The number of tetrahedral voids present in 0. | vedclass

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(D) In a crystal lattice,the number of tetrahedral voids $(TV)$ is twice the number of atoms $(N)$.For an $hcp$ unit cell,the number of atoms per unit

Vedclass · Accepted Answer

$6.02 	imes 10^{23}$

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(D) In a crystal lattice,the number of tetrahedral voids $(TV)$ is twice the number of atoms $(N)$.For an $hcp$ unit cell,the number of atoms per unit cell is $Z = 6$.Therefore,the number of $TV$ per unit cell is $2 	imes 6 = 12$.In $1 \ mole$ of atoms,the number of $TV$ is $2 	imes N_A$.For $0.5 \ mole$ of $hcp$ crystal structure,the number of atoms is $0.5 	imes N_A$.Number of $TV = 2 	imes (0.5 	imes N_A) = 1 	imes N_A = 6.022 	imes 10^{23}$.Wait,re-evaluating: The question asks for the total number of tetrahedral voids in $0.5 \ mole$ of the crystal structure. If we consider $0.5 \ mole$ of the $hcp$ lattice,it contains $0.5 	imes N_A$ atoms. Since $TV = 2N$,the number of $TV = 2 	imes (0.5 	imes N_A) = N_A = 6.022 	imes 10^{23}$.

The number of tetrahedral voids present in $0.5 \ mole$ of $hcp$ crystal structure is:

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