At 300 K two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg. In a equimolar liquid mixture of A and B, the mole fraction of B in the vapour phase above the solution at this temperature is:

Options

A. `0.6`
B. `0.5`
C. `0.8`
D. `0.4`

Answer

D

Explanation

In equimolar liquid mixture
`x_(A)=0.5`
`x_(B)=0.5`
So, `P=0.5xx150+0.5xx100=125`
Now let `y_(B)` be the mole fraction of vapour B then
`y_(B)=(x_(B)p_(B)^(o))/(P)=(0.5xx100)/(125)=0.4`.