Question

The density of a face-centred cubic element (atomic mass = 60.2) is 6.25 g cm⁻¹. Calculate the length of the edge of the unit cell.

Answer 1

`rho = (Z xx M)/(a^3 xx N_A)`   

Where:

ρ = density = 6.25 g/cm³

Z = number of atoms per unit cell = 4 (for fcc)

M = molar mass = 60.2 g/mol

a = edge length of the unit cell (in cm)

N_A​ = Avogadro’s number = 6.022 × 10²³ mol⁻¹

∴ `a^3 = (Z xx M)/(rho xx N_A)`

∴ `a^3 = (4 xx 60.2)/(6.25 xx 6.022 xx 10^(23))`

∴ `a^3 = (240.8)/(37.59 xx 10^(23))`

∴ `a^3 = (240.8)/(3.759 xx 10^(24))`

∴ a³ = 6.4 × 10⁻²³ cm³

a = `root3(6.4 xx 10^(-23)) ≈ 4.00 xx 10^(-8)` cm

a = 4.00 × 10⁻⁸ × 10¹⁰

a = 400 pm