Answer: Density of platinum obtained from the calculation = 19.80 g/cm3
Explanation:
Density of a cubic structure = ((number of atoms per unit cell) × (Atomic weight of Platinum in g/mol))/((volume of the unit cell) × (Avogadro's constant i.e. number of atoms per mol)) = (nA)/((V)(Na))
For FCC, number of atoms per unit cell, a = 4atoms per unit cell.
Atomic weight of platinum, A = 195.1g/mol (from literature)
Volume of unit cell = (lattice constant)^3; lattice constant = a = 0.403nm = (40.3 × (10^-9))cm. Volume of the unit cell = (40.3 × 10^-9)^3 = (6.545 × (10^-23))cm3
Avogadro's constant = (6.022 × (10^23)) atoms/mol
Density = (4 × 195.1)/(6.545 × 6.022) = 19.80 g/cm3
