A Levels Chemistry (9701)•9701/13/M/J/20
Explanation
Molar volume of ideal gas at room conditions Steps:
- Room conditions mean 25°C and 1 atm (RTP), with molar volume of 24 dm³ mol⁻¹ for gases.
- For A, helium volume is 24 dm³, so moles n = V / 24 dm³ mol⁻¹ = 1 mol.
- For B, CaCO₃ molar mass is 100 g mol⁻¹, so n = 100 g / 100 g mol⁻¹ = 1 mol, but mass-based, not volume under RTP.
- For C, H₂SO₄ volume is 0.4 dm³, so n = 0.250 mol dm⁻³ × 0.4 dm³ = 0.1 mol.
Why A is correct:
- At RTP, 24 dm³ of any ideal gas contains exactly 1 mole by definition of molar volume.
Why the others are wrong:
- B: Mass gives 1 mol CaCO₃, but does not use RTP volume measurement for the substance.
- C: Solution contains 0.1 mol H₂SO₄ per concentration-volume formula.
- D: Not enough information.
Final answer: A
Topic: Atoms, molecules and stoichiometry
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