Lithium forms a b.c.c. lattice. If the lattice constant is 3.50 x 10^-10 m and the experimental density is 5.30 x 10² kg m^-3, the percentage occupancy of Li metal is (Li = 7).

We have,
theoretical density = $\frac{\mathrm{zM}}{\mathrm{NV}}=\frac{\mathrm{zM}}{\mathrm{N}\left({\mathrm{a}}^3\right)}$
For a b.c.c. lattice: z = 2 and given that a : $3.50\times {10}^{-10}$m
$\begin{array}{l}\mathrm{M}=7\times {10}^{-3}\mathrm{kg}/\mathrm{mole}\\ {\mathrm{d}}_{\mathrm{cal}}=\frac{2\times 7\times \left({10}^{-3}\right)}{6.022\times {10}^{23}\times {\left(3.50\times {10}^{-10}\right)}^3}\\ =5.42\times {10}^2{\mathrm{kgm}}^{-3}\end{array}$
$\therefore \text{ percentage occupancy }=\frac{{\mathrm{\rho }}_{\exp }}{{\mathrm{\rho }}_{\mathrm{cal}}}\times 100$
$=\frac{5.30\times {10}^2}{5.42\times {10}^2}\times 100=97.78\mathrm{\%}$