Qualitative Analysis - Result Question 4
4. In Carius method of estimation of halogens $250$ $ mg$ of an organic compound gave $141$ $ mg$ of $AgBr$. The percentage of bromine in the compound is (atomic mass $Ag=108, Br=80$ )
(2015 Main)
(a) $24$
(b) $36$
(c) $48$
(d) $60$
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Answer:
Correct Answer: 4. (a)
Solution:
Given, Weight of organic compound $=250$ $ mg$
Weight of $AgBr=141$ $ mg$
$\therefore$ According to formula of $\text{%}$ of bromine by Carius method
$\text{%}$ of $Br=\frac{\text { Atomic weight of } Br}{\text { Molecular weight of } AgBr} \times \frac{\text { Weight of } AgBr}{\text { Weight of organic bromide }} \times 100 $
$\therefore \text{%} \text { of } Br=\frac{80}{188} \times \frac{141}{250} \times 100 =\frac{1128000}{47000}=24 \text{%}$
BETA
