A compount `AB` has a rock type structure with `A:B=1:1`. The formula weight of `AB` is `6.023 Y am u` and the closed `A-B` distance is `Y^(1//3) nm.` `(i)` Find the density of lattice. `(ii)` If the density of lattice is found to be `20 kg m^(-3)`, then predict the type of defect.
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`AB` has rock salt structure `A:B=1:1 i.e., f.c.c.` structure `(z=4)` and formula weight of `AB` is `6.023 Y g` having closest distance `(A-B)Y^(1//3) nm. ` Therefore edge length of unit cell `=2(A^(+)+B^(-))=2xxY^(1//3)xx10^(-9)m` `(i)` Density of `AB=(zxxm)/(Av. no . xxV)=(zxxm)/(Av. no . xxa^(3))` `=(4xx6.023Yxx10^(-3))/(6.023xx10^(23)xx(2Y^(1//3)xx10^(-9))^(3))` `=5.0kg m^(-3)` `(ii)` The observed density of `Ab` is `20 kg m^(-3)`, which is higher than calculated density `5 kg m^(-3)` and thus `AB` has either interstitial impurity defect or substitutional impurity defect.
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