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Question: The Van’t Hoff factor of \(0.005\) M aqueous solution of KCl is \(1.95\) the degree of ionization of...

The Van’t Hoff factor of 0.0050.005 M aqueous solution of KCl is 1.951.95 the degree of ionization of KCl is

A

0.950.95

B

0.970.97

C

0.940.94

D

0.960.96

Answer

0.950.95

Explanation

Solution

KCl K++Cl−(n=2)K^{+} + Cl^{-}(n = 2)

α=i−1n−1=1.95−12−1=0.95\alpha = \frac{i - 1}{n - 1} = \frac{1.95 - 1}{2 - 1} = 0.95