Suppose the edge of the unit cell = $a$ pm Number of atoms present per unit cell = $Z$ Atomic mass of the element = $M$ $\therefore\;$ Volume of the unit cell $=(a \pm)^3$$= (a \times 10^{-10} \mathrm{cm})^{3}$$=a^{3} \times 10^{-30} \mathrm{cm}^{3}$Density of the unit cell = $\frac{\text{Mass of the unit cell}}{\text{Volume of the unit cell}}$Mass of the unit cell$\;\;\;\;$= Number of atoms in the unit cell × Mass of each atom = $Z \times m$where $m =$ mass of each atom$m=\frac{\text{Atomic mass}}{\text{Avogadro's number}}=\frac{M}{N_{0}}$$\therefore\;$ Density of the unit cell, $\rho = \frac{Z \times M}{a^{3} \times N_{0} \times 10^{-30}}$ $g$ / $\mathrm{cm}^{3}$where edge, a is in pm and molar mass, $M$ in $g$ mol$^{-1}$.Atomic mass can be calculated by using the expression$M = \frac{\rho \times a^{3} \times N_{0} \times 10^{-30}}{Z} \ \mathrm{g\ mol}^{-1}$