An element with molar mass

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,

$d=\frac{z, M}{a^{3} \cdot \mathrm{N}_{\mathrm{A}}}$

$z=\frac{d \cdot a^{3} \mathrm{~N}_{\mathrm{A}}}{M}$

$=\frac{2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3} \times\left(4.05 \times 10^{-10} \mathrm{~m}\right)^{3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}}{2.7 \times 10^{-2} \mathrm{~kg} \mathrm{~mol}^{-1}}$

$=4.004$

$=4$

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).