JEE Main · 2025 · Shift-IIhardMOLE-212

In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K…

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is: (Aqueous tension at 300 K = 15 mm Hg)

Options
  1. a

    1.257

  2. b

    20.87

  3. c

    18.67

  4. d

    12.57

Correct Answerd

12.57

Detailed Solution

🧠 Same Dumas trick, different sample mass Just like Dumas with any sample — subtract aqueous tension first, find moles of \ceN2\ce{N2} from PV=nRTPV = nRT, then take the percentage in the original 0.5g0.5\,\text{g} sample.

🗺️ Step by step Dry pressure =71515=700mm Hg=700760atm= 715 - 15 = 700\,\text{mm Hg} = \frac{700}{760}\,\text{atm}.

Moles of \ceN2\ce{N2}: n=(700/760)(0.060)0.0821×3002.244×103mol.n = \frac{(700/760)(0.060)}{0.0821 \times 300} \approx 2.244 \times 10^{-3}\,\text{mol}.

Mass of N =2.244×103×280.0628g= 2.244 \times 10^{-3} \times 28 \approx 0.0628\,\text{g}.

%N=0.06280.50×10012.57%.\%\,\text{N} = \frac{0.0628}{0.50} \times 100 \approx 12.57\%.

⚠️ Common mistake This is exactly MOLE-208 with sample mass 0.5g0.5\,\text{g} instead of 0.4g0.4\,\text{g}. Same setup, just a bigger denominator — so %\%N is smaller.

Answer: (d)\boxed{\text{Answer: (d)}}

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