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The vapour pressure of pure liquids A an...

The vapour pressure of pure liquids `A` and `B` is 450 and `700mm Hg`, respectively, at `350K. ` Find out the composition of the liquid mixture if the total vapour pressure is `600mm Hg`. Also find the composition of the vapour phase.

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`P_(A)^(@)=450mm,P_(B)^(@)=700mm,P_("total")=600mm`
As `P_("total")=P_(A)+P_(B)`
`=x_(A)P_(A)^(@)+x_(B)P_(B)^(@)` (from raoult's law)
`=x_(A)P_(A)^(@)+(1-x_(A))P_(B)^(@)`
`=P_(B)^(@)+(P_(A)^(@)-P_(B)^(@))x_(A)`
`implies600=700+(450-700)x_(A)`
or `x_(A)=0.40`
`thereforex_(B)=1-x_(A)=1-0.40=0.60`
`thereforeP_(A)=x_(A)P_(A)^(@)=0.40xx450=180mm`
`thereforeP_(B)=x_(B)P_(B)^(@)=0.60xx700=420mm`
`therefore` Mole fraction of A in vapoure phase
`=(P_(A))/(P_(A)+P_(B))=(180)/(180+420)=0.30`
and, mole fraction of B in vapour phase
`=1-0.30=070`.
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