Question

KBr is doped with ${10}^{-5}$ mole per cent of $SrB{r}_2$. The number of cationic vacancies in 1 g of KBr crystal is $\times {10}^{14}$. (Round off to the Nearest Integer).


$[AtomicMass:K=39.1u,Br=79.9u,N_A=6.023\times {10}^{23}]$

Answer 1

For every $S{r}^{+2}$ ion, 1 cationic vacancy is created. Hence, no. of $S{r}^{+2}$ ion = Number of cationic vacancies
1 mole KBr (= 119 gm) has $\frac{{10}^{-5}}{100}$ moles of $SrB{r}_2$
which is equal to ${10}^{-7}$ cationic vacancies.
The number of cationic vacancies in 1 g of KBr crystal:
$\frac{{10}^{-7}\times 6.023\times {10}^{23}}{119}=5.06\times {10}^{14}\approx 5\times {10}^{14}$
Correct answer is 5