KBr is doped with \(10^{-5}\) mole percent of \(\mathrm{SrBr_2}\). The number of cationic vacancies in 1g of KBr crystal is ____________ \(10^{14}\). (Round off to the Nearest Integer).
[Atomic Mass : K : 39.1 u, Br : 79.9 u \(N_A = 6.023 \times 10^{23}\)]
[Atomic Mass : K : 39.1 u, Br : 79.9 u \(N_A = 6.023 \times 10^{23}\)]
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Correct Answer: 5
Solution and Explanation
Moles of KBr in 1 g = 1/119 = 8.40 x 10-3 mol.
\(10^{-5}\) mole percent of SrBr2 means \(10^{-5}\) mol of SrBr2 per 100 mol of KBr, i.e. a mole fraction of \(10^{-7}\).
Moles of Sr2+ = \(10^{-7} \times 8.40 \times 10^{-3} = 8.40 \times 10^{-10}\) mol.
Each Sr2+ replaces one K+ and creates one cationic vacancy (to keep neutrality).
Vacancies = \(8.40 \times 10^{-10} \times 6.023 \times 10^{23} = 5.06 \times 10^{14}\) = 5 x 1014.
Answer: 5
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