Question

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

• A. $1\times {10}^{-3}mol$%
• B. $2\times {10}^{-3}mol$%
• C. $3\times {10}^{-3}mol$%
• D. $4\times {10}^{-3}mol$%

Answer 1

The correct option is A $1\times {10}^{-3}mol$%

Each $S{r}^2+$ ion introduces one cation vacancy
$\Rightarrow$ number of cation vacancies = % mole of $S{r}^{2+}={10}^{-3}mol$%