Aluminium metal has a density of 2.72 g cm^ | vedclass

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(B) The density formula is given by $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Given: $d = 2.72 \ g \ cm^{-3}$,$M = 27 \ g \ mol^{-1}$,$a = 404 \ pm = 40

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It forms $fcc$ unit cell

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(B) The density formula is given by $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Given: $d = 2.72 \ g \ cm^{-3}$,$M = 27 \ g \ mol^{-1}$,$a = 404 \ pm = 404 	imes 10^{-10} \ cm$,$N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Substituting the values: $2.72 = \frac{Z 	imes 27}{6.022 	imes 10^{23} 	imes (404 	imes 10^{-10})^3}$.Calculating $Z$: $Z = \frac{2.72 	imes 6.022 	imes 10^{23} 	imes 6.60 	imes 10^{-23}}{27} \approx 4$.Since $Z = 4$,the unit cell is $fcc$ (face-centered cubic).In an $fcc$ lattice,the coordination number is $12$.

Aluminium metal has a density of $2.72 \ g \ cm^{-3}$ and crystallizes in a cubic lattice with an edge length of $404 \ pm$. Which of the following is correct?

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