The vapour pressures of pure liquids A B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the compositon of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition in the Vapour phase.

Explanation

Vapour pressure of pure liquid A `(P_(A)^(@))=450mm`
Vapour pressure of pure liquid B `(P_(B)^(@))`=700mm
Total vapour pressure of the solution (P)=600mm
According to Raoult's Law, P = `P_(A)^(@)x_(A)+P_(B)^(@)x_(B)=P_(A)^(@)x_(A)+P_(B)^(@)(1-x_(A))`
`(600mm)=450mmxxx_(A)(450-700)mm`
`=700mm x_(A) (250mm)`
`x_(A)=((600-700)mm)/(-(250mm))=0.40`
Mole fraction of A`(x_(A)=0.40`
Mole fraction of B `(x_(B))`=1-0.40=0.60
`P_(A)=P_(A)^(@)x_(A)=(450mm)xx0.40=180mm`
`P_(B) = P_(B)^(@)x_(B)=(700m)xx0.60=420mm`
Mole fraction of A in the vapour phase = `P_(A)/(P_(A)+P_(B))=((180mm))/((180+420)mm)=0.30`
Mole fraction of B in the vapour phase = `P_(B)/(P_(A)+P_(B))=((420mm))/((180+420)mm)=0.70`