Steam distillation concept check:\nAssume 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?

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:

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