Some Basic Concepts of Chemistry 2 Question 46
46. A sample of hydrazine sulphate $\left(\mathrm{N}{2} \mathrm{H}{6} \mathrm{SO}_{4}\right)$ was dissolved in $100 \mathrm{~mL}$ of water, $10 \mathrm{~mL}$ of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it, required $20 \mathrm{~mL}$ of $\mathrm{M} / 50$ potassium permanganate solution. Estimate the amount of hydrazine sulphate in one litre of the solution.
Reaction $4 \mathrm{Fe}^{3+}+\mathrm{N}{2} \mathrm{H}{4} \longrightarrow \mathrm{N}_{2}+4 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+}$
$$ \mathrm{MnO}{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}{2} \mathrm{O} $$
(1988, 3M)
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Solution:
- $\mathrm{Meq}$ of $\mathrm{MnO}_{4}^{-}$required $=20 \times \frac{1}{50} \times 5=2$
$\Rightarrow$ Meq of $\mathrm{Fe}^{2+}$ present in solution $=2$
$\Rightarrow$ millimol of $\mathrm{Fe}^{2+}$ present in solution $=2(n$-factor $=1)$
Also,
$\because 4$ millimol of $\mathrm{Fe}^{2+}$ are formed from 1 millimol $\mathrm{N}{2} \mathrm{H}{4}$
$\therefore 2$ millimol $\mathrm{Fe}^{2+}$ from $\frac{1}{4} \times 2=\frac{1}{2}$ millimol $\mathrm{N}{2} \mathrm{H}{4}$
Therefore, molarity of hydrazine sulphate solution
$$ =\frac{1}{2} \times \frac{1}{10}=\frac{1}{20} $$
$\Rightarrow$ In $1 \mathrm{~L}$ solution $\frac{1}{20} \mathrm{~mol} \mathrm{~N}{2} \mathrm{H}{6} \mathrm{SO}_{4}$ is present.
$\Rightarrow$ Amount of $\mathrm{N}{2} \mathrm{H}{6} \mathrm{SO}_{4}=\frac{1}{20} \times 130=6.5 \mathrm{gL}^{-1}$
