If a 0.614 g sample of a gas maintains a pressure of 238 mm Hg when contained in a 1.0 L flask at 0.0°C,what is the gas's molecular weight?

Question

Quizplus · Accepted Answer

To find the molecular weight, use the ideal gas law: PV = nRT. Convert pressure to atm (238 mm Hg = 0.313 atm) and temperature to Kelvin (0°C = 273 K). Solve for n (moles): n = PV/RT = (0.313 atm)(1.0 L) / (0.0821 L·atm/mol·K)(273 K) = 0.014 mol. Molecular weight = mass/moles = 0.614 g / 0.014 mol = 44 g/mol.

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