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.

Answer

0.73

Explanation

Molar volume of liquid benzene`=78/(0.877)"mL"`
Molar volume of benzene vapour at `20^(@)"C"=2750xx(78)/(0.877)"mL"=(2750xx78)/(877)"litre"`
`"P"_("benzene")^(0)=("nRT")/("V"_("benzene"))`
`"P"_("benzene")^(0)=(1xx0.082xx293xx877)/(250xx75)"atm"`
`=0.09823"atm"`
`=74.65"torr"`
Molar volume of liquid toluene `=(92)/(0.867)"mL"`
Molar volume of toluene vapours at `20^(@)"C"=7720xx(92)/(0.867)"mL"=(7720xx92)/(867)"litre"`
Now `P_("toluene")^(0)=(1xx0.082xx293xx867)/(7720xx92)"atm"`
`=0.02932 "atm"`
`=22.29"torr"`
`"P"_("total")="X"_("benzene")("P"_("benzene")^(0)-"P"_("toluene")^(0))+"P"_("toluene")^(0)`
`46="X"_("benzene")(74.65-22.29)+22.29`
`"X"_("benzene")=0.4528`
Mole fraction of benzene in vapour phase `("Y")_("b")=("X"_("benzene")xx"P"_("benzene")^(0))/("P"_("total"))=(0.4528xx74.65)/(46)=0.73`