Question

An aqueous solution containing 28% by weight of liquid A (mol. wt. 140) has a vapour pressure of 160 mm at 37ºC. Find the vapour pressure of the pure liquid A (the vapour pressure of water at 37ºC is 150 mm).

(A) 3.583 mm 

(B) 35.83 mm 

(C) 358.3 mm 

(D) None

Answer 1

Correct Option (C) 358.3 mm

Explanation:

 We know that for a mixture of two miscible liquids, P_total = P^o_A .x_A + P^o_B .x_B where x_A and x_B are mole fractions of liquids A and B, and P^o_A and P^oB are their vapour pressure in pure state.

Moles of liquid ‘A’ = 28/140 = 0.2

Moles of water (liquid B) = 72/18 = 4

Total number of moles in the solution = 4 + 0.2 = 4.2

Moles fraction of A = 0.2/4.2 - 0.048

Mole fraction of water (B) = 1 – 0.048 = 0.952

P_total = Vapour pressure of the solution = 160 mm(given)