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Question

If NaCl is doped with 104 mol% SrCl2, the concentration of cation vacancies will be (NA=6.02×1023mol1)


A

6.02×1014mol1

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B

6.02×1015mol1

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C

6.02×1016mol1

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D

6.02×1017mol1

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Solution

The correct option is D

6.02×1017mol1


Number of moles of cationic vacancies

104102=106mole

Number of cationic vacancies =106×6.02×1023=6.02×1017 mol1


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