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)

$\frac{P_{A^{o}}}{P_{B^{o}}}=\frac{1}{3}$
$\frac{\text{n}_{\text{A}}}{\text{n}_{\text{B}}} = \frac{1}{3}$
$X_{A}=\frac{1}{4}$
$X_{B}=\frac{3}{4}$
$P_{A}=P_{A}^{0}X_{A}=P_{A}^{0}\times \frac{1}{4};P_{B}=P_{B}^{0}X_{B}=P_{B}^{0}\times \frac{3}{4}$
$\frac{P_{A}^{0}}{P_{B}^{0}}=\frac{1}{3}\Rightarrow 3P_{A}^{0}=P_{B}^{0}$
$P=P_{A}^{0}X_{A}+P_{A}^{0}X_{B}=P_{A}^{0}\times \frac{1}{4}+\frac{9}{4}P_{A}^{0}\left[\because P_{B}^{0}\right]=3P_{A}^{0}$
$P=\frac{10}{4}P_{A}^{0}X_{A}=\frac{\frac{1}{4 P_{A}^{0}}}{\frac{10}{4} P_{A}^{0}}=\frac{1}{10}=0.1$