$Λ°_m$ for $NaCl, HCl$, and $NaAc$ are 126.4, 425.9 and $91.0\, S\, cm^2\, mol^{-1}$, respectively. $Λ_m$ for HAc is

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

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

To find the molar conductivity ($\Lambda_m^\circ$) for acetic acid ($HAc$) at infinite dilution, we use Kohlrausch's Law of Independent Migration of Ions.

1. Kohlrausch's Law Principle

The law states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its anions and cations.

$\Lambda_m^\circ(HAc) = \lambda^\circ(H^+) + \lambda^\circ(Ac^-)$

Since we cannot measure the molar conductivity of a weak electrolyte like acetic acid directly by extrapolation, we use values from strong electrolytes ($NaCl, HCl, NaAc$) that contain the same ions.

2. Formulating the Equation

We want to combine the given electrolytes to leave only $H^+$ and $Ac^-$ ions.

• Keep $H^+$: Use $HCl$ ($\Lambda_m^\circ = \lambda^\circ(H^+) + \lambda^\circ(Cl^-)$)
• Keep $Ac^-$: Use $NaAc$ ($\Lambda_m^\circ = \lambda^\circ(Na^+) + \lambda^\circ(Ac^-)$)
• Remove $Na^+$ and $Cl^-$: Subtract $NaCl$ ($\Lambda_m^\circ = \lambda^\circ(Na^+) + \lambda^\circ(Cl^-)$)

The mathematical relationship is:

$\Lambda_m^\circ(HAc) = \Lambda_m^\circ(HCl) + \Lambda_m^\circ(NaAc) - \Lambda_m^\circ(NaCl)$

3. Calculation

Substitute the given values into the equation:

$\Lambda_m^\circ(HAc) = 425.9 + 91.0 - 126.4$

$\Lambda_m^\circ(HAc) = 516.9 - 126.4$

$\Lambda_m^\circ(HAc) = 390.5 \text{ S cm}^2 \text{ mol}^{-1}$