Difficulty: Easy
Correct Answer: 22.4 litres
Explanation:
Introduction / Context:The molar volume at S.T.P. is a benchmark figure for gas calculations, quick estimates of moles from volumes, and stoichiometry in reaction engineering. Though different standards exist (e.g., IUPAC’s 0°C, 1 bar), many textbooks still reference 22.414 L/mol at 0°C and 1 atm.
Given Data / Assumptions:
Concept / Approach:Using PV = nRT with n = 1 mol, P = 1 atm, T = 273.15 K, and R = 0.082057 L·atm·mol^-1·K^-1 gives V ≈ 22.414 L. This is typically rounded to 22.4 L for quick work.
Step-by-Step Solution:
Step 1: Write PV = nRT.Step 2: Substitute P = 1 atm, n = 1 mol, R = 0.082057 L·atm·mol^-1·K^-1, T = 273.15 K.Step 3: Compute V ≈ 22.414 L.Step 4: Round appropriately to ~22.4 L.Verification / Alternative check:If using 1 bar instead of 1 atm, the molar volume is ~22.71 L/mol. The answer here follows the 1 atm convention consistent with many problem sets.
Why Other Options Are Wrong:
Common Pitfalls:Mixing definitions of S.T.P. or mixing units (L, mL, c.c.). Always check the stated standard before using the molar volume constant.
Final Answer:22.4 litres
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