Question

NaCl is crystallized in the presence of small quantity of SrCl$_2$. The formula of crystallized solid is Na$_{0.9998}$Sr$_{0.0001}$Cl. The number of cationic vacancies per mole of this solid is (N=6$\times10^{23}$ mol$^{-1}$)

• A. $6\times10^{-23}$
• B. $6\times10^{19}$
• C. $6\times10^{18}$
• D. $6\times10^{17}$

Hint:

Doping with higher-valent cation creates vacancies for charge balance. Use formula subscripts to find dopant moles. Multiply by Avogadro’s number for particles. Verify neutrality in lattice.

Answer 1

The Correct Option is B

1. SrCl$_2$ doping in NaCl introduces Sr$^{2+}$ ions, replacing two Na$^+$ ions to maintain charge neutrality, creating one cationic vacancy per Sr$^{2+}$.
2. Formula Na$_{0.9998}$Sr$_{0.0001}$Cl indicates 0.0001 moles of Sr$^{2+}$ per mole of solid.
3. Each Sr$^{2+}$ causes 1 Na$^+$ vacancy. Thus, vacancies = 0.0001 moles $\times 6\times10^{23}$ = $6\times10^{19}$ vacancies/mol.
4. The crystal remains neutral as Cl$^-$ balances charges.
5. Thus, the answer is (2) $6\times10^{19}$.