KBr is doped with 10^–5 mole percent of SrBr₂. The number of cationic vacancies in 1 g of KBr crystal is ____ 1014.
(Atomic Mass: K : 39.1u, Br : 79.9u; N_A = 6.023 × 10²³)

There is one cationic vacancy produced for every Sr⁺² ion. Sr⁺² ion number therefore equals the number of cationic vacancies.
Given that the ratio of SrBr₂ doped moles to total KBr moles is 10^-5.
Number of cationic vacancies 

 $=\frac{{10}^{-5}}{100}\times \frac{1}{119}\times {\mathrm{N}}_{\mathrm{A}} = \frac{1}{119}\times {10}^{-7}\times 6.022\times {10}^{23}$
$\begin{array}{l}=5\times {10}^{-2}\times {10}^{-7}\times {10}^{23}\\ =5\times {10}^{14}\\ =5\end{array}$

Therefore, the correct answer is 5