If `NaCI` is droped with `10^(-3)` mole of `SrCI_(2)` then number of cationic vacancies is

Explanation

The number of cation valencies in the lattice of `NaCl` is equal to the number of divalent `Sr^(2+)` ions added.
The concentration of `Sr^(2+)` ion `= 10^(-3) mol` percent
`(10^(-5) "mol")/(100) = 10^(-5) mol`
` :.` mol of `Sr^(2+)` ions contains `6.023 xx 10^(23) Sr^(2+)` ions
`:. 10^(-5)` mol of `Sr^(2+)` ions contains
`= (6.023 xx 10^(23)) xx 10^(-5) = 6.023 xx 10^(18) Sr^(2+)` ions
Therefore, the number of cation vacancies in `NaCl` crystal is `6.023 xx 10^(18)`, when it is doped with `10^(-3)` mole% of `SrCl_(2)`