KBr is doped with 10 ^ - 5 mole percent of SrBr _ 2 The number of...

$1moleKBr ( = 119gm )$ have $\dfrac{{10}^{- 5}}{100}$ moles ${SrBr}_{2}$ and hence, ${10}^{- 7}$ moles cation vacancy (as $1{Sr}^{2 +}$ will result $1$ cation vacancy)

$∴$ Required number of cation vacancies

$= \dfrac{{10}^{- 7} \times 6 . 023 \times {10}^{23}}{119} = 5 . 06 \times {10}^{14} ≃ 5 \times {10}^{14}$