Silver has a cubic unit cell with a cell edge of 408 pm. Its density is 10.6 g cm^–3. How many atoms of silver are there in the unit cell?

Length of the edge of unit cell = 408 pm
Volume of unit cell
= (408pm)³ = 67.92×10–24 cm³
Mass of unit cell = Density x Volume
= 10.6 g cm^–3 ×67.92 ×10^–24 cm3 = 7.20
Mass of unit cell = Number of atoms in unit
cell × Mass of each atom
Now, mass of each atom
$=\frac{\text{ Atomic mass }}{\text{ Avogadro number }}=\frac{108}{6.023\times {10}^{23}}=1.79\times {10}^{-22}\mathrm{g}$
Let the unit cell contain ‘Z’ atoms, so that
Mass of unit cell = Z×1.79×10^–22 = 7.20×10^–22
$\mathrm{Z}=\frac{7.20\times {10}^{-22}}{1.79\times {10}^{-22}}=4.02$
$\therefore$number of atoms present in a unit cell = 4