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 mol%`
B. `34 mol%`
C. `48mol%`
D. `50 mol%`

Answer

D

Explanation

At boiling point `P_(sys) = P_(sorr')`
`760 = 520 xx x_(A) + 1000 (1-x_(A))`
`x_(A) = (240)/(480) = 0.50` or `50%`