The equivalent weight of FeS₂ in the following reaction is ${\mathrm{FeS}}_2+{\mathrm{O}}_2\to {\mathrm{Fe}}^{+3}+{\mathrm{SO}}_2$

${\mathrm{FeS}}_2⟶{\mathrm{Fe}}^{2+}\text{ and }{\mathrm{S}}_2^{2-}$
i.e., Sulphur is in -1 state
$\therefore \text{ For }1\text{ mole of }{\mathrm{FeS}}_2,\mathrm{\Delta O}\cdot \mathrm{N}.=+11$
$\Rightarrow \mathrm{n}\text{ factor }=11$
$\mathrm{\Delta O}.\mathrm{N}.=\text{ change in oxidation number }.$