Question

$NaCl$ is doped with $2\times {10}^{-3}$ mol percent $SrC{l}_2$, the concentration of cation vacancies is:

• A. $3.01\times {10}^{18}$ mol${}^{-1}$
• B. $12.04\times {10}^{18}$ mol${}^{-1}$
• C. $6.02\times {10}^{18}$ mol${}^{-1}$
• D. $12.04\times {10}^{20}$ mol${}^{-1}$

Answer 1

The correct option is B $12.04\times {10}^{18}$ mol${}^{-1}$

$2\times {10}^{-3}$ mol of NaCl is doped with $SrC{l}_3=\frac{2\times {10}^{-3}}{100}=2\times {10}^{-5}$ mol

As each $S{r}^{+2}$ ion introduces one cation vacancy, the concentration of cation vacancies

=$2\times {10}^{-5}\times 6.023\times {10}^{23}$ mol${}^{-1}$ of NaCl $=12.04\times {10}^{18}$ mol${}^{-1}$

Hence, the correct option is $\text{B}$