Question

$Ga$ (atomic mass $70\,u$ ) crystallizes in a hexagonal close packed structure. The total number of voids in $0.581\,\,g$ of $Ga$ is _______.$\times 10^{21} .$ (Round off to the Nearest Integer).

Answer 1

$HCP$ structure : Per atom, there will be one octahedral void $(OV)$ and two tetrahedral voids $(TV)$.

Therefore total three voids per atom are present in $HCP$ structure.

$\rightarrow$ therefore total no of atoms of $Ga$ will be-

$=\frac{\text { Mass }}{\text { Molar Mass }} \times N _{ A }=\frac{0.581 g }{70 g / mol } \times 6.023 \times 10^{23}$

$\rightarrow$ Now, total Number of voids $=3 \times$ total no. of atoms

$=3 \times \frac{0.581}{70} \times 6.023 \times 10^{23}=14.99 \times 10^{21}$

$\simeq 15 \times 10^{21}$