An element crystallising in body centred cubic lattice has an edge length of $500pm$. If density is $4gc{m}^{-3},$ the atomic mass of the element $\left(ingmo{l}^{-1}\right)$ is :
Take $N_A=6\times {10}^{23}$

The correct option is B 150
$\text{Density}\rho =\frac{Z\times M}{a^3\times N_A}$
where, Z = total number of atoms present in 1 unit cell
M = molar mass of particle
a = edge length



$\rho =4g/c{m}^3,a=500pm=500\times {10}^{-10}cm$
For a BCC structure:
$Z=2$

$\therefore M=\frac{\rho \times a^3\times N_A}{Z}$
$M=\frac{4\times (500\times {10}^{-10}{)}^3\times 6\times {10}^{23}}{2}$
$M=150gmo{l}^{-1}$