If NaCl is doped with 10^{-3} , mol % of SrCl | vedclass

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(A) It is given that $NaCl$ is doped with $10^{-3} \, mol \%$ of $SrCl_{2}$.This means that $100 \, mol$ of $NaCl$ is doped with $10^{-3} \, mol$ of $

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$6.022 	imes 10^{18} \, mol^{-1}$

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(A) It is given that $NaCl$ is doped with $10^{-3} \, mol \%$ of $SrCl_{2}$.This means that $100 \, mol$ of $NaCl$ is doped with $10^{-3} \, mol$ of $SrCl_{2}$.Therefore,$1 \, mol$ of $NaCl$ is doped with $\frac{10^{-3}}{100} = 10^{-5} \, mol$ of $SrCl_{2}$.Since each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality,one $Sr^{2+}$ ion creates one cation vacancy.Thus,the number of cation vacancies per $mol$ of $NaCl$ is equal to the number of $Sr^{2+}$ ions per $mol$ of $NaCl$.Concentration of cation vacancies $= 10^{-5} 	imes 6.022 	imes 10^{23} \, mol^{-1} = 6.022 	imes 10^{18} \, mol^{-1}$.

If $NaCl$ is doped with $10^{-3} \, mol \%$ of $SrCl_{2}$,what is the concentration of cation vacancies?

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