The vapour presure 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.

Answer

`CHMi_(A)^(l)=0.40,CHMi_(B)^(l)=0.60`
`CHMi^(V)``_(A)=0.030,CHMi^(V)``_(B)=0.70`

Explanation

Given `:`
`p_(A)^(@)=450mm,p_(B)^(@)=700mm, P_(tot al)=600mm`
Applying Raoult's law,
`p_(A)=p_(A)^(@)xxCHMi_(A)`
`p_(B)=p_(B)^(@)xxCHMi_(B)=p_(B)^(@)xx(1-CHMi_(A))`
`p_(t o t al)=p_(A)+p_(B)=CHMi_(A)p_(A)^(@)+p_(B)^(@)(1-CHMi_(A))`
`=p_(B)^(@)+(p_(A)^(@)-p_(B)^(@))CHMi_(A)....(i)`
Substituting the given values in `Eq. (i)`, we get
`600=700+(450-700)CHMi_(A)`
or `250CHMi_(A)=100`
or `CHMi_(A)=(100)/(250)=0.40`
`:. CHMi_(B)=1-0.40=0.60`
`p_(A)=p_(B)^(@)CHMi_(A)=450mmxx0.40=180mm`
`p_(B)=p_(B)^(@)CHMi_(B)=700mmxx0.60=420mm`
`P_(t otal)=(180+420)=600mm)` This is also given `)`
`CHMi_(A)^(V)(mol` fraction of `B` in vapour phase `):`
`=1-CHMi_(A)^(V)=1-0.30=0.70`