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.

 Solution:
We have given that

Unit cell length,
a = 0.40806 nm  = 4.086 x 10^-10 m

  If fcc lattice the number of atoms per unit cell,
i.e. Z = 4

        M for Ag = 107.88 g mol<sup>-1

                      $=1.0788\times {10}^{-1} \mathrm{kg} {\mathrm{mol}}^{-1}$

              ${\mathrm{N}}_{\mathrm{A}} = 6.023 \times {10}^{23}$</sup>

For bcc structure
        

Solution:
We have given that
Density = 10 gm^-3
Mass = 100g
edge of unit cell ,a= 4 since it is a Fcc crystal

we have to find total number of atom, So by following relation we can get the result,

Total No. of atoms = $\frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{a}}^3.\mathrm{d}}$

Therefore, number of atoms

$$\begin{gathered} =\frac{4\times 100 \mathrm{g}}{(100\times {10}^{-12}\mathrm{m}{)}^3\times 10 {\mathrm{gm}}^{-3}} \\ = 4\times {10}^{31} \mathrm{atoms} \end{gathered}$$
 
thus the number of atoms is 4 x 10³¹