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 solution of 'A' and 'B' boils at `80^(@)C` and 1 atm pressure, the amount of 'A' in the mixture is (1 atm `= 760 mm Hg)`

Options

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

Answer

C

Explanation

`P_(A)^(@)=520 p_(B)^(@)=1000`
`p_(T)=p_(A)^(@)x_(A)+p_(B)^(@)x_(B)`
`760=520 x_(A)+1000(1-x_(A))`
`760=520 x_(A)[phone] x_(A)`
`480 x_(A)=240, x_(A)=(1)/(2)`.