From the given reaction:2 KMnO_4 + 3 H_2 SO_4 | vedclass

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(D) According to the law of equivalence,the number of gram equivalents of $H_2 O_2$ must equal the number of gram equivalents of $KMnO_4$.Gram equival

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

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(D) According to the law of equivalence,the number of gram equivalents of $H_2 O_2$ must equal the number of gram equivalents of $KMnO_4$.Gram equivalents $= N 	imes V 	ext{ (in Litres)} = M 	imes n_{factor} 	imes V 	ext{ (in Litres)}$.For $KMnO_4$,the $n_{factor}$ (change in oxidation state of $Mn$ from $+7$ to $+2$) is $5$.Thus,$N_{H_2 O_2} 	imes V_{H_2 O_2} = M_{KMnO_4} 	imes n_{factor} 	imes V_{KMnO_4}$.Given: $V_{H_2 O_2} = 20 \ mL$,$M_{KMnO_4} = 0.02 \ M$,$V_{KMnO_4} = 16 \ mL$,$n_{factor} = 5$.$N_{H_2 O_2} 	imes 20 = 0.02 	imes 5 	imes 16$.$N_{H_2 O_2} = \frac{0.02 	imes 5 	imes 16}{20} = \frac{0.1 	imes 16}{20} = \frac{1.6}{20} = 0.08 \ N$.$N_{H_2 O_2} = 8 	imes 10^{-2} \ N$.

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