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 ca

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))
Answer» `6.02xx10^14 "mol"^(-1)` `6.02xx10^15 "mol"^(-1)` `6.02xx10^16 "mol"^(-1)` `6.02xx10^17 "mol"^(-1)` Solution :For each `Sr^(2+)` ion INTRODUCED, one cation 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`. `therefore SrCl_2` doped per mole of NaCl=`10^(-4)//100` =`10^(-6)` mole =`10^(-6)xx (6.02xx10^23) Sr^(2+)` ions `=6.02xx10^17 Sr^(2+)` ions Hence, concentration of cation vacancies =`6.02xx10^17 mol^(-1)`