Question

If NaCl is doped with ${10}^{-3}\mathrm{mol}\mathrm{percent}$ of${\mathrm{SrCl}}_2$. What is the concentration of cation vacancies?

Answer 1

Step: 1 Calculating the number of moles of NaCl

The total number of moles of the solution: 100 mol.

Number of moles of NaCl = 100 – moles of SrCl₂

Since the number of moles of SrCl₂ is negligible, we can ignore it

Then, the number of moles of NaCl becomes 100

Step: 2 Calculating the concentration of Cationic Vacancies

Concentration of ${\mathrm{SrCl}}_2$ = ${10}^{-3}\mathrm{mol}\mathrm{percent}$

1 mol of NaCl is doped with = $\frac{{10}^{-3}}{100}\mathrm{moles}\mathrm{of}{\mathrm{SrCl}}_2=={10}^{–5}\mathrm{mol}\mathrm{of}{\mathrm{SrCl}}_2$

The cation vacancies per mole of NaCl = ${10}^{–5}\times 6.022\times {10}^{23}==6.02\times {10}^{18}$

Final answer

Therefore, the concentration of cationic vacancies created by doping NaCl crystals with SrCl₂ is $6.02\times {10}^{18}$.