Total number of voids in 0.5 moles of a compound forming a hexagonal close-packed structure is:
Answers
Number of close-packed particles = 0.5×6.022×1023= 3.011×1023
Therefore, number of octahedral voids = 3.011×1023
And, number of tetrahedral voids = 2×3.011×1023= 6.022×1023
Therefore, total number of voids = 3.011×1023+ 6.022×1023= 9.033×1023
Dear Student,
◆ Answer -
9.033×10^23 voids
● Explaination -
No of atoms present in 0.5 moles of compound are -
No of atoms = 0.5 × Avogadro's number
No of atoms = 0.5 × 6.022×10^23
No of atoms = 3.011×10^23 atoms
Octahedral voids are same no as of atoms in HCP.
Octahedral voids = no of atoms
Octahedral voids = 3.011×10^23 voids
Tetrahedral voids are twice the no as of atoms in HCP.
Tetrahedral voids = 2 × no of atoms
Tetrahedral voids = 2 × 3.011×10^23
Tetrahedral voids = 6.022×10^23 voids
Total no of voids -
Total voids = tetrahedral voids + octahedral voids
Total voids = 6.022×10^23 + 3.011×10^23
Total voids = 9.033×10^23 voids
Therefore, 9.033×10^23 voids will be present in 0.5 mol of compound.
Thanks dear.