If NaCl is doped with `10^(-3)` mol percent of `SrCI_(2)`, what is the concentration of cation vacancy?

Explanation

One cation of `Sr^(2+)` would create one cation vacancy in NaCl. Therefore , the number of cation vacancies created in the lattice of NaCl is equal to the number of divalent `Sr^(2+)` ions added.
`:.` Concentration of cation vacancy on being doped with `10^(-3) mol% SrCl_2`
`=10^(-3) mol% = (10^(-3))/(100) = 10^(-5) mol`
No. of `Sr^(2+)` ions in `10^(-5)` mol = `10^(-5) xx 6.023 xx 10^(23)`
`= 6.023 xx 10^(18) Sr^(2+)` ions
No. of cation vacancies = `6.023 xx 10^(18)`