If the equivalent conductance of 1 M benzoic | vedclass

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(B) The equivalent conductance at infinite dilution for benzoic acid is calculated using Kohlrausch's law: $\Lambda_{eq}^o (C_6H_5COOH) = \Lambda_{eq}

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$3.9$

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(B) The equivalent conductance at infinite dilution for benzoic acid is calculated using Kohlrausch's law: $\Lambda_{eq}^o (C_6H_5COOH) = \Lambda_{eq}^o (C_6H_5COO^-) + \Lambda_{eq}^o (H^+) = 42 + 288.42 = 330.42 \ \Omega^{-1} \ cm^2 \ eq^{-1}$. The degree of dissociation $(\alpha)$ is given by the ratio of equivalent conductance at a given concentration $(\Lambda_{eq}^c)$ to the equivalent conductance at infinite dilution $(\Lambda_{eq}^o)$: $\alpha = \frac{\Lambda_{eq}^c}{\Lambda_{eq}^o} = \frac{12.8}{330.42} \approx 0.03874$. To express this as a percentage: $\alpha \% = 0.03874 	imes 100 = 3.874 \% \approx 3.9 \%$.

If the equivalent conductance of $1 \ M$ benzoic acid is $12.8 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ and the equivalent conductances of benzoate ion and $H^{+}$ ion are $42$ and $288.42 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively,its degree of dissociation is ............. $\%$

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