CLASS – 12TH

CBSE 2024

Solutions
APNI KAKSHA NOTES

APNI KAKSHA 1
Solutions
Solutions are homogeneous mixtures of two or more than two components.
Homogeneous mixture means Composition and properties are uniform throughout the
mixture. Generally, the components that is present in the largest quantity is known as
Solvent. Components present in the solution other than solvent, is known as solute.
Binary Solution = Solvent + Solute
Types of solution Solute Solvent Common examples
Gas Gas Mixture of oxygen and
Gaseous Liquid Gas nitrogen gases Chloform
solutions Solid Gas mixed with nitrogen gas
camphor in nitrogen gas
Gas Liquid Oxygen dissolved in water
Liquid Solutions Liquid Liquid Ethanol dissolved in water
Solid Liquid Glucose dissolved in water

Gas Solid Solution of hydrogen in

Solid Solutions palladium
Liquid Solid Amalgam of mercury with
sodium
Solid Solid Copper dissolved in gold

Expressions for concentration of solution

1. Mole fraction [X]
Mole fraction of a component = No. of moles of component
Total no. of modes of all components
for a binary mixture (having component A and B), if the no. of moles of A and B are hA
and hB respectively. Then mole fraction of A will be XA = hA /hA + hB

APNI KAKSHA 2
 mole fraction of B =XB= hB/hA + hB
For binary mixture XA + XB = 1 [ Sum of all the mole fractions is unity]
Example:-A mixture having two gaseous component N2(s) and O2(g) with no. of moles of
10 and 20 respectively Then what is the male fraction of both component
hn2 10 1
xn2   
Ans:- hn2  no2 10  20 3
1 2
xo2  1  
3 3

2. Mass Percentage [W/w] :-

Mass % of a component = Mass of the component in solution × 100
Total mass of solution
10% Glucose in water means 10 g of glucose is dissolved in 90 g of water resulting in 100
g solution.
3. Volume Percentage [V/v] :-
Volume Percentage of component = Volume of the Component
Total volume of Solution
10% Ethanol in water means that 10 ml of ethanol is dissolved in water such that total
volume of solution is 100 ml.
4. Mass by volume Percentage [m/v] :- It is mars of solute dissolved in 100 ml of the
solution.
5. Parts per million [ppm] :- It is used when solute quantity is very very small.
Parts per million = No. of parts of components × 10 6
Total no. of parts of all components
of the solution
Concentration in parts per million can also be expressed as mass to mass (W/w) volume
to volume (V/v) and mass to volume (m/v)
% w/v = Weight of solute(g)×100
Volume of solution (ml)
Similarly: ppm (w/v)= Weight of solute(g)×106
Volume of solution (ml)
5.8 ppm O2 is dissolved in sea:- means 5.8 g of O2 is dissolved in 106 g of sea water.
The concentration of pulutants in water and atmosphere is expressed in term of ppm
Q. Calculate the mole fraction of ethylene glycol ( C2H6O2) in a solution containing 20% of
C2H6O2 by mass?
Ans 20% of C2H6O2 means 20 g of C2H6O2 is present in 80 g of water Molar mass of C2H6O2
=2×12 + 1×6+16×2=62 g/mol

APNI KAKSHA 3
 Moles of C2H6O2 = 20 g/62 g mol-1 =0.322mol = h glycol
Moles of water = 80 g/ 18 g mol-1 =4.444 mol=h water
Mole fraction of glycol x glycol = h glycol/ h glycol + h water =0.322/0.322+ 4.444 = 0.062
Similarly x water =1- x glycol =1-0.068 =0.932
6. Molarity [M]:- It is defined as number moles of solute dissolved in one litre
of solution
Molarity = Moles of solute
Volume of solution (in litre)
Unit=mol/litre[M]
0.25 M solution of NaoH means that 0.25 mol of NaoH has been dissolved in 1 litre
Q. Calculate the molarity of a solution containing 5 g of NaoH in 450 ml solution ?
Ans Moles of NaoH = 5g /40 g mol-1 =0.125 mol ,
volume of solution in litre =450/100=0.45 L
Molarity = 0.125 mol/ 0.45 L =0.278 mol L -1 =0.278 mol dm-3, 1 litre =1 dm3

Solve this after reading concept of molality

Q. A solution of glucose in water is labeled as 10% by weight .What would be the
molality of a solution ?[ Molar mass of glucose = 180 g mol -1
Ans 10% of glucose solution by weight means 10 g glucose is present in 100 g solution. Then
weight of water = 100-10 = 90 g
Molality = Moles of solute
Mass of solvent (kg)
(10/180) mol
(90/1000) kg
=0.61 mol kg -1 = 0.61 m

7. Molality [m]:- it is defined as no. of moles of the solute per kg of solvent.

Molality = Moles of solute
Mass of solvent (kg)
Unit = mol/kg
1 m solution of kcl means that 1 mol of kcl (74.5 g) is dissolved in 1 kg of water.

