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(B) The equivalent conductance at concentration $c$ is given as $\lambda_{eq}^{c} = 12.8 \ ohm^{-1} \ cm^{2} \ eq^{-1}$.According to Kohlrausch's law,

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$3.9$

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(B) The equivalent conductance at concentration $c$ is given as $\lambda_{eq}^{c} = 12.8 \ ohm^{-1} \ cm^{2} \ eq^{-1}$.According to Kohlrausch's law,the equivalent conductance at infinite dilution is the sum of the ionic conductances of the constituent ions:$\lambda_{eq}^{\infty}(C_{6}H_{5}COOH) = \lambda_{eq}^{\infty}(C_{6}H_{5}COO^{-}) + \lambda_{eq}^{\infty}(H^{+})$Substituting the given values:$\lambda_{eq}^{\infty} = 42 + 288.42 = 330.42 \ ohm^{-1} \ cm^{2} \ eq^{-1}$.The degree of dissociation $\alpha$ is calculated as:$\alpha = \frac{\lambda_{eq}^{c}}{\lambda_{eq}^{\infty}} = \frac{12.8}{330.42} \approx 0.03874$.To express this as a percentage:$\% \alpha = 0.03874 	imes 100 \approx 3.874 \% \approx 3.9 \%$.

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