Question

KBr is doped with 10^–5 mole percent of SrBr₂ . The number of cationic vacancies in 1 g of KBr crystal is ..... 10^14. (Round off to the Nearest Integer).

[Atomic Mass : K : 39.1 u, Br : 79.9 u, N_A = 6.023 × 10²³]

Answer 1

1 mole KBr (= 119 gm) have \(\frac{10^{-5}}{100}\) moles SrBr₂ and hence, 10^–7 moles cation vacancy (as 1 Sr²⁺ will result 1 cation vacancy)

∴ Required number of cation vacancies