A certain metal with a molar mass of 48 g/mol crystallizes in a...

The formula for the density of a cubic unit cell is:
$d = \dfrac{Z \times M}{N_A \times a^3}$

Given:
$d = 20 \text{ g/cm}^3$
$M = 48 \text{ g/mol}$
$N_A = 6 \times 10^{23} \text{ mol}^{-1}$
$a = 200 \text{ pm} = 2 \times 10^{-8} \text{ cm}$

Volume of the unit cell, $a^3 = (2 \times 10^{-8})^3 = 8 \times 10^{-24} \text{ cm}^3$

Rearranging the density formula to solve for the number of atoms per unit cell ($Z$):
$Z = \dfrac{d \times N_A \times a^3}{M}$
$Z = \dfrac{20 \times 6 \times 10^{23} \times 8 \times 10^{-24}}{48}$
$Z = \dfrac{20 \times 48 \times 10^{-1}}{48} = 2$

Since $Z = 2$, the metal crystallizes in a body-centered cubic (BCC) lattice.
The coordination number of an atom in a BCC lattice is $8$.

Answer: $8$