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JEE Main 2022 (27 Jun, Shift 1) · Gas Laws & Ideal Gas Equation

Gas Laws & Ideal Gas Equation question from JEE Main 2022 (27 Jun, Shift 1)

Q56JEE Main 2022 · 27 JUN · SHIFT 1Medium

Gas Laws & Ideal Gas Equation

A mixture of hydrogen and oxygen has volume 2000 cm3, temperature 300 K, pressure 100 kPa and mass 0.76 g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be:

[Take gas constant R = 8.3 JK−1{}^{-1}mol−1{}^{-1}]

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Answer: (b)

  1. Total moles n=PVRT=105×2×10−38.3×300=0.0803n = \frac{PV}{RT} = \frac{10^5 \times 2 \times 10^{-3}}{8.3 \times 300} = 0.0803 mol.
  2. Let xx = moles of H₂, yy = moles of O₂: x+y=0.0803x + y = 0.0803, 2x+32y=0.762x + 32y = 0.76.
  3. Solving: 30y=0.76−0.161⇒y≈0.02030y = 0.76 - 0.161 \Rightarrow y \approx 0.020, x≈0.060x \approx 0.060.
  4. xy=31\frac{x}{y} = \frac{3}{1}, so the answer is (b).
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