Question

An element has a face centred cubic structure with a cell edge of 0.2nm. calculate its density in gm/cm3 if 400gm of this element contains 4.8*10^24 atoms

Answer 1

Each unit cell has:

8 corner atoms, each shared by 8 unit cells = (1/8)(8) = 1 atom
6 face atoms, each shared by 2 faces = (1/2)(6) = 3 atoms
Total: 4 atoms per unit cell

1nm=10^-9 m

V = a³ = (0.2 x 10^-9 m)³ = 8 x 10^-30 m³ = 8 x 10^-24 cm³

Now do a proportion to get the mass of the 4 atoms in a unit cell:

400 g → 4.8 x 10²⁴ atoms

400/4.8 x 10²⁴g → 1 atoms

x g→ 4 atoms



x = 4 · 400 / (4.8 x 10²⁴) g ≅ 3.33 x 10^-22 g

ρ = m/V = (3.33 x 10^-22 g) / (8 x 10^-24 cm³) ≅ 41.625 g/cm³