If two substances A and B have \(P^o_A : P^o_B = 1 : 2\) and have mole fraction in the ratio \(1 : 2\), then mole fraction of \(A\) in vapours is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.2
For an ideal solution, Raoult’s law gives the partial vapour pressure of each component as its mole fraction in the liquid times its pure vapour pressure, pA = xAp°A. By Dalton’s law, the mole fraction of A in the vapour is its share of the total pressure, yA = pA/(pA + pB).
Let p°A = p and p°B = 2p. The mole fractions in the liquid are in the ratio 1 : 2:
\( x_A = \frac{1}{3},\quad x_B = \frac{2}{3} \)
\( p_A = \frac{1}{3}\times p = \frac{p}{3} \)
\( p_B = \frac{2}{3}\times 2p = \frac{4p}{3} \)
\( \displaystyle y_A = \frac{p/3}{p/3 + 4p/3} = \frac{1}{5} = 0.2 \)
So the mole fraction of A in the vapour is 0.2. The vapour is richer in B, the more volatile component.
Shortcut: yA : yB = (1 × 1) : (2 × 2) = 1 : 4, so yA = 1/5.