What is the degree of dissociation of CH_3COO | vedclass

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(C) The degree of dissociation $(\alpha)$ is given by the formula: $\alpha = \frac{\wedge_m^c}{\wedge_m^{\circ}}$.First,calculate the molar conductivi

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$5 	imes 10^{-2}$

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(C) The degree of dissociation $(\alpha)$ is given by the formula: $\alpha = \frac{\wedge_m^c}{\wedge_m^{\circ}}$.First,calculate the molar conductivity at infinite dilution $(\wedge_m^{\circ})$ for $CH_3COOH$ using Kohlrausch's law:$\wedge_m^{\circ}(CH_3COOH) = \wedge^{\circ}(CH_3COO^{-}) + \wedge^{\circ}(H^{+}) = 50 + 350 = 400 \ S \ cm^2 \ mol^{-1}$.Given the molar conductivity at concentration $c$ $(\wedge_m^c)$ is $20 \ S \ cm^2 \ mol^{-1}$.Now,calculate $\alpha$:$\alpha = \frac{20}{400} = \frac{1}{20} = 0.05 = 5 	imes 10^{-2}$.

What is the degree of dissociation of $CH_3COOH$ if $\wedge^{\circ}(CH_3COO^{-}) = 50 \ S \ cm^2 \ mol^{-1}$,$\wedge^{\circ}(H^{+}) = 350 \ S \ cm^2 \ mol^{-1}$ and the molar conductivity of $5 \times 10^{-2} \ M \ CH_3COOH$ is $20 \ S \ cm^2 \ mol^{-1}$?

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