Q. Calculate molality of 2.5 g of ethanoic acid in 75 g of benzene?

Ans Molar mass = CH3-COOH =12×2 + 1×4 + 16 ×2 =60 g mol -1
Molality of CH3-COOH = Moles of CH3-COOH

APNI KAKSHA 4
 Kg of benzene
(2.5/60)/(75/100)=0.556 mol/kg

Main advantage of molality over molarity:- molality does not change with change
with temperature while. Molarity decreases with rise in temperature.
This is because volutre depends on temperature ( V  T ) but mass does not.

Relation between Density, Molarity , and Molality :- If d is density of solution , m is

molality and M is molarity, M2 is molar mass of solute . then
Molality m = M×1000
1000×d – M× M2
(No need to memorise this formula)

Q. Calculate the molarity of 9.8 % w/w solution of H2SO4 if the density of the solution
is 1.02 g ml-1 [Molar mass of H2SO4 = 98 g mol-1
Ans 9.8 % solution means 9.8 g of H2SO4 is present in 100 g of the solution.
Density of solution = 1.02 g ml-1
Then volume of solution = mass of solution
Density of solution
=100/1.02 =0.098
Moles of H2SO4 =9.8 g /98 g mol-1 =0.1 mol
Molarity = 0.1 mol/0.098 =1.019 mol-1 =1.02 M

Q. A solution of glucose (molar mass = 180 g mol-1) in water is labeled as 10% (mass).
What would be the molality and molarity of the solution?
Density of solution = 1.2 g mol-1
Ans 10% glucose solution (by mass) means that
Mass of glucose = 10 g
Mass of solution = 100 g
Mass of solvent = 90 g (molar mass of glucose =180 g/mol)
Density of solution = 1.2 g mol-1
Volume of solution = m/d =100 g/ 1.2 g mol-1 =1000/12 ml =1/12 litre
Now molarity = moles of glucose
Volume of solution
(10/180)/(1/12)=0.667 M
Molality = moles of glucose
APNI KAKSHA 5
 Mass of solvent (in Kg)
(10/180)/(90/1000)=0.617 m

Solubility
Solubility of a substance is its maximum amount that can be dissolved in a fix abount of
solvent.
When a solid solute is added to solvent, some solute dissolves and its concentration
increases in solution. This process is known as Dissolution.
Some solute particles in solution collide with the solid solute particles and get separated
out of solution. This process is known as Crystallization.
At point of equilibrium, no. of solute particles going into solution will be equal to solute
particles separating out . Solute + Solvent = Solution
At this dyamic equilibrium concentration of solute in solution will remain constant at
given temperature and pressure.
Such a solution in which no more solute can be dissolved at the same temperature and
pressure is called a saturated solution.
Unsaturated solution: A solution in which more solute can be dissolved.

Effect of temperature:- if in a nearly saturated solution, the dissolution process is

endothermic ( H > O ) then on increase of temperature = solubility increase
For exothermic ( H < O ) then on increase of temperature = solubility decrease
Pressure does not have any significant effect on solubility of solids in liquids. Because solid
and liquid both are incompressible

Solubility of gas in liquid :-

Factor affecting solubility of a gas in liquid :
1. Nature of gas
2. Nature of liquid
3. Temperature
4. Pressure

APNI KAKSHA 6
Henry Law [ Effect of pressure on solubility]:-
“The solubility of a gas in a liquid at a given temperature is directly
proportional to the pressure at which it is dissolved.”
If we use mole fraction (x) of a gas in solution as a measure of its solubility, then
Partial pressure of a gas (p) = KH . x Henry’s law constant
partial pressure of HCI/torr

1000

500

0
0.010 0.020
Mole fraction of HCI in its
solution in cyclohexane

i. KH is a function of the nature of a gas means that different gases have different gases
have different KH value.
ii. Higher the value of KH – lower is the solubility because at constant pressure = KH 
1/x
iii. Unit of KH = unit of pressure = atm/torr
KH is proportional to temperature [ KH  T and KH  1/x = T 1/x ]
Means that solubility of a gas increase with decrease of temperature.
Due to above reason O2 is getting more dissolve in cold water than in warm water. So
aquatic species are more comfortable in cold water than warm water.

