Aluminium crystallises in a cubic close-packed structure. Its metallic radius is $125 \ pm$.
$(i)$ What is the length of the side of the unit cell?
$(ii)$ How many unit cells are there in $1.00 \ cm^3$ of aluminium?

$(i)$ For a cubic close-packed $(ccp)$ structure, the relationship between the edge length $(a)$ and the metallic radius $(r)$ is given by $a = 2\sqrt{2}r$.
Substituting the value of $r = 125 \ pm$:
$a = 2 \times 1.414 \times 125 \ pm = 353.55 \ pm \approx 354 \ pm$.
$(ii)$ The volume of one unit cell is $a^3 = (354 \ pm)^3 = (354 \times 10^{-10} \ cm)^3 = 4.436 \times 10^{-23} \ cm^3$.
The number of unit cells in $1.00 \ cm^3$ is calculated as:
$\text{Number of unit cells} = \frac{1.00 \ cm^3}{4.436 \times 10^{-23} \ cm^3} = 2.254 \times 10^{22}$.