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?

Similar Questions

Select the correct statement :-

In which of the following reactions does $H_2O_2$ act as an oxidising agent (either in acidic,alkaline,or neutral medium)?
$(i)$ $2Fe^{2+} + H_2O_2 \rightarrow$
$(ii)$ $2MnO_4^{-} + 6H^{+} + 5H_2O_2 \rightarrow$
$(iii)$ $I_2 + H_2O_2 + 2OH^{-} \rightarrow$
$(iv)$ $Mn^{2+} + H_2O_2 \rightarrow$

Which of the following cannot be oxidised by $H_2O_2$?

Strength of a $H_2O_2$ solution is labelled as $1.79 \, N$. Its strength can also be expressed as closest to $... \, \text{volume}$.

The product$(s)$ formed when $H_2O_2$ reacts with disodium hydrogen phosphate is

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