The ammonia (NH$$_3$$) released on quantitative reaction of 0.6 g urea (NH$$_2$$CONH$$_2$$) with sodium hydroxide (NaOH) can be neutralized by

First we recall the balanced chemical reaction between urea and aqueous sodium hydroxide, which furnishes ammonia as one of the products:

$$$\mathrm{NH_2CONH_2 + 2\,NaOH \;\longrightarrow\; Na_2CO_3 + 2\,NH_3}$$$

This equation tells us that

$$$1\;\text{mol urea} \;\xrightarrow{\phantom{2\,NaOH}}\; 2\;\text{mol NH}_3$$$

Now we calculate how many moles of urea are present in the given 0.6 g sample. The molar mass of urea is obtained by adding the individual atomic masses:

$$$M(\mathrm{NH_2CONH_2}) = 2\times14 \;(\text N) + 4\times1 \;(\text H) + 12 \;(\text C) + 16 \;(\text O) = 28 + 4 + 12 + 16 = 60\;\text{g mol}^{-1}$$$

Hence,

$$$n(\text{urea}) = \frac{m}{M} = \frac{0.6\;\text g}{60\;\text{g mol}^{-1}} = 0.01\;\text{mol}$$$

Using the stoichiometric ratio obtained earlier, the moles of ammonia liberated are

$$$n(\mathrm{NH_3}) = 2 \times n(\text{urea}) = 2 \times 0.01 = 0.02\;\text{mol}$$$

Next, ammonia is a base that reacts with hydrochloric acid according to the simple 1 : 1 neutralisation equation,

$$\mathrm{NH_3 + HCl \;\longrightarrow\; NH_4Cl}$$

Therefore, the number of moles (or equivalents, because HCl is monoprotic) of HCl needed equals the moles of NH$$_3$$ present:

$$n(\mathrm{HCl\;required}) = 0.02\;\text{mol}$$

Let us now examine each option by converting its volume and normality into moles of HCl supplied, using the relation $$n = N \times V$$ with $$V$$ in litres.

For Option A: $$$n = 0.4\;\text N \times 0.200\;\text L = 0.080\;\text{mol}$$$

For Option B: $$$n = 0.2\;\text N \times 0.200\;\text L = 0.040\;\text{mol}$$$

For Option C: $$$n = 0.2\;\text N \times 0.100\;\text L = 0.020\;\text{mol}$$$

For Option D: $$$n = 0.1\;\text N \times 0.100\;\text L = 0.010\;\text{mol}$$$

The requirement is 0.02 mol, and only Option C furnishes exactly this amount of acid.

Hence, the correct answer is Option C.