In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ______________. (Nearest integer)
[Atomic mass : Ag = 108, Br = 80]
Answer (integer)
40
Solution
n<sub>AgBr</sub> = ${{0.188g} \over {188g/mol}}$ = 10<sup>$-$3</sup> mol<br><br>$\Rightarrow$ n<sub>Br</sub> = n<sub>AgBr</sub> = 0.001 mol<br><br>$\Rightarrow$ mass<sub>Br</sub> = (0.001 $\times$ 80) gm = 0.08 gm<br><br>$\Rightarrow$ mass% = ${{0.08 \times 100} \over {0.2}} = 40\%$
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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