An element has a body centred cubic (bcc) structure with a cell edge of 444 pm. The density of the element is 7.2 g/cm³. How many atoms are present in 404 g of the element?

Volume of the unit cell = (444 pm)³
                                    =(444x10^-10cm)³
                                    =8.75x10^-23cm³
Volume of 404 g of the element $=\frac{\mathrm{mass}}{\mathrm{density}}$
                                                   $=\frac{404 \mathrm{g}}{7.2 \mathrm{g} {\mathrm{cm}}^{-3}}=56.11 {\mathrm{cm}}^3$
$\therefore$Number of unit cells in this volume
                              $=\frac{56.11{\mathrm{cm}}^3}{8.75\times {10}^{-23}{\mathrm{cm}}^3/\mathrm{unit} \mathrm{cell}}=6.41\times {10}^{23} \mathrm{unit} \mathrm{cells}$