# When 10 mL of an aqueous solution of KMnO4 was titrated in acidic medium, equal volume of 0.1 M

Question:

When $10 \mathrm{~mL}$ of an aqueous solution of $\mathrm{KMnO}_{4}$ was titrated in acidic medium, equal volume of $0.1$

$\mathrm{M}$ of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of $\mathrm{KMnO}_{4}$ in grams per litre is__________________

$\times 10^{-2}$. (Nearest integer)

[Atomic mass of $\mathrm{K}=39, \mathrm{Mn}=55, \mathrm{O}=16$ ]

Solution:

Let molarity of $\mathrm{KMnO}_{4}=\mathrm{x}$

$\mathrm{KMnO}_{4}+\mathrm{FeSO}_{4} \rightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{Mn}^{2+}$

$\mathrm{n}=5 \quad \mathrm{n}=1$

(Equivalents of $\mathrm{KMnO}_{4}$ reacted) $=$ (Equivalents of

$\mathrm{FeSO}_{4}$ reacted)

$\Rightarrow(5 \times \mathrm{x} \times 10 \mathrm{ml})=1 \times 0.1 \times 10 \mathrm{ml}$

$\Rightarrow \mathrm{x}=0.02 \mathrm{M}$

Molar mass of $\mathrm{KMnO}_{4}=158 \mathrm{gm} / \mathrm{mol}$

$\Rightarrow$ Strength $=(\mathrm{x} \times 158)=3.16 \mathrm{~g} / \ell$