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

molar volume of benzene `=(78)/(0.877)ml`
molar volume of toluene `=(92)/(0.867)ml`
At `20^@C` volume of vapour of benzene and toluene
`V_B =(78)/(0.877)xx2750=244.58 ml`
`P_B^0=(RT)/(V_B), P_T^0=(RT)/(V_T), P_("total")=P_B^@chi_B +P_T^@chi_T`
`rArr (46)/(760)=chi_B((RT)/(V_B))+chi_T((RT)/(V_T))`
Substitute for `V_B and V_T and chi_T =1-chi_B`
`rArr chi_B=0.452`
Now find the vapour fraction of benzene.
`Chi_B^V =(P_B^0chi_B)/(P_("total"))=(0.452)/(46//760)((RT)/(V_B))=0.734`