Chemistry, asked by Wazowski8998, 1 year ago

Total number of voids in 0.5 moles of a compound forming a hexagonal close-packed structure is:

Answers

Answered by anirudhajith112
275

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


anirudhajith112: 10^23 there mate my bad forgot the ^ symbol
Answered by gadakhsanket
190

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.

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