Liquids `A` and `B` form an ideal mixture, in which the mole fraction of `A` is `0.25`. At temperature `T`, a small quantity of the vapour in equilibrium with the liquid is collected and condensed. This process is repeated for a second time with first condensate. The second condensate now contains 0.645 mole fraction of `A`. Calculate the ratio `(P_(A)^(@)//P_(B)^(@))`. What will be the mole fraction of `B` in the third condensate?

Explanation

Mole fraction of `A (chi_(A))=0.25`
Mole fraction of `B (chi_(B))=0.75`
After first condensation
Partial presssure of `A (p_(A))=P_(A)^(@)chi_(A)=P_(A)^(@)xx0.25`
Partial presssure of `B (p_(B))=P_(B)^(@)xx0.75`
`:. p_(A)/p_(B)=P_(A)^(@)/P_(B)^(@)xx(0.25)/(0.75)`
After second condensation
Mole fraction of `A (chi_(A))=0.645`
Mole fraction of `B(chi_(B))=1-0.645=0.355`
Ratio of partial pressures
`p_(A)^"/p_(A)^"=(p_(A)^(@)/p_(B)^(@))^(2)xx0.25/0.75=0.645/0.345`
`:.(p_(A)^(@)/p_(B)^(@))^(2)=0.645/0.345xx0.75/0.25`
`:.P_(A)^(@)/P_(B)^(@)=2.33`
After third condensation
`p_(A)^"'/p_(A)^"'=(p_(A)^(@)/p_(B)^(@))^(3)xx0.25/0.75=(2.33)^(3)xx0.25/0.75=4.242`
`:.`Mole fraction of B
`p_(B)^"'/p_(B)^"'+p_(B)^"'=1/5.42=0.191`=Mole fraction of `B`