Question

If $NaCl$ is doped with $10^{-4}mol\%$ of $SrC{l}_2,$ the concentration of cation vacancies will be
$(N_A=6.023\times 10^{23}mo{l}^{-1})$

• A. $6.023\times 10^{15}mo{l}^{-1}$
• B. $6.023\times 10^{16}mo{l}^{-1}$
• C. $6.023\times 10^{17}mo{l}^{-1}$
• D. $6.023\times 10^{14}mo{l}^{-1}$

Answer 1

Answer: (C)

Doping of $NaCl$ with $10^{-4}mol\%ofSrC{l}_2$ means,
$100$ moles of $NaCl$ are doped with $10^{-4}molofSrC{l}_2$
$\therefore$ $1$ mol of $NaCl$ is doped with
$SrC{l}_2=\frac{10^{-4}}{100}=10^{-6}mole$
As each $S{r}^{2+}$ ion introduces one cation vacancy.
$\therefore$ Concentration of cation vacancies
$=10^{-6}$ mol of $NaCl$
$=10^{-6}\times 6.023\times 10^{23}mo{l}^{-1}$
$=6.023\times 10^{17}mo{l}^{-1}$