Question

The vapour pressures of pure liquids A and B are 400 and 600 mm Hg, 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 fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are:

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

Answer 1

The correct option is A 500 mm Hg; 0.4,0.6

Given, the mole fraction of B in liquid $X_B=0.5$
So the mole fraction of A in liquid $X_A=0.5$
According to Raoult's law,
$P_{total}=X_A{P}_A^o+X_B{P}_B^0=0.5\times 400+0.5\times 600=500mmHg$
Now, mole fraction of A in vapour $Y_A=\frac{P_A}{P_{total}}=\frac{0.5\times 400}{500}=0.4$
So, mole fraction of B in vapour $Y_B=1-0.4=0.6$