• Example:
If 10 g NaOH (i.e. 0.25 g eq) is dissolved in 500 mL
solution, 0.5 N solution is obtained.
• Normality of Dilution:
Before dilution
N1V1
=
After dilution
N2V2
• Normality of Mixing:
Let V1 mL solution of N1 normality is mixed with V2 mL solution
of N2 normality.
Then the normality the new solution will be:
N
N1V1
V1
N 2 V2
V2
Note: Learn at Home:
 What is equivalent weight?
 What are the Eq. wts. of:
NaOH, H2SO4, Na2CO3, H2C2O4?
 What will be normality of solution containing 4 g
NaoH dissolved in 250 ml solution?
Molarity
• Definition:
No. of gram moles per liter of solution
• How to Calculate:
molarity
gram moles
1000
ml of solution
OR
molarity
grams of solute 1000
molecular wt. ml of solution
• 1 M Solution:
1 gram mole dissolved in 1 L solution
• Gram Mole or Mole:
Molecular wt. expressesed in grams.
gram moles
grams of solute
molecular wt.
• Example:
If 49 g H2SO4 (i.e. 0. 5 mole) is dissolved in 250 mL
solution, 2 M solution is obtained.
• Molarity of Dilution:
Before dilution
M1V1
=
After dilution
M2V2
• Molarity of Mixing:
Let V1 mL solution of M1 molarity is mixed with V2 mL solution of
M2 molarity. Then the molarity the new solution will be:
M
M1V1
V1
M 2 V2
V2
• Relation between Molarity and Normality:
Normality = n x molarity
Where, n is the acidity of base or basicity of acid.
Note: Normality is equal to or greater than molarity.
Note: Learn at Home:
 What is molecular weight?
 What is acidity of a base and basicity of an acid?
 Give acidity or basicity of the following:
NaOH, Ca(OH)2, NH4OH, Al(OH) 3
H2SO4, CH3COOH, H3PO4, HCN
 What are the M. wts. of:
NaOH, H2SO4, Na2CO3, urea, glucose and H2C2O4?
 How much amount of urea will be needed to prepare
2 Lit of its 0.2 M solution?
Molality
• Definition:
No. of gram moles per 1000 g (1 kg) of solvent
OR
Or Number of moles of solute in 1 kg solvent
• How to Calculate:
molality
gram moles
1000
wt. of solvent
OR
molality
grams of solute 1000
molecular wt. wt. of solvent
• 1 m Solution:
1 gram mole dissolved in 1 kg solvent
• Example:
•
•
•
•
If 45 g glucose (i.e. 0.25 mole) is dissolved in 500 g water,
0.5 m solution is obtained.
This unit of concentration does not involve volume of
solution or solvent.
Therefore this unit of concentration is independent of
temperature
The concentration terms involving V are dependent of T
And the terms involving w only are independent of T
Relation between molality (m) and molarity (M):
M/m =
W/V
Where:
W is the weight of solvent in gm
V is the volume of solution in mL
When the density of the solution is given:
m
M
d M M2
Where:
d is the density of the solution
M2 is the molecular wt. of the solute
Note: Learn at Home:
Relationship between E. wt. and M. wt.?
Relationship between normality and molarity?
What is the molality of a solution containing 0.2
mole sugar dissolved in 500 g water?
How much oxalic acid is needed to prepare 0.05 m,
0.5 L solution?
Calculate the molality of pure water.
