The vapour pressures of two liquids A and B in their pure states are in the ratio 1 : 2. A binary solution of A and B contains A and B in the mole ratio 1 : 2. The mole fraction of A in the vapour phase of the solution will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.2
By Raoult’s law, the partial vapour pressure of each component equals its pure vapour pressure times its mole fraction in the solution. By Dalton’s law, the mole fraction of a component in the vapour is its partial pressure divided by the total pressure:
\( \displaystyle p_A=p_A^{\circ}\,x_A \qquad y_A=\frac{p_A}{p_{\text{total}}} \)
Let \( p_A^{\circ}=p \), so \( p_B^{\circ}=2p \). The mole ratio A : B = 1 : 2 gives \( x_A=\frac{1}{3} \) and \( x_B=\frac{2}{3} \).
\( \displaystyle p_A=p\times\frac{1}{3}=\frac{p}{3} \)
\( \displaystyle p_B=2p\times\frac{2}{3}=\frac{4p}{3} \)
\( \displaystyle p_{\text{total}}=\frac{p}{3}+\frac{4p}{3}=\frac{5p}{3} \)
\( \displaystyle y_A=\frac{p/3}{5p/3}=\frac{1}{5}=0.2 \)
The mole fraction of A in the vapour is 0.2, lower than its 0.33 in the liquid, because the vapour is richer in the more volatile component B.