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Q.Ionic conductance at infinite dilution of Al3+\text{Al}^{3+} and SO42−\text{SO}_4^{2-} are 189 and 160 mho cm² equiv⁻¹ respectively. Calculate the equivalent and molar conductance of the electrolyte Al2(SO4)3\text{Al}_2(\text{SO}_4)_3 at infinite dilution.

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Step 1. Al2(SO4)3\text{Al}_2(\text{SO}_4)_3 dissociates as Al2(SO4)3→2Al3++3SO42−\text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+}+3\text{SO}_4^{2-}.

Step 2. Equivalent conductance at infinite dilution: since the given ionic conductances (189 and 160 mho cm² equiv⁻¹) are already on a per-equivalent basis, they add directly regardless of charge: Λeqo=λo(Al3+)+λo(SO42−)=189+160=349\Lambda_{eq}^{o} = \lambda^{o}(\text{Al}^{3+}) + \lambda^{o}(\text{SO}_4^{2-}) = 189+160 = 349 mho cm² equiv⁻¹.

Step 3. Molar conductance requires converting to a per-mole basis. The total number of equivalents per mole of Al2(SO4)3\text{Al}_2(\text{SO}_4)_3 (its n-factor) is found from either ion: 2 mol Al³⁺ × charge 3 = 6, or 3 mol SO₄²⁻ × charge 2 = 6 — consistently 6. …

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