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If NaCl is doped with $10^{-3}$ mole percentage of $SrCl_2$,what is the concentration of cation vacancy?

$\begin{array}{1 1}6.023\times 10^{18}\\5.023\times 10^{18}\\6.023\times 10^{15}\\4.023\times 10^{18}\end{array} $

1 Answer

Introduction of one $Sr^{2+}$ ion introduces one cation vacancy because one $Sr^{2+}$ replaces two $Na^+$ ions.Therefore,introduction of $10^{-3}$ mole of $SrCl_2$ per 100 moles of NaCl would introduce $10^{-3}$ mole cation vacancies in 100 moles of NaCl
No. of vacancies per mole of NaCl =$\large\frac{10^{-3}}{100}$$=10^{-5}$
$\Rightarrow 10^{-5}\times 6.023\times 10^{23}=6.023\times 10^{18}$
answered Aug 1, 2014 by sreemathi.v