In order to prepare one litre of a 1 N (norm | vedclass

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(B) The normality $(N)$ of a solution is given by the formula: $N = \frac{	ext{Mass of solute in grams}}{	ext{Equivalent weight} 	imes 	ext{Volume

Vedclass · Accepted Answer

$31.60$

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(B) The normality $(N)$ of a solution is given by the formula: $N = \frac{	ext{Mass of solute in grams}}{	ext{Equivalent weight} 	imes 	ext{Volume in litres}}$.In an acidic medium,the reduction half-reaction for $KMnO_4$ is: $MnO_4^- + 8H^+ + 5e^- 	o Mn^{2+} + 4H_2O$.The change in oxidation state of $Mn$ is from $+7$ to $+2$,so the $n$-factor is $5$.The equivalent weight of $KMnO_4 = \frac{	ext{Molecular weight}}{n	ext{-factor}} = \frac{158}{5} = 31.6 \ g/eq$.For $1 \ L$ of $1 \ N$ solution: $	ext{Mass} = N 	imes 	ext{Equivalent weight} 	imes 	ext{Volume} = 1 	imes 31.6 	imes 1 = 31.6 \ g$.

Isotope	Natural abundance (%)	Molar mass $(u)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

In order to prepare one litre of a $1 \ N$ (normal) solution of $KMnO_4$,how many grams of $KMnO_4$ are required if the solution is to be used in an acidic medium for oxidation?

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