Some Basic Concepts of Chemistry - Result Question 40 — JEE PYQ Chapterwise

Integer Answer Type Questions

41. A 1.0 g sample of $\mathrm{Fe}_2 \mathrm{O}_3$ solid of $55.2 %$ purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made up to 100.0 mL. An aliquot of 25.0 mL of this solution requires for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration.

(1991, 4M)

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Answer:

Correct Answer: 41. $\left(1.04 \times 10^4\right)$

Solution:

Mass of $\mathrm{Fe}_2 \mathrm{O}_3=0.552 \mathrm{~g}$

millimol of $\mathrm{Fe}_2 \mathrm{O}_3=\frac{0.552}{159.69} \times 1000=3.45$

During treatment with Zn dust, all $\mathrm{Fe}^{3+}$ is reduced to $\mathrm{Fe}^{2+}$, hence

millimol of $\mathrm{Fe}^{2+}($ in 100 mL$)=3.45 \times 2=6.90$

$\Rightarrow$ In 25 mL aliquot, $\frac{6.90}{4}=1.725$ millimol $\mathrm{Fe}^{2+}$ ions.

Finally $\mathrm{Fe}^{2+}$ is oxidised to $\mathrm{Fe}^{3+}$, liberating one electron per $\mathrm{Fe}^{2+}$ ion. Therefore, total electrons released by the reducing agent.

$\begin{aligned} & =1.725 \times 10^{-3} \times 6.023 \times 10^{23} \\ & =1.04 \times 10^{21} \end{aligned}$

Some Basic Concepts of Chemistry - Result Question 40 — JEE PYQ Chapterwise

Answer:

Solution:

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