If NaCl is doped with 10−4 mol% SrCl2, the concentration of cation vacancies will be (NA=6.02×1023mol−1)
6.02×1014mol−1
6.02×1015mol−1
6.02×1016mol−1
6.02×1017mol−1
Number of moles of cationic vacancies
10−4102=10−6mole
⇒ Number of cationic vacancies =10−6×6.02×1023=6.02×1017 mol−1