The Solid State

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

Solution:
We have given that 

Gram atomic mass of Cr(M) = 52.0 g mol^-1
   Edge length of unit cell (a) = 289 pm
   Density of unit cell $\left(\mathrm{\rho }\right)$ = 7.2 g cm^-3
   Avogadro's Number $\left({\mathrm{N}}_0\right) = 6.022\times {10}^{23} {\mathrm{mol}}^{-1}$       

     $\mathrm{\rho } = \frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}\times {10}^{-30}}$

or              $\mathrm{Z}=\frac{\mathrm{\rho }\times {\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}\times {10}^{-30}}{\mathrm{M}}$
   
   $\mathrm{Z}=\frac{(7.2 \mathrm{g} {\mathrm{cm}}^{-3}) \times (289{)}^3\times (6.022\times {10}^{23} {\mathrm{mol}}^{-3}) \times ({10}^{-30} {\mathrm{cm}}^3)}{(52.0 \mathrm{g} {\mathrm{mol}}^{-1})}=2$

Since the unit cell has 2 atoms, it is body centre in nature.

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.

Solution:
we have given that
Atomic mass of element = 60.
Number of atoms per fcc unit cell = 4
Density of the element = 6.23 g/cm³

                              $= \frac{6.23}{{10}^{23}} \mathrm{g}/(\mathrm{pm}{)}^3$

Density,         $\mathrm{d}=\frac{\mathrm{N}\times \mathrm{M}}{{\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3}$