If NaClis doped with 10^(-4) mol% "of" SrCl_(2)the concentration of cation vacancies will be (N_(A)=6.02times10^(23)mol^(-1))

If NaClis doped with 10^(-4) mol% "of" SrCl_(2)the concentration of ca

1.	If NaClis doped with 10^(-4) mol% "of" SrCl_(2)the concentration of cation vacancies will be (N_(A)=6.02times10^(23)mol^(-1))
Answer» `6.02times10^(23) mol^(-1)` `6.02times10^(15) mol^(-1)` `6.02times10^(16) mol^(-1)` `6.02times10^(17) mol^(-1)` Solution :For each `Sr^(2+)`ion introduced, one vacancy is createdbecause `2Na^(+)`ions are REMOVED and one of the twovacant sites is OCCUPIED by `Sr^(2+)`. Doping with `10^(-4)` mol % `SrCl_(2)` means 100 moles of NaCl are doped with `10^(-4)` mole of `SrCl_(2)`. `therefore SrCl_(2)" droped per mole of " NaCl = 10^(-4)//100` =`10^(-6) "mole" = 10^(-6)times(6.02times10^(23))Sr^(2+) Sr^(2+)ions` `therefore"concentration of cation VACANCIES"=6.02times10^(17) mol^(-1)`