Application of Henry Law:-

1. To increase the solubility of co2 in soft drinks and soda water, the bottle is sealed under
high pressure.
2. At higher altitudes the partial pressure of oxygen is less than that at ground level this
leads to low concentration of oxygen in blood. Low blood oxygen causes climbers to
become weak and unable to think clearly, symtoms of a condition known as anoxia.

Q. The partial pressure of ethane over a saturated solution containing 6.56 × 10 -2 g of

ethane is 1 bar. If solution contains 5.0 ×10 -2 of ethane, then what will be partial
pressure of the gas ?
Ans. According to Henry’s law:-
I. Mass of ethane of partial pressure [ m1 = KH . P2 ] = 6.56 × 10 -2 g = KH .1 bar
KH =6.56 × 10 -2 g/bar
APNI KAKSHA 7
 II. M2 = KH P2 Partial pressure of gas (P2) = M2/ KH =5×10-2 g/ 6.56 × 10 -2 g/bar
=0.762 bar
Q. If N2 gas is bubbled through wate at 293 k, how many millimoles of N 2 gas would
dissolve in 1 L of water ? Assume that N2 exerts a partial pressure of 0.987 bar.
Given= 76.48 k bar at 293 k.
Ans According to Henry’s law:-
pn2  kH .xn2
pn2 0.987bar
xn2    1.29  103
kH 76.48  10 bar
3

If h moles of N2 are present in 1 L or 1000 g of water then moles of water = h water

=1000/18 =55.5 moles
Mole fraction of N2 = h/h + h water = h/ 55.5

xn2 (55.5) =h = 55.5 × 1.29 × 10-5=71.595 × 10-5 moles

h = 0.716 × [10-3 moles] 0.716 milimoles

Vapour pressure of liquid solutions:-

Defination of vapour pressure:- The partial pressure of vapours in
equilbrium with pure solid or pure liquid at a given temperature.
At equilibrium :- rate of evaporation = rate of condensation
For example :- H2O (l) H2O (g)

pH2O(g )
Equilibrium constant (kp) =
pH2O(l )

For pure H2O (l) : pH2O(l ) =1 then kp= pH2O (g )

For liquid vapour : vapour pressure is an equilibrium constant . So it depends only on

a. Nature of liquid
b. Temp. of liquid
If magnitude of intermolecular forces in the liquid is smaller than we observe high
vapour pressure because loosely held molecules escope more easily into vapour phase.

APNI KAKSHA 8
 Boinling point :- The temperature at which vapour pressure of a liquid is equal to
external pressure.
A liquid having high vapour pressure boils at low temperature [B.P  1/V.P]

Vapour pressure of liquid-liquid solutions:-

Raoult’s law:- For a solution of volatile liquids, the partial vapour pressure of each
component in the solution is directly proportional to its mole fraction.
For a binary liquid-liquid solution if p1 and p2 are partial pressure of liquid-1 and liquid
-02 respectively and x1 and x2 are moles fraction of liquid-1 and 2 in solution respectively
For liquid-1: p1  x1
p1 = p1x1…(i)
p1= vapour pressure of pure liquid 1
For liquid-1: p1  x2
P2 = p2x2…(ii)
P2= vapour pressure of pure liquid 2
Vapour pressure of
pure solvent
vapour pressure

0 1
Mole fraction of solvent
According to Dalton;s law of partial pressures :- Total pressure over solution phase in
the container will be the sum of the partial pressure of components of the solution.
ptotal = p1 + p2 = p1x1 + p2x2
ptotal = (1- x2) p1 + p2x2 = p1 + (p2 - p1)x2 [x1 + x2 ]
y = mx + c
linear graph between total v.p and x2
For line II: ptotal = p2
Initially x1 = 0
x2 =1

APNI KAKSHA 9
 vapour pressure
III
p
= p 1+ 2
P
P total
II
p0
1

P2

P1
I
X1 = 1 x1=0
x 2 =1
x 2= 0 Mole fraction
x2

If y1 and y2 are the mole fraction of the liquid -1 and liquid -2 respectively in vapour phase
then, using Dalton’s law of partial pressures
p1= y1 ptotal
p2= y2 ptotal

