KBr is doped with 105 mole percent of SrBr2 The number of ca

Question

KBr is doped with 10⁵ mole percent of SrBr₂. The number of cationic vacancies in 1g of KBr crystal is ____________ 10¹⁴. (Round off to the Nearest Integer).

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Accepted Answer

For every Sr⁺² ion, 1 cationic vacancy is created. Hence, no. of Sr⁺² ion = Number of cationic vacancies

Since mole percentage of SrBr₂ dropped is 10^-5 to that of total moles of KBr.

Hence,

No. of cationic vacancy = 10^-5 100 1 119 N_A...

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