Equivalent mass of a substance may be calculated as, Equivalent mass=("Molecular mass")/("n-factor")=("Atomic mass")/(n-factor") n-factor= Basicity of acid or acidity of base n-factor= Number of moles of electrons gainted or lost per mole of oxidising or reducing agents n-factor= Total positive or negative valency of a salt n-factor= Valency of an ion. Concept of n-factor is very important for redox as well as for non-redox reactions. When KMnO_(4) is titrated against ferrous ammonium sulphate in acid medium then equivalent mass of KMnO_(4) will be :

Equivalent mass of a substance may be calculated as, Equivalent mass=

1.	Equivalent mass of a substance may be calculated as, Equivalent mass=("Molecular mass")/("n-factor")=("Atomic mass")/(n-factor") n-factor= Basicity of acid or acidity of base n-factor= Number of moles of electrons gainted or lost per mole of oxidising or reducing agents n-factor= Total positive or negative valency of a salt n-factor= Valency of an ion. Concept of n-factor is very important for redox as well as for non-redox reactions. When KMnO_(4) is titrated against ferrous ammonium sulphate in acid medium then equivalent mass of KMnO_(4) will be :
Answer» `("Molecular mass")/(10)` `("molecular mass")/(5)` `("Molecular mass")/(3)` `("molecular mass")/(2)` Solution :`MnO_(4)^(-)+8H^(+)+5E^(-) to Mn^(2+) +4H_(2)O`n-factor=5