20. At 300 K two pure liquids A and B have vapour pressure respectively 150 mm Hg. In an equimolar liquid mixture of A and B, the mole fraction of B in the vapour mixture at this temperature is
a. 0.6
b. 0.5
c. 0.8
d. 0.4
Ans. d
Explanation : In equimolar liquid mixture
x = 0.5
x = 0.5
So, P = 0.5 x 150 + 0.5x 100 = 125
Now let y be the mole fraction of vapour B then
y = xB p°B /P = 0.5 x 100 /125 = 0.4
(19)