Question

A mixture of xenon and oxygen gases, in a 8.22 L flask at 76 °C, contains 17.0 grams of xenon and 3.94 grams of oxygen. The p
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Answer #1

1)

Molar mass of O2 = 32 g/mol

mass(O2)= 3.94 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(3.94 g)/(32 g/mol)

= 0.1231 mol

Given:

V = 8.22 L

n = 0.1231 mol

T = 76.0 oC

= (76.0+273) K

= 349 K

use:

P * V = n*R*T

P * 8.22 L = 0.1231 mol* 0.08206 atm.L/mol.K * 349 K

P = 0.4289 atm

Answer: 0.429 atm

2)

1st find partial pressure of Xe

Molar mass of Xe = 131.3 g/mol

mass(Xe)= 17.0 g

use:

number of mol of Xe,

n = mass of Xe/molar mass of Xe

=(17 g)/(1.313*10^2 g/mol)

= 0.1295 mol

Given:

V = 8.22 L

n = 0.1295 mol

T = 76.0 oC

= (76.0+273) K

= 349 K

use:

P * V = n*R*T

P * 8.22 L = 0.1295 mol* 0.08206 atm.L/mol.K * 349 K

P = 0.451 atm

Now use:

Total pressure = p(Xe) + p(O2)

= 0.451 atm + 0.429 atm

= 0.880 atm

Answer: 0.880 atm

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