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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?

1 Answer

1 vote

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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