AT equilibrium , vapour phase will be always rich in the component which is more
volatile.
Q. The vapour pressure of pure liquids A and B are 450 mm and 700 mm of Hg
respectively at 350 k. find out the composition of the liquid mixture if total vapour
pressure is 600 mm of Hg. Also find the composition in vapour phase?
Ans PA =450 mm of Hg PB= 700 mm of Hg
PT = 600 mm of Hg XA/ XB/ YA/ YB ?
PT = PB + (PA - PB) XA = 600 = 700 + (450 – 700 ) XA
Mole fraction of A in liquid XA =100/250 =0.40
Mole fraction of B in liquid phase XB =1- XA =1-0.40=0.60
PA = XA. PA =(0.40) (450) =180 mm of Hg
PB = XB. PB =(0.60) (700) =420 mm of Hg
PA = YA PT = YA = mole fraction of A in vapour phase = PA/ PT =180/600=0.30
PB = YB PT = YB = PB/ PT =420/600 =0.70
Note :- in above question : YB> YA [so vapour phase is more rich is component (l) and
B=B is more volatile liquid than A

Raoult’s law as a special case of Henry’s law:-

According to raoult’s law , vapour pressure of volatile component in a given
solution p=x p …(i)
In the solution of a gas in liquid if one of the components is so volatile that
it exists as a gas is directly proportional to mole fraction then its solubility
p= KH.x..(ii)[Henry’s law]

APNI KAKSHA 10
 When KH becomes equal to p [Both are constant for a given liquid at constant

temperature then raoult’s law bescomes special case fo Henry’s law.

Ideal and Non-ideal solution:- liquid-liquid solution can be classified into

ideal and non-ideal solution on the basis of raoults law.
Ideal solution Non-ideal solution
1. The solution which obey Raoults law over When a solution does not obey raoults law
entire range of concentration are known as over entire range of concentration, then it is
ideal solution called non-ideal solution.
2. PA= XA. PA PA# XA. PA
PB= XB. PB PB# XB. PB
3. Component A + Component B – solution AB Component A + Component B (solution
forces of attration between A-B are exactly components AB) fA-B # fA-A # fB-B
same as A-A and B-B. fA-B = fA-A = fB-B
4. On mixing , there is no enthalpy and On mixing , there is enthalpy and volume
volume change,Hmix =0 Vmix=0 change,Hmix #0 Vmix#0
Example:- H2O and ethanol H2O and CH3 COOH
H-Hexane and h-heptane bromoethane and CH3Cl and CCl4
chloroethane benzene and toleene

Two types of Non-ideal solution:

1. Positive deviation from Raoult’s law:- A-B interaction are weaker than A-
A and B-B interaction. Due to this vapour pressure increase which results in
positive deviation.
(ptotal)exp > (XA. PA+ XB. PB)
Liquid A + Liquid B= solution AB Hmix =+ve, Vmix =+ve
Example:- H2O +CH3OH, Ethanol + Acetone
H2O +C2H5OH, CS2 + Acetone, CHCl3 + CCl4

2. Negative deviation from raoult’s law:- A-B interactions are stronger than
A-A and B-B interactions. Due to this, vapour pressure decrease which
results in negative deviation.

APNI KAKSHA 11
 (PTotal) exp < XA. P+ XB P
Hmix =-ve and Vmix =-ve
O
O

H O + CH OH H O + H C O - H
Example :- 2 3
C
2
Phenol + aniline
CHCl3 + Acetone
O

CH C CH
A solution of CHCl3 (chloroform) and 3 3 (Acetone) shows –ve deviation?
Why?
In pure CHCl3 and pure acetone , there is no hydrogen bonding but after mixing there is
hydrogen bonding between hydrogen of CHCL3 and oxygen of acetone. Due to strong
interaction between solution molecules, vapour pressure decrease which results in –ve
deviation
Cl
C H O=
Cl
Cl

Azotropes:-Binary mixtures having same composition in liquid and vapour

phase boil at a constant temperature. [XA.=YA and XB = YB]

Vapour pressure of solution

vapour pressure

P1 P2

X =0 X1=1
1
X2 =1 X 2=0
Mole fraction
X1
X2

APNI KAKSHA 12
 Vapour pressure of solution

vapour pressure

P2 P1

X =0 X1=1
1
X2 =1 X 2=0
Mole fraction
X1
X2

Azeotropes cant be separated into its components by fractional distillations.

