JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization
\(0.400\,g\) of an organic compound \(( X )\) gave \(0.376\,g\) of \(AgBr\) in Carius method for estimation of bromine. \(\%\) of bromine in the compound \(( X )\) is \(.........\).(Given: Molar mass \(AgBr =188\,g\,mol ^{-1} Br =80\,g\) \(\left.mol ^{- I }\right)\)
- A \(20\)
- B \(30\)
- C \(50\)
- D \(40\)
Answer & Solution
Correct Answer
(D) \(40\)
Step-by-step Solution
Detailed explanation
mole of \(AgBr =\frac{0.376}{188}\) mole of \(Br ^{-}=\)mole of \(AgBr =\frac{0.376}{188}\) mass of \(Br ^{-}=\frac{0.376}{188} \times 80\) \(\%\) of \(Br ^{-}=\frac{0.376 \times 80}{188 \times 0.4} \times 100=40 \%\)
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