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.

Explanation

(a)As AB has rock salt structure, i.e., it has fcc structure , Z=4
Edge of the unit cell = `2(r_(A^+)+r_(B^-))=2xxd_(A-B)=2xxy^(1//3) nm =2xxy^(1//3) xx10^(-9)`m
Calculated density `(rho)=(ZxxM)/(a^3xxN_0) =(4xx(6.023 y xx 10^(-3) "kg mol"^(-1)))/((2y^(1//3)xx10^(-9) m)^3 (6.023xx10^23 "mol"^(-1)))=5.0 kg m^(-3)`
(b)As observed density of AB is `20 "kg m"^(-3)` which is higher than the calculated density `(5.0 "kg m"^(-3))` , this shows that AB has either interstitial impurity defect or substitutional impurity defect.