Question

In: Chemistry

Part A A mixture of He, Ar, and Xe has a total pressure of 2.70 atm...

Part A

A mixture of He, Ar, and Xe has a total pressure of 2.70 atm . The partial pressure of He is 0.450 atm , and the partial pressure of Ar is 0.450 atm . What is the partial pressure of Xe?

Express your answer to three significant figures and include the appropriate units.

Part B

A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He. The temperature of the mixture is 0 ∘C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

Express your answer to three significant figures and include the appropriate units.

Solutions

Expert Solution

A)

Use Dalton law of partial pressure:

Ptotal = p(He) + p(Ar) + p(Xe)

2.70 atm = 0.450 atm + 0.450 atm + p(Xe)

p(Xe) = 1.80 atm

Answer: 1.80 atm

B)

1st calculate the total mol of gas.

Given:

P = 1.0 atm

V = 18.0 L

T = 0.0 oC

= (0.0+273) K

= 273 K

find number of moles using:

P * V = n*R*T

1 atm * 18 L = n * 0.08206 atm.L/mol.K * 273 K

n = 0.8035 mol

Now use:

total mol of gas = mol of N2 + mol of O2 + mol of He

0.8035 mol = 0.250 mol + 0.250 mol + mol of He

mol of He = 0.3035 mol

Molar mass of He = 4.003 g/mol

use:

mass of He,

m = number of mol * molar mass

= 0.3035 mol * 4.003 g/mol

= 1.215 g

Answer: 1.22 g


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