Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of $$5.0$$ $$\text{\AA}$$. If density of the FeO in the crystal is $$4.0$$ g cm$$^{-3}$$, then the number of FeO units present per unit cell is ______ (Nearest integer)
Given: Molar mass of Fe and O is 56 and 16 g mol$$^{-1}$$ respectively.
N$$_A = 6.0 \times 10^{23}$$ mol$$^{-1}$$

We use the density formula for a crystal unit cell:

$$d = \frac{Z \times M}{N_A \times a^3}$$

where $$Z$$ is the number of formula units per unit cell, $$M$$ is the molar mass, $$N_A$$ is Avogadro's number, and $$a$$ is the edge length.

We have $$d = 4.0$$ g cm$$^{-3}$$, $$M = 56 + 16 = 72$$ g mol$$^{-1}$$ (for FeO), $$a = 5.0$$ $$\text{\AA}$$ $$= 5.0 \times 10^{-8}$$ cm, and $$N_A = 6.0 \times 10^{23}$$ mol$$^{-1}$$.

Now, $$a^3 = (5.0 \times 10^{-8})^3 = 125 \times 10^{-24} = 1.25 \times 10^{-22}$$ cm$$^3$$.

So $$N_A \times a^3 = 6.0 \times 10^{23} \times 1.25 \times 10^{-22} = 75$$.

From the density formula:

$$Z = \frac{d \times N_A \times a^3}{M} = \frac{4.0 \times 75}{72} = \frac{300}{72} \approx 4.17$$

Rounding to the nearest integer gives $$Z = 4$$.

So, the answer is $$4$$.