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 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. 50 mol per cent
B. 52 mol per cent
C. 34 mol per cent
D. 48 mol per cent

Answer

A

Explanation

`p_("total")=p_(A)^(@)x_(A) + p_(B)^(@)x_(B)`
`760=520x_(A)+1000(1-x_(A))`
`760=520 x_(A)[phone]x_(A)`
`480x_(A)=240`
`x_(A)=0.5`
`:.` moles of A `=50%`