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

Answer

`6.022xx10^(18)`

Explanation

Due to the addition of strontium chloride `(SrCl_(2))`, each `Sr^(2+)` ion replaces two `Na^(+)` ions but occupies only one lattice point in place of `Na^(+)` ion. This leads to one cation vacancy.
Number of moles of cation vacancies is 100 mol of NaCl = `10^(-3)`
Number of moles of cation vacancies in 1 mol `=(10^(-3))/(100)=10^(-5)mol`
Total number of cation vacancies `=10^(-5)xxN_(0)`
`=(10^(-5)mol)xx(6.022xx10^(23)mol^(-1)`
`=6.022xx10^(18)`