If NaCl is doped with 10^{-4} mol % of SrCl_ | vedclass

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(B) Doping $NaCl$ with $10^{-4} \ mol \%$ of $SrCl_2$ means that $100 \ mol$ of $NaCl$ contains $10^{-4} \ mol$ of $SrCl_2.$Therefore,$1 \ mol$ of $Na

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$6.02 	imes 10^{17} \ mol^{-1}$

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(B) Doping $NaCl$ with $10^{-4} \ mol \%$ of $SrCl_2$ means that $100 \ mol$ of $NaCl$ contains $10^{-4} \ mol$ of $SrCl_2.$Therefore,$1 \ mol$ of $NaCl$ contains $10^{-4} / 100 = 10^{-6} \ mol$ of $SrCl_2.$Each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality,creating one cation vacancy per $Sr^{2+}$ ion introduced.Concentration of cation vacancies $= (10^{-6} \ mol \ SrCl_2 / mol \ NaCl) 	imes (N_A \ ions / mol) 	imes (1 \ 	ext{vacancy} / Sr^{2+} \ 	ext{ion}).$$= 10^{-6} 	imes 6.02 	imes 10^{23} \ 	ext{vacancies} \ mol^{-1} = 6.02 	imes 10^{17} \ mol^{-1}.$

If $NaCl$ is doped with $10^{-4} \ mol \%$ of $SrCl_2,$ the concentration of cation vacancies will be $(N_A = 6.02 \times 10^{23} \ mol^{-1}).$

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