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Elden
27 days ago
13

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. Wh

at is the mole fraction of nitrogen in this mixture?
Chemistry
1 answer:
eduard [2.6K]27 days ago
3 0

Answer:

The mole fraction of nitrogen is   0.52

Explanation:

Provided information:

Temperature =  31.2 °C

Pressure = 870.2 mmHg

Volume = 15.1 L

Mass of the mixture = 24.1 g

Mole fraction of nitrogen =?

Process:

Converting pressure:

870.2 / 760 = 1.12 atm

Converting temperature:

31.2 + 273 = 304.2 K

Total moles calculation:

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

Let x be the number of moles of nitrogen.

Thus, the moles of CO₂ = 0.676 - x

The mass of nitrogen = x mol × 28 g/mol, and for CO₂ the mass = 44 g/mol (0.676 - x)

We have 24.1  = 28x + (29.7 - 44x)

Rearranging gives: 24.1 - 29.7  =  28x  - 44x

-5.6 = -16 x

Therefore, x = 0.35

Mole fraction of nitrogen:

Mole fraction of nitrogen = moles of nitrogen / total moles

Mole fraction of nitrogen =   0.35  mol / 0.676 mol

Mole fraction of nitrogen =   0.52

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Explanation:

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