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Example · Example 21

Q.A gas has a density of 1.964 g L−11.964\ \text{g L}^{-1} at STP (0∘C0^\circ\text{C}, 1 atm1\ \text{atm}). Calculate its molar mass and identify a common gas it could be.

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The ideal gas equation, rewritten in terms of density ρ=m/V\rho = m/V, gives M=ρRTPM = \dfrac{\rho RT}{P}. At STP, T=0∘C=273 KT = 0^\circ\text{C} = 273\ \text{K} and P=1 atmP = 1\ \text{atm}. Substituting ρ=1.964 g L−1\rho = 1.964\ \text{g L}^{-1}, R=0.0821 L atm K−1mol−1R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}: M=1.964×0.0821×2731≈44.0 g mol−1M = \frac{1.964\times0.0821\times273}{1} \approx 44.0\ \text{g mol}^{-1} A molar mass of 44.0 g mol−144.0\ \text{g mol}^{-1} matches carbon dioxide, CO2\text{CO}_2 ($12.0 + 2\times16.0 = 44.0\ \text{g mol}^{ …

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