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QJEEM21Ch73a8: JEE Main 2021 Chemistry - Class 12 JEEM21Ch73a8 Solution: KBr is doped with 10 5 mole percent of SrBr 2 . The number of…

KBr is doped with 10 −- 5 mole percent of SrBr2. The number of cationic vacancies in 1g of KBr crystal is ____________ 1014. (Round off to the Nearest Integer).

[Atomic Mass : K : 39.1 u, Br : 79.9 u NA = 6.023 ×\times 1023]

Step-by-Step Verified Solution

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 =10−5100×1119×NA= {{{{10}^{-5}}} \over {100}} \times {1 \over {119}} \times {N_A}

=1119×10−7×6.022×1023= {1 \over {119}} \times {10^{ - 7}} \times 6.022 \times {10^{23}}

=5×10−2×10−7×1023=5×1014= 5 \times {10^{ - 2}} \times {10^{ - 7}} \times {10^{23}} = 5 \times {10^{14}}

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