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Which of the following expressions correctly represents the equivalent conductance at infinite dilution of $Al_2(SO_4)_3$? Given that $\Lambda^\circ_{Al^{3+}}$ and $\Lambda^\circ_{SO_4^{2-}}$ are the equivalent conductances at infinite dilution of the respective ions.
A
$2\Lambda^\circ_{Al^{3+}} + 3\Lambda^\circ_{SO_4^{2-}}$
B
$\Lambda^\circ_{Al^{3+}} + \Lambda^\circ_{SO_4^{2-}}$
C
$(\Lambda^\circ_{Al^{3+}} + \Lambda^\circ_{SO_4^{2-}})\times6$
D
$\frac{1}{3}\Lambda^\circ_{Al^{3+}} + \frac{1}{2}\Lambda^\circ_{SO_4^{2-}}$
Detailed Solution
Kohlrausch's law: at infinite dilution each ion makes its own contribution to the conductance, independent of the other ion.
In terms of equivalent conductances, one equivalent of any electrolyte contains one equivalent of the cation and one equivalent of the anion.
So $\Lambda^\circ_{eq}(\text{electrolyte}) = \Lambda^\circ_{eq}(\text{cation}) + \Lambda^\circ_{eq}(\text{anion})$, with no stoichiometric multipliers.
The ionic values given are already equivalent conductances, so: $\Lambda^\circ_{eq}[Al_2(SO_4)_3] = \Lambda^\circ_{Al^{3+}} + \Lambda^\circ_{SO_4^{2-}}$
(The form $2\Lambda^\circ + 3\Lambda^\circ$ would apply only if molar ionic conductances were given and the molar conductance were asked.)
In terms of equivalent conductances, one equivalent of any electrolyte contains one equivalent of the cation and one equivalent of the anion.
So $\Lambda^\circ_{eq}(\text{electrolyte}) = \Lambda^\circ_{eq}(\text{cation}) + \Lambda^\circ_{eq}(\text{anion})$, with no stoichiometric multipliers.
The ionic values given are already equivalent conductances, so: $\Lambda^\circ_{eq}[Al_2(SO_4)_3] = \Lambda^\circ_{Al^{3+}} + \Lambda^\circ_{SO_4^{2-}}$
(The form $2\Lambda^\circ + 3\Lambda^\circ$ would apply only if molar ionic conductances were given and the molar conductance were asked.)