Initially total H₂SO₄ present = 100 mL of 0.1 M = $${{{100} \over {1000}} \times 0.1}$$ mole = 0.01 mole

2NaOH + H₂SO₄ $$ \to $$ Na₂SO₄ + 2H₂O
Let in this reaction H₂SO₄ required n mole

$$\therefore$$ $${{{{20} \over {1000}} \times 0.5} \over 2}$$ = $${n \over 1}$$

$$ \Rightarrow n$$ = 0.005 mole

So now remaining H₂SO₄ = 0.01 - 0.005 = 0.005 mole . Those remaining H₂SO₄ will react with NH₃.

2NH₃ + H₂SO₄ $$ \to $$ (NH₄)₂SO₄
Let no moles of NH₃ produce through this reaction is = $$x$$

$$\therefore$$ $${x \over 2}$$ = $${{0.005} \over 1}$$

$$ \Rightarrow x$$ = 0.01

In NH₃ no of N atom is 1 and H atom is 3. So no of moles of N atom in NH₃ = 0.01$$ \times $$1 = 0.01 mole

This 0.01 mole or 0.01$$ \times $$14 gm N is produced from 0.3 gm unknown organic compound.

$$\therefore$$ % of N in unknown compound is

= $${{0.01 \times 14} \over {0.3}} \times 100$$

= 46.6

% of N in urea [(NH₄)₂CO] = $${{14 \times 2} \over {60}} \times 100$$ = 46.6 %
[ Mol weight of urea = 60]

% of N in benzamide [C₆H₅CONH₂] = $${14 \over {121}} \times 100$$ = 11.5 %
[ Mol weight of benzamide [C₆H₅CONH₂] = 121]

% of N in acetamide [CH₃CONH₂] = $${14 \over {59}} \times 100$$ = 23.4 %
[ Mol weight of acetamide [CH₃CONH₂] = 59]

% of N in thiourea [NH₂CONH₂] = $${{14 \times 2} \over {76}} \times 100$$ = 36.8 %
[ Mol weight of thiourea [NH₂CSNH₂] = 76]

$$\therefore$$ compound is urea.