If one mole of AgCl is dopped with 10^–5 mole of CaCl₂, then number of Ag⁺ ions lost from the lattice

2 cations and 1 can leave 1 vacancy for Ag⁺.
-2x10^-5 vacancies of Ag⁺ in 10^-5 moles of cations
= Loss of 2x10^-5 moles of Ag⁺
number of Ag+ ions = Mol × Avogadro's number

$\begin{array}{l}=2\times {10}^{-5}\times 6.023\times {10}^{23}\\ =12\times {10}^{18}\\ =1.2\times {10}^{19}\end{array}$

Therefore, the correct option is C 1.2x10¹⁹