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

$\begin{array}{1 1}6.02\times 10^{18}\\5.02\times 10^{18}\\3.02\times 10^{18}\\4.02\times 10^{18}\end{array} $

1 Answer

Doping of NaCl with $10^{-3}$mol % $SrCl_2$ means that 100 moles of NaCl are doped with $10^{-3}$ mol of $SrCl_2$
$\therefore$ 1 mole of NaCl is doped with $SrCl_2=\large\frac{10^{-3}}{100}$$\times 6.02\times 10^{23}$
$\Rightarrow 6.02\times 10^{18}$
answered Jul 31, 2014 by sreemathi.v