[happyanimesh]

Question:

20ml of 0.2 M MnSO₄ are completely oxidised by 16ml ok KMnO₄ of unknown normality each forming Mn⁺⁴ oxidation state. Find out the normality and molarity of KMnO₄ solution.

My Attempt

I know that M₁V₁=M₂V₂.
So,
    20*.2=M₂*16
Solving this equation, i get:
M₂=0.25M

I also know that Normality=Molarity * valency factor
So, Normality=0.25*4                (Due to the presence of Mn⁺⁴ ion)
Hence, Normality of KMnO4 solution is 1.

But the answer given at the back is:
Normality = 0.5
Molarity = 0.167

Kindly Help...
Is there any mistake in my calculation?   

[Schrödinger]

My Attempt

I know that M₁V₁=M₂V₂.
So,
    20*.2=M₂*16

When you do this, you are actually equating moles. But it is equivalents that need to be equated. So, N₁V₁ = N₂V₂

Normality = Molarity * valency factor
Valency factor is calculated from the change in oxidation state.
For MnSO₄ to Mn⁴⁺ the factor is 4 - 2 = 2
For KMnO₄ to Mn⁴⁺, the factor is 7 - 4 = 3

[happyanimesh]

Thanx a lot!

I was really confused in the concept of molarity and normaity..
U cleared it all!!