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

Answer» Doping of `NaCl` with `10^(-2) mol% SrCl_(2)` means that `100` mol of `NaCl` are doped with `10^(-2)` mol of `SrCl_(2)`.
`:. 1 mol` of `NaCl` is doped with `SrCl_(2) = (10^(-2))/(100) = 10^(-4) mol`
As each `Sr^(2+)` ion introduces one cation vacancy, therefore, the concentration of cation vacancies
`= 10^(-4)` mol/mol of `NaCl`
`= 10^(-4) xx 6.023 xx 10^(23) mol^(-1)`
`= 6.02 xx 10^(19) mol^(-1)`


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