Equivalent weight = n-factor is very important in redox as well as non-redox reactions…

Equivalent weight =
n-factor is very important in redox as well as non-redox reactions. With the help of n-factor we can predicts the molar ratio of the reactant species taking part in reactions. The reciprocal of n-factor’s ratio of the reactants is the molar ratio of the reactants.
In general n-factor of acid/base is number of moles of
H⁺/OH^– furnished per mole of acid/base. n-factor of a reactant is no. of moles of electrons lost or gained per mole of reactant.
Example 1 :
1 . In acidic medium : \(\mathrm{KMnO}_{4}(\mathrm{n}=5) \rightarrow \mathrm{Mn}^{2+}\) 
2 . In neutral medium : \(\mathrm{KMnO}_{4}(\mathrm{n}=3) \rightarrow \mathrm{Mn}^{2+}\)
3 . In basic medium : \(\mathrm{KMnO}_{4}(\mathrm{n}=1) \rightarrow \mathrm{Mn}^{6+}\) 
Example 2 :
\(\mathrm{FeC}_{2} \mathrm{O}_{4} \rightarrow \mathrm{Fe}^{3+}+2 \mathrm{CO}_{2}\) 
Total no. of moles of e^– lost by 1 mole of FeC₂O₄
= 1 + 1 × 2 \(\text { ⇒ }\) 3
\(\text { ∴ }\) n-factor of FeC₂O₄ = 3
For the reaction, Fe_0.95 O (molar mass: M)\(\text { → }\)Fe₂O₃. What is the eq. wt. of Fe_0.95 O ?