Question

$3.92g$ of ferrous ammonium sulphate crystal are dissolved in $100mL$ of water, $20mL$ of this solution requires $18mL$ of potassium permanganate during titration for complete oxidation. The weight of $KMn{O}_4$ present in one litre of the solution is:

• A. $34.76g$
• B. $12.38g$
• C. $1.238g$
• D. $3.476g$

Answer 1

The correct option is D $3.476g$

Molecular weight of Ferrous ammonium sulphate ($\left(N{H}_4{)}_{2}Fe\right(S{O}_4{)}_2.6H_{2}O$) = 392 g/mol

No. of moles of Ferrous ammonium sulphate ($\left(N{H}_4{)}_{2}Fe\right(S{O}_4{)}_2.6H_{2}O$) = $\frac{3.92}{392}=0.01$ moles

Normality of Ferrous ammonium sulphate ($\left(N{H}_4{)}_{2}Fe\right(S{O}_4{)}_2.6H_{2}O$) solution = $\frac{0.01\times 1000}{100}=0.1N$

${Mn{O}_4}^{-}+8H^{+}+5e^{-}\to {Mn}^{2+}+4H_{2}O$

${Fe}^{2+}\to {Fe}^{3+}+e^{-}$

From the equations n-factor of $\left(N{H}_4{)}_{2}Fe\right(S{O}_4{)}_2.6H_{2}O=1$

and n-factor for $KMn{O}_4=5$

20 ml of Ferrous ammonium sulphate ($\left(N{H}_4{)}_{2}Fe\right(S{O}_4{)}_2.6H_{2}O$) neutralised 18 ml of $KMn{O}_4.$

$=>$ $N_1{V}_1=N_2{V}_2$

$=>$ $0.1\times 20=N_2\times 18$

$=>$ $N_2=\frac{1}{9}N$

Equivalent weight of $KMn{O}_4=\frac{Mol.Wt.}{5}=\frac{158}{5}=31.6$ $g{mol}^{-1}{eq}^{-1}$.

Weight of $KMn{O}_4$ present in 1 L of solution = $\frac{1\times 31.6}{9}\approx 3.476$ g