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Question

Calculate the normality of a solution containing 15.8 g of KMnO4 in 50 mL of acidic solution.

A
1 N
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B
10 N
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C
100 N
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D
0.5 N
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Solution

The correct option is B 10 N
Normality is the number of gram or mole equivalents of solute present in one litre of a solution.

Normality (N) =W×1000E×V (in mL)
where,
W = Weight of the solute
E = Equivalent mass of solute
V = Volume of solution in mL

Equivalent mass of solute=Molecular massValence factor
Valence factor is the number of electron released or absorbed.

Reaction of KMnO4 in acidic solution,
MnO4+8H++5eMn2++4H2O

Here, Mn+7 is reduced to Mn+2. Hence, it gains 5 e so valence factor is 5
Molecular mass of KMnO4 is 158 g mol1

E=Molar mass of KMnO4Valence factor=158/5=31.6

Normality (N) =15.8×100031.6× 50

So, Normality of the solution = 10 N

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