Solutions and Their Concentration — Part 2: Raoult’s law, non-ideal solutions and azeotropes: visual lesson, ISC Class 12 Chemistry

Lesson

Vapour pressure of solutions

Part 2 — how the vapour above a mixture of liquids behaves, and why some mixtures cannot be separated by distillation.

In this lesson you will

Apply Raoult’s law to a mixture of two volatile liquids and to a solution of a non-volatile solute
Compare ideal and non-ideal solutions in terms of intermolecular forces, enthalpy and volume of mixing
Distinguish minimum-boiling and maximum-boiling azeotropes with examples and graphs
Explain why an azeotrope cannot be separated by fractional distillation

Concept

Raoult’s law for two volatile liquids

In a mixture of volatile liquids A and B, each partial vapour pressure is proportional to the mole fraction of that liquid. By Dalton’s law the total pressure is the sum. The vapour is richer in the more volatile component.

p_\text{A} = p^\circ_\text{A}\, x_\text{A}, \qquad p_\text{B} = p^\circ_\text{B}\, x_\text{B}

p_\text{total} = p^\circ_\text{A} + (p^\circ_\text{B} - p^\circ_\text{A})\, x_\text{B}

y_\text{A} = \dfrac{p_\text{A}}{p_\text{total}}

Diagram

Raoult’s law on a graph

Partial pressure of A
Partial pressure of B
Total vapour pressure

Illustrative values: pure A 24 kPa, pure B 64 kPa at the same temperature.

For an ideal solution, both partial pressures and the total pressure are straight lines against composition.

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