The density of the unit cell of Li is 0.539 | vedclass

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(A) Given:Density $(ho)$ = $0.539 \ g/cm^3$Edge length $(a)$ = $3.5 \ \mathring{A} = 3.5 	imes 10^{-8} \ cm$Atomic mass $(M_A)$ = $6.94 \ g/mol$Avo

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$2$

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(A) Given:Density $(ho)$ = $0.539 \ g/cm^3$Edge length $(a)$ = $3.5 \ \mathring{A} = 3.5 	imes 10^{-8} \ cm$Atomic mass $(M_A)$ = $6.94 \ g/mol$Avogadro constant $(N_A)$ = $6.022 	imes 10^{23} \ mol^{-1}$Formula for density of unit cell:$ho = \frac{Z 	imes M_A}{N_A 	imes a^3}$Rearranging for $Z$:$Z = \frac{ho 	imes N_A 	imes a^3}{M_A}$$Z = \frac{0.539 	imes 6.022 	imes 10^{23} 	imes (3.5 	imes 10^{-8})^3}{6.94}$$Z = \frac{0.539 	imes 6.022 	imes 10^{23} 	imes 42.875 	imes 10^{-24}}{6.94}$$Z = \frac{139.22}{6.94} \approx 2$Therefore,$2$ atoms of lithium are present in a unit cell.

The density of the unit cell of $Li$ is $0.539 \ g/cm^3$ and the edge length of the unit cell is $3.5 \ \mathring{A}$. How many $Li$ atoms are present in a unit cell of lithium? (Atomic mass of $Li = 6.94 \ g/mol$)

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