Iodometric titration is based on the following reaction ${\mathrm{I}}_{2( \mathrm{s})}+{\mathrm{I}}^{-}⟶{\mathrm{I}}_3^{-}$.The equivalent weight of iodine in reaction is

Given, Chemical reaction is

${\mathrm{I}}_{2( \mathrm{s})}+{\mathrm{I}}^{-}⟶{\mathrm{I}}_3^{-}$

The oxidation state of I(terminal) changes from 0 to –1

The oxidation state of I(central) changes from -1 to 0

valence factor=3

$\text{ Equivalent weight }=\frac{\text{ Mol. }wt}{\text{ valence factor }}$

So, $\text{ Equivalent weight }=\frac{M}{3}$

Therefore, option C is correct.