$2 {K M n O}_{4} + 3 H_{2} {S O}_{4} + 5 H_{2} O_{2} \to K_{2} {S O}_{4} + 2 {M n S O}_{4} + 8 H_{2} O + 5 O_{2}$
Find the normality of $H_{2} O_{2}$ solution, if $20 m L$ of it is required to react completely with 16 $m L$ of $0.02 M, {KMnO}_{4}$ solution. (molar mass of $= {K M n O}_{4} = 158 g {m o l}^{- 1})$

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$6 \times 10^{- 2} N$

$8 \times 10^{- 2} N$

$4 \times 10^{- 2} N$

$2 \times 10^{- 2} N$

answer is D.

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Detailed Solution

Given, Chemical reaction

$2 {K M n O}_{4} + 3 H_{2} {S O}_{4} + 5 H_{2} O_{2} \to K_{2} {S O}_{4} + 2 {M n S O}_{4} + 8 H_{2} O + 5 O_{2}$

Molar mass of ${KMnO}_{4} = 158 {gmole}^{- 1}$

Molarity of ${K M n O}_{4} = 0.02$

Volume of $H_{2} O_{2} = 20 m l$

According to the given reaction $H_{2} O_{2}$ work as a reducing agent

Oxidation state of Mn reduces from +7 to +2

Milliequivalents = normality volume

Normality = Molarity $\times$ n-factor

No. Of milliequivalents of ${K M n O}_{4} = n o . O f m i l l i e q u i v a l e n t s o f H_{2} O_{2}$

$0.02 \times 5 \times 16 = N_{H_{2} O_{2}} \times 20$

$N_{H_{2} O_{2}} = 8 \times 10^{- 2}$

Therefore, option D is correct.

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