If NaCl is doped with `10^(-2)` mol percentage of strontium chloride , what is the concentration of cation vacancy ?

Explanation

Given concentration of `SrCl_(2) = 10^(-3) ` mol%
concentration is in percentage so that take total 100 mol of solution.
No. of mol of NaCl = 100- moles of `SrCl_(2)`
Mol of `SrCl_(2)` is very negligible as compare to total moles.
So No. of mole of NaCl = 100
1 mol of NaCl is dipped with
`= 10^(-3) //100` moles of `SrCl_(2)`
`= 10^(-5)` mole of `SrCl_(2)`
So cation vacancies per mole of NaCl = `10^(-5) ` mol
1 mole `= 6.022 xx 10^(23) ` particles .
So cation vacancies per mole of NaCl `=10^(-5) `mol
`= 10^(-5) xx 6.022 xx10^(23)`
`= 6.02 xx 10^(18)`
So, that the concentration of cation vacancies created by `SrCl_(2)` is `6.022 xx 10^(18)` per mole of NaCl.