Molar conductivities Λm∘ at infinite dilution of NaCl, HCI and CH3COONa are 126.4,425.9 and 91.0 S cm2mol-1 respectively. Λm∘ for CH3COOH will be

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425.5 S cm2mol−1

180.5 S cm2 mol−1

290.85 S cm2 mol−1

390.5 S cm2 mol−1

answer is B.

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Detailed Solution

CH3COONa+HCl⟶NaCl+CH3COOHΛmCH3COOH∘=ΛmCH3COONa∘+Λm(HCl)∘−Λm(NaCl)0ΛmCH3COOH=(91.0+425.9−126.4)Scm2mol−1ΛmCH3COOH=390.5Scm2mol−1

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