The amount of KMnO_4 required to prepare 100 | vedclass

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(A) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent as follows:$MnO_4^- + e^- 	o MnO_4^{2-}$Here,the change in oxidation state of $Mn$ is f

Vedclass · Accepted Answer

$1.58$

Vedclass · Answer

(A) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent as follows:$MnO_4^- + e^- 	o MnO_4^{2-}$Here,the change in oxidation state of $Mn$ is from $+7$ to $+6$,so the $n$-factor is $1$.Therefore,the equivalent weight of $KMnO_4 = \frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{158}{1} = 158 \ g/eq$.Using the formula for normality: $	ext{Normality} = \frac{	ext{Mass}}{	ext{Equivalent mass} 	imes 	ext{Volume in } L}$.$0.1 = \frac{	ext{Mass}}{158 	imes (100/1000)}$.$	ext{Mass} = 0.1 	imes 158 	imes 0.1 = 1.58 \ g$.

The amount of $KMnO_4$ required to prepare $100 \ mL$ of $0.1 \ N$ solution in alkaline medium is ............... $g$.

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