Question

The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600mmHg$, respectively at $298K$. 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 fraction of components $A$ and $B$ in vapour phase, respectively are -

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

Answer 1

Answer: (D)

$P_{\text{total }}=X_A.P_A^0+X_B\cdot P_B^0=0.5\times 400+0.5\times 600$

$=500mmHg$

Now, mole fraction of A in vapour,

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

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