Question

1/ 0.987 mol sample of xenon gas at a temperature of 19.0 °C is found to occupy a volume of 28.2 liters. The pressure of this gas sample is .... mm Hg.

2/ A sample of nitrogen gas collected at a pressure of 477 mm Hg and a temperature of 278 K has a mass of 20.0 grams. The volume of the sample is ...... L.

3/ A 4.97 gram sample of carbon dioxide gas has a volume of 856 milliliters at a pressure of 3.92 atm. The temperature of the CO₂ gas sample is ...... °C.

4/ A mixture of xenon and hydrogen gases, in a 9.38 L flask at 77 °C, contains 11.1 grams of xenon and 0.436 grams of hydrogen. The partial pressure of hydrogen in the flask is

(......) atm and the total pressure in the flask is (.......) atm.

5/  A mixture of nitrogen and neon gases is maintained in a 5.23 L flask at a pressure of 3.02 atm and a temperature of 59 °C. If the gas mixture contains 8.18 grams of nitrogen, the number of grams of neon in the mixture is ...... g.

6/ The stopcock connecting a 3.15 L bulb containing carbon dioxide gas at a pressure of 9.69 atm, and a 4.15 L bulb containing xenon gas at a pressure of 2.59 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is .... atm.

7/ A mixture of xenon and hydrogen gases, at a total pressure of 647 mm Hg, contains 10.3 grams of xenon and 0.403 grams of hydrogen. What is the partial pressure of each gas in the mixture?

P_Xe = (.......) mm Hg
P_H2 = (.......) mm Hg

8/ A mixture of nitrogen and neon gases contains nitrogen at a partial pressure of 437 mm Hg and neon at a partial pressure of 429 mm Hg. What is the mole fraction of each gas in the mixture?

X_N2 = .....
X_Ne = .....

9/ A mixture of carbon dioxide and xenon gases at a total pressure of 973 mm Hg contains carbon dioxide at a partial pressure of 426 mm Hg. If the gas mixture contains 7.58 grams of carbon dioxide, how many grams of xenonare present?

= (......) g Xe

Answer 1

1)

Given:

V = 28.2 L

n = 0.987 mol

T = 19.0 oC

= (19.0+273) K

= 292 K

use:

P * V = n*R*T

P * 28.2 L = 0.987 mol* 0.08206 atm.L/mol.K * 292 K

P = 0.8387 atm

= 0.8387 * 760 mmHg

= 637 mmHg

Answer: 637 mmHg

2)

Molar mass of N2 = 28.02 g/mol

mass(N2)= 20.0 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(20 g)/(28.02 g/mol)

= 0.7138 mol

Given:

P = 477.0 mm Hg

= (477.0/760) atm

= 0.6276 atm

n = 0.7138 mol

T = 278.0 K

use:

P * V = n*R*T

0.6276 atm * V = 0.7138 mol* 0.08206 atm.L/mol.K * 278 K

V = 25.9 L

Answer: 25.9 L

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