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Q.Calculate the limiting molar conductivity of Cl−\text{Cl}^{-} ion by using the data :
λCa2+∘=119.0 S cm2 mol−1\lambda^{\circ}_{\text{Ca}^{2+}} = 119.0\,\text{S cm}^2\,\text{mol}^{-1} and Λm∘\Lambda^{\circ}_m for CaCl2=271.6 S cm2 mol−1\text{CaCl}_2 = 271.6\,\text{S cm}^2\,\text{mol}^{-1}.

Karnataka PUCKarnataka II PUC Board 2026Subjective· 3mImportance★★★★★
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Kohlrausch's law of independent migration gives λCl−∘=(271.6−119.0)/2=76.3 S cm2 mol−1\lambda^\circ_{\text{Cl}^-} = (271.6 - 119.0)/2 = 76.3\,\text{S cm}^2\,\text{mol}^{-1}.

By Kohlrausch's law, the limiting molar conductivity of CaCl2\text{CaCl}_2 is the sum of the ionic contributions (with the appropriate stoichiometric numbers):

Λm∘(CaCl2)=λCa2+∘+2 λCl−∘\Lambda^\circ_m(\text{CaCl}_2) = \lambda^\circ_{\text{Ca}^{2+}} + 2\,\lambda^\circ_{\text{Cl}^-}

Substituting the given data:

271.6=119.0+2 λCl−∘271.6 = 119.0 + 2\,\lambda^\circ_{\text{Cl}^-} …

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