Aluminium (Atomic mass = 27) crystallises in a cubic system with edge length of 4 $A^{\circ }$. Its density is 2.7 g $c{m}^{-3}.$ The number of aluminium atoms present per unit cell is

The correct option is B 4
Density of the unit cell, $\rho =\frac{Z\times M}{a^3\times N_A}$
Given : $\rho =2.7g/c{m}^3,M=27,a=4A^{\circ }=4\times {10}^{-8}cm$
$N_A=6.023\times {10}^{23}$
$Z=\frac{\rho \times a^3\times N_A}{M}$
$Z=\frac{2.7\times (4\times {10}^{-8}{)}^3\times (6.023\times {10}^{23})}{27}Z=3.85\approx 4$
Hence option (b) is correct.