Question

A 1.00 L flask is filled with 1.20 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.450 atm . What is the partial pressure of argon, PAr, in the flask?

Answer 1

A 1.00 L flask is filled with 1.20 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.450 atm . What is the partial pressure of argon, PAr, in the flask?

Solution :-

Lets first calculate the moles of the Argon

moles of Ar = 1.20 g * 1 mol / 39.948 g = 0.03004 mol Ar

now lets calculate the initial pressure of the Ar

PV= nRT

P= nRT/V

= 0.03004 mol * 0.08206 L atm per mol K * 298 K / 1.0 L

= 0.735 atm

Total pressure = 1.450 atm

Then partial pressure of the ethane = 1.450 atm -0.735 atm

                                                                   = 0.715 atm

Partial pressure of Ar = 0.735 atm