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. BrO_(3)^(-) ion reacts with Br^(-) to form Br_(2), in acid medium. The equivalent mass of Br_(2) in this reaction is :

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. BrO_(3)^(-) ion reacts with Br^(-) to form Br_(2), in acid medium. The equivalent mass of Br_(2) in this reaction is :
Answer» `(4 M.w.)/(6)` `(3M.w.)/(5)` `(5M.w.)/(3)` `(5M.w.)/(8)` Solution :`2BrO_(3)^(-)+12H^(+)+10e^(-) to Br_(2)+6H_(2)O "" n_(1)=10` `2Br^(-) to Br_(2)+6e^(-) "" n_(2)=2` N-factor`=(n_(1)xxn_(2))/(n_(1)+n_(2))=(10xx2)/(10+2)=(20)/(12)=(5)/(3)` `E.w.=(M.w.)/(5//3)=(3M.w.)/(5)`