When NaCl (molten) is doped with 10^{-5} , mo | vedclass

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(D) The doping concentration is $10^{-5} \, mol \, \%$,which means $10^{-5} \, mol$ of $AlCl_3$ per $100 \, mol$ of $NaCl$.Number of moles of $AlCl_3$

Vedclass · Accepted Answer

$12.04 	imes 10^{16} \, mol^{-1}$

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(D) The doping concentration is $10^{-5} \, mol \, \%$,which means $10^{-5} \, mol$ of $AlCl_3$ per $100 \, mol$ of $NaCl$.Number of moles of $AlCl_3$ per mole of $NaCl = \frac{10^{-5}}{100} = 10^{-7} \, mol$.Since each $Al^{3+}$ ion replaces one $Na^+$ ion and creates vacancies to maintain electrical neutrality,one $Al^{3+}$ ion replaces three $Na^+$ ions. This results in $2$ cation vacancies per $Al^{3+}$ ion.Total cation vacancies per mole of $NaCl = 2 	imes (	ext{moles of } Al^{3+}) 	imes N_A$.$= 2 	imes 10^{-7} 	imes 6.022 	imes 10^{23} \, mol^{-1}$.$= 12.044 	imes 10^{16} \, mol^{-1}$.

When $NaCl$ (molten) is doped with $10^{-5} \, mol \, \%$ of $AlCl_3$,what is the concentration of cation vacancies?

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