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 boil at `80^(@)C` and 1 atm pressure, than the amount of A in the mixture is (! Atm `= 760 mm Hg)`

Options

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

Answer

B

Explanation

`p_("solution")=p_(A)^(@)x_(A)+p_(B)^(@)x_(B)`
`760=520xx x_(A)+1000(1-x_(A))`
`480xxx_(A)=240`
`x_(A)=240//48=0.5`
`:.` mole fraction of A in solution = 0.5 or amount of A in solution = 50 mol percent.