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22 May, 08:46

A mixture containing nitrogen and hydrogen weighs 3.48 g and occupies a volume of 7.47 L at 296 K and 1.02 atm. Calculate the mass percent of these two gases. Assume ideal-gas behavior.

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  1. 22 May, 09:02
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    there is 2% of hydrogen and 98% of nitrogen (mass percent)

    Explanation:

    assuming ideal gas behaviour

    P*V=n*R*T

    n = P*V / (R*T)

    where P = pressure=1.02 atm, V=volume=7.47 L, T=absolute temperature = 296 K and R = ideal gas constant = 0.082 atm*L / (mole*K)

    thus

    n = P*V / (R*T) = 1.02 atm*7.47 L / (296 K * 0.082 atm*L / (mole*K)) = 0.314 moles

    since the number of moles is related with the mass m through the molecular weight M

    n=m/M

    thus denoting 1 as hydrogen and 2 as nitrogen

    m₁+m₂ = mt (total mass)

    m₁/M₁+m₂/M₂ = n

    dividing one equation by the other and denoting mass fraction w₁ = m₁/mt, w₂ = m₂/mt, w₂ = 1 - w₁

    w₁/M₁+w₂/M₂ = n/mt

    w₁/M₁ + (1-w₁) / M₂ = n/mt

    w₁ * (1/M₁ - 1/M₂) + 1/M₂ = n/mt

    w₁ = (n/mt - 1/M₂) / (1/M₁ - 1/M₂)

    replacing values

    w₁ = (n/mt - 1/M₂) / (1/M₁ - 1/M₂) = (0.314 moles/3.48 g - 1 / (14 g/mole)) / (1 / (1 g/mole) - 1 / (14 g/mole)) = 0.02 (%)

    and w₂ = 1-w₁ = 0.98 (98%)

    thus there is 2% of hydrogen and 98% of nitrogen
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