KBr is doped with 10⁻⁵ mole percent of SrBr₂. The number of cationic vacancies in 1g of KBr crystal is ____________ 10¹⁴. (Round off to the Nearest Integer).
[Atomic Mass : K : 39.1 u, Br : 79.9 u N A = 6.023 × 10²³]

Solution

Chemistry ma’am · 4 steps

Each Sr²⁺ that replaces two K⁺ leaves one cation site empty, so count the Sr²⁺ ions.

1. Why vacancies form. One Sr²⁺ takes the place of two K⁺ ions to keep the crystal neutral. It occupies one of the two sites and leaves the other empty. So one Sr²⁺ gives one cationic vacancy.
2. Moles of KBr. Molar mass of KBr = 39.1 + 79.9 = 119 g mol⁻¹. So 1 g is 1/119 = 8.40 × 10⁻³ mol.
3. Mole percent to fraction. 10⁻⁵ mole percent means 10⁻⁵ mol of SrBr₂ per 100 mol, which is a fraction of 10⁻⁷. Multiply this fraction by the moles of KBr and by N_A.

   N_(vacancies) = 1/119 × 10⁻⁷ × 6.023 × 10²³ = 5.06 × 10¹⁴

4. So. There are about 5.06 × 10¹⁴ cationic vacancies in 1 g, which rounds to 5.

The common slip: 506. Using 10⁻⁵ as the fraction and forgetting to divide the percent by 100 gives 5.06 × 10¹⁶, which is 506 in units of 10¹⁴.

The idea to keep. When a divalent cation dopes a crystal of monovalent cations, every dopant ion creates exactly one cation vacancy. Number of vacancies = number of dopant ions.

Answer from the NTA final answer key. Solution by PrepWiser.