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Reviewed CSCA Chemistry question · Standard

A 0.500 g sample of an unknown gas occupies 0.250 L at 27°C and 1.20 atm. What is the molar mass of the gas? (R = 0.0821 L·atm·mol⁻¹·K⁻¹)

  1. 41.0 g/mol
  2. 20.5 g/mol
  3. 82.0 g/mol
  4. 10.3 g/mol
Show the answer and explanation

Correct answer

A. 41.0 g/mol

Principle or equation

The ideal gas equation is PV = nRT, where n = m/M. Rearranging gives M = mRT/(PV). Temperature must be in kelvin.

Why this answer is correct

Convert 27°C to 300 K. Use M = mRT/(PV) = (0.500 g)(0.0821 L·atm·mol⁻¹·K⁻¹)(300 K) / (1.20 atm × 0.250 L) = (12.315) / (0.300) = 41.05 g/mol, approximately 41.0 g/mol.

Example

If 0.500 g of gas occupies 0.250 L at 300 K and 1.20 atm, the molar mass is 41.0 g/mol.

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