Question

What is the density of KCl at 25.00 °C if the edge length of its fcc unit cell is 628 pm?

Answer 1

First we need to find out how many atoms per unit cell. For all unit cells except hexagonal, atoms on the faces contribute 1/2 atom to each unit cell, atoms on the edges contribute 1/4 atom to each unit cell, and atoms on the corners contribute 1/8 atom to each unit cell. K-1/4 x 12+ 1x 14 K atoms per unit cell CI- 1/8 x 81/2 x 6-4 CI atoms per unit cell Density = Mass of unit cell / volume of unit cell Mass of unit cell [(39.10+35.45) g/mol x (1 mol/6.022 x 1023 ions) x (4 ion units/unit cell)] 4.952 x 1022 g Volume of unit cell (Edge length) -(6.28 x 10-8 cm)3 = 2.477 x 10-22 cm3 Density Mass of unit cell / volume of unit cell -4.952 x 1022 g /2.477 x 1022 cm3 1.999 g/cm3 Density 2.00 g/cm3