Question

The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600mm$ of $Hg$ 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 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 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)

Let $X_A$ and $X_B$ be the mole fraction of liquid $A$ and $B$ in the mixture.
Given, $P_A^{\circ }=400mmHg,P_B^{\circ }=600mmHg$
$P_{\text{Total }}=X_A\cdot P_A^{\circ }+X_B\cdot P_B^{\circ }$
$=0.5\times 400+0.5\times 600$
$=500mmHg$
Now, mole fraction of $A$ in vapour,
$Y_A=\frac{P_A}{P_{\text{total }}}=\frac{0.5\times 400}{500}=0.4\because \left[P_A=X_A{P}_A^{\circ }\right]$
and mole fraction of B in vapour, $Y_B=1-0.4=0.6$