Iron crystallizes in FCC with an edge length | vedclass

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(A) In a solid with Schottky defects, atoms are missing from their lattice positions, which decreases the effective number of atoms per unit cell.For

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$5.8$

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(A) In a solid with Schottky defects, atoms are missing from their lattice positions, which decreases the effective number of atoms per unit cell.For an $FCC$ lattice, the number of atoms per unit cell is $Z = 4$.With $0.1 \%$ Schottky defects, the effective number of atoms per unit cell is:$Z_{effective} = 4 	imes (1 - \frac{0.1}{100}) = 4 	imes 0.999 = 3.996$.Using the density formula $d = \frac{Z_{effective} 	imes M}{N_A 	imes a^3}$:Given $M = 56 \ g/mol$, $N_A = 6.022 	imes 10^{23} \ mol^{-1}$, and $a = 400 \ pm = 4 	imes 10^{-8} \ cm$.$d = \frac{3.996 	imes 56}{(6.022 	imes 10^{23}) 	imes (4 	imes 10^{-8})^3} \approx \frac{223.776}{6.022 	imes 10^{23} 	imes 64 	imes 10^{-24}} \approx \frac{223.776}{38.54} \approx 5.81 \ g/cm^3$.

Iron crystallizes in $FCC$ with an edge length of $400 \ pm$. If it contains $0.1 \%$ Schottky defects, calculate its approximate density $[AW \text{ of } Fe = 56 \ g/mol]$. (in $g/cm^3$)

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