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What is the concentration of cation vaca...

What is the concentration of cation vacancies if NaCl is doped with `10^(-3)` mol percent of `MgCl_(2)`?

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Doping of NaCl with `10^(-3)` moles of `MgCl_(2)` means that 100 moles of NaCl are doped with `10^(-3)` mol of `MgCl_(2)`.
Therefore, mols of `MgCl_(2)` in 1 mol of NaCl = `10^(-3)/100=10^(-5)` mol
As each `Mg^(2+)` ion introduces one cation vacancy, the number of cation vacancies = `10^(-5)` mol/mol of NaCl.
= `10^(-5)xx6.023xx10^(23)"mol"^(-1)`
= `6.023xx10^(18)"mol"^(-1)`
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