At `80^@C`, the vapour pressure of pure liquid `A` is `520 mm` Hg and that of pure liquid `B` is `1000 mm Hg`. If a mixture of solution `A` and `B` boils at `80@C` and `1 atm` pressure, the amount of `A` in the mixture is `(1 atm =760 mm Hg)`
a. `50 mol %` , b.`52 mol %` ,c.`34 mol%` ,d.`48 mol %`

Options

A. `52`mol percent
B. `34`mol percent
C. `48`mol percent
D. `50`mol percent

Answer

D

Explanation

`P_(m)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)`
`760=520X_(A)+P_(B)^(@)(1-X_(A))`
`:.X_(A)=0.5=50%`