Fe crystallizes in a bcc lattice. If the edge | vedclass

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(A) For a $bcc$ unit cell, the number of atoms per unit cell $Z = 2$. The density formula is given by $d = \frac{Z 	imes M}{a^3 	imes N_A}$. Here, $

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$6.022 	imes 10^{23} \,mol^{-1}$

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(A) For a $bcc$ unit cell, the number of atoms per unit cell $Z = 2$. The density formula is given by $d = \frac{Z 	imes M}{a^3 	imes N_A}$. Here, $d = 7.874 \,g/cm^3$, $Z = 2$, $M = 55.845 \,g/mol$, and $a = 286.65 \,pm = 286.65 	imes 10^{-10} \,cm$. Rearranging for $N_A$: $N_A = \frac{Z 	imes M}{d 	imes a^3}$. $N_A = \frac{2 	imes 55.845}{7.874 	imes (286.65 	imes 10^{-10})^3}$. $N_A = \frac{111.69}{7.874 	imes 23.556 	imes 10^{-24}}$. $N_A \approx 6.022 	imes 10^{23} \,mol^{-1}$.

$Fe$ crystallizes in a $bcc$ lattice. If the edge length is $286.65 \,pm$ and the density is $7.874 \,g/cm^3$, calculate the Avogadro number. $(Fe = 55.845 \,u)$

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