A metal of molar mass M g mol^ -1 crystallizes in a body-centered...

For a body-centered cubic (bcc) lattice, the number of atoms per unit cell is $Z = 2$.

The relationship between the edge length $a$ and the atomic radius $r$ for a bcc lattice is:

$\sqrt{3}a = 4r \implies a = \dfrac{4r}{\sqrt{3}}$

Since the radius is given in picometers (pm), we must convert it to centimeters (cm) to find the density in $\text{g cm}^{-3}$:

$a = \dfrac{4r}{\sqrt{3}} \times 10^{-10}\text{ cm}$

The volume of the unit cell is:

$a^3 = \left(\dfrac{4r}{\sqrt{3}} \times 10^{-10}\right)^3 = \dfrac{64r^3}{3\sqrt{3}} \times 10^{-30}\text{ cm}^3$

The formula for density is:

$\rho = \dfrac{Z \times M}{N_A \times a^3}$

Substitute the values of $Z$ and $a^3$ into the density formula:

$\rho = \dfrac{2 \times M}{N_A \times \left(\dfrac{64r^3}{3\sqrt{3}} \times 10^{-30}\right)}$

$\rho = \dfrac{2 \times 3\sqrt{3} \times M}{64 N_A r^3 \times 10^{-30}}$

$\rho = \dfrac{6\sqrt{3} M \times 10^{30}}{64 N_A r^3}$

$\rho = \dfrac{3\sqrt{3} M \times 10^{30}}{32 N_A r^3}$

Answer: $\dfrac{3\sqrt{3} M \times 10^{30}}{32 N_A r^3}$