Calculate the mole fraction of toluence in the vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50. The vapour pressure of pure benzene is 199 torr, that of toluene is 37 torr at the same temperature.

Answer

0.237

Explanation

`C_(6)H_(6)"-A",C_(7)H_(8)"-B"`
`X_("B")=0.50,X_("A")=1-0.5`
=0.5
`Pt=X_("A")P_("B")^(0)+X_("B")P_("B")^(0),P_("B")=X_("B")P_("B")^(0)`
`Y_("B")=(P_("B"))/(Pt)`
`(X_("B")P_("B")^(0))/(X_("A")P_("A")^(0)+X_("B")P_("B")^(0))=(0.5xx37)/(0.5xx119+0.5xx37)=37/156=0.237`