At `88^(@)C` benzene has a vapour pressure of 900 torr and toluene has vapour pressure of 360 torr. What is the mole fraction of benzene in the mixture with toluene that will be boil at `88^(@)C` at 1 atm pressure, benzene- toluene form an idean solution.

Explanation

`P_(S) = 760` torr, because solution boils at `88^(@)C`
`:. 760 = 900 a + 360 (1-a)`
`a = 0.74` where 'a' is mole fration `C_(6)H_(6)`.
(ii) For solid -liquid solution:
Let us assume `A =` non volatile solid % `B =` volatile liquid
According to Raoult's law-
`:' P_(s) = X_(A) P_(A)^(0) +X_(B)P_(B)^(0)`
for `A, P_(A)^(0 = 0`
`:. P_(s) = X_(B)P_(B)^(0)` .....(5)
Let `P_(B)^(0) = P^(0) =` Vapour pressure of pure state of solvent,
here `X_(B)` is mole fraction of solution
`P_(s) = (n_(B))/(n_(A) +n_(B)) P^(0)`
`P_(S) prop (n_(B))/(n_(A)+n_(B))` i.e vapour pressure of solution `prop` mole fraction of solvent
`rArr P_(S) = X_(B)P_(B)^(0) rArr P_(S) = (1- X_(A)) P_(B)^(0) rArr P_(S) = P_(B)^(0) - X_(A) P_(B)^(@) rArr (P_(B)^(@) -P_(S))/(P_(B)^(@)) = X_(A)`
or `(P^(@) -P_(S))/(P^(0)) = X_(A)` ....(7), or `(P^(0)-P_(S))/(P^(0)) =(n_(A))/(n_(A)+n_(B))` ..(8)
or `(P^(0))/(P^(0)-P_(S)) = (n_(A)+n_(B))/(n_(A))` or `(P^(0))/(P^(0)-P_(S)) = 1 +(n_(B))/(n_(A))` or `(P^(0))/(P^(0)-P_(S)) -1 =(n_(B))/(n_(A))` or `(P_(S))/(P^(0)-P_(S)) = (n_(B))/(n_(A))`
`(P^(0)-P_(S))/(P_(S)) =(n_(A))/(n_(B)) = (w_(A)m_(B))/(m_(A)w_(B))` ..(9)