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. Equivalentmass of H_(3)PO_(2) when it undergoes disporportionation to PH_(3) " and " H_(3)PO_(3) 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. Equivalentmass of H_(3)PO_(2) when it undergoes disporportionation to PH_(3) " and " H_(3)PO_(3) will be :
Answer» `M.w.//2` `M.w. //4` `M. w. //24` `3M.w. //4` Solution :`H_(3)PO_(2)+4H^(+) +4e^(-) to PH_(3)+2H_(2)O "" N_(1)=4` `H_(3)PO_(2)+H_(2)O to H_(3)PO_(3)+2H^(+)+2e^(-) "" n_(2)=2` N-factor`=(n_(1)xxn_(2))/(n_(1)+n_(2))=(2xx4)/(2+4)=(8)/(6)=(4)/(3)` E.w. `M.w.//(4)/(3)=(3M.w.)/(4)`