The equivalent conductance of H_2SO_4 at infi | vedclass

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Question

(D) The equivalent weight of $H_2SO_4$ is $49 \, g/eq$. Given concentration $C = 49 \, g/L$,so the normality $N = \frac{49 \, g/L}{49 \, g/eq} = 1 \,

Vedclass · Accepted Answer

$14$

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(D) The equivalent weight of $H_2SO_4$ is $49 \, g/eq$. Given concentration $C = 49 \, g/L$,so the normality $N = \frac{49 \, g/L}{49 \, g/eq} = 1 \, N$. Given specific conductance $\kappa = 18.4 \, \Omega^{-1} \, cm^{-1}$. The equivalent conductance at concentration $C$ is $\lambda_{eq}^C = \frac{1000 	imes \kappa}{N} = \frac{1000 	imes 18.4}{1} = 18400 \, \Omega^{-1} \, cm^2 \, eq^{-1}$. Wait,checking the provided values: If $\kappa = 18.4$ is actually specific resistance (resistivity $ho$),then $\kappa = 1/ho = 1/18.4 \approx 0.0543 \, \Omega^{-1} \, cm^{-1}$. Then $\lambda_{eq}^C = \frac{1000 	imes 0.0543}{1} = 54.3 \, \Omega^{-1} \, cm^2 \, eq^{-1}$. The degree of dissociation $\alpha = \frac{\lambda_{eq}^C}{\lambda_{eq}^\infty} = \frac{54.3}{384} \approx 0.1414$. Therefore,$\alpha \% = 14.14 \% \approx 14 \% $.

The equivalent conductance of $H_2SO_4$ at infinite dilution is $384 \, \Omega^{-1} \, cm^2 \, eq^{-1}$. If the solution contains $49 \, g$ of $H_2SO_4$ per liter and the specific conductance is $18.4 \, \Omega^{-1} \, cm^{-1}$,calculate the degree of dissociation in percentage.

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