NaCl is doped with mole % , the concentration of cation vacancies is: — Solid State Chemistry Question

💡 Solution & Explanation

Doping NaCl with $SrCl_2$ introduces $Sr^{2+}$ ions. Each $Sr^{2+}$ replaces two $Na^+$ ions to maintain electrical neutrality, creating one cation vacancy. Thus, the number of cation vacancies equals the number of $Sr^{2+}$ ions added. $2 \times 10^{-3}$ mole % means there are $2 \times 10^{-3} / 100 = 2 \times 10^{-5}$ moles of $SrCl_2$ per mole of NaCl. The number of cation vacancies per mole of NaCl is $2 \times 10^{-5} \times 6.022 \times 10^{23} = 1.2044 \times 10^{19} \text{ mol}^{-1}$. Therefore, correct answer is B.