The equivalent conductance of 1 , M benzoic a | vedclass

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Question

(B) The molar conductivity at infinite dilution for benzoic acid is calculated using Kohlrausch's law:$\Lambda^0_m (C_6H_5COOH) = \Lambda^0_m(C_6H_5CO

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$3.9$

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(B) The molar conductivity at infinite dilution for benzoic acid is calculated using Kohlrausch's law:$\Lambda^0_m (C_6H_5COOH) = \Lambda^0_m(C_6H_5COO^-) + \Lambda^0_m(H^+) = 42 + 288.42 = 330.42 \, \Omega^{-1} \, 	ext{cm}^2 \, 	ext{mol}^{-1}$.The degree of dissociation $(\alpha)$ is given by the ratio of equivalent conductance at a given concentration to the equivalent conductance at infinite dilution:$\alpha = \frac{\Lambda_m}{\Lambda^0_m} = \frac{12.8}{330.42} \approx 0.0387$.To express this as a percentage:$\alpha \% = 0.0387 	imes 100 = 3.87 \% \approx 3.9 \% $.

The equivalent conductance of $1 \, M$ benzoic acid is $12.8 \, \text{mho} \, \text{cm}^2 \, \text{g-eq}^{-1}$. If the molar conductivities of benzoate ion and $H^+$ ion are $42$ and $288.42 \, \Omega^{-1} \, \text{cm}^2$ respectively,then the degree of dissociation is .......... $\%$.

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