Question

The vapour pressures of pure liquids $A $ and $B$ are $400$ and $600\, mm$ of $Hg$ 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 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, $P_A^{\circ}=400 \,mm \,Hg , P_B^{\circ}=600 \,mm\, Hg$
$P _{\text {Total }}= X _{ A } \cdot P _{ A }^{\circ}+ X _{ B } \cdot P _{ B }^{\circ}$
$=0.5 \times 400+0.5 \times 600$
$=500 \,mm\, Hg$
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, $Y_B=1-0.4=0.6$