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?

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