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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: happyanimesh on July 02, 2010, 08:19:35 AM

Title: Ques. on Molarity and Normality
Post by: happyanimesh on July 02, 2010, 08:19:35 AM
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?   :-\


Title: Re: Ques. on Molarity and Normality
Post by: Schrödinger on July 02, 2010, 10:04:50 AM
Quote from: happyanimesh on July 02, 2010, 08:19:35 AM

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
Title: Re: Ques. on Molarity and Normality
Post by: happyanimesh on July 02, 2010, 10:59:13 AM
Thanx a lot!

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