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

JEE Chemistry 2025 Question with Solution

In Dumas' method 292 mg292 \, \text{mg} of an organic compound released 50 mL50 \, \text{mL} of nitrogen gas (N2N_2) at 300 K300 \, \text{K} temperature and 715 mm Hg715 \, \text{mm Hg} pressure. The percentage composition of 'N' in the organic compound is _____ % (Nearest integer) (Aqueous tension at 300 K300 \, \text{K} = 15 mm Hg15 \, \text{mm Hg})

Answer

Correct answer:18

Step-by-step solution

Standard Method

Given: Mass of compound = 292 mg292 \, \text{mg}, volume of N2N_2 = 50 mL50 \, \text{mL}, temperature = 300 K300 \, \text{K}, total pressure = 715 mm Hg715 \, \text{mm Hg}, aqueous tension = 15 mm Hg15 \, \text{mm Hg}.

Find: Percentage of nitrogen in the organic compound.

First, calculate the pressure of dry nitrogen gas:

Pdry N2=Ptotal−aqueous tension=715 mm Hg−15 mm Hg=700 mm HgP_{\text{dry } N_2} = P_{\text{total}} - \text{aqueous tension} = 715 \, \text{mm Hg} - 15 \, \text{mm Hg} = 700 \, \text{mm Hg}

Convert the measured volume of N2N_2 to STP using the combined gas law:

VSTP=Pdry N2×Vmeasured×TSTPPSTP×Tmeasured=700×50×273760×300≈41.91 mLV_{\text{STP}} = \frac{P_{\text{dry } N_2} \times V_{\text{measured}} \times T_{\text{STP}}}{P_{\text{STP}} \times T_{\text{measured}}} = \frac{700 \times 50 \times 273}{760 \times 300} \approx 41.91 \, \text{mL}

Now calculate the moles of N2N_2 at STP:

Moles of N2=41.9122400≈0.001871 mol\text{Moles of } N_2 = \frac{41.91}{22400} \approx 0.001871 \, \text{mol}

Mass of N2N_2 is:

Mass of N2=0.001871×28≈0.05239 g=52.39 mg\text{Mass of } N_2 = 0.001871 \times 28 \approx 0.05239 \, \text{g} = 52.39 \, \text{mg}

Therefore, percentage of nitrogen in the compound is:

% N=52.39292×100≈17.94%\% \, \text{N} = \frac{52.39}{292} \times 100 \approx 17.94\%

Rounding to the nearest integer, the percentage of nitrogen is 18%18\%. Therefore, the required numerical answer is 18.

Ideal Gas Equation Method

Given: Mass of compound = 292 mg=0.292 g292 \, \text{mg} = 0.292 \, \text{g}, volume of N2N_2 = 50 mL=0.05 L50 \, \text{mL} = 0.05 \, \text{L}, temperature = 300 K300 \, \text{K}.

Find: Percentage of nitrogen in the compound.

Correct the pressure for aqueous tension:

P=715−15=700 mm Hg=700760 atm≈0.921 atmP = 715 - 15 = 700 \, \text{mm Hg} = \frac{700}{760} \, \text{atm} \approx 0.921 \, \text{atm}

Use the ideal gas equation:

PV=nRTPV = nRT

So,

n(N2)=PVRT=0.921×0.050.0821×300≈0.00187 moln(N_2) = \frac{PV}{RT} = \frac{0.921 \times 0.05}{0.0821 \times 300} \approx 0.00187 \, \text{mol}

Mass of nitrogen gas:

Mass of N2=n×M=0.00187×28=0.05236 g\text{Mass of } N_2 = n \times M = 0.00187 \times 28 = 0.05236 \, \text{g}

Now compute percentage of nitrogen:

% of N=0.052360.292×100=17.94%≈18%\% \, \text{of N} = \frac{0.05236}{0.292} \times 100 = 17.94\% \approx 18\%

Therefore, the percentage of nitrogen is 18%18\%, so the numerical answer is 18.

Common mistakes

  • Students often forget to subtract the aqueous tension from the total pressure. This is wrong because the collected gas is moist, so the observed pressure is not the pressure of dry N2N_2 alone. Always use Pdry=Ptotal−PaqueousP_{\text{dry}} = P_{\text{total}} - P_{\text{aqueous}}.

  • A common mistake is using 715 mm Hg715 \, \text{mm Hg} directly in the gas law or STP conversion. This gives more moles of nitrogen than actually present. First correct the pressure, then proceed with either the combined gas law or PV=nRTPV = nRT.

  • Some students calculate moles correctly but forget that the gas is N2N_2, not atomic NN. Using molar mass 1414 instead of 28 g/mol28 \, \text{g/mol} gives half the correct mass. Use the molar mass of the gas actually measured, which is N2N_2.

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