A sample of H2O2 is x% by mass. X ml of KMnO4 is required to oxidise 4gm of H2O2 sample. Find Normality of KMnO4.
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Given, 1g of sample of H
2
O
2
has x% H
2
O
2
by mass
This means 100g sample has x gm of H
2
O
2
⇒1g sample has
100gm
1g×xgm
H
2
O
2
⟶(1)
Now, Volume of KMnO
4
used for neutralization=xcm
3
We know,
At equivalence,
No. of gm equivalents of H
2
O
2
=No. of gm equivalents of KMnO
4
⟶(2)
⇒
H
2
O
2
N
1
V
1
=
KMnO
4
N
2
V
2
⇒
100×17
x
=
1000
N
i
X
[∵Normality=
Eq.Wt
Wt
1000Vinml]
[EquivalentWeightofH
2
O
2
=
2
34
=17]
⇒N
2
=
17
10
=0.588N
∴ Normality of KMnO
4
=0.588N
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