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JEE Main 2023 · Chemistry · Solutions, Solid State and Surface Chemistry · easy

Liquids A and B form an ideal solution. Their vapour pressures in the pure state at a given temperature are 80 and 60 torr respectively. A solution…

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Liquids A and B form an ideal solution. Their vapour pressures in the pure state at a given temperature are 8080 and 60 torr60\ \mathrm{torr} respectively. A solution is made from 2 mol2\ \mathrm{mol} of A and 3 mol3\ \mathrm{mol} of B. The mole fraction of A in the vapour in equilibrium with this solution is
  1. A. 0.400.40
  2. B. 0.470.47
  3. C. 0.570.57
  4. D. 0.530.53
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Answer: (B) 0.470.47

Liquid mole fractions: xA=0.4x_A=0.4, xB=0.6x_B=0.6. Partial pressures (Raoult): pA=0.4×80=32p_A=0.4\times80=32 torr, pB=0.6×60=36p_B=0.6\times60=36 torr. Total =68=68 torr. Dalton: yA=pA/ptotal=32/68=0.47y_A=p_A/p_{\mathrm{total}}=32/68=0.47. The vapour is richer in the more volatile component A than the liquid (0.400.40), so the answer must exceed 0.400.40; 0.400.40 is the liquid composition, not the vapour composition.
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