Question:
$\mathrm{KBr}$ is doped with $10^{-5}$ mole percent of $\mathrm{SrBr}_{2}$. The number of cationic vacancies in $1 \mathrm{~g}$ of $\mathrm{KBr}$ crystal is ___________________
$10^{14}$. (Round off to the Nearest Integer). [Atomic Mass : $\mathrm{K}: 39.1 \mathrm{u}, \mathrm{Br}: 79.9 \mathrm{u}$$\left.\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}\right]$
Solution:
(5)
1 mole $\mathrm{KBr}(=119 \mathrm{gm})$ have $\frac{10^{-5}}{100}$ moles $\mathrm{SrBr}_{2}$
and hence, $10^{-7}$ moles cation vacancy $\left(\right.$ as $1 \mathrm{Sr}^{2+}$ will result 1 cation vacancy)
$\therefore$ Required number of cation vacancies
$=\frac{10^{-7} \times 6.023 \times 10^{23}}{119}=5.06 \times 10^{14} \simeq 5 \times 10^{14}$