Question

Two liquids A and B have $P_A^o$ and $P_B^o$ the ratio of 1 : 3. If the ratio of number of moles of A and B are 1 : 3, the mole fraction of ‘A’ in vapour phase in equilibrium with the solution is equal to -

• A. 0.1
• B. 0.2
• C. 0.5
• D. 1.0

Answer 1

Answer: (A)

$\frac{P_{A^o}}{P_{B^o}}=\frac{1}{3}$

$\frac{{\text{n}}_{\text{A}}}{{\text{n}}_{\text{B}}}=\frac{1}{3}$

$X_A=\frac{1}{4}$

$X_B=\frac{3}{4}$

$P_A=P_A^0{X}_A=P_A^0\times \frac{1}{4};P_B=P_B^0{X}_B=P_B^0\times \frac{3}{4}$

$\frac{P_A^0}{P_B^0}=\frac{1}{3}\Rightarrow 3P_A^0=P_B^0$

$P=P_A^0{X}_A+P_A^0{X}_B=P_A^0\times \frac{1}{4}+\frac{9}{4}{P}_A^0\left[\because P_B^0\right]=3P_A^0$

$P=\frac{10}{4}{P}_A^0{X}_A=\frac{\frac{1}{4P_A^0}}{\frac{10}{4}{P}_A^0}=\frac{1}{10}=0.1$