SOLUTION:
Step 1: Calculate the edge length of the unit cell.
[tex]\begin{aligned} a & = \frac{4}{\sqrt{3}} \times r \\ & = \frac{4}{\sqrt{3}} \times \text{128 pm} \\ & = \text{295.6 pm} \\ & = \text{295.6 pm} \times \frac{1 \times 10^{-12} \: \text{m}}{\text{1 pm}} \times \frac{\text{100 cm}}{\text{1 m}} \\ & = 2.956 \times 10^{-8} \: \text{cm} \end{aligned}[/tex]
Step 2: Calculate the volume of the unit cell.
[tex]\begin{aligned} V & = a^3 \\ & = (2.956 \times 10^{-8} \: \text{cm})^3 \\ & = 2.5829 \times 10^{-23} \: \text{cm}^3 \end{aligned}[/tex]
Step 3: Calculate the mass the unit cell (m).
Note that in a body-centered cubic crystal structure, there are 2 atoms in a unit cell.
[tex]\begin{aligned} m & = \text{2 atoms} \times \frac{\text{1 mol}}{6.0221 \times 10^{23} \: \text{atoms}} \times \frac{\text{51.996 g}}{\text{1 mol}} \\ & = 1.7268 \times 10^{-22} \: \text{g} \end{aligned}[/tex]
Step 4: Calculate the density of rhodium.
[tex]\begin{aligned} d & = \frac{m}{V} \\ & = \frac{1.7268 \times 10^{-22} \: \text{g}}{2.5829 \times 10^{-23} \: \text{cm}^3} \\ & = \boxed{\text{6.69 g/cm}^3} \end{aligned}[/tex]
Hence, the density of chromium is 6.69 g/cm³.
[tex]\\[/tex]
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