An element has a body – centred cubic structure with cell edge of 288pm. The density of element is 7.2 g/cc. How many atoms are present in 208g of the element?

$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{unit} \mathrm{cell} ={(288 \mathrm{pm})}^3={(288\times {10}^{-10}\mathrm{cm})}^3 \\ =2.389\times {10}^{-23} {\mathrm{cm}}^3 \\ \mathrm{Volume} \mathrm{of} 208 \mathrm{g} \mathrm{of} \mathrm{the} \mathrm{element} =\frac{\mathrm{Mass}}{\mathrm{Density}}= \frac{208}{ 7.2}=28.89{\mathrm{cm}}^3 \\ \mathrm{Number} \mathrm{of} \mathrm{unit} \mathrm{cells} =\frac{\mathrm{Total} \mathrm{Volume}}{\mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{unit} \mathrm{cell}}= \frac{28.89}{2.389\times {10}^{-23}} \\ =12.09\times {10}^{23} \\ \mathrm{For} \mathrm{a} \mathrm{BCC} \mathrm{structure}, \mathrm{number} \mathrm{of} \mathrm{atoms} \mathrm{per} \mathrm{unit} \mathrm{cell} =2 \\ \therefore \mathrm{Number} \mathrm{of} \mathrm{atoms} \mathrm{present} \mathrm{in} 208 \mathrm{g} = \\ \mathrm{No}. \mathrm{of} \mathrm{atoms} \mathrm{per} \mathrm{unit} \mathrm{cell} \times \mathrm{No}. \mathrm{of} \mathrm{unit} \mathrm{cells}=2\times 12.09\times {10}^{23} \\ =24.18\times {10}^{23} \end{gathered}$$