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If equivalent conductance of $1 \mathrm{M}$ benzoic acid is $12.8 \mathrm{ohm}^{-1} \mathrm{~cm}^2$ and if the conductance of benzoate ion and $\mathrm{H}^{+}$ion are 42 and $288.42 \mathrm{ohm}^{-1} \mathrm{~cm}^2$ respectively. its degree of dissociation is
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$3.9 \%$
$A_{m\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right)}^o=A_{\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right)}^o+A_{\left(\mathrm{H}^{+}\right)}^o$
$=42+288.42=330.42$
$\alpha=\frac{\Lambda_m^c}{\Lambda_m^o}=\frac{12.8}{330.42}=3.9 \%$
$=42+288.42=330.42$
$\alpha=\frac{\Lambda_m^c}{\Lambda_m^o}=\frac{12.8}{330.42}=3.9 \%$
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