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JEE Advanced 2023 Paper 1 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

Two volatile liquids A and B form an ideal solution. At a given temperature the vapour pressures of pure A and B are 400 torr and 200 torr…

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Two volatile liquids A\mathrm{A} and B\mathrm{B} form an ideal solution. At a given temperature the vapour pressures of pure A\mathrm{A} and B\mathrm{B} are 400 torr400\ \mathrm{torr} and 200 torr200\ \mathrm{torr} respectively. An equimolar liquid mixture of A\mathrm{A} and B\mathrm{B} is placed in a closed vessel and the vapour above it is completely condensed into a second liquid. This second liquid is then allowed to come to equilibrium with its own vapour. If the mole fraction of A\mathrm{A} in the vapour above the second liquid is yy, then the value of 10y10y is
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Answer: 8

Use Raoult's law and Dalton's law at each stage: yA=xAPA∘xAPA∘+xBPB∘y_A=\dfrac{x_AP_A^\circ}{x_AP_A^\circ+x_BP_B^\circ}.
Stage 1 (liquid with xA=0.5x_A=0.5): yA=0.5×4000.5×400+0.5×200=200300=23y_A=\dfrac{0.5\times400}{0.5\times400+0.5\times200}=\dfrac{200}{300}=\dfrac23.
Complete condensation of this vapour gives the second liquid with the same composition, xA=23x_A=\tfrac23, xB=13x_B=\tfrac13.
Stage 2 (vapour above the second liquid): y=23×40023×400+13×200=266.7266.7+66.7=0.80y=\dfrac{\tfrac23\times400}{\tfrac23\times400+\tfrac13\times200}=\dfrac{266.7}{266.7+66.7}=0.80.
Hence 10y=810y=8. The enrichment in the more volatile component A\mathrm{A} at each stage (0.50→0.667→0.800.50\to0.667\to0.80) is the principle of fractional distillation. A common slip is to use the mole fraction of the liquid (0.5) directly in the second stage, which gives 0.6670.667 and the wrong answer 6.676.67.
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