If `NaCI` is droped with `10^(-3)` mole of `SrCI_(2)` then number of cationic vacancies is

If NaCI is droped with 10^(-4) "mole"% of SrCI_(2) then number of cationic vacancies will be

I mol of NaCl is doped with 10^(-5) mole of SrCl_(2) .The number of cationic vacancies in the crystal lattice will be

If 1 mole of NaCl is doped with 10^(-3) mole of SrCl_(2) . What is the number of cationic vacancies per mole of NaCl ?

NaCZ is doped with 2 xx 10^(-2) mole% of SrCl_(2) then the number of cation vacancies per mole

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) = 6.022 xx 10^(23) mol^(-1))

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))