25 A 1.00 L flask is filled with 1.00 g of argon at 25 degree C. A sample of ethane vapor is added to the same flask until the total pressure is 1.250 atm.

Question

Question:

25 A 1.00 L flask is filled with 1.00 g of argon at 25 degree C. A sample of ethane vapor is added to the same flask until the total pressure is 1.250 atm.

What is the partial pressure of argon, PAr, in the flask?

What is the partial pressure of ethane, Pethane, in the flask?

Answer and Explanation:

The partial pressure of argon gas is calculated using ideal gas equation as shown below
PV=nRT
N=1.25g Ar x 1mol/39.95g = 0.0313 moles
P = Nrt/V = (0.0313 mol) (0.0821 Latm/Kmol) (298.15K)/1.00L
=0.766 atm = PAr

Second Step
Total Pressure = PAr + Pethane
Pethene = 1.450 atm – 0.766 atm
=0.684 atm =Pethene

0

Answer ( 1 )

Leave an answer