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If NaCI is doped with 10^(-3)"mol"%SrCI(...

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

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If NaCl is doped with 10^(-4) mol % of SrCl_(2) , the concentration of cation vacancies will be ( N_(A) = 6.022 xx 10^(23) mol^(-1))

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