If AgCl is doped with 2 moles of CdCl2,how many cation vacancies would be produced?

1. One Cd²⁺ replaces two Ag⁺ from their positions in crystal lattice.
2. 2 moles of CdCl₂ has  $2 \times 6.023 \times {10}^{23} = 12.046 \times {10}^{23}$ Cd²⁺ ions.
3. So,number of cation vacancies =  $2 \times 12.046 \times {10}^{23} = 24.092 \times {10}^{23}$