JEE Mains · Chemistry · STD 12 - p -Block elements - ll
The ammonia \((NH_{3} )\) released on quantitative reaction of \(0.6\; \mathrm{g}\) urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) with sodium hydroxide (\(NaOH\)) can be neutralized by
- A \(100\; \mathrm{ml}\) of \(0.1\; \mathrm{N} \;\mathrm{HCl}\)
- B \(200\; \mathrm{ml}\) of \(0.4\; \mathrm{N} \;\mathrm{HCl}\)
- C \(100\; \mathrm{ml}\) of \(0.2\; \mathrm{N} \;\mathrm{HCl}\)
- D \(200\; \mathrm{ml}\) of \(0.2\; \mathrm{N} \;\mathrm{HCl}\)
Answer & Solution
Correct Answer
(C) \(100\; \mathrm{ml}\) of \(0.2\; \mathrm{N} \;\mathrm{HCl}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{NH}_{2} \mathrm{CONH}_{2}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{NH}_{3}\) \(10\; moles \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 20 \mathrm{moles}\) Hence, \(\mathrm{NH}_{3}\) will require \(20 \mathrm{moles}\)
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