How many millilitres of 20 volume H_2O_2 solu | vedclass

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(C) The reaction between $KMnO_4$ and $H_2O_2$ in acidic medium is based on the principle of equivalence: $N_1 V_1 = N_2 V_2$. Given,Normality of $KMn

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$140$

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(C) The reaction between $KMnO_4$ and $H_2O_2$ in acidic medium is based on the principle of equivalence: $N_1 V_1 = N_2 V_2$. Given,Normality of $KMnO_4$ $(N_1)$ = $1 \ N$. Volume of $KMnO_4$ $(V_1)$ = $500 \ mL$. The normality of $H_2O_2$ solution is related to its volume strength by the formula: $N = \frac{	ext{Volume strength}}{5.6}$. So,Normality of $H_2O_2$ $(N_2)$ = $\frac{20}{5.6} \ N$. Applying the law of equivalence: $1 	imes 500 = (\frac{20}{5.6}) 	imes V_2$. $V_2 = \frac{500 	imes 5.6}{20} = 25 	imes 5.6 = 140 \ mL$.

How many millilitres of $20$ volume $H_2O_2$ solution is needed to react completely with $500 \ mL$ of acidified $1 \ N \ KMnO_4$ solution?

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