If NaCl is doped with 10–3 mol% of SrCl2, calculate the concentration of cation vacancies


Correct option is

6.02 × 1018 mol–1


Due to addition of SrCl2, each Sr2+ ion replace to Na+ ions, but occupies only one Na+ lattice point. This makes one cationic vacancy.

Number of moles of cation vacancies in 100 mol of  NaCI = 10-3  

Number of moles of cation vacancies in 1 mol = 

Total cationic vacancies = 10-5 × NA  

                                        = 10-5 × 6.023 × 1023

                                                              = 6.023 × 1018 mol-1



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