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The degree of dissociation $(\alpha)$ of a weak electrolyte $A_xB_x$ is related to Van't hoff factor (i) by the experiment

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

1 Answer

Solution :
Hoff factor :
$A_xB_y \leftrightharpoons xA^{y+}+yB^{x-}$
$1- \alpha \qquad \alpha x \qquad \alpha y $
Total amount of spcies $=1+(x+y-1)\alpha$
van't hoff factor $= \large\frac{1+(x+y-)}{1}$
$\alpha = \large\frac{i-1}{x+y-1}$
answered Jul 18, 2014 by sreemathi.v
edited Nov 17, 2017 by meena.p
 

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