Question

Total number of voids in $0.5mol$ of a compound forming hexagonal closed packed structure are

• A. $3.011\times {10}^{23}$
• B. $9.033\times {10}^{23}$
• C. $6.022\times {10}^{23}$
• D. $4.516\times {10}^{23}$

Answer 1

The correct option is B $9.033\times {10}^{23}$

In hcp, there are 6 atoms per unit cell.
$\therefore N=6$
$\text{Number of octahedral voids}=N=6$
$\text{Numer of tetrahedral voids}=2N=2\times 6=12$
$\therefore$
$\text{Total number of voids per unit cell}=18$
$\text{Total number of voids per atom}=\frac{18}{6}=3$

Thus, the total no. of voids per mole will be $3\times 6.023\times {10}^{23}$
$\therefore$
In $0.5mol$ compound,
$\text{Total no. of voids}=3\times 0.5\times 6.023\times {10}^{23}=9.034\times {10}^{23}$