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

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

Solution:

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^-3 moles of SrCl₂

Moles of SrCl₂ is very negligible as compare to total moles so

percentagealways taken on100 so that

so 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 SrCl2 is 6.022 × 10⁸ per mol of NaCl.