$Λ_m^o$ for $NaCl, HCl$ and $CH_3COONa$ are 126.4, 425.9 and $91.0\, S\, cm^2 mol^{-1}$ respectively. Calculate $Λ_m^o$ for $CH_3COOH$

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

The correct answer is Option (4) → $390.5\, S\, cm^2 mol^{-1}$

We can calculate the limiting molar conductivity (Λ⁰ₘ) of CH₃COOH using Kohlrausch’s law of independent migration of ions:

$\Lambda_m^0(\text{CH₃COOH}) = \Lambda_m^0(\text{CH₃COONa}) + \Lambda_m^0(\text{HCl}) - \Lambda_m^0(\text{NaCl})$

Given:

• Λ⁰ₘ(NaCl) = 126.4 S cm² mol^-1
• Λ⁰ₘ(HCl) = 425.9 S cm² mol^-1
• Λ⁰ₘ(CH₃COONa) = 91.0 S cm² mol^-1

Step 1: Apply Kohlrausch’s law

$\Lambda_m^0(\text{CH₃COOH}) = 91.0 + 425.9 - 126.4$

$\Lambda_m^0$(CH₃COOH) = 390.5 S cm² mol^-1