1. Chemistry

2. Solution - II

3. Session objectives Colligative properties
Relative lowering of vapour pressure
Elevation in boiling point
Depression in freezing point
Osmosis and Osmotic pressure
Abnormal molecular mass

4. 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

5. Relative lowering of vapour pressure
From Raoult’s law,
n moles of solute N moles of solvent

6. Relative lowering of vapour pressure
We can write

7. Illustrative Example
Calculate the relative lowering of vapour pressure of a solution of 6g of urea in 90 g of water.
Solution:

8. Illustrative Example
The vapour pressure of pure water at 0°C is mm of Hg. A solution of lactose containing 8.45g of lactose in 100 g of water, has a vapour pressure of mm at the same temperature. Calculate molecular mass of lactose.
Solution :

9. 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.

10. Solution
[\ MB = 90 amu]

11. Elevation in boiling point
The boiling point of the solution (TbK) is higher than that of pure solvent (T0 K). A B D C
Atmospheric pressure
Vapour pressure E F
Solvent
Solution T0 Tb
Temperature

12. 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 ]

13. Solution = 0.36°C = MB = 253 g mol–1 Atomicity = = 8
Hence,Sulphur exists as S8 molecule.

14. Depression in freezing point
The freezing point of a solution (Tf) is less than that of pure solvent (T0).
Vapour pressure A B
Liquid solvent E D
Solution C
Solid solvent Tf T T0
Temperature

15. 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.

16. Osmotic Pressure
The excess hydrostatic pressure on the solution which prevent the movement of solvent molecules through semipermeable membrane.
p = Osmotic pressure
C = Molar conc. (mol/L)
T = Temperature (K)
R = Solution constant
Isotonic solutions :
Equimolar solutions having the same osmotic pressure.

17. 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.

18. Illustrative Example
A solution containing 8.6 g/l of urea (molecular was 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 :

19. 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.

20. Abnormal molecular mass
When the solution is non-ideal i.e. the solution is not dilute.
When the solute undergoes association/dissociation .
van’t Hoff suggested correction factor (i)

21. Dissociation of molecule

22. Association of molecules

23. Illustrative Example HX H++ X-
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 H++ X-
Total no. of moles =

24. 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 :

25. 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.

26. Illustrative Example
How many grams of NaBr must be added to 270 g of water to lower the vapour pressure by mmHg at temperature at which vapour pressure of water is 50 mmHg (Na = 23, Br = 80) ? Assume 100 % ionisation of NaBr.

27. Solution
NaBr taken = 0.5 mol = 0.5 x 103 = 51.5 g

28. 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]

29. Solution Five per cent by mass of urea contains
10 per cent by mass of glucose contains
\ Total moles of solute = 1.389
= 2.58
\ Freezing point of solution = 0 – = –2.58° C or K

30. Thank you