Two liquids A and B have vapour pressure in the ratio PAo  :  PBo=1:2 at a certain temperature. Suppose that we have an ideal solution of A and B in the mole fraction ratio of A : B = 1 : 2, the mole fraction of A in the vapour phase in equilibrium with the solution at a given temperature is?

To determine the mole fraction of liquid A in the vapour phase, we apply Raoult’s Law and Dalton’s Law of Partial Pressures.

Given Data:
Vapour pressure ratio:
PAo : PBo = 1 : 2
Let PAo = P and PBo = 2P.

Mole fraction in liquid phase:
XA : XB = 1 : 2
Since the total mole fraction must sum to 1,
XA = 1/3 and XB = 2/3.

Step 1: Calculate Partial Pressures in Solution
Using Raoult’s Law, the partial pressure of each component is:

PA = XA * PAo = (1/3) * P = P/3
PB = XB * PBo = (2/3) * (2P) = 4P/3

Total vapour pressure, Ptotal = PA + PB = (P/3) + (4P/3) = 5P/3.

Step 2: Mole Fraction in Vapour Phase
Using Dalton’s Law, the mole fraction of A in the vapour phase is:

YA = PA / Ptotal
= (P/3) / (5P/3)
= 1/5.

Final Answer:
The mole fraction of A in the vapour phase is 1/5 or 0.2.