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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

Grade:12

2 Answers

Arun
25750 Points
4 years ago
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 
 
Khimraj
3007 Points
4 years ago
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
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

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