Question

Two liquids $\text{A}$ and $\text{B}$ have ${P^{\circ }}_A:{P^{\circ }}_B=1:3$ at a certain temperature. If the mole fraction ratio of $x_A:x_B=1:3$, the mole fraction of $\text{A}$ in vapour in equilibrium with the solution at a given temperature is

• A. 0.1
• B. 0.2
• C. 0.5
• D. 1

Answer 1

The correct option is A 0.1

From Raoult's law, Vapour pressure of A is
$P_A=x_A{P}_A^0$
and Vapour pressure of B is
$P_B=x_B{P}_B^0$
Now mole fraction of A in vapour is
$y_A=\frac{P_A}{P_A+P_B}=\frac{x_A{P}_A^0}{x_A{P}_A^0+x_B{P}_B^0}$
$y_A=\frac{1}{1+9}=0.1$