Question

The degree of dissociation $(\alpha)$ of a weak electrolyte $A_xB_y$ is related to van’t Hoff factor $(i)$ by the expression

• A. $\alpha=\frac{i-1}{x+y-1}$
• B. $\alpha=\frac{x+y-1}{i-1}$
• C. $\alpha=\frac{x+y+1}{i-1}$
• D. $\alpha=\frac{i-1}{x+y+1}$

Answer 1

Answer: (A)

$A_xB_y \rightarrow xA^{y+} + yB^{x-}$
$\alpha = \frac{i-1}{n-1}$
where $n$ is the number of ions produced.
$\alpha = \frac{i -1}{(x+y-1)}$