At 25^{circ} C,the molar conductance of 0.007 | vedclass

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(D) Degree of dissociation,$\alpha = \frac{\Lambda_{m}^{c}}{\Lambda_{m}^{\circ}} = \frac{150}{500} = 0.3$Given,$C = 0.007 \ M$Hydrofluoric acid dissoc

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$9 	imes 10^{-4} \ M$

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(D) Degree of dissociation,$\alpha = \frac{\Lambda_{m}^{c}}{\Lambda_{m}^{\circ}} = \frac{150}{500} = 0.3$Given,$C = 0.007 \ M$Hydrofluoric acid dissociates as follows:$HF ightleftharpoons H^{+} + F^{-}$Initially: $C, 0, 0$At equilibrium: $C(1 - \alpha), C\alpha, C\alpha$Dissociation constant,$K_{a} = \frac{[H^{+}] [F^{-}]}{[HF]} = \frac{C \alpha \cdot C \alpha}{C(1 - \alpha)} = \frac{C \alpha^{2}}{(1 - \alpha)}$On substituting the values,we get:$K_{a} = \frac{0.007 	imes (0.3)^{2}}{(1 - 0.3)} = \frac{0.007 	imes 0.09}{0.7} = 0.01 	imes 0.09 = 9 	imes 10^{-4} \ M$

At $25^{\circ} C$,the molar conductance of $0.007 \ M$ hydrofluoric acid is $150 \ \text{mho} \ cm^{2} \ mol^{-1}$ and its $\Lambda_{m}^{\circ} = 500 \ \text{mho} \ cm^{2} \ mol^{-1}$. The value of the dissociation constant of the acid at the given concentration at $25^{\circ} C$ is

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