Air is a mixture of 78.1 percentage Nitrogen (N_2), 20.95 percentage Oxygen (O_2

Question

Air is a mixture of 78.1 percentage Nitrogen (N_2), 20.95 percentage Oxygen (O_2), and 0.9 percentage Argon (Ar), and 0.05% carbon dioxide (CO_2). What is the mole fraction of nitrogen in air? 22.6 On a given day, the atmospheric pressure of the air in Utah was 0.895 atm. What was the partial pressure of nitrogen in this air? In Death Valley, temperatures can reach up to 53.9 degree C, with air pressures as high as 767.6 mmHg. How many grams of oxygen gas are in 1.35 L of this air, assuming the composition given in part 1? 21

Explanation / Answer

Mole fraction of nitrogen=

Total percent of air= 78.1+ 20.95+ 0.9 +0.05= 100

Fraction of nitrogen= 0.781

Mole fraction of nitrogen= 0.781 * 28= 21.868

We know that air contains 78% of Nitrogen (ideal behavior)

This means, Mole fraction of nitrogen is (78100) = 0.78

Therefore,

Partial pressure of Nitrogen in air is = 0.895* 0.78= 0.698 atm.

Temperature,

53.9 degrees C + 273 = 326.9 K

767.6 / 760 mmHg = 1.01 atm

PV=nRT

1.01 atm x 1.35 L = n ( .08206 Latm / molK) x 326.9K

n = 0.05083 moles

In grams= 0.05083*15.99= 0.813 g

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