A compound forms a hexagonal close-packed (HC | vedclass

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(B) In a close-packed structure,the number of octahedral voids is equal to the number of atoms present in the lattice.Given,number of moles of the com

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$1.8066 	imes 10^{23}$

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(B) In a close-packed structure,the number of octahedral voids is equal to the number of atoms present in the lattice.Given,number of moles of the compound $= 0.3 \, mol$.Number of atoms $= 0.3 	imes 6.022 	imes 10^{23} \, 	ext{atoms/mol} = 1.8066 	imes 10^{23} \, 	ext{atoms}$.Since the number of octahedral voids is equal to the number of atoms,the number of octahedral voids $= 1.8066 	imes 10^{23}$.

$A$ compound forms a hexagonal close-packed $(HCP)$ structure. How many octahedral voids are present in $0.3 \, mol$ of it?

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