If 10^{-4} % SrCl_2 is added to a crystal of | vedclass

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(B) The addition of $Sr^{2+}$ ions to $NaCl$ creates cationic vacancies because each $Sr^{2+}$ ion replaces two $Na^+$ ions,leaving one vacancy.Number

Vedclass · Accepted Answer

$6.022 	imes 10^{17}$

Vedclass · Answer

(B) The addition of $Sr^{2+}$ ions to $NaCl$ creates cationic vacancies because each $Sr^{2+}$ ion replaces two $Na^+$ ions,leaving one vacancy.Number of cationic vacancies = Number of $Sr^{2+}$ ions.$10^{-4} \% SrCl_2$ means $10^{-4} \, g$ of $SrCl_2$ in $100 \, g$ of $NaCl$,or in terms of moles,$10^{-6} \, mol$ of $Sr^{2+}$ per mole of $NaCl$.Number of $Sr^{2+}$ ions = $10^{-6} 	imes 6.022 	imes 10^{23} \, mol^{-1} = 6.022 	imes 10^{17}$.

If $10^{-4} \% SrCl_2$ is added to a crystal of $NaCl$,find the number of cationic vacancies created in the crystal.

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