In NaCl is doped with10^(-4) mol % ofScCl_(2) , the concentration of cation vacancies will be(N_(A) = 6.02 xx 10^(23) mol^(-1))

In NaCl is doped with10^(-4) mol % ofScCl_(2) , the concentration of c

1.	In NaCl is doped with10^(-4) mol % ofScCl_(2) , the concentration of cation vacancies will be(N_(A) = 6.02 xx 10^(23) mol^(-1))
Answer» ` 6.02 xx 10^(14) mol^(-1)` `6.02 xx 10^(15) mol^(-1)` ` 6.02 xx 10^(16) mol^(-1)` ` 6.02 xx 10^(17)mol^(-1) ` Solution :For each ` Sr^(2+)`ion INTRODUCED, one catio vacancy is created because` 2 Na^(+)`ions are REMOVED and one vacant site is occupied by ` Sr^(2+)` . Doping with ` 10^(-4) "mol"%" of " SrCl_(2)`means 100 moles of NaCl are doped with ` 10^(-4)` mole of ` SrCl_(2)` `SrCl_(2)` doped per mole of NaCl = `10^(-4) //100` ` 10^(-6)` mole =`10^(-6) xx ( 6.02 xx 10^(23)) Sr^(2+)` ions ` = 6.02 xx 10^(17) Sr^(2+)` ions Hence, concetration of cation vacancies ` 6.02 xx 10^(17) "mol"^(-1)`