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

Let moles of benzene vaproises, at `20^(@)C=n_(1)`
volume of `n_(1)` mole of benzene `(f)=(78n_(1))/(0.877)`
Volume of `n_(1)` mole of benzene `(g)=2750 ((78n_(1))/(0.877))`
`PV=nRT`
`P^(@)=((78n_(1))/(0.877))((2750)/(1000))=n_(1)xx0.082xx293`
`P_("Benzene")=0.0982` atm.
`=74.63 mm Hg`
`P_("Benzene")^(@)` at `27^(@)C` can be calculated as
`log((P_(2))/(P_(1)))=(DeltaH_("vap"))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log ((P_(2))/(74.63))=(394.57xx78)/(2.303xx8.314)[(7)/(300xx293)]`
`P_("Benzene")^(@) ("at"27^(@)C)=100.2` mm Hg
Molality of the solution `=(X_("solvent")xx1000)/(X_("solvent")xx78)=(((100.2-98.88)/(98.88))xx10000)/(0.98xx78)=0.17`
`Deltat_(f)=K_(f)m=5xx0.17=0.85`
We known that `K_(f), =[(RT_(f)^(2))/(1000DeltaH_(f))]M`
`5=((8.314xxT_(1)^(2))/(1000xx10060))78, T_(f)-278.5K`