Question

The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fractional of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fraction of components A and B in vapour phase, respectively are

• A. 500 mm Hg, 0.5, 0.5
• B. 450 mm Hg, 0.4, 0.6
• C. 450 mm Hg, 0.5, 0.5
• D. 500 mm Hg, 0.4, 0.6

Answer 1

Answer: (D)

Let $X_A$ and $X_B$ be the mole fraction of liquid A and B in the mixture.

Given, $P_A^{{}^{\circ }}=400$ $mmHg,P_B^{{}^{\circ }}=600$ $mmHg$

$P_{Total}=X_A.P_A^{{}^{\circ }}+X_B.P_B^{{}^{\circ }}=0.5\times 400+0.5\times 600=500$ $mmHg$

Now, mole fraction of A in vapour,

$Y_A=\frac{P_A}{P_{total}}=\frac{0.5\times 400}{500}=0.4$

Hence $[P_A=X_A{P}_A^{{}^{\circ }}]$

and mole fraction of B in vapour, $Y_B=1-0.4=0.6$