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Q.
The non stoichiometric compound $Fe _{0.94} O$ is formed when $x$ $\%$ of $Fe ^{2+}$ ions are replaced by as many $\frac{2}{3} F e^{3+}$ ions, $x$ is
The number of $Fe ^{3+}$ ions replacing $\times Fe ^{2+}$ ions
$=\frac{2 x}{3}$ vacancies of cations
$=x-\frac{2 x}{3}=x / 3$
But $x / 3=1-0.94=0.06, x=0.06 \times 3$
$=0.18=18 \%$