A 1.00 L flask is filled with 1.30 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.300 atm . Part A What is the partial pressure of argon, PAr, in the flask? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) PArP A r P_Ar = nothing nothing Part B What is the partial pressure of ethane, Pethane, in the flask?

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Alleei Alleei · Answer 1 · 2019-10-10T12:36:17+02:00

Answer :

Part A : The partial pressure of argon is, 0.795 atm

Part B : The partial pressure of ethane is, 0.505 atm

Explanation :

Part A :

First we have to calculate the moles of argon.

Molar mass of argon = 39.95 g/mole

[tex]\text{Moles of argon}=\frac{\text{Mass of argon}}{\text{Molar mass of argon}}=\frac{1.30g}{39.95g/mol}=0.0325mole[/tex]

Now we have to calculate the partial pressure of argon.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of argon = ?

V = Volume of argon = 1.00 L

n = number of moles of argon = 0.0325 mole

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of argon = [tex]25^oC=273+25=298K[/tex]

Putting values in above equation, we get:

[tex]P_{Ar}=\frac{nRT}{V}[/tex]

[tex]P_{Ar}=\frac{(0.0325mol)\times (0.0821L.atm/mol.K)\times (298K)}{1.00L}=0.795atm[/tex]

The partial pressure of argon is, 0.795 atm

Part B :

Now we have to calculate the partial pressure of ethane.

As we know that,

Total pressure = Partial pressure of argon + Partial pressure of ethane

1.300 atm = 0.795 atm + Partial pressure of ethane

Partial pressure of ethane = 1.300 - 0.795

Partial pressure of ethane = 0.505 atm

The partial pressure of ethane is, 0.505 atm