If `NaCl` is doped with `10^(-3)` mol% of `SrCl_(2)`, what is the concentration of cation vacancies?

Explanation

Let moles of NaCl=100
`therefore` Moles of `SrCl_(2)` doped `=10^(-3)`
Each `Sr^(2+)` will repalce two `Na^(+)` ions. To maintain electrical neutrality it occupies one positon and thus creates one cation vacancy
`therefore` Moles of cation vacancy in 100 moles `NaCl=10^(-3)`
Moles of cation vacancy in one mole
Moles of cation vancancy in one mole
NaCl`=10^(-3)xx10^(-2)=10^(-5)`
`therefore `Number of cation vacancies
`=10^(-5)xx6.022xx10^(23)=6.022xx10^(18)"mol"^(-1)`