The molar volume of liquid benzene (density `0.877 g mL^(-1))` increases by a factor of `2750` as it vapourises at `20^(@)C` and that of liquid toluene(density `0.867 g mL)` increases by a factor of `7720`at `20^(@)C`. A solution of benzene and toluene at `20^(@)C` has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution.

Explanation

The volume of 1 mole of `C_(6)H_(6)` (liquid) `=(78)/(0.877)mL`
The volume of 1 mole of `C_(6)H_(6)` (vapour) `=(78)/(0.877)xx2750mL`
`=244.58` litres
for 1 mole of `C_(6)H_(6)` vapour at `20^(@)C`
`p_(b)^(0)V=RT`
`p_(b)^(0)=(0.0821xx293)/(244.58)`atm`=74.75mm`
Similarly for 1 mole of `C_(6)H_(5)CH_(3)` vapour at `20^(@)C`
`p_(t)^(0)=(0.0821xx293)/(((92)/(0.867)xx(7720)/(1000)))"atm"=22.32mm`
Now,
`(1)/(p)=(x_(b).)/(p_(b)^(0))+((1-x_(b).)/(p_(t)^(0)))`
`(1)/(46)=(x_(b).)/(74.75)+(1-x_(b).)/(22.32)`
`x_(b)^(.)=0.7336`