Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 19th Sep 2024
Total views: 413.1k
Views today: 12.13k
Verified
413.1k+ views
Hint: 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.

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 ${\text{6}}{\text{.023 \times 1}}{{\text{0}}^{{\text{17}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$.

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.
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.