Question

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

Answer 1

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