solution: 
we have given a= 2.88A⁰
Density = 7.20g cm^-3
mass = 100g

Volume of the unit cell = $(2.88 \mathrm{A}°{)}^3$
                          $$\begin{gathered} = (2.88\times {10}^{-8}\mathrm{cm}{)}^3 \\ = 23.9 \times {10}^{-24} {\mathrm{cm}}^3 \end{gathered}$$

Volume of 100g of the metal

                 $$\begin{gathered} = \frac{\mathrm{Weight}}{\mathrm{Density}}=\frac{100}{7.20} \\ = 13.9 {\mathrm{cm}}^3 \end{gathered}$$

No. of unit cell in $13.9 {\mathrm{cm}}^3$
   
                   $$\begin{gathered} = \frac{13.9 {\mathrm{cm}}^3}{23.9 \times {10}^{-24} {\mathrm{cm}}^3} \\ = 5.82 \times {10}^{23}. \end{gathered}$$

solution:
we have given

Density of element,  $\mathrm{\rho }=10.3 \mathrm{g} {\mathrm{cm}}^{-3}$
Cell edge  a= 314pm or 3.14 x 10^-10 cm          

     ${\mathrm{N}}_{\mathrm{A}} =6.023\times {10}^{23} {\mathrm{mol}}^{-1}$

Atomic mass $= 96 \mathrm{g} {\mathrm{mol}}^{-1}$

                 $$\begin{gathered} \mathrm{P} = \frac{\mathrm{Z}\times \mathrm{M}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{Z} = \frac{\mathrm{P}\times {\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}} \end{gathered}$$

$= \frac{10.\mathrm{g} {\mathrm{cm}}^{-3}\times (3.14{)}^3\times {10}^{-30}{\mathrm{cm}}^3\times 6.023\times {10}^{23} {\mathrm{mol}}^{-1}}{96 \mathrm{g} {\mathrm{mol}}^{-1}}=2$

The structure of the crystal lattice is B.C.C.

solution:
we have given that
mass = 200g
density = 7.2 g cm^-3
N_A = 4.12 x 10²⁴
Z = 4 
by using the following equation,
 
$\mathrm{\rho } = \frac{\mathrm{z}\times \mathrm{M}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}}$
                     

$\Rightarrow$               ${\mathrm{a}}^3 = \frac{\mathrm{Z}\times \mathrm{M}}{\mathrm{d}\times \mathrm{N}}$

Solution:
we have given that  
density, d= 7.2g cm^-3
edge = 288 x 10^-10 cm 
N_A = 6.02 x 10²³
for Bcc structure Z= 2

Atomic mass = ?

 
$$\begin{gathered} \mathrm{d} = \frac{\mathrm{ZM}}{{\mathrm{a}}^3{\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{M} = \frac{\mathrm{d}\times {\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}}{\mathrm{Z}} \\ \mathrm{M}=\frac{7.25\times (2.88\times {10}^{-10})\times 6.02\times {10}^{23}}{2}\mathrm{g}/\mathrm{mol} \\ \mathrm{M}= 51.77 \mathrm{g}/\mathrm{mol} \mathrm{of} \mathrm{atoms} \end{gathered}$$
        
Thus, atomic mass of element is =51.77 g/mol