At 298 K,the limiting molar conductivity of | vedclass

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(C) The degree of dissociation is given by $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.For the initial solution,$\alpha = \frac{y 	imes 10^2}{4 	im

Vedclass · Accepted Answer

$0.21, 0.86$

Vedclass · Answer

(C) The degree of dissociation is given by $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.For the initial solution,$\alpha = \frac{y 	imes 10^2}{4 	imes 10^2} = \frac{y}{4}$.The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1-\alpha} = \frac{C (\Lambda_m / \Lambda_m^\circ)^2}{1 - (\Lambda_m / \Lambda_m^\circ)} = \frac{C \Lambda_m^2}{\Lambda_m^\circ(\Lambda_m^\circ - \Lambda_m)}$.For the initial solution: $K_a = \frac{C (y 	imes 10^2)^2}{4 	imes 10^2 (4 	imes 10^2 - y 	imes 10^2)} = \frac{C y^2}{4(4-y)}$.After $20$ times dilution,the concentration becomes $C' = C/20$ and molar conductivity becomes $\Lambda_m' = 3y 	imes 10^2$.$K_a = \frac{(C/20) (3y 	imes 10^2)^2}{4 	imes 10^2 (4 	imes 10^2 - 3y 	imes 10^2)} = \frac{C (9y^2)}{20 	imes 4 (4-3y)} = \frac{9 C y^2}{80(4-3y)}$.Equating the two expressions for $K_a$: $\frac{C y^2}{4(4-y)} = \frac{9 C y^2}{80(4-3y)}$.$\frac{1}{4-y} = \frac{9}{20(4-3y)}$ $\Rightarrow 20(4-3y) = 9(4-y)$ $\Rightarrow 80 - 60y = 36 - 9y$.$44 = 51y \Rightarrow y = \frac{44}{51} \approx 0.86$.Then $\alpha = \frac{y}{4} = \frac{44/51}{4} = \frac{11}{51} \approx 0.2156 \approx 0.21$.

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