The molar conductances of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91 ohm^-1 cm² mol^-1 respectively. The molar conductance of CH₃COOH at infinite dilution is

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698.28 ohm^-1cm² mol^- 1

390.71 ohm^-1cm² mol^- 1

540.48 ohm^-1cm² mol^- 1

201.28 ohm^-1cm² mol^- 1

answer is B.

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Detailed Solution

$\land_{m}^{0} ({CH}_{3} COOH) - \land^{0} ({CH}_{3} COONa) + \land^{0} (HCl) - \land^{0} (NaCl) = 91 + 426.16 - 126.45 = 390.71 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$

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