JEE Main · 2025 · Shift-ImediumMOLE-221

At the sea level, the dry air mass percentage composition is given as nitrogen gas: 70.0, oxygen gas: 27.0 and argon…

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

Question

At the sea level, the dry air mass percentage composition is given as nitrogen gas: 70.0, oxygen gas: 27.0 and argon gas: 3.0. If total pressure is 1.15 atm, then calculate the ratio of followings respectively:

(i) partial pressure of nitrogen gas to partial pressure of oxygen gas

(ii) partial pressure of oxygen gas to partial pressure of argon gas

(Given: Molar mass of N, O and Ar are 14, 16, and 40 g mol−1^{-1} respectively)

Options
  1. a

    4.26, 19.3

  2. b

    2.59, 11.85

  3. c

    5.46, 17.8

  4. d

    2.96, 11.2

    ✓
Correct Answerd

2.96, 11.2

Detailed Solution

🧠 Mole fraction = pressure fraction At a fixed total pressure, partial pressure ratios equal mole ratios. So convert the mass percentages to moles, then take the requested ratios.

🗺️ Step by step Per 100 g100\,\text{g} of dry air:

  • Moles of \ceN2\ce{N2} =70/28=2.5= 70/28 = 2.5.
  • Moles of \ceO2\ce{O2} =27/32=0.844= 27/32 = 0.844.
  • Moles of Ar =3/40=0.075= 3/40 = 0.075.

Ratio (i): \ceN2/\ceO2=2.5/0.844≈2.96\ce{N2}/\ce{O2} = 2.5/0.844 \approx 2.96. Ratio (ii): \ceO2/Ar=0.844/0.075≈11.25\ce{O2}/\text{Ar} = 0.844/0.075 \approx 11.25.

Closest match is option (d).

⚠️ Common mistake Don't use the mass ratio 70/27=2.5970/27 = 2.59 directly. Mass ratios are not pressure ratios — you must divide by molar masses first.

Answer: (d)\boxed{\text{Answer: (d)}}

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