Question

1.A mixture of argon and xenon gases contains argon at a partial pressure of 450 mm Hg and xenonat a partial pressure of 523 mm Hg. What is the mole fraction of each gas in the mixture?

X_Ar =
X_Xe =

2.
A mixture of nitrogen and oxygen gases at a total pressure of 912 mm Hg contains nitrogen at a partial pressure of 340 mm Hg. If the gas mixture contains 3.95 grams of nitrogen, how many grams of oxygen are present?

_____g O₂

_3.

The stopcock connecting a 3.99 L bulb containing nitrogen gas at a pressure of 8.46 atm, and a 7.52L bulb containing oxygen gas at a pressure of 2.15 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is_____ atm.

Answer 1

1.

Partial Pressure of Argon = 450 mm Hg

Partial Pressure of Xenon = 523 mm Hg

Total pressure of the mixture = 523 + 450 = 973 mm Hg

Mole fraction of Argon = 450/973 = 0.46

Mole fraction of Xenon = 523/973 = 0.54

2.

Total pressure of the mixture = 912 mm Hg

Partial Pressure of Nitrogen = 340 mm Hg

Partial Pressure of Oxygen = 912-340 mm Hg = 572 mm Hg

Mole fraction of Nitrogen = 340/912 = 0.37

Mole fraction of Oxygen = 572/912 = 0.63

No. of moles of Nitrogen = mass/molar mass = 3.95/28 = 0.14

Mole fraction of Nitrogen = No. of moles of Nitrogen/Total no. of moles

0.37 = 0.14/(0.14+x)      where x is no. of moles of oxygen

0.0518 + 0.37x = 0.14

x = 0.24

Mass of oxygen = no. of moles of oxygen molar mass of oxygen = 0.24 32 = 7.63 g O₂

3.

Let the temperature be 300 K.

Let 1 be nitrogen gas and 2 be oxygen gas

P₁V₁ = n₁RT

8.46 3.99 = n₁ 0.0821 300

n₁ = 1.37

P₂V₂ = n₂RT

2.15 7.52 = n₂ 0.0821 300

n₂ = 0.656

P_totalV_total = n_totalRT

P_total (7.52 + 3.99) = (1.37 + 0.656) 0.0821 300

P_total = 4.34 atm