The limiting molar conductivities for $NaCl, HCl$ and $CH_3COONa$ are 126.4, 425.9 and $91.0\, S\, cm^2\, mol^{-1}$, respectively. Calculate the limiting molar conductivity for acetic acid.

$390.5\, S\, cm^2\, mol^{-1}$

The correct answer is Option (1) → $390.5\, S\, cm^2\, mol^{-1}$

Use Kohlrausch’s law of independent migration of ions.

$\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{HCl}) + \Lambda_m^\circ(\text{CH}_3\text{COONa}) - \Lambda_m^\circ(\text{NaCl})$

Substitute the given values:

$=425.9+91.0−126.4$

$= 390.5\ \text{S cm}^2\text{ mol}^{-1}$