An unknown gas at 71.1 °C and 1.00 atm has a molar mass of 28.01 g/mol. What would the density of the gas be?

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Question:

An unknown gas at 71.1 °C and 1.00 atm has a molar mass of 28.01 g/mol. What would the density of the gas be?

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Answer:0.992 g/L Step-by-Step Explanation:We can use the Ideal Gas Law to calculate the density of the gas.    pV = nRT       n = m/M           Substitute for n    pV = (m/M)RT     Multiply both sides by M pVM = mRT            Divide both sides by V   pM = (m/V) RT      ρ = m/V            Substitute for m/V  pM = ρRT             Divide each side by RT     ρ = (pM)/(RT) =============== p = 1.00 atm M = 28.01 g/mol R = 0.082 06 L·atm·K⁻¹mol⁻¹ T = 71.1 °C = 344.25 K            Solve for ρ =============== ρ = (1.00 × 28.01)/(0.082 06 × 344.25) ρ = 28.01/28.25 ρ = 0.992 g/L

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