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

JEE Chemistry 2025 Question with Solution

In Dumas' method for estimation of nitrogen 0.4 g0.4 \, \text{g} of an organic compound gave 60 mL60 \, \text{mL} of nitrogen collected at 300 K300 \, \text{K} temperature and 715 mm Hg715 \, \text{mm Hg} pressure. The percentage composition of nitrogen in the compound is (Given : Aqueous tension at 300 K300 \, \text{K} = 15 mm Hg15 \, \text{mm Hg})

  • A

    15.71%15.71\%

  • B

    20.95%20.95\%

  • C

    17.46%17.46\%

  • D

    7.85%7.85\%

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Mass of organic compound = 0.4 g0.4 \, \text{g}, volume of nitrogen collected = 60 mL=60×10−3 L60 \, \text{mL} = 60 \times 10^{-3} \, \text{L}, temperature = 300 K300 \, \text{K}, pressure of wet nitrogen = 715 mm Hg715 \, \text{mm Hg}, aqueous tension = 15 mm Hg15 \, \text{mm Hg}.

Find: Percentage composition of nitrogen in the compound.

First, subtract the aqueous tension to get the pressure of dry nitrogen:

P=715 mm Hg−15 mm Hg=700 mm HgP = 715 \, \text{mm Hg} - 15 \, \text{mm Hg} = 700 \, \text{mm Hg}

Convert this pressure into atmosphere:

P=700760 atmP = \frac{700}{760} \, \text{atm}

Using the ideal gas equation:

PV=nRTPV = nRT

So,

n=PVRT=(700760)×(60×10−3)0.0821×300n = \frac{PV}{RT} = \frac{\left(\frac{700}{760}\right) \times (60 \times 10^{-3})}{0.0821 \times 300} n=42000×10−3760×0.0821×300=42760×24.63=4218718.8≈0.002244n = \frac{42000 \times 10^{-3}}{760 \times 0.0821 \times 300} = \frac{42}{760 \times 24.63} = \frac{42}{18718.8} \approx 0.002244

Now calculate the mass of nitrogen gas evolved:

Mass of N2=n×28=0.002244×28≈0.06283 g\text{Mass of } N_2 = n \times 28 = 0.002244 \times 28 \approx 0.06283 \, \text{g}

Hence, percentage of nitrogen in the compound is:

% Nitrogen=0.062830.4×100≈15.7075%\%\, \text{Nitrogen} = \frac{0.06283}{0.4} \times 100 \approx 15.7075\%

Rounding to two decimal places, the percentage composition of nitrogen is 15.71%15.71\%. Therefore, the correct option is A.

Stepwise Calculation

Given: Wet nitrogen is collected over water, so water vapour pressure must be subtracted from the total pressure.

Find: Percentage of nitrogen in the sample.

  1. Pressure due to nitrogen alone:
PN2=715 mm Hg−15 mm Hg=700 mm HgP_{N_2} = 715 \, \text{mm Hg} - 15 \, \text{mm Hg} = 700 \, \text{mm Hg}
  1. Convert pressure to atmosphere:
PN2=700760 atmP_{N_2} = \frac{700}{760} \, \text{atm}
  1. Convert volume to litre:
V=60×10−3 LV = 60 \times 10^{-3} \, \text{L}
  1. Apply the gas equation:
n=P×VR×T=(700760)×60×10−30.0821×300n = \frac{P \times V}{R \times T} = \frac{\left(\frac{700}{760}\right) \times 60 \times 10^{-3}}{0.0821 \times 300}
  1. This gives approximately:
n≈0.00229 moln \approx 0.00229 \, \text{mol}
  1. Mass of nitrogen gas:
mass of N2=n×28≈0.06412 g\text{mass of } N_2 = n \times 28 \approx 0.06412 \, \text{g}
  1. Percentage of nitrogen:
(0.064120.4)×100≈15.71%\left(\frac{0.06412}{0.4}\right) \times 100 \approx 15.71\%

Therefore, the percentage composition of nitrogen in the compound is 15.71%15.71\% and the correct option is A.

Common mistakes

  • Using the total pressure 715 mm Hg715 \, \text{mm Hg} directly is incorrect because the gas was collected over water. The aqueous tension must be subtracted first. Always use the dry nitrogen pressure 700 mm Hg700 \, \text{mm Hg} in the gas law.

  • Treating 60 mL60 \, \text{mL} as 60 L60 \, \text{L} is a unit-conversion error. The volume must be converted to 60×10−3 L60 \times 10^{-3} \, \text{L} before applying PV=nRTPV = nRT.

  • Using atomic mass 1414 instead of molar mass 28 g mol−128 \, \text{g mol}^{-1} for N2N_2 is wrong because the evolved gas is molecular nitrogen. Use the molar mass of N2N_2, not of a single nitrogen atom.

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