If the molar conductivities at infinite dilution for $CH_3COOK, K_2SO_4$ and $H_2SO_4$ are respectively, z, y and x ($S\, cm^2\, mol^{-1}$), the molar conductivity of $CH_3COOH$ at infinite dilution will be:

$(x - y + 2z)/2$

The correct answer is Option (1) → $(x - y + 2z)/2$

Given:

• Λ₀(CH₃COOK) = z
• Λ₀(K₂SO₄) = y
• Λ₀(H₂SO₄) = x

Step 1: Express Λ₀ using ionic contributions

CH₃COOK → CH₃COO⁻ + K⁺

$\Lambda_0(CH_3COOK) = \lambda_0(CH_3COO^-) + \lambda_0(K^+) = z$

K₂SO₄ → 2 K⁺ + SO₄²⁻

$\Lambda_0(K_2SO_4) = 2 \lambda_0(K^+) + \lambda_0(SO_4^{2-}) = y$

H₂SO₄ → 2 H⁺ + SO₄²⁻

$\Lambda_0(H_2SO_4) = 2 \lambda_0(H^+) + \lambda_0(SO_4^{2-}) = x$

Step 2: Express Λ₀(CH₃COOH → CH₃COO⁻ + H⁺)

$\Lambda_0(CH_3COOH) = \lambda_0(CH_3COO^-) + \lambda_0(H^+)$

Step 3: Express λ₀(K⁺) and λ₀(SO₄²⁻) from K₂SO₄

$\lambda_0(SO_4^{2-}) = y - 2 \lambda_0(K^+)$

$\lambda_0(CH_3COO^-) = z - \lambda_0(K^+)$

Also from H₂SO₄:

$\lambda_0(H^+) = \frac{x - \lambda_0(SO_4^{2-})}{2} = \frac{x - (y - 2 \lambda_0(K^+))}{2} = \frac{x - y + 2 \lambda_0(K^+)}{2}$

Step 4: Λ₀(CH₃COOH)

$\Lambda_0(CH_3COOH) = \lambda_0(CH_3COO^-) + \lambda_0(H^+) = (z - \lambda_0(K^+)) + \frac{x - y + 2 \lambda_0(K^+)}{2}=\frac{x - y + 2z}{2}$