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

Explanation

Doping of NaCl with `10^(-3)` mol % `SrCl_(2)` means that 100 100 moles of NaCl are doped with `10^(-3)` mol of `SrCl_(20`.
`:.` 1 mole of NaCl is doped with `SrCl_(2)`
`= ( 10^(-3))/( 100) xx 6.02xx 10^(23) = 6.02 xx 10^(18)`