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

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(A) When $NaCl$ is doped with $SrCl_2$,each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality. One $Sr^{2+}$ ion creates one ca

Vedclass · Accepted Answer

$6.02 	imes 10^{18} \, mol^{-1}$

Vedclass · Answer

(A) When $NaCl$ is doped with $SrCl_2$,each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality. One $Sr^{2+}$ ion creates one cation vacancy. Given doping is $10^{-3} \, mol \% \, SrCl_2$,which means $10^{-3} \, mol$ of $SrCl_2$ in $100 \, mol$ of $NaCl$. So,$Sr^{2+}$ ions per mole of $NaCl = \frac{10^{-3}}{100} = 10^{-5} \, mol$. Since each $Sr^{2+}$ ion creates one cation vacancy,the concentration of cation vacancies is $10^{-5} \, mol \, per \, mole$ of $NaCl$. To find the number of vacancies per mole,we multiply by Avogadro's number $(N_A = 6.022 	imes 10^{23})$: Number of vacancies $= 10^{-5} 	imes 6.022 	imes 10^{23} = 6.022 	imes 10^{18} \, mol^{-1}$.

If $NaCl$ is doped with $10^{-3} \, mol \% \, SrCl_2$,what will be the concentration of cation vacancies?

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