Normality
Molarity
w 1000
E v
w 1000
M v
Molality
w2
M
1000
w1
• Ratio of the no of moles of a constituent to
the total number of moles of all constituents
• Mole fraction of SOLUTE: X2
X2
n2
n1
n2
• Mole fraction of SOLVENT: X1
X1
n1
n1
n2
• For a 2 component-solution:
X1 + X 2 = 1
• This unit of concentration does not involve volume of solution or
solvent
• Therefore this unit of concentration is independent of temperature
• Mole fraction is a relative quantity
• It does not give the actual quantity of the components
• It only gives the quantity of a component relative to the total
quantity of all components
• If a mole fraction is multiplied by 100 it will give the % mole of the
component
• Total of the % mole is equal to 100
Percent Weight
Grams of solute dissolved per 100 g of solution
percent by weight
grams of solute
grams of solution
100
For example:
 5% glucose solution
 It means 5 g glucose dissolved in 100 g glucose solution
 It does not mean 5 g glucose in 100 g water
Note
 % by wt. means gms of solute per 100 g of solution
 % by vl. means gms of solute per 100 mL of solution
Relation between normality, molarity and wt %:
1. Normality and % by wt.:
normality
% by wt. d 10
E
2. Normality and % by vl.:
normality
% by vl. 10
E
3. Molarity and % by wt.:
molarity
% by wt. d 10
M
4. Molarity and % by vl.:
molarity
% by vl. 10
M
Property of dilute solutions which depends on the no.
of particles in the solution
• It depends only on the no. of particles of solute in the
solution
• It does not depend on the nature of particles of the
solute
• There are four Colligative properties which are related
to each other
• They are used to calculate molecular wt. of the solute
There are following four Colligative Properties:
Lowering of vapor
pressure
Elevation
of
boiling
point
Osmotic
pressure
Depression of
freezing point
A. LOWERING OF VAPOR PRESSURE
Vapor pressure:
–
–
–
–
Equilibrium pressure exerted by the vapors of the liquid on it
VP increases exponentially with the increase in temperature
VP is inversely proportional to the intermolecular forces
With rise in T:
•
•
•
•
Intermolecular forces breakdown
Kinetic energy of molecules increases
Molecules at the surface pass into vapor state
Ultimately VP increases
– Liquids with high VP (Low BP) are volatile (eg: ether, benzene)
– Liquids with low VP (high BP) are non-volatile (eg: water, oil)
– Relation between VP and T is derived by Clausus and Clapeyron
• Lowering of vapor pressure:
 On addition of non-volatile solute, the VP of liquid decreases
 P = Po – P
 Po > P
If more solute is added, lowering of VP will increase
• Relative lowering of vapor pressure:
– Ratio of Lowering of VP to the VP of pure solvent
Po
P
Po
– The relation between the RL of VP and amount of solute is
given by Raoult
• If more solute is added, lowering of VP will increase
• Raoult’s Law gives the relation between the RL of VP and
amount of solute added to the solution (concentration)
“Relative lowering of VP is equal to the mole fraction of
the solute”
Po P
Po
n2
n2
n1
Alternate Statement:
“Partial VP of liquid over its solution is proportional to its mole fraction”
VP1 α X1
or
VP1 = Po x X1
Limitations of Raoult’s Law
It is applicable only to
very dilute
solutions
non-volatile
solutes
unionized
solutes
unassociated
solutes
Ostwald and Walker Method
• Used to calculate Relative lowering of VP
• Lowering of VP is used to calculate M Wt. of solute
• Limitations of the Method:
– All limitations of Raoult’s Law
• Only for non-volatile solutes
• Applicable for very dilute solutions
• Solute should be ionised or associated
– Uniform temperature is needed
– Low air pressure is necessary
 The Relative Lowering of Vapor Pressure is calculated using the
following formula
 This formula is derived from the observations of the above
experiment
o
P
P
o
P
Where:
m2
m2 m 1
m2 is loss in wt of solvent
m2 + m1 is gain in wt of CaCl2
Calculation of M. Wt. of solute from Relative L of VP
o
P
Po
P
w 2 M1
w 1 M2
In the above formula:
 The Relative Lowering of Vapor Pressure is calculated from
the Ostwald-Walker Method.
 w2 is the wt. of solute.
 w1 is the wt. of solvent.
 M1 is the Mol. wt. of the solvent.