There are two types of azeotropes.
1. Minimum boiling point azeotrope:- The solution which show large positive deviation
from Raoult’s law form minimum boiling azeotrope at a specific composition
For example :- C2H5OH + H2O [with 95% by volume of ethanol]
2. Maximum boiling azeotrope:- The solution that show large negative deviation from
Raoult’s law form maximum boiling azeotrope at a specific composition.
For example:- HNO3 + H2O [ 68% nitric acid + 32 % water by mass]

Colligative Properties
The properties of solution which depend only on the number of solute particle.
Colligative properties  no. of particles in solution  1/molar mass of solute
Four colligative properties:-
1. Relative lowering of vapour pressure[RLVP]
2. Depression of freezing point of solvent
3. Elevation of boiling point of solvent
4. Osmotic pressure of solution

Relative lowering of vapour pressure [RLVP]:-

When a non-volatile solute is dissolved in a solvent, v.p of the solution becomes lower
than that of pure solvent which is known as RLVP
If X1 : mole fraction of solvent and X2 : mole fraction of solute
Then vapour pressure of solution p = X1P [P = v.p of pure solvent]
The reaction in the v.p of solvent (p) = P - P = P - X1P = (1- X1) P
p = X2P = RLVP p/ P = X2 means that RLVP = mole fraction of solute
If h1 and h2 are number of moles of solvent and solute. Then
RLVP = P - P/ P = X2 = h2/ h2+ h1

APNI KAKSHA 13
 For dilute solution h2 <<< = h1
P - P/ P = h2/ h1=(W2/M2) (M1/W1)
Here W1 and W2 are the masses and M1 and M2 are the molar masses of solvent and solute
respectively.

Q. A solution is prepared by dissolving 10 g of non – volatile solute in 200 g of water.

It has a vapour pressure of 31.84 mm of Hg at 308 k. Calculate the molar mass of
solute. [v.p of pure water = 32 mm of Hg]
Ans. W2 = 10 g W1 = 200 g p = 31.84 mm of Hg
M2 = ? M1=18 g (H2O) p= 32 mm of Hg
P - P/ P = h2/ h1=(W2/M2) (W1/M1)
32-31.84/32 = (10/ M2)/(200/18)
M2 = 180 g / mol
Elevation of boiling point :-
As we know B.P of a liquid is a temperature at which its V.P becomes equal to the
atmospheric pressure.
The B.P of a solution is always higher than the B.P of pure solvent in which the solution
is prepared. This is because V.P of solution at a given temperature is found to be lower
than V.P of pure solvent at the same temperature.
In a pure liquid the entire surface is occupied by the molecules of liquid. If a non- volatile
solute is added to pure liquid to give a solution. Then surface has both and solvent
molecules, so the fraction of surface covered by solvent molecules gets reduced. So the
no. of solvent molecules escaping from surface is reduced. Thus the v.p is also reduced.

! mol of solvent
Pure Solvent 1 mol of solute

a Solvent Solute
b

APNI KAKSHA 14
 Boiling point of
Solvent Solution
1.013 bar
or 1 Atm t
lv en
So
n
lu tio
So

vapour pressure

Tb
Tb

Tb

Temperature/k

If Tb is B.P of pure solvent and Tb is the B.P of solution, then elevation in B.P
 T b = T b - T b
For dilute solutions  Tb  m [here m is molality or molal concentration of solute in a
solution]
 Tb = Kb . m [Kb : B.P elevation constant / molal elevation constant]
As unit of m = mol/kg , so unit of Kb = k kg mol-1
If W2 g of solute of molar mass M2 is dissolved in W1 g of solvent , then molality of solution
m = (W2/M2)mole/ (W1/1000)kg
Then,  Tb = Kb 1000 W2/M2.W1
And molar mass of solute M2 = 1000 W2. Kb/ Tb.W1
Q. 18g of glucose (Molar mass = 180 g mol-1) is dissolved in 1 kg of water in a source
pan. At what temperature will this solution boil? [ k b for water = 0.52 k kg mol-1,
B.P of pure water = 373.15 k)
Ans W1 = 1 kg = 1000 g Tb = 373.15
W2 = 18 g Kb= 0.52 k kg mol-1 M2 = 180 g mol-1
 Tb = Kb .m= kb 1000 W2/M2.W1 =0.52×1000×18/180/1000 = 0.052 k
 Tb = Tb - Tb= Tb – 373.15 = 0.052 k
Tb = 373.15 + 0.052 = 373.202 k
Tb = 373.2 k

