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 = x_B p°_{B /}P = 0.5 x 100 /125 = 0.4

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