The vapour pressures of pure liquids A and B are 450 mm Hg and 700 mm Hg respectively at 350 K. Find the composition of the liquid mixture if its total vapour pressure is 600 mm Hg. (Assume that the solution is ideal.)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → xA = 0.4, xB = 0.6
For an ideal solution of two volatile liquids, Raoult’s law gives each partial vapour pressure as p = x p0, and the total vapour pressure is their sum. With xB = 1 − xA:
\( \displaystyle p_{total}=x_A\,p_A^0+(1-x_A)\,p_B^0 \)
\( \displaystyle 600=450\,x_A+700\,(1-x_A) \)
\( \displaystyle 250\,x_A=100 \qquad x_A=0.4 \)
So xA = 0.4 and xB = 1 − 0.4 = 0.6. Check: 450 × 0.4 + 700 × 0.6 = 180 + 420 = 600 mm Hg.