Q13 / 90 · KVPY 2011 (SB/SX) · Chemistry
At $300 \mathrm{~K}$ the vapour pressure of two pure liquids, $\mathrm{A}$ and $\mathrm{B}$ are 100 and $500 \mathrm{~mm} \mathrm{Hg}$, respectively. If in a mixture of $\mathrm{A}$ and $\mathrm{B}$, the vapour pressure is $300 \mathrm{~mm} \mathrm{Hg}$, the mole fractions of $\mathrm{A}$ in the liquid and in the vapour phase, respectively, are-
- A$1 / 2$ and $1 / 10$
- B$1 / 4$ and $1 / 6$
- C$1 / 4$ and $1 / 10$
- Correct answerD$1 / 2$ and $1 / 6$
Solution
$\mathrm{Y}_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}}^{0} \mathrm{X}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^{0} \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^{0} \mathrm{X}_{\mathrm{B}}}=\frac{100 \cdot \frac{1}{2}}{100 \cdot \frac{1}{2}+500 \cdot \frac{1}{2}}$
$=\frac{50}{50+250}$
$=\frac{50}{300}$
$=1 / 6$
$\because \mathrm{PS}=\left(\mathrm{p}_{\mathrm{A}}^{0}-\mathrm{p}_{\mathrm{B}}^{0}\right) \mathrm{X}_{\mathrm{A}}+\mathrm{p}_{\mathrm{B}}^{0}$
$300=(-400) \mathrm{X}_{\mathrm{A}}+500$
$\mathrm{X}_{\mathrm{A}}=\frac{1}{2}$
$\therefore \mathrm{X}_{\mathrm{B}}=\frac{1}{2}$
Hence, the correct answer is option: 'D'.