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If NaCI is droped with 10^(-3) mole of S...

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

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Doping of NaCl with `10^(-3)` mol % `SrCl_(2)`, means that 100 moles of NaCl are doped with `10^(-3)` mol of `SrCl_(2)`.
1 mole of NaCl is doped with `SrCl_(2)= (10^(-3))/(100) "mole" = 10^(-5)` mole
As each `Sr^(2+)` ion introduces one cation vacancy, therefore, concentration of cation
vacancies = `10^(5)` mol/mol of NaCl = `10^(5)xx 6.02 xx 10^(23) "mol"^(-1) = 6.02 xx 10^(18) "mol"^(1)`
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