The molar conductivity of a 0.02 M solution of propanoic acid is 31.2 S cm ^2 mol ⁻¹ . What is the percentage dissociation of the acid in this solution? [Given: Limiting molar conductivities of H ^+ and C ₂ H ₅ COO ^- are 349.6 S cm ^2 mol ⁻¹ and 40.4 S cm ^2 mol ⁻¹ respectively]

Step-by-step solution

The limiting molar conductivity of propanoic acid is given by Kohlrausch's law: _ m ^ ( C ₂ H ₅ COOH ) = ^ ( H ⁺) + ^ ( C ₂ H ₅ COO ⁻) Substituting the given values: _ m ^ = 349.6 + 40.4 = 390.0 S cm ² mol ⁻¹ The degree of dissociation is the ratio of molar conductivity at a given concentration to the limiting molar conductivity: = _ m _ m ^ = 31.2 390.0 = 0.08 The percentage dissociation is: 100 = 0.08 100 = 8.0 % Answer: 8.0 %