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NVAEasyJEE 2025Quantitative Analysis (C, H, N…)

JEE Chemistry 2025 Question with Solution

In Dumas' method for estimation of nitrogen 1 g1 \, \text{g} of an organic compound gave 150 mL150 \, \text{mL} of nitrogen collected at 300 K300 \, \text{K} temperature and 900 mm Hg900 \, \text{mm Hg} pressure. The percentage composition of nitrogen in the compound is _____ % (nearest integer).

(Aqueous tension at 300 K300 \, \text{K} = 15 mm Hg15 \, \text{mm Hg})

Answer

Correct answer:20

Step-by-step solution

Standard Method

Given: Mass of organic compound = 1 g1 \, \text{g}, volume of N2\mathrm{N_2} collected = 150 mL=0.150 L150 \, \text{mL} = 0.150 \, \text{L}, temperature = 300 K300 \, \text{K}, total pressure = 900 mm Hg900 \, \text{mm Hg}, aqueous tension = 15 mm Hg15 \, \text{mm Hg}.

Find: Percentage composition of nitrogen in the compound.

First, calculate the pressure of dry nitrogen gas:

PN2=900−15=885 mm HgP_{\mathrm{N_2}} = 900 - 15 = 885 \, \text{mm Hg}

Converting into atmosphere:

PN2=885760 atm≈1.164 atmP_{\mathrm{N_2}} = \frac{885}{760} \, \text{atm} \approx 1.164 \, \text{atm}

Using the ideal gas law:

PV=nRTPV = nRT

So,

n=PVRT=1.164×0.1500.0821×300n = \frac{PV}{RT} = \frac{1.164 \times 0.150}{0.0821 \times 300} n≈0.0071 moln \approx 0.0071 \, \text{mol}

Now calculate the mass of nitrogen gas:

Mass of N2=0.0071×28.02 g\text{Mass of } \mathrm{N_2} = 0.0071 \times 28.02 \, \text{g} Mass of N2≈0.198762 g\text{Mass of } \mathrm{N_2} \approx 0.198762 \, \text{g}

Percentage of nitrogen in the compound:

% N=0.1987621×100≈19.8762%\%\,\text{N} = \frac{0.198762}{1} \times 100 \approx 19.8762\%

Nearest integer:

20%20\%

Therefore, the percentage composition of nitrogen is 20%20\%.

The final numerical answer is 20.

Direct Calculation

Given: Dry nitrogen pressure = 885 mm Hg885 \, \text{mm Hg}, volume = 0.150 L0.150 \, \text{L}, temperature = 300 K300 \, \text{K}.

Find: Percentage of nitrogen.

Compute moles of nitrogen directly from the corrected pressure and then convert to mass:

n=(885760)×0.1500.0821×300≈0.0071 moln = \frac{\left(\frac{885}{760}\right) \times 0.150}{0.0821 \times 300} \approx 0.0071 \, \text{mol} Mass of nitrogen=0.0071×28≈0.1988 g\text{Mass of nitrogen} = 0.0071 \times 28 \approx 0.1988 \, \text{g} % N=0.19881×100≈19.85%≈20%\%\,\text{N} = \frac{0.1988}{1} \times 100 \approx 19.85\% \approx 20\%

Therefore, the answer is 20.

Common mistakes

  • Subtracting no aqueous tension from the total pressure is incorrect because the collected gas is moist. Always use dry gas pressure: 900−15=885 mm Hg900 - 15 = 885 \, \text{mm Hg} before applying the gas law.

  • Using 150 mL150 \, \text{mL} directly in the ideal gas equation is wrong because R=0.0821R = 0.0821 is in L atm mol−1K−1\text{L atm mol}^{-1} \text{K}^{-1}. Convert volume to 0.150 L0.150 \, \text{L} first.

  • Forgetting to convert pressure from mm Hg\text{mm Hg} to atm\text{atm} leads to inconsistent units in PV=nRTPV=nRT. Use P=885760 atmP = \frac{885}{760} \, \text{atm}.

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