When NaCl is dopped with `10^(-5) "mole % of" SrCl_(2)`, what is the no. of cationic vacanies?

Explanation

Number of moles of cationic vacancies = 100.
Mole of NaCl `= 10^(-3)`
`therefore` Number of moles of cationic vacancies per mole of NaCl `= (10^(-3))/(100)=10^(-5)`
`therefore` Total number of cationic vacancies `= 10^(-5)xx N_(0)=6.02xx10^(18)`