Two liquids `A` and `B` have `P_(A)^(@)` and `P_(B)^(@)` in the ratio of `1:3` and the ratio of number of moles of `A` and `B` in liquid phase are `1:3` then mole fraction of `A` in vapour phase in equilibrium with the solution is equal to :

Options

A. 0.1
B. 0.2
C. 0.5
D. `1.0`

Answer

A

Explanation

`X_(A)=(1)/(4),X_(B)=(3)/(4)`
`P_("total")=P_(A)^(@)X_(A)+P_(B)^(o)X_(B)`
` =1xx(1)/(4)+3xx(3)/(4)`
` =(1)/(4)+(9)/(4)=(10)/(4)=(5)/(2)`
`Y_(A)=(P_(A)^(@)X_(A))/(P_("total"))=(4)/((5)/(2))=(2)/(20)=(1)/(10)=0.1`