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The degree of dissociation $(\alpha)$ of a weak electrolyte $A_xB_y$ is related to van't Hoff factor (i) by the expression

$\begin{array}{1 1} \alpha= \frac{i-1}{(x+y-1)} \\ \alpha= \frac{i-1}{(x+y+1)} \\ \alpha= \frac{x+y-1}{(i-1)} \\ \alpha= \frac{x+y+1}{(i-1)} \end{array} $

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Answer :$\alpha= \frac{i-1}{(x+y-1)}$
The correct expression is $\alpha= \frac{i-1}{(x+y-1)}$
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