The specific conductance (kappa) of 0.02 M a | vedclass

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(A) Given,specific conductance $(\kappa) = 1.65 	imes 10^{-4} \ S \ cm^{-1}$ and molarity $(M) = 0.02 \ M$. Molar conductance $(\lambda_{m})$ is calc

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$0.021$

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(A) Given,specific conductance $(\kappa) = 1.65 	imes 10^{-4} \ S \ cm^{-1}$ and molarity $(M) = 0.02 \ M$. Molar conductance $(\lambda_{m})$ is calculated as: $\lambda_{m} = \frac{1000 	imes \kappa}{M} = \frac{1000 	imes 1.65 	imes 10^{-4}}{0.02} = 8.25 \ S \ cm^{2} \ mol^{-1}$. Molar conductance at infinite dilution $(\lambda_{m}^{\infty})$ is: $\lambda_{m}^{\infty} = \lambda_{m(H^{+})}^{\infty} + \lambda_{m(CH_{3}COO^{-})}^{\infty} = 349.1 + 40.9 = 390 \ S \ cm^{2} \ mol^{-1}$. The degree of dissociation $(\alpha)$ is given by: $\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\infty}} = \frac{8.25}{390} \approx 0.0211$.

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