15 g Ba(MnO4)2, sample containing inert impurity is completely reacting with 100 mL of 11.2 V H2O2. What will be the % purity of Ba(MnO4)2 in the sample?

Engineering

Chemistry

Law of Chemical Equivalence

Titration

Question

15 g Ba(MnO₄)₂, sample containing inert impurity is completely reacting with 100 mL of 11.2 V H₂O₂. What will be the % purity of Ba(MnO₄)₂ in the sample?

None of the above

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Normality of $H_{2} O_{2} = \frac{11.2}{5.6} N$
Milliequivalents Ba(MnO₄)₂ reacted = milli equivalents of H₂O₂ reacted = 2 × 100 = 200 meq. = 0.2 gram eq.
Moles of Ba(MnO₄)2 = $\frac{0.2}{10} = 0.02$
∴ Weight of Ba(MnO₄)₂ = 0.02 × 375
Percentage purity of Ba(MnO₄)₂ = $\frac{375 \times 0.02}{15} \times 100 = 50 %$ .