What is the molar conductivity of CH_3COOH at | vedclass

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Question

(A) According to Kohlrausch's law, the molar conductivity at infinite dilution $(\Lambda^0_m)$ can be expressed as the sum of the ionic conductivities

Vedclass · Accepted Answer

$\frac{x - y}{2} + z$

Vedclass · Answer

(A) According to Kohlrausch's law, the molar conductivity at infinite dilution $(\Lambda^0_m)$ can be expressed as the sum of the ionic conductivities.Given:$\Lambda^0_m(H_2SO_4) = 2\lambda^0(H^+) + \lambda^0(SO_4^{2-}) = x$$\Lambda^0_m(K_2SO_4) = 2\lambda^0(K^+) + \lambda^0(SO_4^{2-}) = y$$\Lambda^0_m(CH_3COOK) = \lambda^0(CH_3COO^-) + \lambda^0(K^+) = z$We need to find $\Lambda^0_m(CH_3COOH) = \lambda^0(H^+) + \lambda^0(CH_3COO^-)$.From the given equations:$\frac{\Lambda^0_m(H_2SO_4) - \Lambda^0_m(K_2SO_4)}{2} = \frac{(2\lambda^0(H^+) + \lambda^0(SO_4^{2-})) - (2\lambda^0(K^+) + \lambda^0(SO_4^{2-}))}{2} = \lambda^0(H^+) - \lambda^0(K^+) = \frac{x - y}{2}$.Adding this to $\Lambda^0_m(CH_3COOK)$:$\frac{x - y}{2} + z = (\lambda^0(H^+) - \lambda^0(K^+)) + (\lambda^0(CH_3COO^-) + \lambda^0(K^+)) = \lambda^0(H^+) + \lambda^0(CH_3COO^-) = \Lambda^0_m(CH_3COOH)$.Thus, the correct option is $A$.

What is the molar conductivity of $CH_3COOH$ at infinite dilution if the molar conductivities of $H_2SO_4$, $K_2SO_4$, and $CH_3COOK$ at infinite dilution are respectively $x$, $y$, and $z$ $\Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}$?

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