Q. A solution of glycerol [C3H8O3] in water was prepared by dissolving some glycerol

in 500 g of water. This solution has a B.P of 100.42 C What mass of glycerol was
dissolved to make this solution? [kb for water = 0.512 k kg mol-1]
Ans M2 = 92 g mol-1 W1 = 500 g W2 = ? Kb= 0.512 k kg mol-1
 Tb = Tb - Tb=100.42 C - 100C = 0.42 k
 Tb = kb (W2/M2) × ( 1000/W1) = 0.42 = 0.512(W2/92) × ( 1000/500)
W2 = 37.73 g = mass of glycerol dissolved.

APNI KAKSHA 15
Depression of freezing point :-
The freezing point of a substance is the temperature at which V.P of the substance in its
liquid phase is equal to its V.P in solid phase.
When a non-Volatile solute is added to a solvent , the freezing point of the solution is
always lower that of pure solvent , as v.p of solvent decrease in presence of non-volatile
solute.
This difference in freezing point is known as depression in freezing point.
t
en
solv
id
liqu
ion
lu t
nt

so
l ve
so
n
ze
fro
vapour pressure

T1

Tt Tt

Temperature/k

 T f = T f - T f
Tf= freezing point of pure solvent.
Tf = freezing point of solution
 Tf  m
 Tf = kf m
 Tf = kf ×(W2/M2) × ( W1/1000)
kf = freezing point depression constant
f=molar depression constant
k =cryoscopic constant

Q. Calculate the freezing point of a solution containing 60 g of glucose (molar mass =

180 g mol-1) in 250 g of water. (kf of water =1.06 k kg mol-1)
Ans W2 =60 g M2 = 180 g mol-1 W1 =250 g
Molality of solution = m = (W2/ M2)×1000/W1 =60g/180 g mol-1 ×1000/250 g mol-1
M = 1.33
 Tf = kf m = 1.86 × 1.33 = 2.47 k
Tf - Tf = 2.47 k [ freezing point of pure water = 273.15 k = 0C
Freezing point of solution (Tf) = 273.15 – 2.47 = 270.68 k

Q. Calculate the mass of compound (molar mass = 256 g mol-1) to be dissolved in 75 g

of benzene to lower its freezing point by 0.48 k. (kf =5.12 k kg mol-1)
Ans W1 =75 g  Tf =0.48 k M2 = 256 g mol-1

APNI KAKSHA 16
 kf =5.12 k kg mol-1
 Tf = kf m = kf ×W2 ×1000/ W1 . M2 =0.48=5.12× W2×1000/75×256
W2 =mass of solute =1.8 g
Q. A 10% solution (by Mass) of Sucrose in water has freezing point of 269.15 k.
Calculate freezing point of 10% glucose in water. If freezing point of pure water is
273.15 k. [ Given molar mass of sucrose = 342 g mol-1 , molar mass of glucose = 180
g mol-1]
Ans 10% of solution by mass means 10 g of sucrose dissolved in 90 g of water.
W2 =10 g W1 =90 g M2 = 342 g mol-1
Then molality m =(W2/ M2)×( W1/1000) =(10/342)/(90/1000)=0.324 mol kg-1
 Tf = Tf - Tf =273.15-269.15 = 4 k
 Tf = kf m = kf = Tf/m =4/0.324 [kf is constant (same) for solvent (water)
# for glucose solution :
W2 =10 g W1 =90 g M2 = 180 g mol-1[then molality m =(10/180)/(90/1000)
M=0.617 mol/kg
 Tf = kf m = 4/0.324×0.617 = 7.617 k
Hence freezing point of glucose solution =273.15 – 7.617 =265.53 k

Osmosis and Osmotic pressure:-

The process of flow of solvent molecules from solution of lower
concentration to solution of higher concentration through semipermeable
membrane is known as osmosis
The hydrostatic pressure which develops on account of osmosis is called osmotic
pressure
The pressure which just stops the flow of solvent is called Osmotic pressure of solution.
Osmotic pressure (π)  Molarity (c)
π =CRT =h2/v RT
Here, V=volume of solution (in L)
h2= moles of solute (=W2/M2)
π = (W2/ M2)/V× RT
M2 = W2RT/πv = osmotic pressure is used to determine molar masses of macromolecules
like proteins, polymers.

