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]
Answer (integer)
5
Solution
For every Sr<sup>+2</sup> ion, 1 cationic vacancy is created. Hence, no. of Sr<sup>+2</sup> ion = Number of cationic vacancies<br><br>Since mole percentage of SrBr<sub>2</sub> dropped is 10<sup>$-$5</sup> to that of total moles of KBr.<br><br>Hence,<br><br>No. of cationic vacancy $= {{{{10}^{-5}}} \over {100}} \times {1 \over {119}} \times {N_A}$<br><br>$= {1 \over {119}} \times {10^{ - 7}} \times 6.022 \times {10^{23}}$<br><br>$= 5 \times {10^{ - 2}} \times {10^{ - 7}} \times {10^{23}} = 5 \times {10^{14}}$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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