1.4 g of a gas occupies 1.23 dm3 at 1 atm and 300 K. What is its molar mass? (R = 0.0821 dm3 atm K−1 mol−1)
- Use M = mRT ÷ PV. The units are atm, dm3 and K, so R = 0.0821.
- RT = 0.0821 × 300 = 24.63.
- M = (1.4 × 24.63) ÷ (1 × 1.23) = 34.48 ÷ 1.23 = 28.0.
Answer: About 28 g mol−1, so the gas could be nitrogen, N2.