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 $a = 400 \ pm = 4 	imes 10^{-8} \ cm$.$d = \frac{3.996 	imes 56}{(6.022 	imes 10^{23}) 	imes (4 	imes 10^{-8})^3} \approx \frac{223.776}{6.022 	imes 10^{23} 	imes 64 	imes 10^{-24}} \approx \frac{223.776}{38.54} \approx 5.81 \ g/cm^3$.

List-$I$ (Defect)	List-$II$ (Examples)
$A$. Frenkel defects	$1$. $FeO$
$B$. Schottky defects	$2$. $NaCl$
$C$. Vacancy defects	$3$. $AgCl$
$D$. Metal deficiency defects	$4$. Crystals with vacant lattice sites

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 $g/cm^3$)

Similar Questions

Match List-$I$ with List-$II$.

List-$I$ (Defect) List-$II$ (Examples)

$A$. Frenkel defects $1$. $FeO$

$B$. Schottky defects $2$. $NaCl$

$C$. Vacancy defects $3$. $AgCl$

$D$. Metal deficiency defects $4$. Crystals with vacant lattice sites

In which crystal is the Schottky defect observed?

In which of the following is impurity defect observed?

If $NaCl$ is doped with $10^{-3} \ mol \ \%$ $SrCl_2$,then the concentration of cation vacancies will be

$KBr$ is doped with $10^{-5}$ mole percent of $SrBr_2$. The number of cationic vacancies in $1 \ g$ of $KBr$ crystal is ........ $\times 10^{14}$. (Round off to the Nearest Integer).
[Atomic Mass : $K = 39.1 \ u$,$Br = 79.9 \ u$; $N_A = 6.023 \times 10^{23} \ mol^{-1}$]

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