if al3+ replaces na+ at the edge center of NaCl lattice then calculate the vacancies in one mole of NaCl
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Na+ the number of atoms => 12*1/4 = 3
ANd 3 Na+ = Al3+
Then for 1 mole that is 1*6*10^23 atoms of Na+ would have => 6*10^23/3 Al3+ atoms => 2*10^23.
ANd 3 Na+ = Al3+
Then for 1 mole that is 1*6*10^23 atoms of Na+ would have => 6*10^23/3 Al3+ atoms => 2*10^23.
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