Search any question & find its solution
Question:
Answered & Verified by Expert
Liquids A and B form an ideal solution in the entire composition range. At $350 \mathrm{~K}$, the vapor pressures of pure A and pure B are $7 \times 10^{3} \mathrm{~Pa}$ and $12 \times 10^{3} \mathrm{~Pa}$, respectively. The composition of the vapour is in equilibrium with a solution containing 40 mole percent of $\mathrm{A}$ at this temperature is:
Options:
Solution:
2538 Upvotes
Verified Answer
The correct answer is:
$x_{\mathrm{A}}=0.28 ; x_{\mathrm{B}}=0.72$
$\mathrm{P}_{\mathrm{A}}^{\circ}=7 \times 10^{3}$
$\mathrm{P}_{\mathrm{B}}^{\circ}=12 \times 10^{3}$
$x_{\mathrm{A}}^{\prime}=0.4 ; x_{\mathrm{B}}^{\prime}=1-0.4$
$x_{\mathrm{B}}^{\prime}=0.6$
$\mathrm{P}_{\text {total }}=\mathrm{P}_{\mathrm{A}^{\circ}}^{x_{\mathrm{A}}^{\prime}}+\mathrm{P}_{\mathrm{B}}^{\circ} x_{\mathrm{B}}^{\prime}$
$=7 \times 10^{3} \times 0.4+12 \times 10^{3} \times 0.6$
$=(7 \times 0.4+12 \times 0.6) \times 10^{3}=10^{4}$
$x_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}^{\circ}}^{x^{\prime}} \mathrm{A}}{\mathrm{P}_{\text {total }}}=\frac{7 \times 10^{3} \times 0.4}{10^{4}}$
$\therefore x_{\mathrm{A}}=0.28, x_{\mathrm{B}}=1-0.28=0.72$
$\mathrm{P}_{\mathrm{B}}^{\circ}=12 \times 10^{3}$
$x_{\mathrm{A}}^{\prime}=0.4 ; x_{\mathrm{B}}^{\prime}=1-0.4$
$x_{\mathrm{B}}^{\prime}=0.6$
$\mathrm{P}_{\text {total }}=\mathrm{P}_{\mathrm{A}^{\circ}}^{x_{\mathrm{A}}^{\prime}}+\mathrm{P}_{\mathrm{B}}^{\circ} x_{\mathrm{B}}^{\prime}$
$=7 \times 10^{3} \times 0.4+12 \times 10^{3} \times 0.6$
$=(7 \times 0.4+12 \times 0.6) \times 10^{3}=10^{4}$
$x_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}^{\circ}}^{x^{\prime}} \mathrm{A}}{\mathrm{P}_{\text {total }}}=\frac{7 \times 10^{3} \times 0.4}{10^{4}}$
$\therefore x_{\mathrm{A}}=0.28, x_{\mathrm{B}}=1-0.28=0.72$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.