The specific conductance (k) of 0.02 M aqueo | vedclass

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(A) Step $1$: Calculate the molar conductivity $(\lambda_m)$ at the given concentration.$\lambda_m = \frac{k 	imes 1000}{C} = \frac{1.65 	imes 10^{-

Vedclass · Accepted Answer

$0.021$

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(A) Step $1$: Calculate the molar conductivity $(\lambda_m)$ at the given concentration.$\lambda_m = \frac{k 	imes 1000}{C} = \frac{1.65 	imes 10^{-4} 	imes 1000}{0.02} = 8.25 \ S \ cm^2 \ mol^{-1}$.Step $2$: Calculate the molar conductivity at infinite dilution $(\lambda_m^{\infty})$ using Kohlrausch's law.$\lambda_m^{\infty}(CH_3COOH) = \lambda_{H^{+}}^{\infty} + \lambda_{CH_3COO^{-}}^{\infty} = 349.1 + 40.9 = 390.0 \ S \ cm^2 \ mol^{-1}$.Step $3$: Calculate the degree of dissociation $(\alpha)$.$\alpha = \frac{\lambda_m}{\lambda_m^{\infty}} = \frac{8.25}{390.0} = 0.02115 \approx 0.021$.

Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$	Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$
$H^{+} : 349.6$	$OH^{-} : 199.1$
$Na^{+} : 50.1$	$Cl^{-} : 76.3$
$K^{+} : 73.5$	$Br^{-} : 78.1$
$Ca^{2+} : 119.0$	$CH_3COO^{-} : 40.9$
$Mg^{2+} : 106.0$	$SO_4^{2-} : 160.0$

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