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.

At 88^(@)C benzene has a vapour pressure of 900 torr and toluene has v

1.	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.
Answer» Solution :`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)