The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Apply Raoult's law `P_(T)=P_(A)^(0)x_(A)+P_(B)^(0)x_(B)=P_(B)^(0)x_(A)+P_(B)^(0)(1-x_(A))` to calculate mole fraction of `A(x_(A))` and `B(x_(B))`.
In vapour phase, partial pressure are used insted of number of moles.

Explanation

Step I : Composition in liquid phase
Vapour pressure of pure liquid A `(P_(A)^(0))=450" mm"`
Vapour pressure of pure liquid B `(P_(B)^(0))=700" mm"`
Total vapour pressure of the solution (P) = 600 mm
According to Raoult's law,
`P=P_(A)^(0)x_(A)+P_(A)^(0)x_(B)=P_(A)^(0)x_(A)+P_(B)^(0)(1-x_(A))`
`(600" mm")=450" mm"xx x_(A)+700" mm"(1-x_(A))`
`=700" mm"+x_(A)+(450-700)" mm"`
`=700-x_(A)(250" mm")`
`x_(A)=((600-700))/(-(250" mm"))=0.40`
Mole fraction of `A(x_(A))=0.40`
Mole fraction of `B(x_(B))=1-0.40=0.60`
Step II : Composition in vapour phase
`P_(A)=P_(A)^(0)x_(A)=(450" mm")xx0.40=180" mm"`
`P_(B)=P_(B)^(0)x_(B)=(700" mm")xx0.60=420" mm"`
Mole fraction of A in vapour phase `=(P_(A))/(P_(A)+P_(B))`
`=((180)" mm")/((180+420)" mm")=0.30`
Mole fraction of B in vapour phase `=(P_(B))/(P_(A)+P_(B))=((420)" mm")/((180+420)" mm")=0.30`
Mole fraction of B in vapour phase `=(P_(B))/(P_(A)+P_(B))=((420)" mm")/((180+420)" mm")=0.70`