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The degree of dissociation (α ) of a weak electrolyte A_xB_y is related to Van't Hoff factor (i) by the expression

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$\large \alpha \; = \;\frac{{i - 1}}{{\left( {x + y + 1} \right)}}$

$\large \alpha \; = \;\frac{{i - 1}}{{\left( {x + y - 1} \right)}}$

$\large \alpha \; = \;\frac{{\left( {x + y - 1} \right)}}{{i - 1}}$

$\large \alpha \; = \;\frac{{\left( {x + y + 1} \right)}}{{i - 1}}$

answer is A.

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Detailed Solution

$\large A_xB_y\rightarrow xA^{+y}+yB^{-x}$

No.of particles in aq. sol. =(x+y)

Let $\large \alpha$ be degree of ionization of the electrolyte

$\large A_xB_y\rightarrow xA^{+y}+yB^{-x}$

Before ionization; 1 0 0

After ionization; 1- $\large \alpha$ x $\large \alpha$ y $\large \alpha$

total no.of particles After ionization =1- $\large \alpha$ + $\large \alpha$ (x+y)

$\large Van't Hoff factor (i) = \frac{observed\;colligative\;property}{normal\;colligative\;property}$

$\large i=\frac{1-\alpha +\alpha (x+y)}{1}$

i = 1+ $\large \alpha$ [(x+y)-1]

$\large \alpha =\frac{i-1}{(x+y)-1}$

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