When Fe$$_{0.93}$$O is heated in presence of oxygen, it converts to Fe$$_2$$O$$_3$$. The number of correct statement/s from the following is _____.
A. The equivalent weight of Fe$$_{0.93}$$O is $$\frac{\text{Molecular weight}}{0.79}$$
B. The number of moles of Fe$$^{2+}$$ and Fe$$^{3+}$$ in 1 mole of Fe$$_{0.93}$$O is 0.79 and 0.14 respectively.
C. 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 Fe$$_{0.93}$$O is 85% and 15% respectively.

We are given that $$Fe_{0.93}O$$ is heated in the presence of oxygen and converts to $$Fe_2O_3$$. We need to find the number of correct statements from A, B, C, D.

We begin by determining the composition of $$Fe_{0.93}O$$. In $$Fe_{0.93}O$$, one mole of the compound contains 0.93 moles of iron and 1 mole of oxygen. The iron ions are a mixture of $$Fe^{2+}$$ and $$Fe^{3+}$$. Let $$x$$ = moles of $$Fe^{2+}$$ and $$y$$ = moles of $$Fe^{3+}$$. From the total iron: $$x + y = 0.93 \quad \cdots (i)$$
From charge balance (total positive charge = total negative charge from $$O^{2-}$$): $$2x + 3y = 2 \quad \cdots (ii)$$

Next, substituting $$x = 0.93 - y$$ from equation (i) into equation (ii) gives: $$2(0.93 - y) + 3y = 2 \implies 1.86 - 2y + 3y = 2 \implies y = 0.14$$
Then $$x = 0.93 - 0.14 = 0.79$$
This shows that there are 0.79 moles of $$Fe^{2+}$$ and 0.14 moles of $$Fe^{3+}$$ per mole of $$Fe_{0.93}O$$.

Now we verify each statement.

Statement A: The equivalent weight of $$Fe_{0.93}O$$ is $$\frac{\text{Molecular weight}}{0.79}$$. When $$Fe_{0.93}O$$ is oxidized to $$Fe_2O_3$$, the 0.79 moles of $$Fe^{2+}$$ are oxidized to $$Fe^{3+}$$, each losing one electron. The total change in oxidation state per mole of $$Fe_{0.93}O$$ is 0.79. Therefore, the n-factor is 0.79, and the equivalent weight = $$\frac{M}{0.79}$$. Statement A is CORRECT.

Statement B: The moles of $$Fe^{2+}$$ and $$Fe^{3+}$$ are 0.79 and 0.14 respectively. This is exactly what we calculated earlier. Statement B is CORRECT.

Statement C: $$Fe_{0.93}O$$ is metal deficient with a lattice comprising cubic close-packed arrangement of $$O^{2-}$$ ions. $$Fe_{0.93}O$$ is a non-stoichiometric compound derived from FeO (which has the NaCl structure with FCC arrangement of $$O^{2-}$$ ions). It is metal-deficient because some $$Fe^{2+}$$ sites are vacant, with charge balance maintained by some $$Fe^{2+}$$ converting to $$Fe^{3+}$$. The $$O^{2-}$$ ions form a cubic close-packed (CCP/FCC) lattice. Statement C is CORRECT.

Statement D: The percentage composition of $$Fe^{2+}$$ and $$Fe^{3+}$$ is 85% and 15% respectively. $$\%Fe^{2+} = \frac{0.79}{0.93} \times 100 \approx 84.95\% \approx 85\%$$
$$\%Fe^{3+} = \frac{0.14}{0.93} \times 100 \approx 15.05\% \approx 15\%$$ Statement D is CORRECT.

All 4 statements are correct. The answer is 4.