An element X (At. wt.= 24) forms FCC lattice. If the edge length of lattice is $4\times {10}^{-8}\mathrm{cm}$ and the observed density is $2.4 \mathrm{x} {10}^3 \mathrm{kg}/{\mathrm{m}}^3$. Then the percentage occupancy of lattice point by element X is: (${\mathrm{N}}_{\mathrm{A}}=6\mathrm{x}{10}^{23}$ )

We know that,

$\mathrm{Density} \mathrm{of} \mathrm{unit} \mathrm{cell}=\frac{\mathrm{Z}·\mathrm{M}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}}$

In FCC H- atom is present.

So, $\text{ Density }=\frac{4\times 24}{64\times 6\times {10}^{23}\times {10}^{-24}}\mathrm{g}/{\mathrm{cm}}^3=0.25\times {10}^1\mathrm{g}/{\mathrm{cm}}^3$

Calculated Density $=2.5\mathrm{g}/{\mathrm{cm}}^3$

$$\begin{gathered} \text{ Percentage occupancy }=\frac{\text{ Observed density }}{\text{ Calculated density }} \\ =\frac{2.4}{2.5}\times 100 \\ =96\mathrm{\%} \end{gathered}$$