Boiling of immiscible liquid mixtures: The boiling point of a heterogeneous mixture of immiscible liquids is __________ the boiling point of any single component.

Introduction / Context:Steam distillation and co-distillation exploit the behavior of immiscible liquid mixtures, which boil when the sum of their vapor pressures equals the external pressure. This allows boiling at a temperature lower than the normal boiling point of either component, enabling gentle recovery of thermally sensitive compounds.

Given Data / Assumptions:

• Two immiscible liquids in equilibrium with their vapors.
• External (usually atmospheric) pressure fixed.
• Ideal or near-ideal additivity of vapor pressures for immiscible systems.

Concept / Approach:For immiscible liquids A and B, the boiling condition at temperature T is p_A^sat(T) + p_B^sat(T) = P_ext. Since each pure-component saturated vapor pressure is less than P_ext at T lower than their individual boiling points, their sum can reach P_ext at a temperature below either pure boiling point. Therefore, the mixture boils at a lower temperature than any component alone.

Step-by-Step Solution:

Verification / Alternative check:Practical steam distillation of essential oils demonstrates boiling of oil–water mixtures well below the normal boiling point of the oil, confirming the rule.

Why Other Options Are Wrong:

• Higher than or equal to: contradicts vapor-pressure additivity for immiscible pairs.
• Either (a) or (b) and undetermined: do not apply; the direction is definite for immiscible systems.

Common Pitfalls:Confusing immiscible with miscible mixtures (Raoult’s law). For miscible systems, behavior depends on activity coefficients; for immiscible systems, the simple pressure-sum rule applies.

Final Answer:lower than