NaCl is doped with 2 times 10^{-3} , mol % , | vedclass

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(B) When $SrCl_2$ is added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality.One $Sr^{2+}$ ion creates one catio

Vedclass · Accepted Answer

$12.04 	imes 10^{18} \, mol^{-1}$

Vedclass · Answer

(B) When $SrCl_2$ is added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality.One $Sr^{2+}$ ion creates one cation vacancy.Concentration of $SrCl_2 = 2 	imes 10^{-3} \, mol \% = \frac{2 	imes 10^{-3}}{100} = 2 	imes 10^{-5} \, mol$ per $100 \, mol$ of $NaCl$.Number of cation vacancies per mole of $NaCl = 2 	imes 10^{-5} 	imes N_A$.$= 2 	imes 10^{-5} 	imes 6.022 	imes 10^{23} \, mol^{-1}$.$= 12.044 	imes 10^{18} \, mol^{-1}$.

$NaCl$ is doped with $2 \times 10^{-3} \, mol \% \, SrCl_2$. What is the concentration of cation vacancies?

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