Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of \(197 \mathrm{pm},\) calculate the density of solid calcium.

In a ccp structure, the atoms are arranged in a face-centered cubic (fcc) arrangement, which means that the atoms are present at the corners and the center of each face of the cube. The relationship between the atomic radius (r) and the unit cell edge length (a) in an fcc structure is given by the following equation: \[a = 2\sqrt{2}r\] We are given the atomic radius of calcium (r) as 197 pm. Now, we can calculate the length of a unit cell (a) for calcium: \[a = 2\sqrt{2}(197 \, \mathrm{pm}) = 556.16 \, \mathrm{pm}\]

Now that we have the edge length (a) of the unit cell, we can calculate the volume (V) of the unit cell using the formula for the volume of a cube: \[V = a^3\] Substituting the edge length (556.16 pm) calculated in Step 1: \[V = (556.16 \, \mathrm{pm})^3 = 1.719 \times 10^8 \, \mathrm{pm}^3\]

In an fcc arrangement, there are 4 atoms per unit cell. This is because each of the 8 corners of the cube contains 1/8 of an atom, and each of the 6 faces contains 1/2 of an atom: \[\text{Number of atoms per unit cell} = 8 \times (\frac{1}{8}) + 6 \times (\frac{1}{2}) = 4\]

To calculate the mass of the atoms in the unit cell, we need to find the molar mass of calcium and convert it to the mass of one atom. The molar mass of calcium is 40.08 g/mol. To find the mass of one atom, we can use Avogadro's number (6.022 x 10^23 atoms/mol): \[\text{Mass of one calcium atom} = \frac{40.08 \, \mathrm{g/mol}}{6.022 \times 10^{23} \, \mathrm{atoms/mol}} = 6.655 \times 10^{-23} \, \mathrm{g}\] Therefore, the mass of the atoms in the unit cell (4 calcium atoms) is: \[\text{Total mass in one unit cell} = 4 \times (6.655 \times 10^{-23} \, \mathrm{g}) = 2.662 \times 10^{-22} \, \mathrm{g}\]

Finally, we can calculate the density (ρ) of solid calcium using the mass (M) and volume (V) of the unit cell: \[\rho = \frac{M}{V}\] Substituting the values obtained in Steps 2 and 4: \[\rho = \frac{2.662 \times 10^{-22} \, \mathrm{g}}{1.719 \times 10^8 \, \mathrm{pm}^3}\] To convert the volume from \(pm^3\) to \(cm^3\), we use the conversion factor \(1\, cm = 1\times 10^{10}\, pm\), and so: \[V = 1.719 \times 10^8 \, \mathrm{pm}^3 \times (\frac{1 \, \mathrm{cm}}{1\times 10^{10} \, \mathrm{pm}})^3 = 1.719 \times 10^{-6} \, \mathrm{cm}^3\] Now we can calculate the density: \[\rho = \frac{2.662 \times 10^{-22}\, \mathrm{g}}{1.719 \times 10^{-6}\, \mathrm{cm}^3} = 1.548 \, \mathrm{g/cm^3}\] Thus, the density of solid calcium is approximately 1.548 g/cm³.