The degree of dissociation $(\alpha)$ of a weak electrolyte, $A_x B_y$ is related to van't Hoff factor (i)…
The degree of dissociation $(\alpha)$ of a weak electrolyte, $A_x B_y$ is related to van't Hoff factor (i) by the expression:
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$\alpha=\frac{i-1}{x+y+1}$
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$\alpha=\frac{x+y-1}{i-1}$
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$\alpha=\frac{x+y+1}{i-1}$
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$\alpha=\frac{i-1}{(x+y-1)}$
Solution
$
\begin{aligned}
& i=1-\alpha+n \alpha=1+\alpha(n-1) \\
& \frac{i-1}{n-1}=\alpha \\
& A_x B_y \rightarrow x A^{+y}+y B^{-x} \\
& n=x+y
\end{aligned}
$
So $\alpha=\frac{i-1}{x+y-1}$
Asked in: JEE Main 2011
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