Question

If an ionic solid XY ( X and Y are monovalent ions) is doped with 10^-2 mole % of another ionic solid AY₃ then the concentration of the cationic vacancies created is
(ans : 1.205 * 10 ²⁰ mol^​-1 )

Answer 1

1. When XY is doped by AB₃, three X⁺ are replaced by one A³⁺. So, number of cationic vacancy per unit cell = 3-1 = 2.

2. By 10^-2 mole % we mean that in 100 moles of XY there are 10^-2 A³⁺

So, 1 mole of XY has = $\frac{{10}^{-2}}{100}={10}^{-4}$ of A³⁺

Number of cation vacancies = 2$\times$6.023​$\times$10²³$\times$10^-4 = 1.205 ​$\times$ 10²⁰.