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

Answer

D

Explanation

`p_(T)=p_(A)^(@)x_(A)+p_(B)^(@)x_(B)`
"Mixture solution boils at 1 atm =760mm=total pressure."
`760=520x_(A)+100(1-x_(A))`
`x_(A)=0.5,mol%ofA=50%`