Question

When $NaCl$ is doped with ${10}^{-5}$ moles % of $SrC{l}_2$, what is the number of cationic vacancies per mole?

• A. ${10}^{-5}\times N_A$
• B. ${10}^{-7}\times N_A$
• C. $2\times {10}^{-7}{N}_A$
• D. None of these

Answer 1

The correct option is B ${10}^{-7}\times N_A$

Given,

NaCl is doped with 10^-5 moles % of SrCl₂.This means, 100 moles of NaCl are doped with 10^-5 moles of SrCl₂.

Therefore, 1mole of NaCl is doped with 10^-5 /100 = 10^-7 moles of SrCl₂ or 10^-7 moles of Sr²⁺.

Now,

One Sr²⁺ ion create one cation vacancy.

Therefore, no. of cation vacancies created by 10^-7 moles Sr²⁺

= 10^-7 moles/mole of NaCl

= 10^-7 x 6.022 x 10²³ / mole of NaCl = ${10}^{-7}\times N_A$