If Nacl is Doped With 10?3 Mol % of Srcl2, What is the Concentrat Exercise Chapter 1 the Solid State Chemistry Class

Question 25: If NaCl is doped with 10⁻³ mol % of SrCl₂, what is the concentration of cation vacancies?

Answer
Given Conetration of SrCl₂ = 10⁻³ mol%
Concentration is in percentage so that take total 100 mol of solution
Number of moles of NaCl     = 100 – moles of SrCl₂
Moles of SrCl₂is very negligible as compare to total moles so
Number of moles of NaCl      = 100
1 mol of NaCl is dipped with  = 10−3/100  moles of SrCl₂
                                         = 10^–5  mol of SrCl₂
So cation vacancies per mole of NaCl =10^–5  mol
1 mol = 6.022 x10²³ particles
So
So cation vacancies per mole of NaCl  = 10–5 x 6.022 x10²³ 
                                                     = 6.02 x10¹⁸ 
So that, the concentration of cation vacancies created by SrCl₂is 6.022 × 10⁸ per mol of NaCl.