The crystal structure of an element has an fc | vedclass

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(C) The effective number of atoms in an $fcc$ unit cell is $Z = 4$. The formula for density is $d = \frac{Z 	imes M}{N_{A} 	imes a^3}$. Given: $d =

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$108.0$

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(C) The effective number of atoms in an $fcc$ unit cell is $Z = 4$. The formula for density is $d = \frac{Z 	imes M}{N_{A} 	imes a^3}$. Given: $d = 11.21 \ g \ cm^{-3}$,$a = 4 \ \mathring{A} = 4 	imes 10^{-8} \ cm$,$N_{A} = 6.023 	imes 10^{23} \ mol^{-1}$. Rearranging for molar mass $(M)$: $M = \frac{d 	imes N_{A} 	imes a^3}{Z}$. $M = \frac{11.21 	imes 6.023 	imes 10^{23} 	imes (4 	imes 10^{-8})^3}{4}$. $M = \frac{11.21 	imes 6.023 	imes 10^{23} 	imes 64 	imes 10^{-24}}{4}$. $M = 11.21 	imes 6.023 	imes 16 	imes 0.1 = 108.0 \ g \ mol^{-1}$.

The crystal structure of an element has an $fcc$ lattice. If the edge length of the crystal is $4 \ \mathring{A}$,what is the atomic weight (in $g \ mol^{-1}$) of the element,if the density of the crystal is $11.21 \ g \ cm^{-3}$ $(N_{A} = 6.023 \times 10^{23} \ mol^{-1})$?

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