JEE Main 2004 — Some Basic Concepts Of Chemistry Question with Solution

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

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

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

$$\Rightarrow n$$ = 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 = $$x$$

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

$$\Rightarrow x$$ = 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

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

= 46.6

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

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

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

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

$$\therefore$$ compound is urea.