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. 48 mol precent
B. 52 mol precent
C. 50 mol precent
D. 34 mol precent

Answer

3

Explanation

According to Raoult's law
`P_(T)=P_(A)^(0) chi_(B)`
At the boiling point, total vapor pressure of solution becomes equal to 1 atm pressure (760 mmHg=(520 mmHg) `chi_(A)+(1000 mmHg) chi_(B)`
`=(520 mmHg) chi_(A)+(1000 mmHg)(1-chi_(A))`
Solving for `chi_(A)`, we get
`chi_(A)=0.5`
Thus, mol % of `A=0.5xx100%`
`=50%`