Question

Potassium crystallizes in a body centred cubic lattice. If, the approximate number of unit cells in $4.0$ g potassium is $x\times {10}^{22}$, then $x$ is ($N_A=6\times {10}^{23}$)

Answer 1

There will be eight atoms at corners of the cube and one atom at the body centre.

$\therefore \text{Number of atoms per unit cell}=(8\times \frac{1}{8})=2$
$\text{Number of atoms in 4.0 g of potassium}=\frac{4}{39}\times 6\times {10}^{23}$
$\therefore \text{Number of unit cells in 4.0 g of potassium}=\frac{4}{39}\times \frac{6\times {10}^{23}}{2}=3.07\times {10}^{22}$