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

It is given that NaCl is doped with 10 −3 mol% of SrCl 2 .This means that 100 mol of NaCl is doped with 10 −3 mol of SrCl 2
Therefore, 1 mol of NaCl is doped with 10^-3/1000 mol of SrCl2
=10^-5
C ation vacancies produced by one Sr 2+ ion = 1
concentration of the cation vacancies produced by 10^-5 mole of SrCl2 =10^-5*6.022*10^23
=6.022*10^18 per mole

Given Conetration of SrCl2 = 10−3 mol%Concentration is in percentage so that take total 100 mol of solutionNumber of moles of NaCl = 100-3 moles of SrCl2Moles of SrCl2 is very negligible as compare to total moles sopercentagealways taken on100 so thatso 1 mol of NaCl is dipped with = 10−3/100 moles of SrCl2 = 10–5 mol of SrCl2So cation vacancies per mole of NaCl =10–5 mol1 mol = 6.022 x1023 particlesSoSo cation vacancies per mole of NaCl = 10–5 x 6.022 x1023 = 6.02 x1018 So that, the concentration of cation vacancies created by SrCl2 is 6.022 × 108 per mol of NaCl.

Hii Prasanjeet,
The solution is as follows
It is given that NaCl is doped with 10 −3 mol% of SrCl 2 .
This means that 100 mol of NaCl is doped with 10 −3 mol of SrCl 2
Therefore, 1 mol of NaCl is doped with 10^-3/1000 mol of SrCl2
=10^-5
C ation vacancies produced by one Sr 2+ ion = 1
concentration of the cation vacancies produced by 10^-5 mole of SrCl2 =10^-5*6.022*10^23
=6.022*10^18 per mole
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