Question

AgCl is doped with 10⁻² mol% of CdCl₂. Find the concentration of cation vacancies.

Answer 1

Given: AgCl is doped with 10⁻² % mol of CdCl₂.

1 mol % = `1/100` mol per mol of host material

Avogadro’s number, N_A = 6.022 × 10²³

⇒ Convert mol% to mol fraction:

10⁻²% = `10^-2/100` = 10⁻⁴ mol CdCl₂​ per mol of AgCl.

Each mol of CdCl₂ introduces one mole of cation vacancies.

∴ If 10⁻⁴ mol of Cd²⁺ is added per 1 mol of AgCl, then

Cation vacancies = 10⁻⁴ mol per mol of AgCl

⇒ Convert moles to number of vacancies:

Number of vacancies = 10⁻⁴ × 6.022 × 10²³

= 6.022 × 10¹⁹ mol⁻¹