Calculate the radius of a palladium (Pd) atom, given that Pd has an FCC crystal structure, a density of \(12.0 \mathrm{~g} / \mathrm{cm}^{3}\), and an atomic weight of \(106.4 \mathrm{~g} / \mathrm{mol}\)

Question: Determine the atomic radius of a palladium (Pd) atom given its density (12.0 g/cm³), atomic weight (106.4 g/mol), and face-centered cubic (FCC) crystal structure. Answer: The atomic radius of a palladium atom can be determined using the given information and following the steps outlined in the provided solution as shown below: 1. Calculate the number of moles per unit cell: moles = 4 / (6.022 × 10^23 atoms/mol) 2. Calculate the mass of one unit cell: mass = moles × 106.4 g/mol 3. Calculate the volume of one unit cell: volume = mass / 12.0 g/cm³ 4. Calculate the side length of the unit cell: a = (volume)^(1/3) 5. Calculate the atomic radius using FCC side length relationship: r = a / (2 √2) Using these steps and the given values, you can solve for the atomic radius of a palladium atom.