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

Explanation

`P_(T)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)implies760=520x_(A)+P_(B)^(@)(1-X_(A))" "impliesX_(A)=0.5`
Thus mole % of `A=50%`