When AgCl is doped with 10^{-4} , text{mole p | vedclass

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(D) When $Ru^{3+}$ ions are doped into $AgCl$,each $Ru^{3+}$ ion replaces $3 \ Ag^+$ ions to maintain electrical neutrality.This creates $2$ cationic

Vedclass · Accepted Answer

$12.046 	imes 10^{17}$

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(D) When $Ru^{3+}$ ions are doped into $AgCl$,each $Ru^{3+}$ ion replaces $3 \ Ag^+$ ions to maintain electrical neutrality.This creates $2$ cationic vacancies for every $Ru^{3+}$ ion added.Given doping is $10^{-4} \, 	ext{mole percent}$,which is $\frac{10^{-4}}{100} = 10^{-6} \, 	ext{moles of } RuCl_3 	ext{ per mole of } AgCl$.Number of vacancies $= 2 	imes 10^{-6} \, 	ext{moles per mole of } AgCl$.Concentration of vacancies $= 2 	imes 10^{-6} 	imes 6.023 	imes 10^{23} \, 	ext{vacancies/mole}$.$= 12.046 	imes 10^{17} \, 	ext{cationic vacancies per mole of } AgCl$.

When $AgCl$ is doped with $10^{-4} \, \text{mole percent}$ $RuCl_3$,what will be the concentration of cationic vacancies per mole of $AgCl$?

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