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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

The correct expression is $\alpha=\large\frac{i-1}{(x+y-1)}$
answered Jul 18, 2014 by sreemathi.v
 

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