Question

Total number of voids in $0.45mol$ of a compound forming a hexagonal close packed structure is:

• A. $6.45\times {10}^{23}$
• B. $3.92\times {10}^{23}$
• C. $8.13\times {10}^{23}$
• D. $5.59\times {10}^{22}$

Answer 1

The correct option is C $8.13\times {10}^{23}$

$\text{In HCP, number of atoms per unit cell}=6$

$\text{Number of octahedral voids}=6$

$\text{Number of tetrahedral voids}=2\times 6=12$

$\text{Total number of voids}=6+12=18$

$\text{Total number of voids per atom}=\frac{18}{6}=3$

$\text{Total number of voids in}1mol=3\times 6.023\times {10}^{23}$

$\text{Total number of voids in}0.45mol=0.45\times 3\times 6.023\times {10}^{23}=8.13\times {10}^{23}$