A metal has a bcc lattice and edge length of the unit cell is $2.37\times {10}^{-10}m$. Thus, atomic volume of the metal is ………..$c{m}^3$

Volume of the cube $=a^3\times N_0$$={(2.37\times {10}^{-10}m)}^3\times 6.023\times {10}^{23}mo{l}^{-1}$

In BCC lattice, number of atoms = 2

Hence volume of one atom $=\frac{8\times {10}^{-6}}{2}{m}^{3}mo{l}^{-1}=4\times {10}^{-6}{m}^3$

$=4\times {10}^{-6}\times {10}^{6}c{m}^3=4.0c{m}^3$