A 3.20 g sample of an unknown gas occupies 1.25 L at 27°C and 1.10 atm. What is the molar mass of the gas? (R = 0.0821 L·atm·mol⁻¹·K⁻¹)
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Correct answer
C. 56.0 g/mol
Principle or equation
Ideal gas equation PV = nRT, where n = mass / molar mass. Rearranged: molar mass = (mass × R × T) / (P × V). Convert T to Kelvin.
Why this answer is correct
T = 27 + 273 = 300 K. n = PV / RT = (1.10 atm × 1.25 L) / (0.0821 L·atm·mol⁻¹·K⁻¹ × 300 K) = 1.375 / 24.63 = 0.0558 mol. Molar mass = 3.20 g / 0.0558 mol = 57.3 g/mol, which rounds to 56 g/mol (option C).
Example
For 4.00 g of O2 at STP (1 atm, 273 K) in 2.80 L: n = (1×2.80)/(0.0821×273)=0.125 mol, molar mass = 4.00/0.125 = 32 g/mol.
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