JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization
In an estimation of bromine by Carius method,\(1.6\, g\) of an organic compound gave \(1.88 \,g\) of \(AgBr\). The mass percentage of bromine in the compound is.......... (Atomic mass, \(Ag =108, Br =80\, g\, mol ^{-1}\) )
- A \(50\)
- B \(55\)
- C \(45\)
- D \(40\)
Answer & Solution
Correct Answer
(A) \(50\)
Step-by-step Solution
Detailed explanation
Carius method \(\%\) of \(Br =\frac{\text { wt of } AgBr }{\text { wt. of organic compound }} \times 100 \times \frac{\text { molar mass of Br }}{\text { AgBr }}\) \(=\frac{1.88}{1.6} \times \frac{80}{188} \times 100=\frac{15040}{300.8}=50 \%\)
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