The correct option is A 50.00 mL
n factor of KMnO4,
Equivalents of KMnO4=5× moles of KMnO4
Normality KMnO4=5× molarity of KMnO4
For Fe(NO3)2, moles and equivalents are equal, as n factor is 1.
At the equivalence point,
Equivalents of KMnO4 = Equivalents of Fe(NO3)2
or VKMnO4× Normality of KMnO4
=VFe(NO3)2× Normality of Fe(NO3)2
For 0.02 M KMnO4 solution.
Normality of KMnO4=(5)(0.02000)=0.1 N
and for 0.20 M Fe(NO3)2 solution
Normality of Fe(NO3)2=0.20 N
putting the values,
VKMnO4=(25.00 mL)(0.20000.1000)=50.00 mL