 Therefore, M2, the Mol. wt. of the Solute can be calculated.
B. ELEVATION IN BOILING POINT
• Boiling Point:
– “The temp. at which VP of liquid equals Atmospheric
Pressure (AP)”
– VP increases exponentially with the increase in T
– With rise in T:
• Intermolecular forces breakdown and the kinetic energy of
molecules increases
• They pass into the vapor phase and VP increases
• When VP equalizes the AP, liquid starts boiling
• This T is the BP of the liquid
– Boiling of liquid takes place throughout the liquid while
vaporization takes place only on the surface
• Elevation of Boiling Point:
– “Increase in the boiling point of liquid on addition of
solute”
–
–
–
–
–
–
On adding non-volatile solute, VP decreases
Therefore, temperature required to equalize VP to AP increases
Ultimately, the B. Pt. of the liquid increases (elevated)
This is called as E of BP
E of BP is the Colligative Property
If equal no. of moles of solute is dissolved in equal wt. of solvent,
the two solutions have same E of BP
– E of BP is proportional to the molality
• Formula:
Tb = T – To
(T > To)
Landsberger and
Walker Method:
• Used to calculate Elevation
of B. Pt.
• Elevation of B. Pt. is used to
calculate M Wt. of solute
o
• First T is determined
• Then
T is determined
• Then is Tb calculated
• Using following Formula
• Tb = T – ToU
• This experiment is
called as Ebullioscopy
Calculation of M. Wt. of solute from Elevation of B. Pt.
Tb
Kb
w 2 1000
w 1 M2
OR
Tb
Kb
m
In the above formulae:





w2 is the wt. of solute.
w1 is the wt. of solvent.
M2 is the Mol. wt. of the solute.
Kb is the molal elevation constant.
Therefore, Tb, the elevation in boiling point can be calculated.
• Molal elevation constant or Ebullioscopic Constt., Kb:
– Elevation in B. Pt. when 1 mole solute is dissolved in 1000 g
solvent
OR
– Elevation in B. Pt. of 1 molal solution
• Unit:
– K m-1
or
K kg mol-1
• Kb of water:
– 0.52 K m-1
– Kb is constant for a solvent
– It is independent of the nature of solute
• Note that:
– Sea water boils at higher temperature than the fresh water
– E of BP is proportional to Molality (concentration)
– It is inversely proportional to the M2
C. DEPRESSION OF FREEZING POINT
• Freezing Point:
– Temperature at which VP of liquid equals VP of its solid
– At FP the liquid and solid co-exist in equilibrium
– At FP the liquid and solid have same temperature
• Depression of Freezing Point:
–
–
–
–
Decrease in the Freezing pt. of liquid on addition of solute
On adding non-volatile solute, VP decreases
Therefore, temperature required to equalise VP to AP decreases
Ultimately, the FP of the liquid decreases (depressed)
• Formula:
Tf = To – T
(To > T)
Beckmann Method:
• Used to calculate
Depression of F. Pt.
• Depression of F. Pt. is used
to calculate M Wt. of solute
o
• First T is determined
• Then
T is determined
• Then is Tf calculated
• Using following Formula
• Tf = To – T
• This experiment is
called as Cryoscopy
Calculation of M. Wt. of solute from Depression of F. Pt.
