To a 25 , mL of H_2O_2 solution,an excess of | vedclass

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(A) The reaction is: $H_2O_2 + 2KI + 2H^+ \rightarrow 2H_2O + I_2$ and $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.Equivalents of $H_2O_2$ = Equ

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(A) The reaction is: $H_2O_2 + 2KI + 2H^+ ightarrow 2H_2O + I_2$ and $I_2 + 2Na_2S_2O_3 ightarrow 2NaI + Na_2S_4O_6$.Equivalents of $H_2O_2$ = Equivalents of $I_2$ = Equivalents of $Na_2S_2O_3$.Normality of $Na_2S_2O_3$ = $Molarity 	imes n	ext{-factor} = 0.3 	imes 1 = 0.3 \, N$.Equivalents of $Na_2S_2O_3 = 0.3 	imes 20 	imes 10^{-3} = 6 	imes 10^{-3}$.Normality of $H_2O_2$ in $25 \, mL$ solution = $\frac{6 	imes 10^{-3}}{25 	imes 10^{-3}} = 0.24 \, N$.Volume strength of $H_2O_2 = 5.6 	imes 	ext{Normality} = 5.6 	imes 0.24 = 1.344 \, mL$.

To a $25 \, mL$ of $H_2O_2$ solution,an excess of acidified solution of $KI$ was mixed. The liberated $I_2$ required $20 \, mL$ of $0.3 \, M$ hypo $(Na_2S_2O_3)$ solution for neutralization. The volume strength of $H_2O_2$ is......$mL$.

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