Question

If ${\text{NaCl}}$ is doped with ${10^{ - 4}}\%$ of ${\text{SrC}}{{\text{l}}_{\text{2}}}$, the concentration of cation vacancies will be: $\left( {{{\text{N}}_{\text{A}}}{\text{ = 6}}{\text{.02 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)$.

Answer 1

Here strontium is present in $+ 2$ oxidation state in strontium chloride and sodium is present in $+ 1$ oxidation state in sodium chloride. So that one strontium cation will have power to replace or remove two sodium cations from the given composition.

Complete answer:
In question it is given that, Sodium chloride (${\text{NaCl}}$) is doped with ${10^{ - 4}}\%$ of Strontium chloride (${\text{SrC}}{{\text{l}}_{\text{2}}}$). It means:
$100$ moles of Sodium chloride (${\text{NaCl}}$) doped with = ${10^{ - 4}}$ moles of Strontium chloride (${\text{SrC}}{{\text{l}}_{\text{2}}}$)
So that one mole of Sodium chloride (${\text{NaCl}}$) doped with = $\dfrac{{{{10}^{ - 4}}}}{{100}} = {10^{ - 6}}$ moles of Strontium chloride (${\text{SrC}}{{\text{l}}_{\text{2}}}$)
-As it is clear that one strontium cation will have the capacity of replacing two sodium cations.
-So each strontium cation will replace one sodium cation and produce one cationic vacancy.
-And concentration of cationic vacancies will be calculated as follow:
Concentration of cation vacancies = ${\text{1}}{{\text{0}}^{{\text{ - 6}}}}{\text{mol}}$ of strontium in per mole of sodium chloride
It is given that, ${{\text{N}}_{\text{A}}}{\text{ = 6}}{\text{.02 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
Concentration of cation vacancies = ${\text{1}}{{\text{0}}^{{\text{ - 6}}}} \times 6.02 \times {10^{23}}$
Hence, concentration of cation vacancies is equal to ${\text{6}}{\text{.023 \times 1}}{{\text{0}}^{{\text{17}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$.

Additional information:
In the given question ${{\text{N}}_{\text{A}}}{\text{ = 6}}{\text{.02 \times 1}}{{\text{0}}^{{\text{23}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ shows the Avogadro number, which means this amount of particles are present in one mole of solution.

Note:
Here some of you may do wrong calculation if you were not familiar with the fact that strontium have oxidation state of $+ 2$ i.e. strontium will remove two electrons from it for bonding with other atoms who will accept those electrons.