If NaCl is doped with10^(-3)mol %SrCl_(2) , what is the concentration of cation , vacancies ?

If NaCl is doped with10^(-3)mol %SrCl_(2) , what is the concentration

1.	If NaCl is doped with10^(-3)mol %SrCl_(2) , what is the concentration of cation , vacancies ?
Answer» Solution :Doping of NACL with` 10^(-3)`mol % ` SrCl_(2)`means that 100 moles of NaCl are doped with`10^(-3) " mol of " SrCl_(2)` 1 mole of NaCl is doped with `SrCl_(2) = ( 10^(-3))/100 " mole"= 10^(-5)`mole As each `SR^(2+)`ion introduces one CATION vancancy, therefore, concentration , of cation vanancies. ` = 10^(-5) `mol/mol of NaCl ` = 10^(-5)xx 6.02 xx 10^(23)" mol"^(-1) = 6.02 xx 10^(18)"mol"^(-1)`