If `Al^(3+)` replaces `Na^(+)` at the edge centre of `NaCl` lattice ,then the cation vacancies in `1` mole of `NaCl` will be

Explanation

1 mole of NaCl contains 1 mole of `Na^+` ions, i.e., `6.023xx10^23 Na^+` ions .
NaCl is fcc arrangement of `Cl^-` ions and `Na^+` ions are present at the edge centes and body centre. As there are 12 edge centres and each centre is shared by 4 unit cells, their contribution per unit cell=`1/4xx12`=3
Contribution of `Na^+` ion at the body-centre=1
Thus, for every 4 `Na^+` ions, the ions present at the edge centres=3. This means that `Na^+` ions which have been replaced =`3/4xx6.023xx10^23 =4.517xx10^23`
1 `Al^(3+)` ion will replace 3 `Na^+` ions to maintain electrical neutrality. One vacancy will be occupied by `Al^(3+)` and the remaining 2 will be vacant.
This means that `1/3` rd of these positions will be occupied by `Al^(3+)` ions and `2/3` rd will remain vacant. Hence, no. of vacancies in 1 mole of NaCl =`2/3xx4.517xx10^23=3.01xx10^23`