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KBr is doped with 10−5 mole percent of SrBr2. The number of cationic vacancies in 1 g of KBr crystal is___×1014(Atomic Mass: K:39.1u,Br:79.9u NA=6.023×1023)(Round off to nearest integer)

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answer is 5.

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Detailed Solution

Sr+2 will result one cation vacany nSrBr2=nSr+2=10−5100=10−7Per mole cationic vacancies =10−7×6.023×1023i.e. 119g→6.023×10161g →?6.023×1016119=5.06×1014

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KBr is doped with 10−5 mole percent of SrBr2. The number of cationic vacancies in 1 g of KBr crystal is___×1014(Atomic Mass: K:39.1u,Br:79.9u NA=6.023×1023)(Round off to nearest integer)