Consider the temperature-composition diagram for water and ethanol: When a 50%/50% mixture of water and ethanol is boiled and a vapour is removed, some of the liquid remains in the distillation flask. That liquid

Question

BlackTom AI · Accepted Answer

Ethanol is more volatile than water (it has a higher vapor pressure and a lower boiling point), so the vapour formed above a liquid mixture is enriched in ethanol relative to the liquid. Using vapor–liquid equilibrium concepts (Raoult’s law and relative volatility) explains why.

Relevant concepts:
- Raoult’s law and vapor composition: for a binary mixture, the vapour mole fraction yA = (α xA) / [1 + xA(α − 1)], where α (relative volatility) = pA*/pB* (>1 if A is more volatile).
- When vapour is removed, the remaining liquid composition shifts opposite to the vapour enrichment (conservation of mass).

Step-by-step reasoning:
1. Start with a liquid mixture x_ethanol = 0.50. Ethanol has higher vapour pressure than water, so α > 1.
2. Plugging into the vapour-composition relation (illustrative algebra): if α = 2 (example to show effect),
   y_ethanol = (2 × 0.50) / [1 + 0.50(2 − 1)] = 1.00 / 1.50 = 0.667.
   This shows the vapour is richer in ethanol (≈66.7%) than the liquid (50%).
3. Removing that vapour preferentially removes ethanol from the flask, so the residual liquid loses more ethanol than water.
4. After some vapour removal, the liquid ethanol fraction decreases (x_ethanol < 0.50) and the liquid becomes richer in water.

Why the correct answer is correct:
- Because the vapour removed is enriched in ethanol, the remaining liquid is depleted in ethanol and therefore richer in water relative to the original 50/50 mixture.

Why the other options are incorrect:
- (a) “is pure water” — incorrect: removing vapour cannot leave pure water unless all ethanol is removed; initial boiling produces a vapour mixture, not complete separation to pure water.
- (c) “is richer in ethanol” — incorrect: vapour is richer in ethanol, but removing it leaves the residue poorer in ethanol.
- (d) “has the same composition… because of an azeotrope” — incorrect: while ethanol–water has a 95.6% ethanol azeotrope, a 50/50 starting mixture is not at that azeotropic composition; the composition of the liquid does change as distillation proceeds.

Conclusion: The liquid remaining in the flask becomes richer in water compared to the starting 50/50 mixture, so option b is correct.

Consider the temperature-composition diagram for water and ethanol: When a 50%/50% mixture of water and ethanol is boiled and a vapour is removed, some of the liquid remains in the distillation flask. That liquid单项选择题

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