An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm^–3. Calculate the number of atoms in 108 g of the element.

$\mathrm{d}=\frac{\mathrm{zm}}{{\mathrm{Na}}^3}$
Where,
d= density of element
z= number of atom per unit cell
m =mass of element
N = number of atoms
a = edge length
Let us put the values in formula
$\begin{array}{r}\mathrm{N}=\frac{108\times 4}{10.8\times 27\times {10}^{-24}}\\ =1.48\times {10}^{24}\text{ atoms }\end{array}$