Ask Question
20 September, 08:36

A 24.1-g mixture of nitrogen and carbon dioxide is found to occupy a volume of 15.1 L when measured at 870.2 mmHg and 31.2oC. What is the mole fraction of nitrogen in this mixture?

+2
Answers (1)
  1. 20 September, 09:09
    0
    Mole fraction of nitrogen = 0.52

    Explanation:

    Given dа ta:

    Temperature = 31.2 °C

    Pressure = 870.2 mmHg

    Volume = 15.1 L

    Mass of mixture = 24.1 g

    Mole fraction of nitrogen = ?

    Solution:

    Pressure conversion:

    870.2 / 760 = 1.12 atm

    Temperature conversion:

    31.2 + 273 = 304.2 K

    Total number of moles:

    PV = nRT

    n = PV/RT

    n = 1.12 atm * 15.1 L / 0.0821 L. atm. mol⁻¹. K⁻¹ * 304.2 K

    n = 16.9 L. atm. / 25 L. atm. mol⁻¹

    n = 0.676 mol

    Number of moles of nitrogen are = x

    Then the number of moles of CO₂ = 0.676 - x

    Mass of nitrogen = x mol. 28 g/mol and for CO₂ Mass = 44 g/mol (0.676 - x)

    24.1 = 28x + (29.7 - 44x)

    24.1 - 29.7 = 28x - 44x

    -5.6 = - 16 x

    x = 0.35

    Mole fraction of nitrogen:

    Mole fraction of nitrogen = moles of nitrogen / total number of moles

    Mole fraction of nitrogen = 0.35 mol / 0.676 mol

    Mole fraction of nitrogen = 0.52
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A 24.1-g mixture of nitrogen and carbon dioxide is found to occupy a volume of 15.1 L when measured at 870.2 mmHg and 31.2oC. What is the ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers