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If the molar conductivity ($\Lambda_m$) of a $0.050\text{ mol L}^{-1}$ solution of a monobasic weak acid is $90\text{ S cm}^2\text{mol}^{-1}$, its extent (degree) of dissociation will be [Assume $\lambda_+^\circ = 349.6\text{ S cm}^2\text{mol}^{-1}$ and $\lambda_-^\circ = 50.4\text{ S cm}^2\text{mol}^{-1}$]:
A
0.115
B
0.125
C
0.225
D
0.215
Explanation
$\Lambda_m^\circ = 349.6 + 50.4 = 400.0\text{ S cm}^2\text{mol}^{-1}$. Degree of dissociation $\alpha = \Lambda_m / \Lambda_m^\circ = 90 / 400 = 0.225$.
Detailed Solution
According to Kohlrausch's law of independent migration of ions, limiting molar conductivity is $\Lambda_m^\circ = \lambda_+^\circ + \lambda_-^\circ = 349.6 + 50.4 = 400.0\text{ S cm}^2\text{mol}^{-1}$. The degree of dissociation $\alpha$ is given by $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{90\text{ S cm}^2\text{mol}^{-1}}{400.0\text{ S cm}^2\text{mol}^{-1}} = 0.225$.