Q.If NaCl is doped with $10^{-3}$ mol % of $SrCl_2$, what is the concentration of cation vacancies?

A $Sr^{2+}$ ion substitutes for two $Na^+$ ions to keep charge balance but occupies only one lattice site, so each $Sr^{2+}$ creates exactly one $Na^+$ vacancy.

$10^{-3}\text{ mol\%}=\frac{10^{-3}}{100}=10^{-5}\text{ mol }SrCl_2\text{ per mol }NaCl.$ …

Step 1 – Charge-balance argument.

When $NaCl$ is doped with $SrCl_2$, each $Sr^{2+}$ ion carries the charge of two $Na^+$ ions. To preserve electrical neutrality, two $Na^+$ sites are affected: one is taken by $Sr^{2+}$ and the other is left vacant. Hence every $Sr^{2+}$ ion introduces one cation ($Na^+$) vacancy.

Step 2 – Convert the doping level to moles.

$10^{-3}$ mol % means $10^{-3}$ mol of $SrCl_2$ per $100$ mol of $NaCl$:

Step 3 – Number of cation vacancies. …

Recognise the aliovalent-doping rule: a divalent $Sr^{2+}$ replaces two monovalent $Na^+$ but occupies one site, so it produces exactly one cation vacancy. Convert the mol% dopin …

• Treating "mol %" as a fraction of 1 instead of per 100 — forgetting the extra factor of $10^{-2}$ (answer off by 100).
• Assuming each $Sr^{2+}$ makes two vacancies; it creates only one (it fills the other site). …