Calculate a value for the density of FCC platinum in grams per cubic centimeter from its lattice constant "a" of 0.403 nm. The answer should have two decimals of accuracy.

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