Question

At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mm Hg, respectively. If in a mixture of a and B, the vapoure is 300 mm Hg, the mole fractions of a in the vapour phase, respectively, are -
A. `0.6`
B. `0.5`
C. `0.8`
D. `0.4`

Answer 1

Correct Answer - D
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)^(@))/(P)=(0.5xx100)/(125)=0.4`.