What is the number of atoms present per unit | vedclass

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(D) The formula for density is $ho = \frac{n 	imes M}{a^{3} 	imes N_{A}}$.Given: $a = 4 \ \mathring{A} = 4 	imes 10^{-8} \ cm$,$ho = 2.7 \ g \

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$4$

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(D) The formula for density is $ho = \frac{n 	imes M}{a^{3} 	imes N_{A}}$.Given: $a = 4 \ \mathring{A} = 4 	imes 10^{-8} \ cm$,$ho = 2.7 \ g \ cm^{-3}$,$M = 27 \ g \ mol^{-1}$,$N_{A} = 6.022 	imes 10^{23} \ mol^{-1}$.Rearranging for $n$: $n = \frac{ho 	imes a^{3} 	imes N_{A}}{M}$.Substituting the values: $n = \frac{2.7 	imes (4 	imes 10^{-8})^{3} 	imes 6.022 	imes 10^{23}}{27}$.$n = \frac{2.7 	imes 64 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}{27} = 0.1 	imes 64 	imes 0.6022 \approx 3.85$.Since the number of atoms per unit cell must be an integer,$n \approx 4$,which corresponds to a face-centered cubic $(FCC)$ unit cell.

What is the number of atoms present per unit cell of aluminium having edge length $4 \ \mathring{A}$? (Given: density of $Al = 2.7 \ g \ cm^{-3}$,atomic mass of $Al = 27 \ g \ mol^{-1}$)

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