Calculate the number of atoms present per unit cell if product of density and volume of unit cell is $1.8 \times 10^{-22} \mathrm{~g}$ [Mass of an atom $=4.5 \times 10^{-23} \mathrm{~g}$]

The density $\rho$ of a unit cell relates atomic parameters through $\rho V = Z m$, where $V$ is volume, $Z$ is atomic count per unit cell, and $m$ is atomic mass.

Given $\rho V = 1.8 \times 10^{-22}$ g and $m = 4.5 \times 10^{-23}$ g, we substitute into the relation:

$1.8 \times 10^{-22} = Z \times (4.5 \times 10^{-23})$

Solving for $Z$:

$Z = \frac{1.8 \times 10^{-22}}{4.5 \times 10^{-23}} = \frac{1.8}{4.5} \times 10^{1} = 0.4 \times 10 = 4$

The unit cell contains 4 atoms, corresponding to choice $\boxed{C}$.