Question

1mole of $NaCl$ is doped with ${10}^{-5}$ mole of $SrC{l}_2.$ The number of cationic vacancies in the crystal lattice will be:

• A. $6.022\times {10}^{18}$
• B. $6.022\times {10}^{15}$
• C. $6.022\times {10}^{23}$
• D. $12.044\times {10}^{20}$

Answer 1

The correct option is A $6.022\times {10}^{18}$

1 mol of $NaCl$ is doped with ${10}^{-5}$ mol of $SrC{l}_2={10}^{-5}$
So for every 2+ ion that comes in, 2 1+ ions must leave to maintain charge neutrality. This means that the cation vacancies produced by one $S{r}^{2+}$ ion = 1, as it fills one of the two spots it creates.
Number of the cation vacancies produced by ${10}^{-5}$ mole of $SrC{l}_2={10}^{-5}\times 6.022\times {10}^{23}=6.022\times {10}^{18}$ per mole