The equivalent conductance of a monobasic aci | vedclass

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(D) Equivalent conductivity at infinite dilution $\Lambda^{\circ}_{eq} = 348 \, \Omega^{-1} \, cm^2 \, eq^{-1}$.Normality $(N)$ of the solution $= \fr

Vedclass · Accepted Answer

$51.7$

Vedclass · Answer

(D) Equivalent conductivity at infinite dilution $\Lambda^{\circ}_{eq} = 348 \, \Omega^{-1} \, cm^2 \, eq^{-1}$.Normality $(N)$ of the solution $= \frac{	ext{mass}}{	ext{equivalent mass} 	imes 	ext{volume in } L} = \frac{15}{49 	imes 1} \approx 0.3061 \, N$.Conductivity $(\kappa) = \frac{1}{	ext{resistivity}} = \frac{1}{18.5} \approx 0.05405 \, \Omega^{-1} \, cm^{-1}$.Equivalent conductivity $(\Lambda_{eq}) = \frac{\kappa 	imes 1000}{N} = \frac{0.05405 	imes 1000}{0.3061} \approx 176.58 \, \Omega^{-1} \, cm^2 \, eq^{-1}$.Degree of dissociation $(\alpha) = \frac{\Lambda_{eq}}{\Lambda^{\circ}_{eq}} = \frac{176.58}{348} \approx 0.5074$.Percentage dissociation $= 0.5074 	imes 100 \approx 50.74 \, \%$,which is closest to $51.7 \, \%$.

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