Application of law of equivalence in simple titrations?

Question

Can any one say me how to find the n factor of a salt.
0.2M KMnO4 solution completely reacts with 0.05M FeSO4 solution under acidic conditions . The volume of FeSO4 used is 50 ml. Wat? volume of KMnO4
was used?
On referring to my txt book its given
0.2*5*V=0.05*50
V=2.5 ml
So how did dis 5 get in between.

Answer 1

This is a redox reaction.
By law of equivalence:
Equivalents of KMnO4=Equivalents of FeSO4

MilliMoles of FeSO4=Molarity of FeSO4*Volume of FeSO4 used.
= 0.05*50
=2.5
Since on oxidation Fe(+2) changes to Fe(+3)
Change of oxidation number = 1
thus n=1
Equivalents of FeSO4=n*millimoles
=2.5

Oxidation no of Mn in KMnO4=7
Reduction of MnO4(-) occurs to Mn(+2)
Change of oxidation number = 7-2=5
n=5
Equivalents of KMnO4= Molarity of KMnO4 * volume of KMnO4* n
=0.2*V*5

Thus 0.2*V*5=0.05*50
V=2.5 ml of KMnO4