JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization
An organic compound is subjected to chlorination to get compound A using \(5.0\, \mathrm{~g}\) of chlorine. When \(0.5\, \mathrm{~g}\) of compound \(\mathrm{A}\) is reacted with \(\mathrm{AgNO}_{3}\) [Carius Method], the percentage of chlorine in compound \(A\) is \(.....\) when it forms \(0.3849\) \(g\) of \(\mathrm{AgCl}\). (Round off to the Nearest Integer) (Atomic masses of \(\mathrm{Ag}\) and \(\mathrm{Cl}\) are \(107.87\) and \(35.5\) respectively)
- A \(19\)
- B \(21\)
- C \(25\)
- D \(80\)
Answer & Solution
Correct Answer
(A) \(19\)
Step-by-step Solution
Detailed explanation
\(\mathrm{n}_{\mathrm{cl}} \text { in compound }=\mathrm{n}_{\mathrm{AgCl}}=\frac{0.3849 \mathrm{~g}}{(107.87+35.5)}\, \mathrm{g} / \mathrm{mol}\) \(\Rightarrow \text { mass of chlorine }=\mathrm{n}_{\mathrm{Cl} } \times 35.5=0.0953\, \mathrm{gm}\)…
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