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

Answer

A

Explanation

We have , `P _(A) ^(@) = 520` mm Hg
`and P _(B ) ^(@)= 1000` mm Hg
Let mole fraction of A in solution ` = x _(A)`
and mole fraction of B in solution `= X _(B)`
Then, at 1 atm pressure i.e., at 760 mm Hg,
`P _(A) ^(@) X _(A) + P _(B) ^(@) X _(B) = 760 mm Hg `
`P _(A) ^(@) X _(A) + P _(B) ^(@) ( 1-x _(A)) = 760` mm Hg
`implies 520 x _(A) + 100 0 - 1000 X _(A)`
` = 760 mm Hg `
`implies X _(A) = (1)/(2) or 50 ` mol % .