A 0.900 g sample of an unknown gas occupies 0.500 L at 127°C and 0.820 atm. What is the molar mass of the gas? (R = 0.0821 L·atm·mol⁻¹·K⁻¹)
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Correct answer
A. 72.0 g/mol
Principle or equation
The ideal gas equation PV = nRT can be rearranged to n = PV/(RT). The molar mass M = mass / n.
Why this answer is correct
Convert temperature to kelvin: T = 127 + 273.15 = 400.15 K (approximately 400 K). n = (0.820 atm × 0.500 L) / (0.0821 L·atm·mol⁻¹·K⁻¹ × 400.15 K) = 0.410 / 32.85 ≈ 0.01248 mol. M = 0.900 g / 0.01248 mol ≈ 72.1 g/mol, which rounds to 72.0 g/mol.
Example
For a 1.00 g gas at 1.00 atm, 1.00 L, 273 K, n = 0.0446 mol, M = 22.4 g/mol.
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