A metal ๐ forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _______ $ร10^{21}$. (Nearest integer). ( Given $N_A=6.02ร10^{23}$)
The number of voids in a hexagonal close-packed (hcp) structure can be determined from the formula for voids in a closed-packed structure: for each atom, there is 1 octahedral void and 2 tetrahedral voids. Given that there are 0.02 mol of the metal M, we use Avogadro's number ($N_A=6.02ร10^{23}$) to find the number of atoms. Number of atoms = $0.02 \text{ mol} \times 6.02ร10^{23} \text{ atoms/mol} = 1.204ร10^{22} \text{ atoms}$. In an hcp structure, the number of total voids (octahedral plus tetrahedral) per atom is 3. So, total voids = Number of atoms ร Voids per atom = $1.204ร10^{22} \text{ atoms} \times 3 = 3.612ร10^{22}$ voids. Finally, we need to express this in terms of $10^{21}$ as stated in the question: $3.612ร10^{22} = 36.12ร10^{21}$. The nearest integer is 36. Hence, the total number of voids in 0.02 mol of the metal is 36 $ร10^{21}$, which falls within the expected range.
