Question

If $NaCl$ is doped with ${10}^{-4}$ mole % of $SrC{l}_2$, the concentration of cation vacancies will be:
$[N_A=6.02\times {10}^{23}mo{1}^{-1}]$

• A. $6.02\times {10}^{14}mo{1}^{-1}$
• B. $6.02\times {10}^{15}mo{1}^{-1}$
• C. $6.02\times {10}^{16}mo{1}^{-1}$
• D. $6.02\times {10}^{17}mo{1}^{-1}$

Answer 1

Answer: (D)

$6.02\times {10}^{17}mo{1}^{-1}$

Given, NaCl is doped with ${10}^{-4}$ mole % of $SrC{l}_2$.

This means ,100 moles of NaCl are doped with ${10}^{-4}$ moles of $SrC{l}_2$

Therefore , 1mole of NaCl is doped with $\frac{{10}^{-4}}{100}={10}^{-6}$ moles of $SrC{l}_2$ or ${10}^{-6}$ moles of $4S{r}^{2+}$

Now, One $S{r}^{2+}$ ion creates one cation vacancy.

Therefore, no.of cations vacancies created by ${10}^{-6}$ moles $S{r}^{2+}$

= ${10}^{-6}$ moles/mole of NaCl

= ${10}^{-6}\times 6.022\times {10}^{23}$ vacancies/mole of NaCl

= $6.022\times {10}^{17}$ vacancies /mole of NaCl.

So the correct answer is option D.