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

Answer

A

Explanation

`p=p_(A)^(0)x_(A)+p_(B)^(0)`
`760=520xxx_(A)+1000(1-x_(A))`
`x_(A)=0.50`, i.e., mole per cent.