Question

If $NaCl$ is doped with ${10}^{-4}$ mole % of ${SrCl}_2$, the concentration of cation vacancies will be:

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

Answer 1

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