Question

The van't Hoff factor of a 0.005 M aqueous solution of $KCl$ is 1.95. The degree of ionization of $KCl$ is:

• A. 0.95
• B. 0.97
• C. 0.94
• D. 0.96

Answer 1

The correct option is A 0.95

$KCl\rightleftharpoons K^{+}+C{l}^{-}(n=2)$

$\alpha =\frac{i-1}{n-1}=\frac{1.95-1}{2-1}=0.95$