When $\mathrm{Fe_{0.93}O}$ is heated in presence of oxygen, it converts to $\mathrm{Fe_2O_3}$. The number of correct statement/s from the following is ________

A. The equivalent weight of $\mathrm{Fe_{0.93}O}$ is ${{\mathrm{Molecular\,weight}} \over {0.79}}$

B. The number of moles of Fe$^{2+}$ and Fe$^{3+}$ in 1 mole of $\mathrm{Fe_{0.93}O}$ is 0.79 and 0.14 respectively

C. $\mathrm{Fe_{0.93}O}$ is metal deficient with lattice comprising of cubic closed packed arrangement of O$^{2-}$ ions

D. The % composition of Fe$^{2+}$ and Fe$^{3+}$ in $\mathrm{Fe_{0.93}O}$ is 85% and 15% respectively

Let the number of $\mathrm{Fe}^{2+}$ ions be $x$ and number of $\mathrm{Fe}^{3+}$ ions be $0.93-x$. <br/><br/>According to the charge neutrality, <br/><br/>$$ \begin{aligned} & 2 x+3(0.93-x)=2 \Rightarrow 2 x-3 x+2.79=2 \Rightarrow x=0.79 \\\\ & \mathrm{Fe}^{2+} \text { ion }=x=0.79 \\\\ & \mathrm{Fe}^{3+} \text { ion }=0.93-x=0.93-0.79=0.14 \\\\ & \% \text { of } \mathrm{Fe}^{2+} \text { ion }=\frac{0.79}{0.93} \times 100=84.95 \% \approx 85 \% \\\\ & \% \text { of } \mathrm{Fe}^{3+} \text { ion }=\frac{0.14}{0.93} \times 100=15.05 \% \end{aligned} $$ <br/><br/>As it is a metal deficiency defect, <br/><br/>so, 1 mol of $\mathrm{Fe}_{0.93} \mathrm{O}$ contains $1 \mathrm{~mol}$ of oxide ions and $0.93 \mathrm{~mol}$ of iron ions as $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$. <br/><br/>So, moles of $\mathrm{Fe}^{2+}=0.79$ <br/><br/>and moles of $\mathrm{Fe}^{3+}=0.14$ <br/><br/>$$ \text { The equivalent weight }=\frac{\text { Molecular weight }}{n \text {-factor }} $$ <br/><br/>$F{e^{2 + }} \to F{e^{3 + }} + {e^ - }$ <br/><br/>For one $\mathrm{Fe}^{2+}, n$-factor $=1$ <br/><br/>$\therefore $ For $0.79 \,\mathrm{Fe}^{2+}, n$-factor $=0.79$ <br/><br/>Out of $0.93 \mathrm{~mol}$, there are $0.79 \mathrm{~mol} \,\mathrm{Fe}^{2+}$ ions are present.