JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization
When \(0.15\, \mathrm{~g}\) of an organic compound was analyzed using Carius method for estimation of bromine, \(0.2397 \,\mathrm{~g}\) of \(AgBr\) was obtained. The percentage of bromine in the organic compound is \(.....\) (Nearest integer) [Atomic mass : Silver \(=108\), Bromine \(=80]\)
- A \(96\)
- B \(12\)
- C \(85\)
- D \(68\)
Answer & Solution
Correct Answer
(D) \(68\)
Step-by-step Solution
Detailed explanation
Moles of \(\mathrm{Br}=\) Moles of \(\mathrm{AgBr}\) obtained \(\Rightarrow \quad\) Mass of \(\mathrm{Br}=\frac{0.2397}{188} \times 80\, \mathrm{~g}\) therefore \(\%\) Br in the organic compound \(=\frac{W_{\text {Br }}}{W_{T}} \times 100\)…
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