The limiting molar conductivities for H^+, Na^+, Cl^-, CH_3COO^- are…

The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

List-I Electrolyte	List-II $Λ°_m$
(A) $HCl$	(III) $425.9\,S\, cm^2\, mol^{-1}$
(B) $NaCl$	(II) $126.4\,S\, cm^2\, mol^{-1}$
(C) $CH_3COONa$	(IV) $91.0\,S\, cm^2\, mol^{-1}$
(D) $CH_3COOH$	(I) $390.5\,S\, cm^2\, mol^{-1}$

To calculate the limiting molar conductivities ($\Lambda^\circ_m$) for each electrolyte, we apply Kohlrausch's Law of Independent Migration of Ions, which states that $\Lambda^\circ_m$ for an electrolyte is the sum of the individual molar conductivities of its constituent ions.

Calculation Steps

Given the ionic conductivities ($\lambda^\circ$) at 298 K:

• $\lambda^\circ(H^+) = 349.6 \text{ S cm}^2 \text{ mol}^{-1}$
• $\lambda^\circ(Na^+) = 50.1 \text{ S cm}^2 \text{ mol}^{-1}$
• $\lambda^\circ(Cl^-) = 76.3 \text{ S cm}^2 \text{ mol}^{-1}$
• $\lambda^\circ(CH_3COO^-) = 40.9 \text{ S cm}^2 \text{ mol}^{-1}$

(A) $\Lambda^\circ_m$ for $HCl$

$\Lambda^\circ_m(HCl) = \lambda^\circ(H^+) + \lambda^\circ(Cl^-) = 349.6 + 76.3 = \mathbf{425.9 \text{ S cm}^2 \text{ mol}^{-1}}$

Matches with (III).

(B) $\Lambda^\circ_m$ for $NaCl$

$\Lambda^\circ_m(NaCl) = \lambda^\circ(Na^+) + \lambda^\circ(Cl^-) = 50.1 + 76.3 = \mathbf{126.4 \text{ S cm}^2 \text{ mol}^{-1}}$

Matches with (II).

(C) $\Lambda^\circ_m$ for $CH_3COONa$

$\Lambda^\circ_m(CH_3COONa) = \lambda^\circ(CH_3COO^-) + \lambda^\circ(Na^+) = 40.9 + 50.1 = \mathbf{91.0 \text{ S cm}^2 \text{ mol}^{-1}}$

Matches with (IV).

(D) $\Lambda^\circ_m$ for $CH_3COOH$

$\Lambda^\circ_m(CH_3COOH) = \lambda^\circ(CH_3COO^-) + \lambda^\circ(H^+) = 40.9 + 349.6 = \mathbf{390.5 \text{ S cm}^2 \text{ mol}^{-1}}$

Matches with (I).