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Question

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

A
6.02×1016mol1
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B
6.02×1017mol1
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C
6.02×1014mol1
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D
6.02×1015mol1
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Solution

The correct option is A 6.02×1017mol1
Moles of SrCl2 in 1 mole NaCl=104100=106 moles

One Sr+2 creates one cationic vacancies.

Concentration of the cationic vacancies produced by 106 moles of SrCl2

=106×6.023×1023

=6.023×1017 mol1

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