Question

Assuming 100% ionisation of $BaC{l}_2\left(aq\right)$, what will be the value of van't Hoff factor?

Answer 1

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


For 100% ionisation, van't Hoff factor is equal to the number of particles in which 1 ​molecule of electrolyte dissociates​.

$BaC{l}_2\to B{a}^{+2}+2C{l}^{-}$
$i=3$