AgCl is doped with 10-2 mol % of CdCl2, find the concentration of cation vacancies
Given in the question is AgCl is doped with 10-210-2 mol % of CdCl2CdCl2. From this we can say that 100 moles of AgCl are doped with 0.01 moles of CdCl2CdCl2. So, we can say that 1 mole of AgCl will be doped with = 0.01/100=10-40. moles of CdCl2 Therefore, 10-4 moles of cation vacancies will be formed.
