Chemistry
Solution - II
Session objectives

Colligative properties





Relative lowering of vapour pressure
Elevation in boiling point
Depression in freezing point
Osmosis and Osmotic pressure
Abnormal molecular mass
Colligative Properties
Properties which depend only on the
number of particles of solute and donot
depend upon the nature of solute particles,
(molecules or ions).
•
Relative lowering of vapour pressure.
•
Elevation in boiling point
•
Depression in freezing point
•
Osmotic Pressure
Relative lowering of vapour pressure
From Raoult’s law,
ps  xsolvent po
ps
o
p
 1  xsolute
xsolute  1 

po  ps
po
ps
po
 Relative lowering of v.p,
po  ps
po  ps
p
p
o

n
nN
o
n
N
 xsolute
moles of solute
moles of solvent
Relative lowering of vapour pressure
We can write
po  ps
o
p

n
N
when n  10%
For calculation purpose
po
po  ps
po
po  ps
ps
po  ps

nN
n
1 

N
n
nN
1
n
po  ps n

ps
N
Illustrative Example
Calculate the relative lowering of vapour
pressure of a solution of 6g of urea in 90 g
of water.
Solution:
Re lative lowering of vapour pressure,
po  ps

6 / 60
n

N  n 90 /18  6 / 60
po  ps

1 10
1


 0.02
10 51 51
po
po
Illustrative Example
The vapour pressure of pure water at 0°C is 4.579 mm of Hg.
A solution of lactose containing 8.45g of lactose in 100 g of
water, has a vapour pressure of 4.559 mm at the same
temperature. Calculate molecular mass of lactose.
Solution :
Let the molar mass of lactose  M
po  ps n


ps
N
8.45
4.579  4.559
 M
100
4.559
18
0.02
8.45  18

4.559
M  100
M 
8.45  18  4.559
0.02  100
 349 gmol1
Illustrative Problem
The vapour pressure of pure A is 10 Torr and at the
same temperature when 1g B is dissolved in 20 g of
A, its vapour pressure is reduced to 9 Torr. If
molecular mass of A is 200 amu, calculate the
molecular mass of B.
Solution
Po  Ps nB

Ps
NA
1
M
10 - 9
= B
20
9
200
1 10
=
9 MB
[ MB = 90 amu]
Elevation in boiling point
The boiling point of the solution (TbK) is higher
than that of pure solvent (T0 K).
Tb  Tb  T0
 RT02M1

Tb  
m
 1000H 
vap 

Kb 
RT02M
103 Hv
B
Tb  Kbm

RT02
103lv
Atmospheric
pressure E
F
A
C
Tb
Hv  Enthalpy of vaporisation
lv  Latent heat of vaporisation
1000 K b W2
M2 
W1Tb
D
T0
Temperature
Tb
Illustrative Example
A solution of 3.8 g of sulphur in 100g of CS2
(Boiling point = 46.30°C) boils at 46.66°C. What is
the formula of sulphur molecule in this solution?
[ Atomic mass of sulphur = 32 g mol –1 Kb
for CS2 = 2.40 K kg mol –1 ]
Solution
Tb  46.66C  46.30C = 0.36°C
MB 
wB  1000  K b
Tb  w1
=
3.8  1000  2.40
0.36  100
MB = 253 g mol–1
Atomicity =
MB
253
=8

gram atomic weight
32
Hence,Sulphur exists as S8 molecule.
Depression in freezing point
The freezing point of a solution (Tf ) is
less than that of pure solvent (T0).
Tf  T0  Tf
Tf  K f m
A


