QJEEM04Ch2771: JEE Main 2004 Chemistry - Class 11 JEEM04Ch2771 Solution: The ammonia evolved from the treatment of 0.30 g of an organic…

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.