Question

$Ga$ (atomic mass = $70u$) crystallizes in a hexagonal close packed structure. The total number of voids in $0.581\text{g}$ of $Ga$ is $\times {10}^{21}$. (Round off to the nearest integer). [Given: $N_A=6.023\times {10}^{23}]$

Answer 1

No. of moles of $Ga=\frac{0.581}{70}$
No. of atoms of $Ga=\frac{0.581}{70}\times N_A$
$\therefore$ Total number of voids $=\frac{0.581}{70}\times N_A\times 3$
$=0.0249\times 6\times {10}^{23}$
$=15\times {10}^{21}$
(As there is one octahedral void and two tetrahedral voids per atom)