Solution:
We have given that
Edge length of the unit cell
= 200 pm = 200 x 10^–10 cm

Vol. of the unit cell
= (200 x 10^–10 cm)³ = 8 x 10^–24 cm³
Therefore, volume of the substance
= Vol. of unit cell x Vol. of 1 unit cell.

Since the element has a fcc unit cell, number of atoms per unit cell = 4.
Total number of atoms = Atoms/unit cell x number of unit cells.
                                     = 24 x 10²³atoms = 4

atoms/unit cell x No. of unit cells.

∴   No. of unit cells = $\frac{24\times {10}^{23} \mathrm{atoms}}{4 \mathrm{atoms}/\mathrm{unit} \mathrm{cell}}$

                               $= 6\times {10}^{23} \mathrm{unit} \mathrm{cells}.$

∴  Volume of the substance
                   $6\times {10}^{23} \mathrm{unit} \mathrm{cell} \times 8\times {10}^{-24 }{\mathrm{cm}}^3/\mathrm{unit} \mathrm{cell} = 4.8 {\mathrm{cm}}^3$


Now,     $$\begin{gathered} \mathrm{Density} = \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ = \frac{200\mathrm{g}}{4.8 {\mathrm{cm}}^3} =4.17 \mathrm{g} {\mathrm{cm}}^{-3}. \end{gathered}$$

Solution;
we have given that    

   $\mathrm{Z} =1, \mathrm{\rho } = 4 \mathrm{g} {\mathrm{cm}}^{-3},$
        $$\begin{gathered} \mathrm{M} = 168.5 {\mathrm{mol}}^{-1}, \\ {\mathrm{N}}_{\mathrm{A}} = 6.02\times {10}^{23}{\mathrm{mol}}^{-1} \\ \mathrm{\rho } = \frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{a}}^3\times \mathrm{N}} \end{gathered}$$

or          $$\begin{gathered} {\mathrm{a}}^3 = \frac{\mathrm{Z}\times \mathrm{M}}{\mathrm{\rho }\times {\mathrm{N}}_{\mathrm{A}}} \\ = \frac{1\times 168.5\mathrm{g} {\mathrm{mol}}^{-1}}{4\times 6.02\times {10}^{23}} \\ = \frac{168.5\times {10}^{-23}}{24.08} \end{gathered}$$
$$\begin{gathered} \log {\mathrm{a}}^3 = \log 168.5 + \log {10}^{-23} - \log 24.08 \\ 3 \log \mathrm{a} = 2.2266 - 23.000-1.3816 \\ 3 \log \mathrm{a} = -22.1550+2-2 \\ 3 \log \mathrm{a} = \overline{24}+(2.1550) \\ \mathrm{a} = \mathrm{Antilog} \overline{8}.7183 = 5.228\times {10}^{-8}\mathrm{cm} \\ = 5.228\times {10}^{-8}\times {10}^{10}\mathrm{pm} = 522.8 \mathrm{pm}. \end{gathered}$$

Volume of the unit cell = (5 A°)
= (5 x 10^–8 cm)³
= 125 x 10^–24 cm³
= 1.25 x 10^–22 cm³
Density of FeO = 4.0 g cm^–3

Therefore, mass of the unit cell
= 1.25 x 10^–22 cm³ x 4 g cm^–3
= 5 x 10^–22g
Mass of all molecules of FeO
$$\begin{gathered} =\frac{72}{6.022\times {10}^{23}} \\ = 1.195 \times {10}^{-22} \mathrm{g} \end{gathered}$$

 Number of FeO molecules/unit cell
$$\begin{gathered} =\frac{5\times {10}^{-22}\mathrm{g}}{1.195\times {10}^{-22}\mathrm{g}} \\ = 4.19\approx 4 \end{gathered}$$

Thus, there are 4 Fe²⁺ ions and 4O^2– ions in each unit cell.