Question

Two moles of pure liquid ‘A’ $\left( {{\text{P}}_{\text{A}}^{\text{0}} = 80{\text{mmHg}}} \right)$ and three moles of pure liquid ‘B’ $\left( {{\text{P}}_{\text{B}}^{\text{0}} = 120{\text{mmHg}}} \right)$ are mixed. Assuming ideal behaviour.
A) Vapour pressure of the mixture is $104{\text{mmHg}}$
B) Mole fraction of liquid ‘A’ in vapour pressure is $0.3077$
C) Mole fraction of ‘B’ in vapour pressure is $0.692$
D) Mole fraction of ‘B’ in vapour pressure is $0.785$

Answer 1

Hint:
To solve this question, you must recall the Raoult’s law and Dalton’s law of partial pressure Dalton’s law of partial pressure states that the total pressure of a mixture of a number of non-reacting gases is equal to the sum of pressures exerted by individual gases.

Formula used:
${{\text{P}}_{\text{T}}} = {{\text{p}}_{\text{1}}} + {{\text{p}}_{\text{2}}} + ........{{\text{p}}_{\text{n}}}$
Where, ${{\text{P}}_{\text{T}}}$ is the total pressure of the mixture of gases
${{\text{p}}_{\text{1}}}{\text{,}}{{\text{p}}_{\text{2}}}{\text{,}}....{{\text{p}}_{\text{n}}}$ are the partial pressures exerted by the individual gases in the mixture.

Complete step by step solution:
The partial pressure of a gas in a mixture is given by Raoult’s Law and it is the product of its vapour pressure in pure liquid form and mole fraction of the gas in the mixture, that is, ${{\text{p}}_{\text{n}}} = \chi \times {{\text{P}}^0}$.
In the question we are given the number of moles of A and B as two and three respectively.
The partial pressure of the pure liquids are given to us as ${\text{P}}_{\text{A}}^{\text{0}} = 80{\text{mmHg}}$ and ${\text{P}}_{\text{B}}^{\text{0}} = 120{\text{mmHg}}$
The total number of moles of gas in the mixture $ = {\text{n}} = {{\text{n}}_{\text{a}}} + {{\text{n}}_{\text{b}}} = 2 + 3 = 5$
So, we can find the mole fractions of the gases in the mixture as,
${\chi _a} = \dfrac{{{{\text{n}}_{\text{a}}}}}{{{{\text{n}}_{\text{a}}}{\text{ + }}{{\text{n}}_{\text{b}}}}} = \dfrac{2}{5} = 0.4$
The mole fraction of A is $0.4$
And, ${\chi _b} = \dfrac{{{{\text{n}}_{\text{b}}}}}{{{{\text{n}}_{\text{a}}}{\text{ + }}{{\text{n}}_{\text{b}}}}} = \dfrac{3}{5} = 0.6$
The mole fraction of B is $0.6$.
So the partial pressures of the gases in the mixture are
${{\text{p}}_{\text{a}}} = \chi \times {\text{p}}_{\text{a}}^0$
${{\text{p}}_a} = 0.4 \times 80 = 32{\text{mmHg}}$
For B:
${{\text{p}}_{\text{b}}} = \chi \times {\text{P}}_{\text{b}}^0$
${{\text{p}}_b} = 0.6 \times 120 = 72{\text{mmHg}}$
The vapour pressure of the mixture is given by the sum of the partial pressures of its constituents as per Dalton's law.
So, the vapour pressure of mixture is, ${{\text{P}}_{\text{T}}} = {{\text{p}}_{\text{a}}} + {{\text{p}}_{\text{b}}} = 32 + 72$
$\therefore {{\text{P}}_{\text{T}}} = 104{\text{ mmHg}}$

Thus, the correct option is A.

Note:
The partial pressure of a gas in a mixture is given by Raoult’s Law and it is the product of its vapour pressure of the pure liquid and mole fraction of the gas in the mixture, that is, ${{\text{p}}_{\text{n}}} = \chi \times {{\text{P}}^0}$. Raoult’s law states that for ideal gases: $\Delta {{\text{G}}_{{\text{mix}}}} = 0{\text{ , }}\Delta {{\text{H}}_{{\text{mix}}}} = 0{\text{ and }}\Delta {{\text{V}}_{{\text{mix}}}} = 0$
Real gases mostly deviate from Raoult’s law and show positive or negative deviations.