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Question

The concentration of cation vacancies when $$NaCl$$ is doped with $$10^{-3}$$ mole percent of $$SrCl_{2}$$ is :


A
6.023×1020
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B
6.023×1023
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C
6.023×1021
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D
6.023×1018
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Solution

The correct option is C $$6.023 \times 10^{18}$$
Given, concentration of $$SrCl_2  = 10^{-3}$$ mol percent

The moles of $$SrCl_2$$ are very negligible as compared to the total moles.

$$1$$ mol of NaCl is doped with $$= \dfrac{ 10^{-3} }{100}$$ $$=10^{-5}$$ moles of $$SrCl_2$$

So, cation vacancies per mole of $$NaCl$$ $$=10^{-5}$$ moles

$$1$$ mol $$= 6.022 \times 10^{23}$$ particles

So, the cation vacancies per mole of $$NaCl$$  $$= 10^{-5} \times 6.022 \times10^{23} = 6.02 \times 10^{18}$$

So, the concentration of cation vacancies created by $$SrCl_2$$ is $$6.022 \times 10^{18}$$ per mol of $$NaCl$$.

Chemistry

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