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. ` 6.023 xx 10^(23) Na^(+)` ions .
NaCl has fcc arragement of `Cl^(-)` ions and ` Na^(+)` ions are present at eh edge centres and body - centre. As there are 12 edge centre and each edge centre is shared by 4 shared by 4 unit cells, their contribution per unit cell ` = 1/4 xx 12 = 3`
Conribution 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 `N^(+)` ions which have been replaced ` = 3/4 xx 6.023 xx 10^(23) = 4.517 xx 10^(23)`
` 1Al^(3+)` ion will replace ` 3Na^(+)` ions to maintain electrical neutrality . one vacancy will be occupied by ` Al^(3+)` ion and the remaining 2 will be vacent.
The 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 vanancies in 1 mole of ` NaCl = 2/3 xx 4.517 xx 10^(23) = 3.01 xx 10^(23)`