The ratio of the vapour pressure of two miscible liquids A and B in pure state is 1 : 3 at a certain temperature. `n_(A)` moles of A and no moles of B are mixed to form an ideal solution. If the ratio of moles of A and B in the vapour phase was found to be 4 : 3, the ratio of moles of A and B in the phase they were mixed is:

Options

A. `4/5`
B. `4`
C. `2/3`
D. `1/3`

Answer

B

Explanation

`P _(M) = P _(A) ^(0) . X _(A) + P _(B) ^(0) . X _(B)`
Also,
`P _(A) . = P _(A) ^(0) . X _(A) = P _(M). X _(A) .`
`P _(B) . = P _(B) ^(0) . X _(B) = P _(M) . X _(B) .`
where, `X _(A) . and X _(B) .` are molar fraction of A and B in vapour phase.
`therefore ( P _(A) .)/( P _(b) .) = ( P _(A) ^(9) . X _(A))/( P _(B) ^(0) . X _(B)) = ( X _(A).)/( X _(B))`
or `(1)/(3) xx (X _(A))/(X _(B)) = (4)/(3)`
`therefore ( X _(A))/( X _(B)) = 4 = ( n _(A))/( n _(B))`