NaCl is doped with 2x10^-2 mole% of SrCl₂ then the number of cation vacancies per mole

Given concentration of SrCl₂ = 2×10⁻² mol%
Concentration is expressed as a percentage, requiring a total of 100 mol of solution.
Number of Mole NaCl = 100-Mole SrCl₂
SrCl₂ moles are extremely small in comparison to total moles.
Therefore, 100 is the number of moles of NaCl.
A dip of 1 mol of NaCl in
$$\begin{gathered} =2\times {10}^{-2}/100\text{ moles of }{\mathrm{SrCl}}_2 \\ =2\times {10}^{-4}\text{ mole of }{\mathrm{SrCl}}_2 \end{gathered}$$

Cation vacancies therefore equal 2x10^-4 mol of NaCl per mole.
6.023 x 10²³ particles per mole
$$\begin{gathered} Cation vacancies therefore equal 2x{10}^{-4} mol of NaCl per mole. \\ \begin{array}{l}=2\times {10}^{-4}\times 6.023\times {10}^{23}\\ =12.046\times {10}^{19}\end{array} \\ Therefore, 12.046x{10}^{19} cation vacancies were produced by SrC{l}_2 for every mole of NaCl. \end{gathered}$$

Therefore, the correct option is B