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Open the NCERT Class 11 Chemistry books and you will find two different molar volumes. The mole chapter says 22.7 L of methane reacts with 45.4 L of oxygen. The organic chapter says 22400 mL of nitrogen at STP weighs 28 g. Neither is a misprint. They use two different definitions of standard pressure. This post explains both, shows that a careful calculation gives the same answer either way, and lists the mixing errors that cause wrong answers.
Where do the two numbers come from?
Both come from pV = nRT at 0 °C (273.15 K). Only the pressure differs:
NIST lists the molar volume of an ideal gas at 273.15 K and 100 kPa (1 bar) as 22.710 954 64 L mol⁻¹.
NIST lists the molar volume at 273.15 K and 101.325 kPa (1 atm) as 22.413 969 54 L mol⁻¹.
The standard atmosphere is exactly 101 325 Pa, so 1 atm is 1.01325 bar. Divide the two volumes and you get 22.711 / 22.414 = 1.01325, the same ratio, because volume is inversely proportional to pressure. The gap is about 1.3%.
Which one does NCERT use, and why?
The current Class 11 chapter Some Basic Concepts of Chemistry (2026-27 reprint) uses the 1 bar value. Its stoichiometry section says "22.7 L of CH₄(g) reacts with 45.4 L of O₂(g) to give 22.7 L of CO₂(g)".
The reason was spelt out in NCERT's earlier States of Matter chapter. An archived 2021-22 copy defines STP as "273.15 K (0°C) temperature and 1 bar (i.e., exactly 10⁵ pascal) pressure", with a molar volume of 22.71098 L mol⁻¹. A footnote adds that the older 1 atm standard, giving 22.413996 L, "is still often used, and applies to all chemistry data more than decade old". That chapter is not in the current rationalised Part I, whose contents run from Unit 1 to Unit 6 (Equilibrium) with no States of Matter unit, but the 22.7 L figure stayed in Unit 1.
The Class 11 organic chemistry chapter still uses the older convention in the Dumas method. It converts the nitrogen volume to STP with 760 mm of mercury and then uses "22400 mL N₂ at STP weighs 28 g".
Worked example 1: NCERT's Dumas problem, both ways
Problem 8.21: 0.3 g of compound gives 50 mL of nitrogen at 300 K and 715 mm, with aqueous tension 15 mm, so the gas pressure is 700 mm.
Using 1 atm (760 mm) and 22400 mL:
V at STP = 273 × 700 × 50 / (300 × 760) = 41.9 mL.
Mass of N₂ = 28 × 41.9 / 22400 = 0.0524 g.
%N = 0.0524 / 0.3 × 100 = 17.46%, NCERT's answer.
Using 1 bar (750.06 mm) and 22711 mL:
1 bar = 760 / 1.01325 = 750.06 mm. V at STP = 273.15 × 700 × 50 / (300 × 750.06) = 42.49 mL.
Mass of N₂ = 28 × 42.49 / 22711 = 0.0524 g.
%N = 17.46%. The same answer.
The gas is the same amount whichever "standard" you convert to, so a consistent calculation cannot change the result.
Mixing them: convert with 760 mm (41.9 mL) but divide by 22711 mL, and you get 28 × 41.9 / 22711 / 0.3 × 100 = 17.22%. That 0.24-point error is the mixing mistake.
Worked example 2: NCERT's welding gas (exercise 1.34)
A hydrocarbon burns to give 3.38 g CO₂ and 0.690 g H₂O, and 10.0 L at STP weighs 11.6 g.
Carbon: 3.38 × 12/44 = 0.922 g, or 0.0768 mol. Hydrogen: 0.690 × 2/18 = 0.0767 g, or 0.0761 mol. The ratio is 1 : 1, so the empirical formula is CH.
Molar mass with 22.7 L: 11.6 × 22.7 / 10.0 = 26.3 g mol⁻¹. With 22.4 L: 26.0 g mol⁻¹.
CH has a mass of 13, so n = 2 either way, and the gas is C₂H₂.
Here the 1.3% gap does not change the formula, because n must be a whole number.
Worked example 3: 4.4 g of CO₂
4.4 g / 44 g mol⁻¹ = 0.100 mol. At 1 bar that is 2.27 L. At 1 atm it is 2.24 L. If the answer options include both, read the conditions in the question carefully.
Common mistakes and how to avoid them
Mixing conventions inside one problem, as in the 17.22% result above.
Using 22.4 L at room temperature. The archived NCERT footnote gives 24.789 L mol⁻¹ at 298.15 K and 1 bar (SATP). Better still, use pV = nRT.
Using molar volume for liquids or solids. It applies only to gases, and ideal ones at that.
Assuming one value is "wrong". Both are correct for their own pressure.
A practice routine
Rework NCERT Problem 8.21 both ways until you can do it without notes.
Solve exercise 1.34 with each molar volume and confirm the formula does not change.
Write five gas-volume questions of your own, each stating a different pressure, and solve them with pV = nRT.
Read our posts on ideal gas problems and common mole concept errors for more practice.
If unit conversions are where your answers drift, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can check each conversion line by line.
Redo the Dumas problem now, once at 1 atm and once at 1 bar, and confirm that both give 17.46%.
