Question

Assuming 100% ionisation, van't Hoff factor for $KCl\left(aq\right)$ and $K_{2}S{O}_4\left(aq\right)$ respectively is:

• A. 2,3
• B. 1,1
• C. 2,2
• D. 3,2

Answer 1

The correct option is A 2,3

$\text{van't Hoff factor}\left(i\right)=\frac{\text{Number of molecules (actual)}}{\text{Number of molecules (expected)}}$

For $KCl$:
$KCl(aq.)\to 1K^{+}+1C{l}^{-}$
$i=2$

$K_{2}S{O}_4(aq.)\to 2K^{+}+1S{O}_4^{2-}$
$i=3$