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 mol of NaCl contains 1 mol of `Na^(+)` ions i.e `6.022 xx 10^(23) Na^(+)` ions.
NaCl has fcc arrangement of `Cl^(-)` ions and `Na^(+)` are present at the edge centres and body centres.
Since there are 12 edge and each edge is shared by 4 unit cells, then contirbution of `Cl^(-)` ions present at the edge centres is `1/4 xx 12 = 3`. Contribution of `Na^(+)` ion present at the body centre = 1 Thus, for every `4 Na^(+)` ions, the ions present at the edge centres
`:. ` The `Na^(+)` ions which have been replaced by `Al^(3+)` ions
`= 3/4 xx 6.022 xx 10^(23) = 4.5165 xx 10^(23)`
To maintain electrical neutrality, `1 Al^(3+)` ion will replace `3 Na^(+)` ions. This means that 1 position will be occupied and remaining 2 will be vacant.
`:.` No. of vacancies in 1 mole of NaCl
`= 2/3 xx 4.5165 xx 10^(23) = 3.011 xx 10^(23)`.