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Chemistry
Solution - II
Colligative properties
Relative lowering of vapour pressure Elevation in boiling point Depression in freezing point Osmosis and Osmotic pressure
Abnormal molecular mass
Session objectives
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
Colligative Properties
Relative lowering of vapour pressure
From Raoult’s law,
os solvent
ssoluteo
p x p
p1 x
p
ssolute o
os
o
px 1
p
p p
p
os
soluteo
Relative lowering of v.p,
p px
p
os
o
p p nn Np
n moles of soluteN moles of solvent
We can write
os
o
p p nNp
For calculation purpose
o
os
p n Nnp p
so
s
p Nnp p
os
s
p p np N
Relative lowering of vapour pressure
when n 10%
o
os
p n N1 1
np p
Illustrative Example
os
o
Relative lowering of vapour pressure,
p p 6 / 60nN n 90 / 18 6 / 60p
os
o
p p 1 10 10.02
10 51 51p
Calculate the relative lowering of vapour pressure of a solution of 6g of urea in 90 g of water.
Solution:
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.
Let the molar mass ofl actose M
os
s
p p np N
8.454.579 4.559 M
1004.55918
0.02 8.45 184.559 M 100
1
8.45 18 4.559M
0.02 100
349gmol
Solution :
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
os B
s A
P P n
P N
B
1M10- 9
=209
200
B
1 10=
9 M[ MB = 90 amu]
The boiling point of the solution (TbK) is higher than that of pure solvent (T0
K).
b b 0 b bT T T T K m
b 22
1 b
1000 K WM
W T
Elevation in boiling point
20 1
bvap
RT MT m
1000 H
2 20 0
b 3 3v v
RT M RTK
10 H 10 l
vH Enthalpy of vaporisation
vl Latent heat of vaporisation T0Tb
A
B
Solv
ent
Vapour
pre
ssu
re
Temperature
Solution
D
C
EF
bT
Atmospheric pressure
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
bT 46.66 C 46.30 C = 0.36°C
MB = 253 g mol–1
B b
Bb 1
w 1000 KM
T w3.8 1000 2.40
0.36 100
=
BM 253
gram atomic weight 32Atomicity = = 8
Hence,Sulphur exists as S8 molecule.
The freezing point of a solution (Tf) is less than that of pure solvent (T0).
f 0 f f fT T T T K m
f 22
1 f
1000 K WM
W T
Depression in freezing point
20 1
ff
RT MT m
1000 H
2 20 0
f 3 3ff
RT M RTK
10 H 10 l
Tf
T T0
A
BLiq
uid
solv
ent
E
D
Solution
CSolid
solve
nt
Vapour
pre
ssu
re
Temperature
OsmosisOsmosis
Semi permeable membraneAllows the passage of solvent molecules but blocks the passage of solute molecules.
Movement of solvent molecules through a semi permeable membrane from a less concentrated solution to a more concentrated solution..
Equimolar solutions having the same osmotic pressure.
Osmotic Pressure
Isotonic solutions :
= Osmotic pressureC = Molar conc. (mol/L)T = Temperature (K)R = Solution constant
nCRT= RT
V
The excess hydrostatic pressure on the solution which prevent the movement of solvent molecules through semipermeable membrane.
Reverse Osmosis
Used for water purification.Used for water purification.
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
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?
Illustrative Example
1Let osmotic pressure of urea=
2osmotic pressure of unknown compound=
1 2 (isotomic)
8.6 5RT RT
60 M
60 50M 348.9
8.6
Solution :
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.
Illustrative Example
Moles of glucose(n1)=10/180 =0.05Moles of urea(n2)=10/60=0.16Moles of sucrose(n3)=10/342=0.02
More the number of moles of solute, higher will be the osmotic pressure.
n n n
p p p
2 1 3
2 1 3
Solution :
Abnormal molecular mass
(1) When the solution is non-ideal i.e. the solution is not dilute.
(2) When the solute undergoes association/dissociation .
Where,
observed colligative propertyi
Theroretical colligative property
Theroretical molar massi
observed molar mass
b b
ff
T iK m
T iK m
icRT
os
soluteo
p pix
p
van’t Hoff suggested correction factor (i)
Dissociation of molecule
1 2 3A A A A ........(n moles)
Ini. C 0
At eqm. C(1 ) Cn
cn C C Cn C C (n 1)
no. of moles at eqm.i
Initial no. of moles
C C (n 1)1 (n 1)
C
i 1 (n 1)
n no. of moles obtained on dissociation
nnA A
Ini. C 0
C `At eqm. C(1 `)
n
n no. of moles associated
` degree of association
e
i
C `C(1 `)n ni
n C
Association of molecules
11 ` 1
n
Note :
i 1 for dissociation
i 1 for association
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).
Illustrative Example
Solution:
HX H++ X-
1
Total no. of moles = 1
ffT iK m
(1 ) 1.86 0.2
1.2 1.86 0.2 ( 0.2)
0.45
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.
Illustrative Example
Solution :
ffT i K m
i 1 since urea does not dissociate
fK 1.86
3x 100.5 1.86
60 (500 128)
Mass ofl iquid water 500 128 372
3
0.5 60 372x
1.86 10
6g
Among equimolal aqueous solution of C6H5NH2Cl, Ca (NO3)2, La(NO3)2, glucose which will have the highest freezing point?
Illustrative Example
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.
Solution
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
1solute
solvent 1 2
n iΔP= X =
P n i+n
1
1
2n3.125=
50 2n +15
1
1
2n +15 50= =16
2n 3.125
1
1n
2
NaBr taken = 0.5 mol = 0.5 x 103 = 51.5 g
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]
Illustrative Example
Solution
Five per cent by mass of urea contains3
35 10= × =0.833 moles in 10 g
60 100
10 per cent by mass of glucose contains
3310 10
= × =0.556 moles in 10 g180 100
Total moles of solute = 1.389
ffΔT =K m=1.86×1.389
= 2.58
Freezing point of solution = 0 – 2.58 = –2.58° C or 270.42 K
Thank you
