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 withfirst 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?

Liquids A and B form an ideal mixture, in which the mole fraction of A

1.	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 withfirst 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?
Answer» <P> Solution :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 secondcondensation 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 thirdcondensation `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`