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(B) For the dissociation of a weak electrolyte $A_{x}B_{y} \rightleftharpoons xA^{y+} + yB^{x-}$,let $\alpha$ be the degree of dissociation.At equilib

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$(\frac{K}{c^{x+y-1} x^{x} y^{y}})^{\frac{1}{x+y}}$

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(B) For the dissociation of a weak electrolyte $A_{x}B_{y} \rightleftharpoons xA^{y+} + yB^{x-}$,let $\alpha$ be the degree of dissociation.At equilibrium,the concentrations are: $[A_{x}B_{y}] = c(1-\alpha)$,$[A^{y+}] = xc\alpha$,and $[B^{x-}] = yc\alpha$.The dissociation constant $K$ is given by: $K = \frac{[A^{y+}]^{x} [B^{x-}]^{y}}{[A_{x}B_{y}]} = \frac{(xc\alpha)^{x} (yc\alpha)^{y}}{c(1-\alpha)}$.Since $A_{x}B_{y}$ is a weak electrolyte,$\alpha \ll 1$,therefore $(1-\alpha) \approx 1$.Thus,$K = \frac{x^{x} c^{x} \alpha^{x} y^{y} c^{y} \alpha^{y}}{c} = c^{x+y-1} x^{x} y^{y} \alpha^{x+y}$.Solving for $\alpha$,we get: $\alpha^{x+y} = \frac{K}{c^{x+y-1} x^{x} y^{y}}$.Therefore,$\alpha = (\frac{K}{c^{x+y-1} x^{x} y^{y}})^{\frac{1}{x+y}}$.

Given is a concentrated solution of a weak electrolyte $A_{x}B_{y}$ of concentration '$c$' and dissociation constant '$K$'. The degree of dissociation is given by:

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