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If NaCI is droped with 10^(-4) "mole"% o...

If `NaCI` is droped with `10^(-4) "mole"% of SrCI_(2)` then number of cationic 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)=6.02xx10^(23)mol^(-1))

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))