163. An element crystallizes in a structure having a fcc unit cell of an edge 200 pm. Calculate its density if 200 g of this element contains 24 x 10²³ atoms.

Solution:
We have given that
Edge length of the unit cell
= 200 pm = 200 x 10^–10 cm

Vol. of the unit cell
= (200 x 10^–10 cm)³ = 8 x 10^–24 cm³
Therefore, volume of the substance
= Vol. of unit cell x Vol. of 1 unit cell.

Since the element has a fcc unit cell, number of atoms per unit cell = 4.
Total number of atoms = Atoms/unit cell x number of unit cells.
                                     = 24 x 10²³atoms = 4

atoms/unit cell x No. of unit cells.

∴   No. of unit cells = $\frac{24\times {10}^{23} \mathrm{atoms}}{4 \mathrm{atoms}/\mathrm{unit} \mathrm{cell}}$

                               $= 6\times {10}^{23} \mathrm{unit} \mathrm{cells}.$

∴  Volume of the substance
                   $6\times {10}^{23} \mathrm{unit} \mathrm{cell} \times 8\times {10}^{-24 }{\mathrm{cm}}^3/\mathrm{unit} \mathrm{cell} = 4.8 {\mathrm{cm}}^3$


Now,     $$\begin{gathered} \mathrm{Density} = \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ = \frac{200\mathrm{g}}{4.8 {\mathrm{cm}}^3} =4.17 \mathrm{g} {\mathrm{cm}}^{-3}. \end{gathered}$$