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Question

If NaCl is doped with $$10^{-3}$$ mol percent of $$SrCl_2$$, what is the number of cation vacancies?


A
3.01×1018 mol1
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B
6.02×1018 mol1
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C
1.2×1019 mol1
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D
18×1018 mol1
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Solution

The correct option is C $$6.02\, \times\, 10^{18}$$ mo$$l^{-1}$$
If NaCl is doped with $$10^{-3}$$ mol percent of $$SrCl_2$$, what is the number of cation vacancies is $$6.02\, \times\, 10^{18}$$ mo$$l^{-1}$$.
$$100$$ moles of $$NaCl$$ contains $$10^{-3}$$ moles of $$SrCl_2$$
1 mole of NaCl will contain $$ \displaystyle 6.02\, \times\, 10^{23} \times \frac {10^{-3}}{100}=6.02\, \times\, 10^{18}$$ molecules of $$SrCl_2$$.
1 molecule of $$SrCl_2$$ will lead to one cation vacancy.
$$6.02\, \times\, 10^{18}$$ molecules of $$SrCl_2$$ will give $$6.02\, \times\, 10^{18}$$ cation vacancies

Chemistry

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