If NaCl is doped with `10^(-3)` mole percent `SrCl_(2)`, what will be the concentration of cation vacancies ? (`N_(A) = 6.02 xx 10^(23) mol^(-1)`)

Explanation

Every `Sr^(2+)` ion causes one cation vacancy (because two `Na^(+)` ions are replaced by one `Sr^(2+)`)
Therefore, introduction of `10^(-3)` moles of `SrCl_(2)` per 100 moles of NaCl would introduce `10^(-3)` mole cation vacancies in 100 moles of NaCl.
No. of vacancies per mole of NaCl
`= (10^(-3))/(10000) xx 6.02 xx 10^(23) = 6.02 xx 10^(18)` vacancies.