APNI KAKSHA 17
 Patm p
atm

SPM

solution solvent

Q. A solution prepared by dissolving 8.95 mg of a gene fragment in 35 ml of water has

an osmotic pressure of 0.335 torr at 25C. Assuming the gene fragment is non-
electrolyte, determine its molar mass.
Ans W2 =8.95 mg =8.95×10-3 g v= 35 ml = 35/1000 L
π = 0.335 torr = 0.335 / 760 atm T = 273+25 =298 k
π =CRT =( W2/ M2)/v×RT
then M2= W2RT/ πv =8.95×10-3(0.0882)(298)/(0.335/760)(35/1000)
M2 = 1.417 ×104 g mol-1
Q. At 300 k , 36 g of glucose (molar mass =180 g mol-1) present per litre in its solution
has an osmotic pressure of 4.98 bar. If the osmotic pressure of another glucose is
1.52 bar at the same temperature. Calculate the concentration of other solution?
Ans π =CRT = W2RT/ M2v
For first solution : 4.98 bar = (36 g)KT/(180 g mol -1).V…(i)
For second solution : 1.52 bar = W2RT/ M2v….(ii)
Dividing (ii) by (i):-
1.52/4.98 = W2RT/ M2v × 180 v/36 RT
W2/ M2 =0.061 mol
Q. Define the following term?
1. Isotonic solutions:-
Two solutions having some osmotic pressure at a given temperature are called isotonic
solutions
2. Hypertonic solutions:-
A solution having lower osmotic pressure than other solution is called hypotonic while a
solution having higher osmotic pressure is called hypertonic.

Example:- The osmotic pressure associated with the fluid inside the blood cell is
equivalent to that of 0.9% (m/v) NaCl solution, called normal saline solution and it is safe
to inject intravenously.

APNI KAKSHA 18
 If we place the cells in a solution containing more than 0.9% (m/v) NaCl , water will flow
out of the cells and they would shrink. Such a solution is called hypertonic
If the salt concentration is less than 0.9 % (m/v) the solution is said to be hypotonic in
this case, water will flow the cells if placed in this solution and they would swell.

Reverse Osmosis :-
If a pressure larger than the osmotic pressure is applied to the solution side,
then the pure solvent flows out of solution through semipermeable
membrane. This phenomenon is called reverse osmosis.
Reverse osmosis is used in desalination of sea water.

pressure>II Piston

Fresh water Salt water

Water out let

SPM

Reverse osmosis occurs when a pressure larger than the osmotic pressure is applied to the
solution.

Important PYQ’s
Q. The molecular masses of polymers are determined by osmotic pressure method
and not by measuring other colligative properties. Give 2 reason?
Ans The osmotic pressure method has the advantage over other colligative properties
because
1. Pressure measurement is around the room temperature and the molarity of the
solution is used instead of molality.
2. Its magnitude is large as compared to other colligative properties even for very dilute
solutions.

Q. Outer hard shells of two eggs are removed. One of the egg placed in pure water
and the other egg is placed in saturated solution of NaCl. What will be observed
and Why?
Ans The egg placed in water swell because the concentration of proteins is high inside the
egg as compared to water. Therefore water diffuses through semipermeable membrane
of egg and egg swells.

APNI KAKSHA 19
 The egg placed in NaCl solution will shrink due to osmosis water will move out of the egg
membrane thereby shrinking the egg.

Q. A 5% solution (by mass) of cane sugar (mw=342 g mol-1) is isotonic with 0.8774
solution of substance x. Find the molecular weight of x.
Ans W cone sugar = 5 g
Wx =0.877 g
M cane sugar = 342 g mol-1
Mx=?
Isotonic means same osmotic pressure
π cane sugar = πx
C cane sugar .RT = Cx .RT
(W cone sugar/ M cane sugar)/V =(Wx / Mx)/v
(5/342) = 0.877/ Mx
Then Mx = 59.9 g mol-1

Q. Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and Why?
Ans 2M glucose has higher B.P because more the concentration, more is the elevation in
boiling point.

Abnormal Molar Mass:-

For the substances undergoing association or dissociation in solution,
molecular mass determined from colligative properties is different [either
lower or higher ] from expected value.
This experimental determined molar mass is known as abnormal molar
mass.

