If NaCl is doped with `10^(-3)` mol percent of `SrCI_(2)`, what is the concentration of cation vacancy?

Explanation

Doping of NaCl with` 10^(-3)` mol % ` SrCl_(2)` means that 100 moles of NaCl are doped with `10^(-3) " mol of " SrCl_(2)`
1 mole of NaCl is doped with `SrCl_(2) = ( 10^(-3))/100 " mole"= 10^(-5)` mole
As each `Sr^(2+)` ion introduces one cation vancancy, therefore, concentration , of cation vanancies.
` = 10^(-5) ` mol/mol of NaCl ` = 10^(-5) xx 6.02 xx 10^(23) " mol"^(-1) = 6.02 xx 10^(18) "mol"^(-1)`