Barium has a body-centered cubic structure. If the atomic radius of barium is \(222 \mathrm{pm}\), calculate the density of solid barium.

The density of solid barium can be calculated using the given atomic radius and following these steps: 1. Calculate the edge length 'a' from the atomic radius (r) using the formula \(4r = \sqrt{3a^2}\), which results in \(a = \sqrt{\frac{(4(222))^2}{3}} \approx 505.77 \mathrm{pm}\). 2. Calculate the volume of the unit cell (V) as \(V = a^3 \approx (505.77)^3 \approx 1.293 \times 10^{8} \mathrm{pm}^3\). 3. In a BCC structure, there are 2 atoms per unit cell, and the molar mass of barium (Ba) is 137.33 g/mol. Find the mass of one Ba atom using Avogadro's Number: \(\frac{137.33 \mathrm{g/mol}}{6.022 \times 10^{23} \mathrm{atoms/mol}} \approx 2.28 \times 10^{-22} \mathrm{g/atom}\). 4. Calculate the mass of atoms in the unit cell: \(2.28 \times 10^{-22} \mathrm{g/atom} \times 2 \approx 4.56 \times 10^{-22} \mathrm{g}\). 5. Calculate the density of solid barium (ρ) using the formula \(ρ = \frac{\text{Mass}}{\text{Volume}}\), which results in \(ρ = \frac{4.56 \times 10^{-22} \mathrm{g}}{1.293 \times 10^{8} \mathrm{pm}^3} \approx 3.53 \mathrm{g/cm^3}\). Therefore, the density of solid barium is approximately \(3.53 \mathrm{g/cm^3}\).