A metal has a cubic lattice of side length 2. | vedclass

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(A) The volume of one unit cell is $V = a^3 = (2.88 	imes 10^{-8} \, cm)^3 = 23.8878 	imes 10^{-24} \, cm^3$. The mass of one unit cell is $m = 	ex

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$4.19 	imes 10^{19}$

Vedclass · Answer

(A) The volume of one unit cell is $V = a^3 = (2.88 	imes 10^{-8} \, cm)^3 = 23.8878 	imes 10^{-24} \, cm^3$. The mass of one unit cell is $m = 	ext{density} 	imes V = 7.2 \, g/cm^3 	imes 23.8878 	imes 10^{-24} \, cm^3 = 171.99 	imes 10^{-24} \, g$. The total mass of the metal is $7.2 \, mg = 7.2 	imes 10^{-3} \, g$. The number of unit cells is $\frac{	ext{Total mass}}{	ext{Mass of one unit cell}} = \frac{7.2 	imes 10^{-3} \, g}{171.99 	imes 10^{-24} \, g} \approx 4.186 	imes 10^{19}$. Thus,the number of unit cells is approximately $4.19 	imes 10^{19}$.

$A$ metal has a cubic lattice of side length $2.88 \, \mathring{A}$. The density of the metal is $7.2 \, g/cm^3$. How many unit cells are present in $7.2 \, mg$ of the metal?

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