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 $$integer)$$

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

Ionic solids

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Question

KBr is doped with $10^{- 5}$ mole percent of $S r B r_{2}$ . The number of cationic vacancies in 1 g of KBr crystal is $___\times 10^{14}$
(Atomic Mass: $K : 39.1 u, B r : 79.9 u$ $N_{A} = 6.023 \times 10^{23}$ )(Round off to nearest integer)

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Solution

$S r^{+ 2}$ will result one cation vacany $n_{S r B r_{2}} = n_{S r + 2} = \frac{10^{- 5}}{100} = 10^{- 7}$
Per mole cationic vacancies = $10^{- 7} \times 6.023 \times 10^{23}$
i.e. $\begin{array}{l} 119 g \to 6.023 \times 10^{16} \\ 1 g \to ? \end{array}$
$\frac{6.023 \times 10^{16}}{119} = 5.06 \times 10^{14}$

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Ionic solids

Unlock the full solution & master the concept.

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)

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 $S r B r_{2}$ . The number of cationic vacancies in 1 g of KBr crystal is $___\times 10^{14}$
(Atomic Mass: $K : 39.1 u, B r : 79.9 u$ $N_{A} = 6.023 \times 10^{23}$ )(Round off to nearest integer)