Aluminum has an atomic radius of 143 \(\mathrm{pm}\) and forms a solid with a cubic closest packed structure. Calculate the density of solid aluminum in \(\mathrm{g} / \mathrm{cm}^{3}\) .

\(a \approx 405.2\, \text{pm}\) #tag_title# Step 3: Calculate the volume of the unit cell #tag_content# Now that we have the edge length, we can calculate the volume of the unit cell (V): \(V = a^3\) Substitute the edge length (a) into the formula: \(V = (405.2\, \text{pm})^3\) Calculating the volume (V): \(V \approx 6.68 \times 10^7\, \text{pm}^3\) #tag_title# Step 4: Convert volume to cm³ #tag_content# To calculate the density in g/cm³, we need to convert the volume from pm³ to cm³: \(1\, \text{pm} = 10^{-10} \, \text{cm}\) \(V \approx 6.68 \times 10^7\, \text{pm}^3 \times (10^{-10}\, \text{cm} / \text{pm})^3\) Calculating the volume in cm³: \(V \approx 6.68 \times 10^{-23}\, \text{cm}^3\) #tag_title# Step 5: Determine the molar mass and Avogadro's number #tag_content# The molar mass of aluminum is 26.98 g/mol. Avogadro's number (N) is approximately \(6.022 \times 10^{23}\) particles/mol. #tag_title# Step 6: Calculate the density #tag_content# The density (ρ) of solid aluminum can be calculated using the formula: \(\rho = \frac{4 \times M_{\text{Al}}}{N \times V}\) Substituting the values found previously: \(\rho \approx \frac{4 \times 26.98\, \text{g/mol}}{6.022 \times 10^{23} \, \text{particles/mol} \times 6.68 \times 10^{-23}\, \text{cm}^3}\) Calculating the density (ρ): \(\rho \approx 2.70\, \text{g/cm}^3\) Thus, the density of solid aluminum is approximately 2.70 g/cm³.