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(D) Degree of dissociation,$\alpha = \frac{\Lambda^{c}}{\Lambda^{\infty}}$Given,$\Lambda^{c} = 8.0 \, 	ext{mho} \, 	ext{cm}^2$ and $\Lambda^{\infty}

Vedclass · Accepted Answer

$1.25 	imes 10^{-5}$

Vedclass · Answer

(D) Degree of dissociation,$\alpha = \frac{\Lambda^{c}}{\Lambda^{\infty}}$Given,$\Lambda^{c} = 8.0 \, 	ext{mho} \, 	ext{cm}^2$ and $\Lambda^{\infty} = 400 \, 	ext{mho} \, 	ext{cm}^2$.$\alpha = \frac{8.0}{400} = 0.02 = 2 	imes 10^{-2}$.For a weak monobasic acid,the dissociation constant $K_{a}$ is given by Ostwald's dilution law:$K_{a} = \frac{C \alpha^{2}}{1 - \alpha}$.Since $\alpha$ is very small $(1 - \alpha \approx 1)$,the formula simplifies to $K_{a} \approx C \alpha^{2}$.Given concentration $C = \frac{1}{32} \, M$.$K_{a} = \frac{1}{32} 	imes (2 	imes 10^{-2})^{2} = \frac{1}{32} 	imes 4 	imes 10^{-4} = \frac{1}{8} 	imes 10^{-4} = 0.125 	imes 10^{-4} = 1.25 	imes 10^{-5}$.

The equivalent conductance of $M/32$ solution of a weak monobasic acid is $8.0 \, \text{mho} \, \text{cm}^2$ and at infinite dilution is $400 \, \text{mho} \, \text{cm}^2.$ The dissociation constant of this acid is

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