Some Basic Concepts of Chemistry - Result Question 40
46. A sample of hydrazine sulphate $\left(\mathrm{N}_2 \mathrm{H}_6 \mathrm{SO}_4\right)$ was dissolved in 100 mL of water, 10 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 mL of $\mathrm{M} / 50$ potassium permanganate solution. Estimate the amount of hydrazine sulphate in one litre of the solution.
$\begin{aligned} & \text { Reaction } 4 \mathrm{Fe}^{3+}+\mathrm{N}_2 \mathrm{H}_4 \longrightarrow \mathrm{N}_2+4 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+} \\ & \mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O} \\ \end{aligned}$
(1988, 4M)
Show Answer
Answer:
Correct Answer: 46. $\left(6.5 \mathrm{gL}^{-1}\right)$
Solution:
- Meq of $\mathrm{MnO}_4^{-}$required $=20 \times \frac{1}{50} \times 5=2$
$\Rightarrow \mathrm{Meq}$ 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 L solution $\frac{1}{20} \mathrm{~mol} \mathrm{~N}_2 \mathrm{H}_6 \mathrm{SO}_4$ is present.
$\Rightarrow$ Amount of $\mathrm{N}_2 \mathrm{H}_6 \mathrm{SO}_4=\frac{1}{20} \times 130=6.5 \mathrm{gL}^{-1}$
BETA
