An element with molar mass 2.7 x 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 x 103 kg m-3, what is the nature of the cubic unit cell?
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 × 10-12m
= 4.05 × 10-10 m
It is known that, Avogadro's number,
NA = 6.022 ×1023 mol-1
Applying the relation,
d = \(\frac{z,M}{a^3.N_A}\)
z = \(\frac{d.a^3N_A}{M}\)
= \(\frac{2.7 \times 10^3kgm^{-3} \times (4.05 \times 10^{-10}m)^3 \times 6.022 \times 10^{23}mol^{-1}}{2.7 \times 10^{-2}kg mol^{-1}}\)
= 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).