Ncert Exercises & Solutions

25. If NaCl is doped with 10⁻³ mol % of SrCl₂, what is the concentration of cation vacancies?

It is given that NaCl is doped with 10⁻³ mol % of SrCl₂.
This means that 100 mol of NaCl is doped with 10⁻³ mol of SrCl₂.

Therefore, 1 mol of NaCl is doped with

\frac{10^{- 3}}{100}

mol of SrCl₂= 10⁻⁵ mol of SrCl₂

Cation vacancies produced by one Sr²⁺ ion = 1

∴

The concentration of the cation vacancies

produced by 10^{- 5} mol of {Sr}^{2 +} ions = 10^{- 5} \times 6.022 \times 10^{23} = 6.022 \times 10^{18} {mol}^{- 1}

Hence, the concentration of cation vacancies created by SrCl₂ is 6.022 × 10⁸ per mol of NaCl.