Q. 253.7( 21 Votes )
If NaCl is doped with 10–3 mol % of SrCl2, what is the concentration of cation vacancies?
We know that two Na 2+ ions are replaced by each of the Sr2+ ions while SrCl2 is doped with NaCl. But in this case only one lattice point is occupied by each other of the Sr2+ ion and produce one cation vacancy. Here 10-3 mole of SrCl2 is doped with 100 moles of NaCl.
Thus, cation vacancies produced by NaCl = 10 -3mol
Therefore, 1 mole of NaCl will produce cation vacancies after doping
= = 10-5 mol
Therefore, total cationic vacancies
= 10-5 × Avogadro’s number
= 10-5 × 6.022 × 10 23
= 6.023 × 10 -18 vacancies.
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