Q74 / 200 · NEET 2022 (Phase 2) · Chemistry
What fraction of $\mathrm{Fe}$ exists as $\mathrm{Fe}$ (III) in $\mathrm{Fe}_{0.96} \mathrm{O}$ ?
(Consider $\mathrm{Fe}_{0.96}$ to be made up of $\mathrm{Fe}$ (II) and $\mathrm{Fe}$ (III) only)
- Correct answerA$\frac{1}{12}$
- B0.08
- C$\frac{1}{16}$
- D$\frac{1}{20}$
Solution
Let's assume $\mathrm{Fe}$ in $\mathrm{Fe}^{2+}=\mathrm{x}$
Let's assume $\mathrm{Fe}$ in $\mathrm{Fe}^{3+}(0.96-x)$
Using the charge conservation formula for
$\mathrm{Fe}_{0.96} \mathrm{O}_{1.00}$
We get: $(x)(+2)+(0.96-x)(+3)-2=0$
$\begin{aligned}
& =2 x+2.88-3 x-2=0 \\
& x=0.88
\end{aligned}$
So, number of ions of $\mathrm{Fe}^{3+}=0.96-x=0.96-0.88=0.08$
$\text {Fraction of } \mathrm{Fe}^{3+} \text { ion } =\frac{0.08}{0.96} =\frac{1}{12}$
Hence, the correct answer is option: 'A'.