Question

The vapour pressure of pure liquid A at `300K` is `577` Torr and that of pure liquid `B` is `390` Torr. These two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a mixture in which the mole fraction of `A` in the vapour is `0.35`. Calculate the total pressure of the vapour and the composition of the liquid mixture.

Answer 1

Correct Answer - `440` Torr
A and B volatile liquids, given `P_(A^(0)=575` Torr, `P_(B^(0)=390` Torr
let mole fraction of `A` in solution `=X_(A)`
hence, `P_("total")=P_(A^(0)X_(A)+P_(B)^(0)(1-X_(A))`
also `X_(A)=` ,p,e fraction of A in the vapour `=0.35`
`X_(A)=(P_(A)^(@)X_(A))/(P_(A)^(@)X_(A)+P_(B)^(@)(1-X_(A))=0.35`
`=(575X_(A))/(575X_(A)+390(1-X_(A))`
this gives `X_(A)=0.27`
hence, total pressure
`P_("total")=575xx0.27+390xx0.73`
`=440` Torr
Composition of liquid mixture,
`A=27` mol `%,B=73` mol `%`
total pressure `=440` Torr