Question

Calculate the normality of a solution containing 15.8 g of $KMn{O}_4$ in 50 mL of acidic solution.

• A. 1 N
• B. 10 N
• C. 100 N
• D. 0.5 N

Answer 1

The correct option is B 10 N

Normality is the number of gram or mole equivalents of solute present in one litre of a solution.

Normality (N) $=\frac{W\times 1000}{E\times V\left(inmL\right)}$
where,
W = Weight of the solute
E = Equivalent mass of solute
V = Volume of solution in mL

$\text{Equivalent mass of solute}=\frac{Molecularmass}{Valencefactor}$
Valence factor is the number of electron released or absorbed.

Reaction of $KMn{O}_4$ in acidic solution,
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

Here, $M{n}^{+7}$ is reduced to $M{n}^{+2}$. Hence, it gains 5 $e^{-}$ so valence factor is 5
$\text{Molecular mass of}KMn{O}_{4}is158gmo{l}^{-1}$

$E=\frac{\text{Molar mass of}KMn{O}_4}{\text{Valence factor}}=158/5=31.6$

Normality (N) $=\frac{15.8\times 1000}{31.6\times 50}$

So, Normality of the solution = 10 N