Q28 / 90 · JEE Main 2021 (27 Aug Shift 1) · Chemistry

When $10 mL$ of an aqueous solution of ${KMnO}_{4}$ was titrated in acidic medium, equal volume of $0.1 M$ of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of ${KMnO}_{4}$ in grams per litre is _______ $\times 10^{- 2} .$ (Nearest integer) [Atomic mass of $K = 39, Mn = 55, O = 16$ ]

Correct answer: 316

Solution

$\underset{vf = 5}{{MnO}_{4}^{-}} + \underset{vf = 1}{{Fe}^{2 +} ⟶} {Mn}^{2 +} + {Fe}^{3 +}$

Mili equ. of ${MnO}_{4} {}^{-}=$ mili equ. of ${Fe}^{2 +}$

$5 [M \times 10] = 1 [0.1 \times 10]$

$M = [\frac{0.1}{5}] = 0.02 \frac{mole}{Lit}$

Strength of ${KMnO}_{4} = 0.02 \times 158 \frac{gram}{Lit} = 3.16 g / L = 316 \times 10^{- 2} g / L$

Hence, the correct answer is 316.