Question

$\mathrm{KBr}$ is doped with 10^-5 mole per cent of SrBr₂. The number of cationic vacancies in 1 g of $\mathrm{KBr}$ crystal is___ 10¹⁴. (Round off to the Nearest Integer).

Answer 1

The atomic mass of potassium =$39.1u$, the mass of bromine =$79u$, and Avagadro's number (N_A) = $6.022\times {10}^{23}$

One cationic vacancy is produced for each ${\mathrm{Sr}}^{2+}$ ion.

Hence, no. of ${\mathrm{Sr}}^{2+}$ ion = Number of cationic vacancies. Since the mole percentage of ${\mathrm{SrBr}}_2$ dopped is ${10}^{-5}$ to that of total moles of $\mathrm{KBr}$.

Hence, the number of cationic vacancies = $\frac{{10}^{-5}}{100}\times \frac{1}{19}\times N_A$

= $\frac{1}{19}\times {10}^{-5}\times 6.022\times {10}^{23}$

= $5\times {10}^{14}$

Hence, the number of cationic vacancies in one gram of $\mathrm{KBr}$ crystal is $5$.