Two liquids A and B have vapour pressures of 0.685 bar and 0.264 bar respectively. In an ideal solution of the two, what is the mole fraction of A at which the two liquids have equal partial pressure?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.278
By Raoult’s law, in an ideal solution the partial vapour pressure of each volatile component equals its mole fraction times the vapour pressure of the pure liquid:
\( \displaystyle p_A=x_A\,p_A^0 \qquad p_B=x_B\,p_B^0=(1-x_A)\,p_B^0 \)
Set the two partial pressures equal:
\( \displaystyle 0.685\,x_A=0.264\,(1-x_A) \)
\( \displaystyle (0.685+0.264)\,x_A=0.264 \)
\( \displaystyle x_A=\frac{0.264}{0.949}=0.278 \)
A is the more volatile liquid, so its mole fraction must be smaller than that of B for the two partial pressures to be equal.