What is the molar conductivity of CH_3-CO_2H | vedclass

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(A) According to Kohlrausch's law of independent migration of ions:$(\Lambda_{m}^{\circ})_{CH_3-COOH} = (\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambd

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$\frac{x_1-x_2}{2} + x_3$

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(A) According to Kohlrausch's law of independent migration of ions:$(\Lambda_{m}^{\circ})_{CH_3-COOH} = (\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambda_{m}^{\circ})_{H^{+}}$We have:$(\Lambda_{m}^{\circ})_{(CH_3-COO)_2Ba} = 2(\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambda_{m}^{\circ})_{Ba^{2+}} = x_1$$(\Lambda_{m}^{\circ})_{BaCl_2} = (\Lambda_{m}^{\circ})_{Ba^{2+}} + 2(\Lambda_{m}^{\circ})_{Cl^{-}} = x_2$$(\Lambda_{m}^{\circ})_{HCl} = (\Lambda_{m}^{\circ})_{H^{+}} + (\Lambda_{m}^{\circ})_{Cl^{-}} = x_3$To get $CH_3-COOH$,we perform the following operation:$\frac{1}{2} [(\Lambda_{m}^{\circ})_{(CH_3-COO)_2Ba} - (\Lambda_{m}^{\circ})_{BaCl_2}] + (\Lambda_{m}^{\circ})_{HCl}$$= \frac{1}{2} [2(\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambda_{m}^{\circ})_{Ba^{2+}} - (\Lambda_{m}^{\circ})_{Ba^{2+}} - 2(\Lambda_{m}^{\circ})_{Cl^{-}}] + (\Lambda_{m}^{\circ})_{H^{+}} + (\Lambda_{m}^{\circ})_{Cl^{-}}$$= (\Lambda_{m}^{\circ})_{CH_3-COO^{-}} - (\Lambda_{m}^{\circ})_{Cl^{-}} + (\Lambda_{m}^{\circ})_{H^{+}} + (\Lambda_{m}^{\circ})_{Cl^{-}}$$= (\Lambda_{m}^{\circ})_{CH_3-COO^{-}} + (\Lambda_{m}^{\circ})_{H^{+}} = (\Lambda_{m}^{\circ})_{CH_3-COOH}$Therefore,the value is $\frac{x_1-x_2}{2} + x_3$.

What is the molar conductivity of $CH_3-CO_2H$ at infinite dilution?
Given that,
$\wedge_{m}^{o}(CH_3-CO_2)_2Ba = x_1 \ S \ cm^2 \ mol^{-1}$
$\wedge_{m}^{o}(BaCl_2) = x_2 \ S \ cm^2 \ mol^{-1}$
$\wedge_{m}^{o}(HCl) = x_3 \ S \ cm^2 \ mol^{-1}$

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What is the molar conductivity of $CH_3-CO_2H$ at infinite dilution?Given that,$\wedge_{m}^{o}(CH_3-CO_2)_2Ba = x_1 \ S \ cm^2 \ mol^{-1}$$\wedge_{m}^{o}(BaCl_2) = x_2 \ S \ cm^2 \ mol^{-1}$$\wedge_{m}^{o}(HCl) = x_3 \ S \ cm^2 \ mol^{-1}$