The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by :
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- 200 ml of 0.02 N HCl
- 100 ml of 0.2 N HCl
- 100 ml of 0.1 HCl
- 200 ml of 0.4 N HCl
The Correct Option is B
Solution and Explanation
Step 2: \(NH_2CONH_2+2NaOH\to Na_2CO_3+2NH_3\), so NH3 formed \(=2\times0.01=0.02\) mol.
Step 3: NH3 + HCl → NH4Cl (1:1), so 0.02 mol HCl (20 meq) is required.
Step 4: Check the options: (A) 200 mL x 0.02 N = 4 meq; (B) 100 mL x 0.2 N = 20 meq; (C) 100 mL x 0.1 = 10 meq; (D) 200 mL x 0.4 N = 80 meq. Only (B) gives 20 meq.
Answer: (B) 100 mL of 0.2 N HCl
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