If NaCl is doped with 10^-3 mol % GaCl₃, the concentration of the cation vacancies is 1.2 $x{10}^{-a}$, the value of a is

 

100 mol of NaCl are doped with 10^-3 mol of GaCl₃.

$\therefore 1\mathrm{mol}\text{ of }\mathrm{NaCl}\text{ is doped with }{\mathrm{GaCl}}_3=\frac{{10}^{-3}}{100}={10}^{-5}\mathrm{mol}$

As one Ga³⁺ ion is introduced, three ${\mathrm{Na}}^{\oplus }$ have to be removed to maintain the electrical neutrality. So as one vacancy is filled by Ga³⁺, two cation vacancies are formed.

$\therefore$ Concentration of cation vacancy

$\begin{array}{l}=2\times {10}^{-5}\mathrm{mol}/\mathrm{mol}\text{ of }\mathrm{NaCl}\\ =2\times {10}^{-5}\times 6.023\times {10}^{23}{\mathrm{mol}}^{-1}\\ =12.046\times {10}^{-18}{\mathrm{mol}}^{-1}\\ =1.2046\times {10}^{-19}{\mathrm{mol}}^{-1}\end{array}$