At `80^(@)"C"`, the vapour pressure of pure liquid 'A' is 250 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 of Hg)"`

Options

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

Answer

D

Explanation

`P_("solution")=P_("A")^(0)"x"_("A")^(0)+P_("B")^(0)(1-x_(A))`
`760=520xxx_("A")+1000(1-x_(A))`
`480x_(A)=1000-760`
`x_(A)=0.5(50%"Mole)`