Jitender Pal
Last Activity: 10 Years ago
(i) Find equivalent wt. of H2O2 and KMnO4.
(ii) Xmol of N.KMnO4 = X ml of N.H2O2
H2O2 → H2O + [O]
2KMnO4 + 3H2SO4→ K2SO4→ K2SO4 + 2MnSO4 + 3H2O + 5 [O]
∴ Equivalent wt. of H2O2 = Mol. wt./2 = 34/2 = 17
Equivalent wt. of KMnO4 = Mol. wt./5 = 158/5 = 31.6
Let normality of KMnO4 solution = N, then
X ml. of N.KMnO4 = X ml. of N.H2O2
w = X * N *17/1000 g of H2O2
According to problem X *17 *N/1000 g of H2O2 is present in
1 g of solution, Also given ……(1)
100 g of H2O2 contain X g of H2O2
1 g of H2O2 contain = X/100 * 1 g of H2O2 …...(2)
Comparing the two relation (1) and (2)
17 * N * X/1000 = X/100
Or N = X/100 * 1000/17 X = 1000/100 * 17 = 10/17
Hence, normality of KMnO4 solution is 10/17 N or 0.588 N
ALTERNATIVE SOLUTION :
The complete oxidation under acidic conditions can be represented as follows:
5H2O2 + 2MnO-4 + 6H+ → 502 + 2Mn2 + 2Mn2+ + 8H2O
Since 34 g of H2O2 = 2000 ml of 1N . H2O2
(∵ Eq. wt or H2O2 = 34/2)
∴ 34 g of H2O2 = 2000 ml of 1 N KMnO4 [∵N1V1 = N2V2]
Or X/100 g of H2O2 = 2000 * X/100 *34 ml of 1 N KMnO4
Therefore the unknown normality = 2000 * X/34 * 100 * X = 10/17 or 0.588 N
