20 mL of KMnO4 solution completely reacts with 10 mL of 1M FeS2 solution and 20 mL of 1 M CuS solution to produce Cu+2, Fe+3, SO2. Normality of KMnO4 solution is

Engineering

Chemistry

Law of Chemical Equivalence

Titration

Question

20 mL of KMnO₄ solution completely reacts with 10 mL of 1M FeS₂ solution and 20 mL of 1 M CuS solution to produce Cu⁺², Fe⁺³, SO₂. Normality of KMnO₄ solution is

57.5 N

11.5 N

6.5 N

8.5 N

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KMnO₄ + FeS₂ + CuS → Cu⁺² + Fe⁺³ + SO₂
n= 11 n = 6
Eq. of KMnO₄ = Eq. of FeS₂ + Eq. of CuS
$\frac{N \times V}{1000} = \frac{{MV}_{{FeS}_{2}}}{1000} \times n + \frac{{MV}_{CuS}}{1000} \times n$

$\frac{N \times 20}{1000} = \frac{10 \times 1}{1000} \times 11 + \frac{20 \times 1 \times 6}{1000}$

N = $\frac{110 + 120}{20} = \frac{230}{20} = 11 .5 N$