Question

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

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

Answer 1

The correct option is C $6.02\times {10}^{17}{mol}^{-1}$

$NaCl$ is doped with ${10}^{-4}$ $mol\%$ of $SrC{l}_2$, i.e., one mole of $NaCl$ will have ${10}^{-6}$ mol of $SrC{l}_2$

$\to S{r}^{+2}$ will replace one cation.

Therefore, concentration of cation vacancy $={10}^{-6}\times 6.022\times {10}^{23}=6.022\times {10}^{17}$ $mo{l}^{-1}$