An element crystallises in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 gcm⁻³. How many atoms are present in 300 g of the element?

The edge length, a = 500 pm = 500 × 10⁻¹² m = 5 × 10⁻⁸ cm
The volume of the unit cell = a³ = (5 × 10⁻⁸ cm)³ = 1.25 × 10⁻²² cm³
For b.c.c unit cell, the number of atoms per unit cell Z = 2
Volume occupied by each atom = 1.25 × 10⁻²² cm³/2 = 6.25 × 10⁻²³ cm³
The density d = 7.5 g/cm³
Mass of sample = 300 g
Volume of sample = mass/density = 300/7.5 = 40 cm³
The number of atoms present in 300 g of the element = 40 cm³ / 6.25 × 10⁻²³ cm³ = 6.4 × 10²³ atoms.