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The non stoichiometric compound $\mathrm{Fe}_{0.94} \mathrm{O}$ is formed when $\mathrm{x} \%$ of $\mathrm{Fe}^{2+}$ ions are replaced by as many $\frac{2}{3} \mathrm{Fe}^{3+}$ ions, $\mathrm{x}$ is
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The correct answer is:
18
The number of $\mathrm{Fe}^{3+}$ ions replacing $\mathrm{x} \mathrm{Fe}^{2+}$ ions $=\frac{2 \mathrm{x}}{3}$ vacancies of cations
$$
\begin{array}{l}
=x-\frac{2 x}{3}=x / 3 \\
\text { But } x / 3=1-0.94=0.06, x=0.06 \times 3 \\
=0.18=18 \%
\end{array}
$$
$$
\begin{array}{l}
=x-\frac{2 x}{3}=x / 3 \\
\text { But } x / 3=1-0.94=0.06, x=0.06 \times 3 \\
=0.18=18 \%
\end{array}
$$
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