If NaCl is doped with 10-4 mol% of SrCl2, the concentration of cation vacancies will be (NA = 6.02 1023 mol-1)
- 6.02 1016 mol-1
- 6.02 1017 mol-1
- 6.02 1014 mol-1
- 6.02 1015 mol-1
Answer
(2) 6.02 1017 mol-1
Solution
As per NCERT, when NaCl is doped with SrCl2, each Sr2+ ion replaces two Na+ ions: it occupies the site of one and the other site remains vacant.
⇒ Number of cation vacancies = number of Sr2+ ions
10-4 mol% means 10-4 mol of SrCl2 in 100 mol of NaCl
⇒ SrCl2 per mole of NaCl = 10-4100 = 10-6 mol
∴ Cation vacancies = 10-6 6.02 1023
= 6.02 1017 mol-1
Answer: (2)