Question

An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m⁻³, what is the nature of the cubic unit cell?

Answer 1

It is given that density of the element, d = 2.7 × 10³ kg m⁻³

Molar mass, M = 2.7 × 10⁻² kg mol⁻¹

Edge length, a = 405 pm = 405 × 10⁻¹² m

= 4.05 × 10⁻¹⁰ m

It is known that, Avogadro’s number, N_A = 6.022 × 1023 mol⁻¹

Applying the relation,

`d = (z,m)/(a^3.N_A)`

`z= (d.a^3N_A)/M`

`=(2.7xx10^3kgm^(-3)xx(4.05xx10^(-10 m))^3 xx6.022xx10^23 mol^(-1))/(2.7xx10^(-2)kg  mol^(-1))`

= 4.004

 = 4

This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).