An element has a BCC structure with a cell edge of 288 pm. The density of the element is 7.2 g cm^-3. The number of atoms present in 208 g of the element?

Volume of unit cell is calculated as:

${\left(288 \mathrm{pm}\right)}^3=2.389\times {10}^{-23} {\mathrm{cm}}^3$

Volume present in 208 g of element is calculated as:

$\mathrm{volume}=\frac{\mathrm{mass}}{\mathrm{density}}=\frac{208}{7.2}=28.89 {\mathrm{cm}}^3$

$\mathrm{Number} \mathrm{of} \mathrm{unit} \mathrm{cell}=\frac{\mathrm{total} \mathrm{volume}}{\mathrm{volume} \mathrm{of} \mathrm{unit} \mathrm{cell}}=\frac{28.89}{2.389\times {10}^{-23}}=12.09\times {10}^{23}$

For BCC, number of atoms per unit cell is 2. Number of atoms present in 208 g:

$2\times 12.09\times {10}^{23}=2.418\times {10}^{24}$