Ionic conductance of Ga^{3+} ions and NO_3^- | vedclass

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(B) The equivalent conductance at infinite dilution for an electrolyte is the sum of the equivalent conductances of its constituent ions: $\Lambda_{eq

Vedclass · Accepted Answer

$90$

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(B) The equivalent conductance at infinite dilution for an electrolyte is the sum of the equivalent conductances of its constituent ions: $\Lambda_{eq}^{\circ} = \lambda_{eq}^{\circ}(Ga^{3+}) + \lambda_{eq}^{\circ}(NO_3^-)$.First,convert molar ionic conductance to equivalent ionic conductance using the relation $\lambda_{eq}^{\circ} = \frac{\lambda_{m}^{\circ}}{n}$,where $n$ is the charge on the ion.For $Ga^{3+}$,$\lambda_{eq}^{\circ}(Ga^{3+}) = \frac{120}{3} = 40 \, \Omega^{-1} \, cm^2 \, eq^{-1}$.For $NO_3^-$,$\lambda_{eq}^{\circ}(NO_3^-) = \frac{50}{1} = 50 \, \Omega^{-1} \, cm^2 \, eq^{-1}$.Therefore,$\Lambda_{eq}^{\circ}(Ga(NO_3)_3) = 40 + 50 = 90 \, \Omega^{-1} \, cm^2 \, eq^{-1}$.

Ionic conductance of $Ga^{3+}$ ions and $NO_3^-$ ions at infinite dilution are $120 \, \Omega^{-1} \, cm^2 \, mol^{-1}$ and $50 \, \Omega^{-1} \, cm^2 \, mol^{-1}$ respectively. Equivalent conductance of $Ga(NO_3)_3$ at infinite dilution will be ........... $\Omega^{-1} \, cm^2 \, eq^{-1}$.

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