A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol-1) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is 10-2. (nearest integer) [Given: Kb of the solvent = 5.0 K kg mol-1] Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Answer
3
Solution
Tb = Kb m
m = 0.55.0 = 0.1 mol kg-1
Moles of X, nX = 0.1 1501000 = 0.015 mol
Moles of Y, nY = 150300 = 0.5 mol
Relative lowering of vapour pressure = p - pp = xX
For a dilute solution, xX nXnY = 0.0150.5 = 0.03 = 3 10-2
(Exact: 0.0150.515 = 2.9 10-2 3 10-2)
Answer: 3