If NaCl is doped with 10^{-3} mol % SrCl_2, | vedclass

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(A) When $SrCl_2$ is added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^+$ ions in the crystal lattice.One $Sr^{2+}$ ion occupies the site of one $Na

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$1 	imes 10^{-3} \ mol \ \%$

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(A) When $SrCl_2$ is added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^+$ ions in the crystal lattice.One $Sr^{2+}$ ion occupies the site of one $Na^+$ ion,while the other $Na^+$ site remains vacant to maintain electrical neutrality.Therefore,the number of cation vacancies created is equal to the number of $Sr^{2+}$ ions added.Given that the concentration of $SrCl_2$ is $10^{-3} \ mol \ \%$,the concentration of cation vacancies will be $1 	imes 10^{-3} \ mol \ \%$.

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

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