Steam distillation concept check: Assume benzene and water are mutually insoluble. Their normal boiling points are 80.1°C (benzene) and 100°C (water). At a total pressure of 1 atm, what is the boiling point of a liquid–liquid mixture of benzene and water?
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A80.1°C
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BLess than 80.1°C
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C100°C
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DGreater than 80.1°C but less than 100°C
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EGreater than 100°C
Answer
Correct Answer: Less than 80.1°C
Explanation
Introduction / Context:Immiscible (mutually insoluble) liquid mixtures such as benzene and water exhibit a special boiling behavior exploited in steam distillation. Instead of following Raoult’s law for ideal, miscible solutions, the total vapor pressure is the sum of the pure-component vapor pressures, allowing the mixture to boil below the normal boiling point of either component.
Given Data / Assumptions:
- Benzene and water are assumed insoluble in each other (two liquid phases).
- Normal boiling points: benzene = 80.1°C; water = 100°C at 1 atm.
- Total system pressure = 1 atm; boiling occurs when total vapor pressure equals 1 atm.
Concept / Approach:For immiscible liquids, the total vapor pressure P_total at a given temperature equals P_benzene^sat(T) + P_water^sat(T). Boiling begins when P_total reaches the external pressure. Because both components contribute to P_total, this condition is met at a temperature lower than the normal boiling point of either pure liquid.
Step-by-Step Solution:
Identify condition: boil when P_benzene^sat(T_boiling) + P_water^sat(T_boiling) = 1 atm.Since each vapor pressure rises with temperature, their sum reaches 1 atm at a temperature below 80.1°C, because neither component alone must achieve 1 atm.Therefore, T_boiling(mixture) < min(80.1°C, 100°C) = 80.1°C.Verification / Alternative check:Typical steam-distillation temperatures for benzene–water at 1 atm are around the high 60s to low 70s °C, clearly below 80.1°C, confirming the principle.
Why Other Options Are Wrong:
- 80.1°C or 100°C: Applies to pure-component boiling, not an immiscible pair.
- Greater than 80.1°C but less than 100°C, or greater than 100°C: Contrary to the additive vapor pressure rule for immiscible mixtures.
Common Pitfalls:Confusing immiscible behavior with ideal-solution Raoult’s law; forgetting that both components contribute independently to total vapor pressure.
Final Answer:Less than 80.1°C