A metallic element has a cubic lattice. Each edge of the unit cell is 2 $2{\mathrm{A}}^{\mathrm{o}}$. The density of the metal is 2.5 g cm^-3. The unit cells in 200 g of metal are

Number of unit cells $=\frac{\mathrm{mass} \mathrm{of} \mathrm{metal}}{\mathrm{mass} \mathrm{of} \mathrm{one} \mathrm{unit} \mathrm{cell} }$
Given, edge length of unit cell $=2{\mathrm{A}}^{\mathrm{o}}=2\times {10}^{-8}\mathrm{cm}$
Mass of metal = 200 g
Density of metal = 2.5 g cm^-3
Volume of unit cell = (edge length)³ 
$\begin{array}{r}={\left(2\times {10}^{-8}\right)}^3\\ =8\times {10}^{-24}{\mathrm{cm}}^3\end{array}$
Mass of one unit cell - volume x density
$=8\times {10}^{-24}\times 2.5=20\times {10}^{-24}$
$\therefore$Number of unit cells in 200 g metal
$\begin{array}{r}=\frac{\text{ mass of metal }}{\text{ mass of one unit cell }}\\ =\frac{200}{20\times {10}^{-24}}=1.0\times {10}^{25}\end{array}$