If equivalent conductance of 1 M benzoic aci | vedclass

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(B) The equivalent conductance at infinite dilution for benzoic acid is calculated using Kohlrausch's law: $\Lambda_{\infty} = \lambda_{\infty}(C_6H_5

Vedclass · Accepted Answer

$3.9$

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(B) The equivalent conductance at infinite dilution for benzoic acid is calculated using Kohlrausch's law: $\Lambda_{\infty} = \lambda_{\infty}(C_6H_5COO^-) + \lambda_{\infty}(H^+)$ $\Lambda_{\infty} = 42 + 288.42 = 330.42 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ The degree of dissociation $\alpha$ is given by the ratio of equivalent conductance at a given concentration to the equivalent conductance at infinite dilution: $\alpha = \frac{\Lambda_c}{\Lambda_{\infty}} = \frac{12.8}{330.42} \approx 0.0387$ To express this as a percentage: $\alpha \% = 0.0387 	imes 100 = 3.87 \% \approx 3.9 \%$

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

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