Question

A compound has hexagonal closed packed structure. What is the total number of voids present in $0.5mol$ of it?

• A. $6.02\times {10}^{23}$
• B. $9.03\times {10}^{23}$
  
• C. $12.04\times {10}^{23}$
• D. $\mathrm{3..01}\times {10}^{23}$

Answer 1

The correct option is B

$9.03\times {10}^{23}$

Total number of atoms in hexagonal closed packing is,
$=0.5\times 6.02\times {10}^{23}=3.01\times {10}^{23}$

In hexagonal close packing,
$\text{Number of atoms present = Number of octahedral voids}$
$\text{Number of atoms present =}2\times \text{Number of tetrahedral voids}$
So,
$\text{Number of Octahedral Voids}=3.011\times {10}^{23}$
$\text{Number of Tetrahedral Voids}=2\times 3.011\times {10}^{23}$
$=6.02\times {10}^{23}$
Hence,
$\text{Total no. of voids in 0.5 mol}=9.03\times {10}^{23}$
HCP: