Question

The Van't Hoff factor ($i$) for ${Na}_{2}S{O}_4$ is:

• A. 3
• B. 2
• C. 1
• D. 0

Answer 1

The correct option is A 3

$N{a}_{2}S{O}_4$ is a strong electrolyte. It dissociates into $N{a}_{2}S{O}_4\to 2N{a}^{+}+S{O}_4^{2-}$

Hence, $i=3.$

So, option A is correct