Question

If NaCl is doped with 10 −3 mol % of SrCl2, what is the concentration of cation vacancies?

Answer 1

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Answer :

The number of cation vacancies created in the lattice of NaCI is equal to the number of divalent $Sr^{2+}$ ions added. The concentration of $Sr^{2+} = 10^{-3}$ mole % = $\frac{10^{-3}}{100}$ = $10^{-5}$ mole.

1 mole of $Sr^{2+}$ contains 6.023 × $10^{23}$ $Sr^{2+}$ ions.

∴ $10^{5}$ mole of $Sr^{2+}$ will contain 6.023 × $10^{23}$ ×$10^{-5}$ = 6.023 ×$10^{18}$ $Sr^{2+}$ ions.

Hence, the concentration of cation vacancies is 6.023 ×$10^{18}$.