Question

If $NaCl$ is doped with ${10}^{-3}$ mol$\%$ of $SrC{l}_2$, the concentration of cation vacancies will be:

$(N_A=6.02\times {10}^{23}mo{l}^{-1}$)

• A. $6.02\times {10}^{15}mo{l}^{-1}$
• B. $6.02\times {10}^{16}mo{l}^{-1}$
• C. $6.02\times {10}^{18}mo{l}^{-1}$
• D. $6.02\times {10}^{14}mo{l}^{-1}$

Answer 1

Answer: (C)

$6.02\times {10}^{18}mo{l}^{-1}$

Given that $1$ $mol$ of $NaCl$ is doped with $\frac{{10}^{-3}}{100}$ $mol$ of $S{r}^{+2}={10}^{-5}$ $mol$

Cation vacancies produced by $S{r}^{2+}$ ion $=1$ [$\because$ 1 $S{r}^{+2}$ can replace 2 $N{a}^{+}$]

So, concentration of cation vacancies produced by ${10}^{-5}$ mole of $SrC{l}_2$

$=6.023\times {10}^{23}\times {10}^{-5}$

$=6.023\times {10}^{18}$ per mole