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)…

Question

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=39.1u, Br=79.9u,N_A=6.023×10^23

MARKS Content Team · Accepted Answer

Correct answer: 5. 1 mole KBr (=119gm) have 10^(-5)100 moles SrBr_2 and hence, 10^(-7) moles cation vacancy (as 1Sr^(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

$KBr$ 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 : $K = 39.1 u, Br = 79.9 u$ , $N_{A} = 6.023 \times 10^{23}]$

Solution

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

Solution

More Solid State Questions

$KBr$ 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 : $K = 39.1 u, Br = 79.9 u$ , $N_{A} = 6.023 \times 10^{23}]$