Question

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

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

Answer 1

Answer: (B)

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

Moles of $SrC{l}_2$ in $1$ mole $NaCl=\frac{{10}^{-4}}{100}={10}^{-6}$ moles

$\therefore$ One $S{r}^{+2}$ creates one cationic vacancies.

Concentration of the cationic vacancies produced by ${10}^{-6}$ moles of $SrC{l}_2$

$={10}^{-6}\times 6.023\times {10}^{23}$

$=6.023\times {10}^{17}$ $mo{l}^{-1}$