Question 9

For a cubic close packed (ccp) lattice, the number of atoms per unit cell is $Z = 4$.

Given:

Edge length ($a$) = $400 \text{ pm} = 400 \times 10^{-10} \text{ cm} = 4 \times 10^{-8} \text{ cm}$.

Atomic mass ($M$) = $105.6 \text{ g mol}^{-1}$. Avogadro's constant ($N_A$) = $6 \times 10^{23} \text{ mol}^{-1}$.

The density ($d$) is calculated as: $d = \frac{Z \times M}{N_A \times a^3}$ $d = \frac{4 \times 105.6}{6 \times 10^{23} \times (4 \times 10^{-8})^3}$

$d = \frac{422.4}{6 \times 10^{23} \times 64 \times 10^{-24}}$

$d = \frac{422.4}{6 \times 6.4 \times 10^{-1}} = \frac{422.4}{38.4} = 11.0 \text{ g cm}^{-3}$

Thus, the density is $11.0$.