Mole fraction question

Question

Consider a solution containing liquids A and B where the mole fraction of B is 0.60. Assuming ideality, calculate the mole fraction of B in the vapour at equilibrium with this solution at 25oC. (The vapour pressures of pure liquid A and B at 25oC are 200 torr and 400 torr, respectively).

Answer 1

Vapour pressure of pure A = 200 torr
Vapour pressure of pure B = 400 torr

In the liquid mixture :
Mole fraction of A = 1-0.6 = 0.4
Mole fraction of B = 0.6

In the vapour of the mixture :
(According to Roault's law :)
Partial pressure of A = 0.4 x 200 = 80 torr
Partial pressure of B = 0.6 x 400 = 240 torr
(According to Dalton's law of partial pressure:)
Total pressure = 80 + 240 = 320 torr
Mole fraction of B in the vapour = 240/320 = 0.75

What syllabus are you studying ?
The above topic has been out of the HKAL Chemistry syllabus since this year.