The edge length of the unit cell of a crystal | vedclass

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(D) The formula for density is $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Given: $d = 7.2 \ g \ cm^{-3}$, $a = 288 \ pm = 2.88 	imes 10^{-8} \ cm$, $M =

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Body-centered cubic

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(D) The formula for density is $d = \frac{Z 	imes M}{a^3 	imes N_A}$.Given: $d = 7.2 \ g \ cm^{-3}$, $a = 288 \ pm = 2.88 	imes 10^{-8} \ cm$, $M = 52 \ g \ mol^{-1}$, $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Substituting the values:$7.2 = \frac{Z 	imes 52}{(2.88 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}$$7.2 = \frac{Z 	imes 52}{23.887 	imes 10^{-24} 	imes 6.022 	imes 10^{23}}$$7.2 = \frac{Z 	imes 52}{14.385}$$Z = \frac{7.2 	imes 14.385}{52} \approx 1.991 \approx 2$.Since the number of atoms per unit cell $Z = 2$, the crystal has a body-centered cubic $(BCC)$ structure.

The edge length of the unit cell of a crystal is $288 \ pm$. If its density is $7.2 \ g \ cm^{-3}$ and the molar mass is $52 \ g \ mol^{-1}$, determine the type of unit cell.

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