1. Find initial moles of sulphuric acid

Given:

• Volume of $\mathrm{H_2SO_4} = 100\ \mathrm{mL} = 0.100\ \mathrm{L}$
• Molarity $= 0.1\ \mathrm{M}$

So, moles of $\mathrm{H_2SO_4}$ initially taken: $n(\mathrm{H_2SO_4}) = M \times V = 0.1 \times 0.100 = 0.010\ \text{mol}$

2. Find moles of NaOH used for neutralizing excess acid

Given:

• Volume of $\mathrm{NaOH} = 20\ \mathrm{mL} = 0.020\ \mathrm{L}$
• Molarity $= 0.5\ \mathrm{M}$

So, $n(\mathrm{NaOH}) = 0.5 \times 0.020 = 0.010\ \text{mol}$

Since neutralization is: $\mathrm{H_2SO_4 + 2NaOH \to Na_2SO_4 + 2H_2O}$

Therefore, moles of excess $\mathrm{H_2SO_4}$ are: $n(\mathrm{H_2SO_4,\ excess}) = \frac{0.010}{2} = 0.005\ \text{mol}$

3. Find moles of sulphuric acid consumed by ammonia

$n(\mathrm{H_2SO_4,\ consumed}) = 0.010 - 0.005 = 0.005\ \text{mol}$

Reaction of ammonia with sulphuric acid: $2\mathrm{NH_3} + \mathrm{H_2SO_4} \to (\mathrm{NH_4})_2\mathrm{SO_4}$

Hence moles of ammonia evolved: $n(\mathrm{NH_3}) = 2 \times 0.005 = 0.010\ \text{mol}$

4. Find mass of nitrogen in the compound

Each mole of $\mathrm{NH_3}$ contains 1 mole of nitrogen atoms.

So moles of nitrogen: $n(\mathrm{N}) = 0.010\ \text{mol}$

Mass of nitrogen: $m(\mathrm{N}) = 0.010 \times 14 = 0.14\ \mathrm{g}$

5. Calculate percentage of nitrogen in the organic compound

Mass of compound taken = $0.30\ \mathrm{g}$

$\%N = \frac{0.14}{0.30} \times 100 = 46.67\%$

6. Match with given compounds

• Urea, $\mathrm{NH_2CONH_2}$

  • Molar mass $= 60$
  • Nitrogen mass $= 28$
  • $\%N = \frac{28}{60} \times 100 = 46.67\%$

• Benzamide, $\mathrm{C_6H_5CONH_2}$

  • Molar mass $= 121$
  • $\%N = \frac{14}{121} \times 100 \approx 11.57\%$

• Acetamide, $\mathrm{CH_3CONH_2}$

  • Molar mass $= 59$
  • $\%N = \frac{14}{59} \times 100 \approx 23.73\%$

• Thiourea, $\mathrm{NH_2CSNH_2}$

  • Molar mass $= 76$
  • $\%N = \frac{28}{76} \times 100 \approx 36.84\%$

Thus the compound is urea.

7. Comparison with stored answer

Derived answer: A: urea

Stored correct answer: A

They match.