Argon has a cubic closest packed structure as a solid. Assuming that argon has a radius of 190. pm, calculate the density of solid argon

A cubic closest packed (ccp) structure is also called a face-centered cubic (fcc) structure. In an fcc structure, there are 8 corner atoms and 6 face atoms totally contributing to just one unit cell.

For the fcc structure, we have: - 8 corner atoms with 1/8 of each inside the unit cell: \(8\cdot \frac{1}{8} = 1\) - 6 face atoms with 1/2 of each inside the unit cell: \(6\cdot \frac{1}{2} = 3\) Therefore, the total number of atoms per unit cell is: \(1+3 = 4\)

In an fcc structure, the lattice constant 'a' can be found by relating it to the atomic radius 'r'. The relationship is given by the equation: \(a = 4r / \sqrt{2}\) We are given the radius of argon, r = 190 pm. Using this, we can calculate the lattice constant (a): \(a = \frac{4 \cdot 190\, \text{pm}}{\sqrt{2}} \approx 269.53\, \text{pm}\)

The volume of a cubic unit cell can be calculated using the formula: \(V = a^3\) Now, we will calculate the volume using the lattice constant (a) obtained in step 3: \(V = (269.53\, \text{pm})^3 \approx 1.955\times 10^7\, \text{pm}^3\)

To proceed to the calculation of density, we need the molar mass of argon which is already known. The molar mass of argon (Ar) = 39.948 g/mol

To calculate the density, we can use the formula: \(\rho = \frac{\text{mass}}{\text{volume}}\) The mass can be calculated using: Mass = (number of atoms per unit cell) x (molar mass of argon) / Avogadro's number The volume will be the volume of the unit cell calculated in step 4. First we find the mass for one unit cell, by dividing the molar mass by Avogadro's number and then multiplying by the number of argon atoms in one unit cell: Mass per unit cell = \(\frac{4\times 39.948\,\text{g/mol}}{6.022\times10^{23}\, \text{ atoms/mol}} \approx 2.660\times10^{-22}\,\text{g}\) Now, we calculate the density by dividing the mass per unit cell by the volume of the unit cell: \(\rho = \frac{2.660\times10^{-22}\,\text{g}}{1.955\times 10^7\, \text{pm}^3} = 1.361\times10^{-29}\, \text{g/pm}^3\) Finally, converting the density from g/pm³ to g/cm³ (1 pm = 1 x 10⁻¹² cm), we get: \(\rho \approx 1.361\times10^{-1}\, \text{g/cm}^3\) Thus, the density of solid argon is approximately 0.1361 g/cm³.