1.18 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? 

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, NA = 6.022 × 1023 mol−1
Applying the relation,
d=zMa3·NA
z=d·a3NAM
=2.7×103 kg m-1×(4.05×10-10 m)3×6.022×1023 mol-12.7×10-2 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).