Tf  
m
1000


H

f

RT02M1
Kf 
RT02M
3
10 Hf

B
RT02
1000 K f W2
M2 
W1Tf
E
3
10 lf
D
Tf
C
T
Temperature
T0
Osmosis
Movement of solvent molecules through a semi
permeable membrane from a less concentrated
solution to a more concentrated solution..
Semi permeable membrane
Allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic Pressure
The excess hydrostatic pressure on
the solution which prevent the
movement of solvent molecules
through semipermeable membrane.
p  CRT=
p=
C=
T=
R=
n
RT
V
Osmotic pressure
Molar conc. (mol/L)
Temperature (K)
Solution constant
Isotonic solutions :
Equimolar solutions having the same
osmotic pressure.
Reverse Osmosis
If the pressure higher than the osmotic pressure is applied on
the solution, the solvent will flow from the solution into the
pure solvent, through the semipermeable membrane
 Used for water purification.
Illustrative Example
A solution containing 8.6 g/l of urea (molecular was
- 60 g mol-1) was found to be isotonic with a 5%
solution of an organic non-volatile solute at 298 K.
What is the molecular mass of the organic solute?
Solution :
Let osmotic pressure of urea=p1
osmotic pressure of unknown compound=p2
 p1  p2 (isotomic)
8.6
5
RT  RT
60
M
M 
60  50
 348.9
8.6
Illustrative Example
What is the relationship between osmotic pressures of
10 grams of glucose(p1),10 grams of urea(p2) and 10
gram of sucrose(p3) which are dissolved in 250 ml of
water respectively at 273 K.
Solution :
Moles of glucose(n1)=10/180 =0.05
Moles of urea(n2)=10/60=0.16
Moles of sucrose(n3)=10/342=0.02
More the number of moles of solute,
higher will be the osmotic pressure.
n2  n1  n3
 p2  p1  p3
Abnormal molecular mass
(1) When the solution is non-ideal i.e. the solution is not dilute.
(2) When the solute undergoes association/dissociation .
van’t Hoff suggested correction factor (i)
Where,
observed colligative property
i
Theroretical colligative property
Theroretical molar mass
i
observed molar mass
po  ps
o
p
 ixsolute
Tb  iKbm
Tf  iK fm
p  icRT
Dissociation of molecule
A
A1  A2  A3  ........(n  moles)
Ini.
C
At eqm. C(1  )
0
Cn
nc  C  C  Cn  C  C(n  1)
i

no. of moles at eqm.
Initial no. of moles
C  C(n  1)
 1  (n  1)
C
 i  1  (n  1)
n  no. of moles obtained on dissociation
Association of molecules
nA
Ini.
An
C
0
At eqm. C(1  `)
C`
n
n  no. of moles associated
` degree of association
i
ne

ni
C(1  `) 
C
C`
n
1

 1  `  1
n

Note : 
i  1 for dissociation
i  1 for association
Illustrative Example
A 0.2 molal aqueous solution of a weak acid (HX) is
20% ionised. What is the depression in freezing point
of the solution (kf for H2O=1.86 K kg mol-1).
Solution:
HX
1
H++ X-


Total no. of moles = 1  
Tf  iK f m
 (1  )  1.86  0.2
 1.2  1.86  0.2
 0.45
(  0.2)
Illustrative Example
A solution of x grams of urea in 500 grams of water is
cooled to -0.5°C .128 grams of ice separated from the
solution. Calculate x, given Kf = 1.86 deg/molal.
Solution :
Tf  i Kf m
i  1 since urea does not dissociate
Kf  1.86
Mass of liquid water  500  128  372
x  103
0.5  1.86 
60  (500  128)
 6g
x 
0.5  60  372
1.86  103
Illustrative Example
Among equimolal aqueous solution of C6H5NH2Cl,
Ca (NO3)2, La(NO3)2, glucose which will have the
highest freezing point?
Solution
Glucose does not ionise. So the number of particles
furnished by glucose in solution will be least
compared to other options.
Hence, the depression in freezing point is minimum.
In other words, the freezing point will be highest for
glucose.
Illustrative Example
How many grams of NaBr must be added to 270
g of water to lower the vapour pressure by 3.125
mmHg at temperature at which vapour pressure
of water is 50 mmHg (Na = 23, Br = 80) ?
Assume 100 % ionisation of NaBr.
Solution
ΔP
Psolvent
= Xsolute =
n1i
n1i + n2
2n1
3.125
=
50
2n1 +15
2n1 +15
50
=
= 16
2n1
3.125
n1 
1
2
NaBr taken = 0.5 mol = 0.5 x 103 = 51.5 g
Illustrative Example
An aqueous solution contains 5% by mass of
urea and 10% by mass of glucose. What will
be its freezing point?
[Kf for H2O is 1.86 K mol–1 kg]
Solution
Five per cent by mass of urea contains
5 103
=
×
= 0.833 moles in 103 g
60 100
10 per cent by mass of glucose contains
10 103
=
×
= 0.556 moles in 103g
180 100
 Total moles of solute = 1.389
 ΔTf =Kfm=1.86×1.389
= 2.58
 Freezing point of solution = 0 – 2.58
= –2.58° C or 270.42 K
Thank you