What is the normality of a KMnO4 solution in which 15.8 g of KMnO4 is present in 250 mL solution and the solution is used as an oxidant, where MnO4− is reduced to Mn2+ ? (Given: molar mass of KMnO4=158 gmol−1 )
2 N
We know that,
Normality = Number of gram equivalents of soluteVolume of solution (in litres)
Equivalent Mass = Molecular massValeny factor
MnO4−→ Mn2+
[Mn7+ + 5 e−→ Mn2+]
→ Valency factor = 5
Molecular mass of KMnO4 = 158
∴ equivalent weight = 1585 = 31.6
No of gram equivalents = 15.831.6
∴ Normality =15.831.6250×10−3=2 N