If NaCl is doped with `10^(-1)mol%` of `SrCl_(2)` ,the concentration of cation vacancies will be "

If NaCl is doped with 10^(-4)mol% of SrCl_(2) the concentration of cation vacancies will be (N_(A)=6.02xx10^(23)mol^(-1))

NaCl is doped with 2xx10^(-3) mol % SrCl_2 , the concentration of cation vacancies is

If NaCI is doped with 10^(-3)"mol"%SrCI_(2) then the concentration of cation vacancies will be: