Question

If NaCl is doped with 10^-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.02 x 10^23 mol^-1)

Answer 1

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

hence, number of cation vacancies will be 6.023 × 10^17

Answer 2

6.02 x 1015 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 1023 x10-8
= 6.02 x 1015 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 1023 x10-8
= 6.02 x 1015 mol-1