Question

The conductivity of a 0.20 M solution of KCl is \(2.48 \times 10^{-2} \, \text{S cm}^{-1}\). Calculate its molar conductivity and degree of dissociation.

Hint:

Molar conductivity \(\Lambda_m\) can be calculated by dividing the conductivity (\(\kappa\)) by the concentration, and degree of dissociation is the ratio of actual molar conductivity to its value at infinite dilution.

Answer 1

The molar conductivity \(\Lambda_m\) is given by the formula: \[ \Lambda_m = \kappa \times \dfrac{1000}{C} \] where \(\kappa = 2.48 \times 10^{-2} \, \text{S cm}^{-1}\) and \(C = 0.20 \, \text{mol L}^{-1}\). \[ \Lambda_m = 2.48 \times 10^{-2} \times \dfrac{1000}{0.20} = 124 \, \text{S cm}^2 \text{mol}^{-1} \] The degree of dissociation \(\alpha\) is given by the relation: \[ \alpha = \dfrac{\Lambda_m}{\Lambda_m^0} \] where \(\Lambda_m^0\) is the molar conductivity at infinite dilution, which for KCl is 142.7 S cm² mol⁻¹. Thus, \[ \alpha = \dfrac{124}{142.7} = 0.87 \]