Q.

If NaCl is doped with 10-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.02 × 1023 mol-1)

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answer is 4.

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Detailed Solution

In NaCl crystal2Na+ replaced by one Sr+2;1 mole of NaCl contain 10-4100=10-6mole of Sr+2so number of cationic vacancies= 10-6 ×6.02×1023 mol-1=6.02×1017 mol-1
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If NaCl is doped with 10-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.02 × 1023 mol-1)