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Question is

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

  • 6.02 x 1015 mol-1

  • 6.02 x 1016 mol-1

  • 6.02 x 1017

  • 6.02 x 1014

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