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Question

$$NaCl$$ is doped with $$2\times10^{-3}$$ mol percent $$SrCl_2$$, the concentration of cation vacancies is:


A
3.01×1018 mol1
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B
12.04×1018 mol1
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C
6.02×1018 mol1
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D
12.04×1020 mol1
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Solution

The correct option is B $$\;12.04\times10^{18}$$ mol$$^{-1}$$

$$2\times10^{-3}$$ mol of NaCl is doped with $$SrCl_3=\displaystyle\frac{2\times10^{-3}}{100}=2\times10^{-5}$$ mol

As each $$Sr^{+2}$$ ion introduces one cation vacancy, the concentration of cation vacancies

 =$$2\times10^{-5}\times6.023\times10^{23}$$ mol$$^{-1}$$ of NaCl $$=12.04\times10^{18}$$ mol$$^{-1}$$

Hence, the correct option is $$\text{B}$$

Chemistry

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