An element germanium crystallises in b c c type crystal structure...

$d=7.2 \mathrm{~g} \mathrm{~cm}^{-3}, a=288 \mathrm{pm}$ and mass of element $=52 \mathrm{~g}$ Hence, $d=\frac{\text{ Mass }}{\text{ Volume }} \Rightarrow 7.2 \mathrm{~g} \mathrm{~cm}^{-3}=\frac{52 \mathrm{~g}}{V} V=\frac{52 \mathrm{~g}}{7.2 \mathrm{~g} \mathrm{~cm}^{-3}}=7.222 \mathrm{~cm}^3$ Edge of unit cell, $a=288 \mathrm{pm}=288 \times 10^{-12} \mathrm{~m}=288 \times 10^{-10} \mathrm{~cm}=2.88 \times 10^{-8} \mathrm{~cm}$ Volume of unit cell, $a^3=\left(2.88 \times 10^{-8} \mathrm{~cm}\right)^3=23.888 \times 10^{-24} \mathrm{~cm}^3$ The number of unit cells in $52 \mathrm{~g}$ of metal $$\begin{aligned} & =\frac{\text{ Total volume }}{\text{ Volume of one unit cell }} \\ & =\frac{7.222 \mathrm{~cm}^3}{23.888 \times 10^{-24} \mathrm{~cm}^3}=0.3023 \times 10^{24}\end{aligned}$$ Germanium has body centred cubic structure. Hence, its unit cell contains 2 atoms. $\therefore$ Total atoms in $3.023 \times 10^{23}$ unit cells are $2 \times 3.023 \times 10^{23}=6.046 \times 10^{23}$ atoms Molar mass $=\frac{52 \mathrm{~g} \times 6.022 \times 10^{23} \text{ atom } \mathrm{mol}^{-1}}{6.046 \times 10^{23} \text{ atom }}$ Hence, atomic mass $=51.8$