KBr is doped with 10 - 5 mole percent of SrBr 2 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 × 10 23

Step-by-step solution

1   mole   KBr   ( = 119 gm ) have 10 - 5 100 moles SrBr 2 and hence, 10 - 7 moles cation vacancy (as 1 Sr 2 + will result 1 cation vacancy) ∴ Required number of cation vacancies = 10 - 7 × 6 . 023 × 10 23 119 = 5 . 06 × 10 14 ≃ 5 × 10 14