Immiscible liquids and boiling: An equimolar, well-agitated mixture of toluene (Tb = 110.6°C) and water (Tb = 100°C) is completely immiscible. At what temperature will the mixture boil at 1 atm total pressure?

Introduction / Context:Immiscible liquid–liquid systems exhibit “steam distillation” behaviour, where the total pressure at boiling equals the sum of the individual vapour pressures. This allows boiling at a temperature below the normal boiling point of either component—a useful principle in separating heat-sensitive organics.

Given Data / Assumptions:

• Toluene normal boiling point: 110.6°C.
• Water normal boiling point: 100°C.
• Liquids are completely immiscible and the system is well agitated so interfacial mass transfer is not limiting.
• Total pressure at boiling is 1 atm.

Concept / Approach:For immiscible liquids A and B, boiling occurs when pA(T) + pB(T) = Ptotal. Because both components contribute vapour pressure, the required temperature to reach 1 atm is lower than the boiling point of either pure component. Composition (e.g., equimolar) of the liquid phase does not directly set the boiling temperature in immiscible systems; the vapour-pressure sum does.

Step-by-Step Solution:

Verification / Alternative check:Classic steam distillation of essential oils follows this: organics distil with steam below their normal boiling points, protecting thermally labile compounds.

Why Other Options Are Wrong:

• 100°C or between 100–110.6°C: Would imply only water’s vapour pressure mattered; incorrect for immiscible pairs.
• 110.6°C: That is toluene’s pure-component boiling point, not the mixture’s.

Common Pitfalls:Confusing immiscible with ideal solutions (Raoult’s law for miscible mixtures). For immiscible liquids, the simple vapour-pressure sum rule applies instead.

Final Answer:Less than 100°C