Van’t Hoff factor :-

It is the ratio of the ratio of the experimental value of colligative property to
the calculated value of the colligative property.
Van’t Hoff factor [i] = Normal molar mass/Abnormal molar mass
= Observed / Experimental colligative property
Calculated colligative property
= total no. of moles of particles after association / dissociation

APNI KAKSHA 20
 Number of moles of particles before association /dissociation
For example :-
1. KCl k+ + Cl- i=2
2. MgSO4 Mg2+ + SO42- i=2 for dissociation i>1
3. K2SO4 2k+ + SO42- i=3
4. CaCl2 Ca+2 + 2 Cl- i=3
5. 2C6H5COOH (C6H5COO)2 i=1/2 [for association i<1]
Dissociation:-
Let the degree fo dissociation is  for Ax BY
Ax BY xAy+ + y B x-
T=0 : C 0 0
Teq : C(1-) xC yC
Total concentration at equilibrium = C(1-) + xC + yC =C [1+(n-1) ]
No. of particles after diss. of one molecule of electrolyte = n= x+4
i= C [1+(n-1) ]/C
then, i=1+(n-1) 

Association:-
hA An
t=0 C 0
teq C(1-) C/n
 = degree of association for dimerization n=2
i= C(1-)+C/n
C
i =1+(1/n-1) 

Note :- Modification in equations for colligative properties.

1. RLVP of solvent P-P/ P =i h2/h1
2. Elevation of B.P Tb = ikb.m
3. Depression of freezing point =Tf = ikf.m
4. Osmotic pressure of solution π = i h2/v ×RT

Q. A 1 molal aq. Solution of trichloroacetic acid is heated to its B.P. the solution has
B.P 100.10 C. Determine Van’t Hoff factor for CCl3COOH. Kb (Water) = 0.512 k kg
mol-1

APNI KAKSHA 21
Ans m=1, Kb=0.512 k kg mol-1
Tb = 100.18 C
 Tb = Tb - Tb =100.18 -100 =0.18C
Tb (water) =100C
 Tb =iKb.m =0.18 k = i×0.512 ×1 then I = 0.35

Q. Calculate the freezing point of an aqueous solution containing 10.5 g of magnesium

bromide in 200 g of water assuming complete dissociation of MgBr2.
Molar mass of MgBr2 =184 g mol-1 , kf (water) = 1.86 k kg mol-1
Ans W2 =10.50 g W1 =200 g, M2(MgBr2)= 184 g mol-1 kf =1.86 k kg mol-1
MgBr2 Mg2+ 2Br- i=3
 Tf = iKf.m =ikf [W2 ×1000/M2×W1] =3×1.86 [10.5×1000/104×200] = 1.592 k
Freezing point of solution = Tf - Tf =273 k – 1.592 k = 273.408 k

Q. Phenol association in benzene to a certain extent to form a dimer. A solution

containing 20 g of phenol in 1 kg of benzene has its freezing point by 0.69 k.
Calculate the fraction of phenol that has dimerised. kf (benzene) = 5.1 k kg mol-1
Ans for dimer on association h=2 W2 =20 g W1 =1kg = 1000 g
 Tf =0.69 k , kf =5.1 k kg mol-1
M2 =(5.1) 20×1000/(0.69)×1000 =147.82 g mol-1 [experimental molar mass]
Calculated molar mass of phenol [C6H5OH]2 degree of association  Then
1 
i  1    1 
n 
i  1 0.635  1
   0.73    73%
1 1
1 1
n 2

Q. An aq. Solution containing 12.48 g of BaCl2 in 1 kg of water at 373.0832 k. Calculate

the degree of dissociation of BaCl2.
[Given kb (H2O) = 0.52 k kg mol-1 , molar mass of BaCl2. =208.34 g mol-1]
Ans W2 =12.48 g , W1 = 1000 g , Tb (solution ) = 373.0832 k
kb (H2O) = 0.52 k kg mol-1, M2 =208.34 g mol-1
Tb = kb W2×1000/M2×W1
Tb=Tb - Tb = 373.0832 -373 =0.0832 k
kbW2  1000 0.52  12.48  1000
M2    78gmol 1 =experimental molar mass
Tb  W1 0.0832  1000

APNI KAKSHA 22
I = M2( calculated)/ M2(experimental) = 208.34/78 = 2.67
For BaCl2 Ba+2 + 2Cl-
I =1+(n-1) d
N= 3 degree of dissociation  =i-1/n-1 =2.67-1/3-1 =0.835
 =83.5%

APNI KAKSHA 23