Grade 12Physical Chemistry
NaCl is doped with 10^-4 moles of SrCl2 . The no of cation vacancies in 117g of NaCl is
NaCl is doped with 10^-4 moles of SrCl2 . The no of cation vacancies in 117g of NaCl is
2 Answers
Arun
NaCl is dropped with 10^-4 mol% of SrCl2 , this means 100 moles of NaCl are dropped in 10^-4 mole of SrCl2
so, 1 mole of NaCl is dropped in 10^-6 mole of SrCl2.
now, here it is clear that one Sr²+ creates one cation vacancy.
so, number cation vacancies = number of mole of SrCl2 × Avogadro's number
= 10^-6 × 6.023 × 10²³
= 6.023 × 10^17
so, 1 mole of NaCl is dropped in 10^-6 mole of SrCl2.
now, here it is clear that one Sr²+ creates one cation vacancy.
so, number cation vacancies = number of mole of SrCl2 × Avogadro's number
= 10^-6 × 6.023 × 10²³
= 6.023 × 10^17
Khimraj
6.02 x 10^15 mol-1
If NaCl is doped with 10^-4 mol% of SrCl2,
2 Na+ ions doped by Sr2+ NA = 6.02 x 10^23
The concentration of cation vacancies
= 6.02 x 10^23 x10^-8
= 6.02 x 10^15 mol-1
2 Na+ ions doped by Sr2+ NA = 6.02 x 10^23
The concentration of cation vacancies
= 6.02 x 10^23 x10^-8
= 6.02 x 10^15 mol-1
If NaCl is doped with 10-4 mol% of SrCl2,
2 Na+ ions doped by Sr2+ NA = 6.02 x 1023
The concentration of cation vacancies
= 6.02 x 10^23 x10^-8
= 6.02 x 10^15 mol-1
2 Na+ ions doped by Sr2+ NA = 6.02 x 1023
The concentration of cation vacancies
= 6.02 x 10^23 x10^-8
= 6.02 x 10^15 mol-1
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