Question

NaCl is doped with 10^-4 mol% SrCl_​​​2 ​​​​​​, the concentration of cation vacancies is - (Concentration of vacancies= 10^-4/100 N_{^{​​​​​​}​​​​​​A} )

Answer 1

Given Conetration of SrCl₂ = 10⁻4 mol%

Concentration is in percentage so that take total 100 mol of solution

Number of moles of NaCl = 100^-4 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 = 10⁻4/100 moles of SrCl₂

= 10^–6 mol of SrCl2

So cation vacancies per mole of NaCl =10^–6 mol

1 mol = 6.022 x10²³ particles

So

So cation vacancies per mole of NaCl = 10^–6x 6.022 x10²³

= 6.02 x10¹7

So that, the concentration of cation vacancies created by SrCl2 is 6.022 × 10^7per mol of NaCl.