The molar conductivity of a 0.02 M solution of propanoic acid is...

The limiting molar conductivity of propanoic acid is given by Kohlrausch's law:

$\Lambda_{m}^{\circ}(\text{C}_{2}\text{H}_{5}\text{COOH}) = \lambda^{\circ}(\text{H}^{+}) + \lambda^{\circ}(\text{C}_{2}\text{H}_{5}\text{COO}^{-})$

Substituting the given values:

$\Lambda_{m}^{\circ} = 349.6 + 40.4 = 390.0 \text{ S cm}^{2} \text{ mol}^{-1}$

The degree of dissociation $\alpha$ is the ratio of molar conductivity at a given concentration to the limiting molar conductivity:

$\alpha = \dfrac{\Lambda_{m}}{\Lambda_{m}^{\circ}}$

$\alpha = \dfrac{31.2}{390.0} = 0.08$

The percentage dissociation is:

$\alpha \times 100 = 0.08 \times 100 = 8.0\%$

Answer: $8.0\%$