If 10^{-4} , text{mole} % of SrCl_2 is added | vedclass

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(C) When $Sr^{2+}$ ions are added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality.One $Sr^{2+}$ ion creates on

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) When $Sr^{2+}$ ions are added to $NaCl$,each $Sr^{2+}$ ion replaces two $Na^+$ ions to maintain electrical neutrality.One $Sr^{2+}$ ion creates one cation vacancy.Given concentration of $SrCl_2 = 10^{-4} \, 	ext{mole} \% = \frac{10^{-4}}{100} = 10^{-6} \, 	ext{mole}$.Number of $Sr^{2+}$ ions = $10^{-6} 	imes N_A = 10^{-6} 	imes 6.02 	imes 10^{23} = 6.02 	imes 10^{17} \, 	ext{mol}^{-1}$.Since each $Sr^{2+}$ ion creates one cation vacancy,the concentration of cation vacancies is $6.02 	imes 10^{17} \, 	ext{mol}^{-1}$.

If $10^{-4} \, \text{mole} \%$ of $SrCl_2$ is added to $NaCl$,the concentration of cation vacancies created is...... $(N_A = 6.02 \times 10^{23} \, \text{mol}^{-1})$

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