The density of potassium bromide crystal is $2.75gc{m}^{-3}$ and the length of a unit cell is $654pm$. What will be the crystal lattice of $KBr$?

$\text{Molar mass of}KBr=119gmo{l}^{-1}$

The correct option is A Face centred cubic
$\text{Density,}\rho =\frac{Z\times M}{N_A\times V}$
where,
$Z$ is number of atoms in unit cell
$N_A$ is avagadro number
$M$ is molar mass of element
$V$ is volume of unit cell

$Z=\frac{\rho \times a^3\times N_A}{M}$
$=\frac{2.75\times (654\times {10}^{-10}{)}^3\times 6.023\times {10}^{23}}{119}=3.89\approx 4$
$\text{Number of atoms per unit cell}=4$
Hence, it has face centre crystal lattice.