85) For a concentrated solution of a weak electrolytic A_x B_y of concentration C, the degree of dissociation is $\alpha$ given by
$$\begin{gathered} A) \alpha =\sqrt{K_{eq}/\left(x+y\right)} \\ B) \alpha =\sqrt{K_{eq}C/\left(x\times y\right)} \\ C) \alpha =K_{eq}/C^{x+y-1} X^x{Y}^Y \\ D) \alpha =K_{eq}{C}^{xy} \\ E) \alpha =K_{eq}/C^{xy} \end{gathered}$$

Dear Student,

When the electrolyte dissociates:

$$\begin{gathered} A_x{B}_y\to x{A}^{y+} + y{B}^{x-} \\ at t=0 c 0 0 \\ at equilibrium c-c\alpha xc\alpha yc\alpha \\ {K}_{eq} = \frac{( xc\alpha {)}^x( yc\alpha {)}^y}{c(1-\alpha )} = \frac{c^{x+y}{\alpha }^{x+y}{x}^x{y}^y}{c (1-\alpha )} \\ {K}_{eq} = c^{x+y-1} {\alpha }^{x+y} x^x{y}^y ( 1-\alpha \approx 1 ) \\ {\alpha }^{x+y} = \frac{K_{eq}}{c^{x+y-1} x^x{y}^y} \\ \alpha = {\left(\frac{K_{eq}}{c^{x+y-1} x^x{y}^y}\right)}^{1/x+y} \end{gathered}$$
The correct option would be the third one but its misprinted as it doesn't have the power mentioned in the answer.

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