Tf
Kf
w 2 1000
w 1 M2
• Molal depression constant or Cryoscopic Constt., Kf:
– Depression in F. Pt. when 1 mole solute is dissolved in 1000
g solvent (i.e. 1 molal solution)
OR
– Elevation in B. Pt. of 1 molal solution
• Unit:
– K m-1
• Kf of water:
– 1.825 K m-1
or
K kg mol-1
• On addition of Non volatile solute to a liquid
– Its VP decreases
– Its BP increases
– Its FP decreases
– More is the LVP, more is the EBP and more is the DFP
• Deviation from Raoult’s Law:
– When VP of liquid is less than calculated from Raoult’s law, it is –ve deviation
– When VP of liquid is more than calculated from Raoult’s law, it is +ve deviation
– Binary solutions with –ve deviation boil at higher T than the components
– Binary solutions with +ve deviation boil at lower T than the components
– At –ve deviation from Raoult’s law: V = +ve and
H = +ve
– At +ve deviation from Raoult’s law: V = –ve and
H = –ve
D. OSMOTIC PRESSURE
OSMOSIS
• Osmosis:
The passage of a solvent through a semipermeable membrane from a dilute to
a more concentrated solution
Osmosis was first observed by Abbe Nollet in 1784
• Diffusion:
Uniform distribution of solute in the solution
• Comparison of Osmosis and Diffusion:
 Osmosis is the spontaneous passage of solvent from low to high conc. through
semipermeable membrane
 Diffusion is the movement of the solute from high to low conc.
 Osmosis can be reversed or stopped by applying opposite pressure
 Diffusion can not be stopped or reversed
 Osmosis occurs in solutions only
 Diffusion occurs in gases also
 For osmosis semipermeable membrane is necessary
 In diffusion semipermeable membrane is not needed
 In osmosis only solvent moves
 In Diffusion solvent and solute both move
 Osmosis is unidirectional
 Diffusion is bidirectional
 Osmosis happens due to differential OP
 Diffusion takes place due to differential concentration
• Semipermeable membrane
 Allows the flow of solvent molecules only
 There are two types of Semipermeable Membrane
1.
2.
Natural: parchment paper, cell wall, gall bladder, egg membrane
Artificial: Cu-ferrocynide membrane, cellulose nitrate membrane
 They serve as molecular filters
• Osmotic Pressure:
Hydrostatic pressure developed within the solution which is
just sufficient to stop the osmosis
OR
External pressure just needed to stop the osmosis
• Reverse Osmosis:
 Flow of solvent from high conc. to low conc. through
semipermeable membrane
 This is possible when external pressure is applied in the
high conc. area
Types of Solutions
(Based on Osmotic Pressure)
Isotonic solns:
Hypertonic soln:
Hypotonic soln:
Having same OP
Having higher OP
Having Lower OP
When a plant cell is put in hypotonic solution, water diffuses in and the cell swells
When a plant cell is put in hypertonic solution, water diffuses out and the cell shrinks
When isotonic solutions are separated by SPM there is no net flow of water
Solvent flows from hypotonic to hypertonic solution
0.91% NaCl solution is isotonic to human blood cells
Blood cells put in 4% NaCl solution will shrink and put in o.1% NaCl will swell
•
•
•
•
•
•
Behaviour of dilute solution resembles a gas
Solute particles behave like gas molecules
OP of a dilute solution resembles gas pressure
Gas laws are applicable to dilute solutions
Equation similar to Ideal gas equation can exist
There are following Laws of OP:
1. Boyle-Van't Hoff Law
2. Pressure-Temperature Law
At constt. Temperature, Osmotic Pressure is
proportional to concentration of solution.
Po
C
But: C = n/V, therefore C = 1/V
Therefore:
Po
1
V
(n = 1)
• At given Concentration, Osmotic Pressure is
proportional to the Temperature of solution.
Van't Hoff Equation for Solutions
• It is similar to Ideal Gas Equation
• It is drawn from the 2 Laws of OP
• R’ is called as Van’t Hoff constant
Po V
nR' T
Calculation of M. Wt. of solute from Osmotic Pressure
Since:
PoV = nR’T
And:
n = w/M
Therefore:
M
wRT
Po V
• When solute gets ionised or associated, actual no. of
particle increases or decreases
• Therefore the OP being collegative property, increases or
decreases
• To consider this fact Van’t of Factor, i , is introduced in the
equation
PoV = i.nR’T
Where:
i
actual number of particles after ionisation
number of particles if there is no ionisation
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