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 mole fraction of nitrogen in this mixture?

Answer 1

Answer:

Mole fraction of nitrogen =   0.52

Explanation:

Given data:

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