In: Chemistry
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?
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