The ammonia (NH₃) released on quantitative reaction of 0.6 g urea (NH₂CONH₂) with sodium hydroxide (NaOH) can be neutralized by :

Solution

Chemistry ma’am · 3 steps

One urea gives two ammonia, so double the urea moles and then match milliequivalents of HCl.

1. Moles of urea. The molar mass of urea NH₂CONH₂ is 60 g mol⁻¹. So 0.6 g is 0.6/60 = 0.01 mol.

   NH₂CONH₂ + 2NaOH → Na₂CO₃ + 2NH₃

2. Ammonia released. Each urea has two N atoms, so it releases two NH₃. That gives 2 × 0.01 = 0.02 mol of NH₃, which is 20 milliequivalents of base. It needs exactly 20 meq of HCl.

   Option	meq of HCl = N × V (mL)	Matches 20 meq?
   (a)	0.02 × 200 = 4	no
   (b)	0.2 × 100 = 20	yes
   (c)	0.1 × 100 = 10	no, half
   (d)	0.4 × 200 = 80	no, four times too much

3. So. Only 100 mL of 0.2 N HCl supplies exactly 20 meq, so the answer is (b).

The idea to keep. For neutralisation, convert everything to milliequivalents (N × V in mL) and match them. Always check how many NH₃ one molecule of the compound gives.

Answer from the NTA final answer key. Solution by PrepWiser.