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 precent
B. 53 mol percent
C. 34 mol percent
D. 48 mo percent

Answer

A

Explanation

`p_("total") = p_(A)^(@)x_(A) + p_(B)^(@) x_(B)`
`760=520 x_(A) + 1000(1-x_(A))`
`760 = 520 x_(A) + 1000 - 1000 x_(A)`
`480 x_(A) = 240`
`480 x_(A) = 240`
`x_(A) = 0.5 ` or Moles of A = 50%