KBr is doped with ${10}^{-5}$ mole percent of ${\mathrm{SrBr}}_2$. The number of cationic vacancies in 1 g of KBr crystal is$\_\_\_\times {10}^{14}$

(Atomic Mass:  $\mathrm{K}=39.1\mathrm{u}, \mathrm{Br}=79.9\mathrm{u}$, ${\mathrm{N}}_{\mathrm{A}}=6.023\times {10}^{23}$)

Atomic mass of potassium is 39.1 u and atomic mass of bromine is 79 u. One cationic vacancy approved by one Sr²⁺ ion. Number of Sr²⁺ ions is equal to the number of cationic vacancies.

Since the mole percentage of SrBr₂ dopped is ${10}^{-5}$ to that of total moles of KBr.

Hence, the number of cationic vacancies:
$\mathrm{Number} \mathrm{of} \mathrm{cationic} \mathrm{vacancies}=\frac{{10}^{-5}}{100}\times \frac{1}{19}\times 6.022\times {10}^{23}=\underset{\mathrm{x}}{\underbrace{5}}\times {10}^{14}$