Ga (atomic mass 70 , u) crystallizes in a hex | vedclass

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(C) In an $HCP$ structure,for every $1$ atom,there is $1$ octahedral void $(OV)$ and $2$ tetrahedral voids $(TV)$.Therefore,the total number of voids

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) In an $HCP$ structure,for every $1$ atom,there is $1$ octahedral void $(OV)$ and $2$ tetrahedral voids $(TV)$.Therefore,the total number of voids per atom is $1 + 2 = 3$.Number of moles of $Ga = \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{0.581 \, g}{70 \, g/mol} = 0.0083 \, mol$.Number of $Ga$ atoms $= 	ext{moles} 	imes N_A = 0.0083 	imes 6.022 	imes 10^{23} \approx 4.998 	imes 10^{21}$ atoms.Total number of voids $= 3 	imes 	ext{Number of atoms} = 3 	imes 4.998 	imes 10^{21} = 14.994 	imes 10^{21}$.Rounding to the nearest integer,we get $15 	imes 10^{21}$.

$Ga$ (atomic mass $70 \, u$) crystallizes in a hexagonal close packed structure. The total number of voids in $0.581 \, g$ of $Ga$ is $\dots \dots \dots \dots \dots \times 10^{21}$. (Round off to the Nearest Integer).

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