NaCl is doped with  then the number of cation vacancies per mole is ____× 10¹⁹ (nearest 2 digits)

Given, NaCl is doped with 2 × 10^–2 mole % of SrCl₂
We need to find the number of cation vacancies per mole.

This indicates that 2x10^-2 moles of SrCl₂ have been added to 100 moles of NaCl.
Consequently, 1 mole of NaCl is doped with =$$\begin{gathered} \frac{2\times {10}^{-2}}{100} = 2\times {10}^{-4} \mathrm{mol}. \mathrm{of} {\mathrm{SrCl}}_2 \\ \mathrm{Cation} \mathrm{vacanies} \mathrm{produced} \mathrm{by}: \\ {\mathrm{Sr}}^{2+}\mathrm{ion} = 1 \mathrm{conc}. \mathrm{of} \mathrm{cation} \mathrm{vacanies} \mathrm{is} \mathrm{product} \\ = 2\times {10}^{-4}\times 6.022\times {10}^{23} \\ = 12.046\times {10}^{+19} \end{gathered}$$

Therefore, the correct answer is 12