Question

Two liquids a and b are mixed in 1:4 mole ratio to form an ideal solution. if pure liquid a and b exert a vapour pressure of 75 mmhg resp, the vapour pressure of liquid a in vapour phase would be?

Answer 1

if both a and b exert 75mmHg of pressure
then pressure exerted by a in vapour phase would be 1/5(75)
= 15mmHg

Answer 2

Hey dear,
There is data missing in question. I added accurate question in image.

● Answer -
pA = 34.5 mm Hg

● Explaination -
From given data,
Mole fraction of A, xA = 1/(1+4) = 0.2
Mole fraction of B, xB = 4/(1+4) = 0.8

According to Rault's law -
Ptotal = pA0.xA + pB0.xB
Ptotal = 75×0.2 + 22×0.8
Ptotal = 32.6 mm Hg

Mole fraction of liquid in vapour phase -
xA = pA / Ptotal
xA = 15 / 32.6
xA = 0.46

Vapour pressure of liquid A in vapor phase is -
pA = 0.46 × 75
pA = 34.5 mm Hg

Therefore, vapour pressure of liquid A is 34.5 mm Hg.

Hope this helps you...