158. The density of chromium is 7.2 g cm^–3. If the unit cell is cubic with edge length of 289 pm, determine the type of the unit cell (Atomic mass of Cr = 52 amu).

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.