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 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)

$P_{\text {total }}=X_{A} . P_{A}^{0}+X_{B} \cdot P_{B}^{0}=0.5 \times 400+0.5 \times 600$

$=500 mmHg$

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, $Y _{ B }=1-0.4=0.6$