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.

Explanation

Here, `p_(A)^(@)="450 mm, "p_(B)^(@)="700 mm, "p_("Total")="600 mm"`
`"Applying Raoult's law, "p_(A)=x_(A)xxp_(A)^(@)," "p_(B)=x_(B)xxp_(B)^(@)=(1-x_(A))p_(B)^(@)`
`" "P_("Total")=p_(A)+p_(B)=x_(A)p_(A)^(@)+(1-x_(A))p_(B)^(@)=p_(B)^(@)+(p_(A)^(@)-p_(B)^(@))x_(A)`
Substituting the given values, we get
`600=700+(450-700)x_(A)" or "250x_(A)=100" or "x_(A)=(100)/(250)=0.40`
Thus, composition of the liquid mixture will be
`x_(A)" (mole fraction of A) = 0.40, "x_(B)" (mole fractiobn of B)"=1-0.40=0.60`
`therefore" "p_(A)=x_(A)xxp_(A)^(@)=0.40xx"450 mm = 180 mm, "p_(B)=x_(B)xxp_(B)^(@)=0.60xx"700 mm = 420 mm"`
`"Mole fraction of A in the vapour phase"=(p_(A))/(p_(A)+p_(B))=(180)/(180+420)=0.30`
Mole fraction of B in the vapour phaes `=1-0.30 = 0.70.`