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85) For a concentrated solution of a weak electrolytic A_{x} B_{y} of concentration C, the degree of dissociation is $\alpha $ given by

$A)\alpha =\sqrt{{K}_{eq}/\left(x+y\right)}\phantom{\rule{0ex}{0ex}}B)\alpha =\sqrt{{K}_{eq}C/\left(x\times y\right)}\phantom{\rule{0ex}{0ex}}C)\alpha ={K}_{eq}/{C}^{x+y-1}{X}^{x}{Y}^{Y}\phantom{\rule{0ex}{0ex}}D)\alpha ={K}_{eq}{C}^{xy}\phantom{\rule{0ex}{0ex}}E)\alpha ={K}_{eq}/{C}^{xy}$

When the electrolyte dissociates:

${A}_{x}{B}_{y}\to x{A}^{y+}+y{B}^{x-}\phantom{\rule{0ex}{0ex}}att=0c00\phantom{\rule{0ex}{0ex}}atequilibriumc-c\alpha xc\alpha yc\alpha \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{K}_{eq}=\frac{(xc\alpha {)}^{x}(yc\alpha {)}^{y}}{c(1-\alpha )}=\frac{{c}^{x+y}{\alpha}^{x+y}{x}^{x}{y}^{y}}{c(1-\alpha )}\phantom{\rule{0ex}{0ex}}{K}_{eq}={c}^{x+y-1}{\alpha}^{x+y}{x}^{x}{y}^{y}(1-\alpha \approx 1)\phantom{\rule{0ex}{0ex}}{\alpha}^{x+y}=\frac{{K}_{eq}}{{c}^{x+y-1}{x}^{x}{y}^{y}}\phantom{\rule{0ex}{0ex}}\alpha ={\left(\frac{{K}_{eq}}{{c}^{x+y-1}{x}^{x}{y}^{y}}\right)}^{1/x+y}\phantom{\rule{0ex}{0ex}}$

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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