Question

If two substances A and B have $P_A^0:P_B^0 = 1:2$ and have mole ratio in solution 1:2 then mole fraction of A in vapours is:
a. 0.33
b. 0.25
c. 0.52
d. 0.2

Answer 1

We can use Raoult’s law and Dalton’s law of partial pressure to solve the following question. Raoult’s law gives us information on vapour pressure and concentration, while Dalton’s law talks about total pressure exerted by the gaseous mixture.

Complete step by step solution:
We have given two substances A and B with the pressure of their pure substances is in the ratio of 1:2.
i.e., $\dfrac{{P_A^0}}{{P_B^0}} = \dfrac{1}{2}$
Also, the mole ratio in solution is given
i.e.,$\dfrac{{{X_A}}}{{{X_B}}} = \dfrac{1}{2}$
We know that Raoult’s gives us the relation between vapour pressure and concentration of the solution. It states that partial vapour pressure of any component of the solution at any given temperature is equal to the mole fraction of that component in the solution and its vapour pressure in the pure state.
${P_A} = P_A^0{X_A}$ and ${P_B} = P_B^0{X_B}$
Where ${P_A}$ and ${P_B}$ is the partial vapour pressure of component A and B.
$P_A^0$ and $P_B^0$ is the vapour pressure in the pure state.
${X_A}$ and ${X_B}$ is its mole fraction.

Now, from Dalton’s pressure when a mixture of two or more non-reacting gases is enclosed in a container then the total pressure exerted by the gaseous mixture is equal to the sum of partial pressures of the individual gases.
Applying that here,
${P_{total}} = {P_A} + {P_B}$
Substituting the values of ${P_A}$ and ${P_B}$, we get
${P_{total}} = P_A^0{X_A} + P_B^0{X_B}$

Also from Dalton’s law, we know that the partial pressure of the gas in a mixture is the pressure which gas would exert when present alone in the same vessel at the same temperature as that in the mixture.
Therefore ${P_A} = {Y_A}{P_{total}}$
${Y_A} = \dfrac{{{P_A}}}{{{P_{total}}}}$
Substituting the values of $P_{total}$ and taking the reciprocate, we get
$\dfrac{1}{{{Y_A}}} = \dfrac{{P_A^0{X_A} + P_B^0{X_B}}}{{P_A^0{X_A}}}$
$\Rightarrow \dfrac{1}{{{Y_A}}} = 1 + \dfrac{{P_B^0{X_B}}}{{P_A^0{X_A}}}$
Now applying the values of given ratio we get
$\Rightarrow \dfrac{1}{{{Y_A}}} = 1 + 2 \times 2 = 5$
$\Rightarrow {Y_A} = \dfrac{1}{5} = 0.2$
The mole fraction of A in vapour pressure is 0.2

**i.e., option (d) is correct

Note:**
We can also assume the values of ${\text{P}}_{\text{A}}^0 = 1\;{\text{and}}\;{\text{P}}_{\text{B}}^0 = 2$ also ${X_A} = 1\;and\;{X_B} = 2$
So we get ${P_{total}} = 1 + 2 \times 2 = 5$ and ${P_A} = 1$
$\Rightarrow {Y_A} = \dfrac{{{P_A}}}{{{P_{total}}}} = \dfrac{1}{5} = 0.2$