An element has a body-centred cubic $\left(bcc\right)$ structure with a cell edge of $288pm$. The density of the element is $7.2gc{m}^{-3}$. How many atoms are present in $208g$ of the element ?

Molar mass of element
Given:

Cell edge, $a=288\times {10}^{-10}cm$, density of
copper, $d=7.2gc{m}^{-3}$

In case of $bcc$ lattice, number of atoms per
unit cell, $z=2$

We know that $N_A=6.022\times {10}^{23}mo{l}^{-1}$
Therfore, $M=\frac{d{N}_A{a}^3}{z}$

By putting the values, we get,
$M=\frac{7.2gc{m}^{-3}\times 6.022\times {10}^{23}mo{l}^{-1}\times (288\times {10}^{-10}cm{)}^2}{2}$

Number of atoms

Given:

Mass of element $m=208g$

We know that,

Number of $\text{mole}=\frac{\text{mass}\left(m\right)}{\text{molar mass}\left(M\right)}=\frac{\text{number of atoms}}{N_A}$

By putting the values, we get
$\frac{\text{number of atoms}}{6.022\times {10}^{23}mo{l}^{-1}}=\frac{208g\times 2}{7.2gc{m}^{-3}\times 6.022\times {10}^{23}mo{l}^{-1}\times (288\times {10}^{-10}cm{)}^2}=24.19\times {10}^{23}\text{atoms}$

Number of atoms present in $208g$ of the element$24.19\times {10}^{23}\text{atoms}$

Final answer: Number of atoms present in
$208g$ of the element is $24.19\times {10}^{23}$.