Question

$3.92 \,g$ of ferrous ammonium sulphate are dissolved in $100\, mL$ water. $20\, mL$ of this solution requires $18 mL$ of potassium permanganate during titration for complete oxidation. The weight of $KMnO _{4}$ present in one litre of the solution is

• A. $34.76\, g$
• B. $12.38\, g$
• C. $1.238\, g$
• D. $3.51 \,g$

Answer 1

Answer: (D)

The redox reaction involving the oxidation of $Fe ^{2+}$

(from ferrous ammonium sulphate) is

$MnO _{4}^{-}+5 Fe ^{2+}+8 H ^{+} \longrightarrow 5 Fe ^{3+}+ Mn ^{2+}+4 H _{2} O$

Mol. wt. of ferrous ammonium sulphate,

$\left( NH _{4}\right)_{2} SO _{4} \cdot FeSO _{4} \cdot 6 H _{2} O =392$

$\therefore $ Molarity of $\left( NH _{4}\right)_{2} SO _{4} \cdot FeSO _{4} \cdot 6 H _{2} O$

$=\frac{W t}{\text { Mol. } \text { wt. }} \times \frac{1000}{\text { Volume }}=\frac{3.92}{392} \times \frac{1000}{100}=0.1$

Applying molarity equation,

$\frac{M_{1} V_{1}}{n_{1}} \text { (Ferrous amm. sulphate) }=\frac{M_{2} V_{2}}{n_{2}}\left( KMnO _{4}\right)$

$\text { or } \frac{0 \cdot 1 \times 20}{5}=\frac{M_{2} \times 18}{1} \text { or } M_{2}=\frac{0 \cdot 1 \times 20}{5 \times 18}=\frac{1}{45}$

Amount of $KMnO _{4}$ present in one litre

$=\text { Molarity } \times \text { Mol. wt. }=\frac{1}{45} \times 158=3 \cdot 51\, g$