Molar volume at NTP – volume occupied by 1 g-mole of an ideal gas At normal temperature and pressure (0 °C and 760 mm Hg), the molecular mass expressed in grams (1 g-mole) of an ideal gas occupies a volume of:

Introduction / Context:
The concept of molar volume connects macroscopic gas volumes to the amount of substance. At widely used reference conditions (NTP/STP in classical teaching), the molar volume of an ideal gas takes a standard value that simplifies quick estimates.

Given Data / Assumptions:

• NTP: T = 273 K, p ≈ 1 atm = 760 mm Hg.
• Ideal-gas behavior.
• Amount of substance = 1 g-mole (i.e., 1 mole when molecular mass is expressed in grams).

Concept / Approach:
From the ideal-gas law, V = nRT/p. Substituting n = 1 mol, R = 0.082057 L·atm/(mol·K), T = 273 K, and p = 1 atm gives V ≈ 22.4 L. This value is used pervasively in stoichiometry and gas-law problems at these reference conditions.

Step-by-Step Solution:
Write V = nRT/p.Insert n = 1, R = 0.082057 L·atm/(mol·K), T = 273 K, p = 1 atm.Compute V ≈ 0.082057 * 273 / 1 ≈ 22.4 L.Select 22.4 litres as the correct molar volume at NTP.

Verification / Alternative check:
Using S.I. units: R = 8.314 kJ/(kmol·K), p = 101.325 kPa, T = 273.15 K, yields V ≈ 22.414 m^3/kmol, i.e., 22.414 L/mol, consistent with 22.4 L for 1 mol.

Why Other Options Are Wrong:

• 0.224 L and 2.24 L: off by factors of 100 and 10 respectively.
• 224 L: off by a factor of 10 too high.
• 11.2 L: corresponds to half a mole at NTP, not one mole.

Common Pitfalls:
Confusing updated metrological definitions of STP/NTP with the classical value used in many exam problems; unless otherwise specified, 22.4 L is expected.

Final Answer:
22.4 litres