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NEET CHEMISTRYMedium

The correct expression that represents the equivalent conductance at infinite dilution of Al2(SO4)3Al_2(SO_4)_3 is: (Given that ΛAl3+∘\Lambda^{\circ}_{Al^{3+}} and ΛSO42−∘\Lambda^{\circ}_{SO_4^{2-}} are the equivalent conductances at infinite dilution of the respective ions)

A

ΛAl3+∘+ΛSO42−∘\Lambda^{\circ}_{Al^{3+}} + \Lambda^{\circ}_{SO_4^{2-}}

B

(ΛAl3+∘+ΛSO42−∘)×6(\Lambda^{\circ}_{Al^{3+}} + \Lambda^{\circ}_{SO_4^{2-}}) \times 6

C

13ΛAl3+∘+12ΛSO42−∘\frac{1}{3}\Lambda^{\circ}_{Al^{3+}} + \frac{1}{2}\Lambda^{\circ}_{SO_4^{2-}}

D

2ΛAl3+∘+3ΛSO42−∘2\Lambda^{\circ}_{Al^{3+}} + 3\Lambda^{\circ}_{SO_4^{2-}}

Step-by-Step Solution

According to Kohlrausch's law of independent migration of ions, the limiting equivalent conductance of an electrolyte is the algebraic sum of the limiting equivalent conductances of its constituent cations and anions. Therefore, for Al2(SO4)3Al_2(SO_4)_3, the equivalent conductance at infinite dilution is Λeq∘=ΛAl3+∘+ΛSO42−∘\Lambda^{\circ}_{eq} = \Lambda^{\circ}_{Al^{3+}} + \Lambda^{\circ}_{SO_4^{2-}}. Note that if the question asked for molar conductance, it would be Λm∘=2Λm∘(Al3+)+3Λm∘(SO42−)\Lambda^{\circ}_{m} = 2\Lambda^{\circ}_{m}(Al^{3+}) + 3\Lambda^{\circ}_{m}(SO_4^{2-}).

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