If NaCl is doped with 10^{-5} mol of AlCl_3, | vedclass

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(D) When $Al^{3+}$ ions are introduced into the $NaCl$ lattice,each $Al^{3+}$ ion replaces one $Na^+$ ion to maintain electrical neutrality. Since $Al

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

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(D) When $Al^{3+}$ ions are introduced into the $NaCl$ lattice,each $Al^{3+}$ ion replaces one $Na^+$ ion to maintain electrical neutrality. Since $Al^{3+}$ has a charge of $+3$ and $Na^+$ has a charge of $+1$,one $Al^{3+}$ ion replaces three $Na^+$ ions. This creates two cation vacancies for every $Al^{3+}$ ion added. Concentration of $Al^{3+}$ = $10^{-5} \ mol$. Number of $Al^{3+}$ ions = $10^{-5} 	imes 6.022 	imes 10^{23} = 6.022 	imes 10^{18} \ ions$. Number of cation vacancies = $2 	imes (6.022 	imes 10^{18}) = 12.044 	imes 10^{18} \ mol^{-1}$.

If $NaCl$ is doped with $10^{-5} \ mol$ of $AlCl_3$,what will be the concentration of cation vacancies?

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