Question

$KBr$ is doped with ${10}^{-5}$ mole percent of $SrB{r}_2$. the number of cationic vacancies in $1g$ of $KBr$ crystal is${10}^{14}$. (Round off the nearest integer). [Atomic mass: $K:39.1u,Br:79.9u,N_A=6.023\times {10}^{23}$]
(JEE Main 2021)

Answer 1

$1S{r}^{2+}$ replaces two $K^{+}$, it occupies $1$ of the position and $1$ void is created. Number of vacancies in 1 mole of $KBr=\frac{{10}^{-5}}{100}\times 6.023\times {10}^{23}$ Moles of KBr given $=\frac{1}{119}$ $\therefore$ Total vacancies in $1g$ of $\text{KBr}=\frac{1}{119}\times \frac{1}{{10}^7}\times 6.023\times {10}^{23}$ $=5.06\times {10}^{14}$ $=5\times {10}^{14}$