Question

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

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

Answer 1

The correct option is A $6.02\times {10}^{17}mo{l}^{-1}$

One cation of Sr²⁺ would create one vacancy in NaCl therefore number of cation vacancies created in the lattice of NaCl is equal to the number of $S{r}^{+2}$ ions added concentration of cation vacancy or being doped with
${10}^{-4}mole\%SrC{l}_2=\frac{{10}^{-4}}{100}={10}^{-6}molNo.ofS{r}^{+2}ionsin{10}^{-6}mole={10}^{-6}\times 6.02\times {10}^{23}=6.02\times {10}^{17}S{r}^{+2}ions,$
No. of cation vacancies = $6.02\times {10}^{17}$