Question

An element has atomic mass 93 g mol⁻¹ and density 11.5 g cm^–3. If the edge length of its unit cell is 300 pm, identify the type of unit cell. (N_A = 6.023 × 10²³ mol⁻¹)

Answer 1

Given: M = 93 g mol⁻¹

ρ = 11.5 g cm⁻³

a = 300 pm = 300 × 10⁻¹⁰ cm = 3 × 10⁻⁸ cm

N_A = 6.023 × 10²³ mol⁻¹

Formula: `rho = (Z xx M)/(N_A xx a^3)`

`Z = (rho xx N_A xx a^3)/M`

`Z = (11.5 xx (3 xx 10^-8)^3 xx6.023 xx 10^23)/93`

`Z = (11.5 xx 27 xx 10^-24 xx 6.023 xx 10^23)/93`

`Z = (11.5 xx 162.621 xx 10^-1)/93`

`Z = (1870.14 xx 10^-1)/93`

`Z = 187.014/930`

Z = 2.01 (approx.)

As the number of atoms present in given unit cells is coming nearly equal to 2, hence the given unit cell is a body-centred cubic unit cell (bcc).