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 at constant volume and temperature until the total pressure is 1.450 atm. What is the partial pressure of argon, Par, in the flask?
Express your answer to three significant figures and include the appropriate units.
To find the partial pressure of argon, we first need to find the moles of argon in the flask.
Using the ideal gas law: PV = nRT, we can find the moles of argon:
n = PV/RT = (1.450 atm)(1.00 L) / (0.0821 L.atm/mol.K)(298 K) ≈ 0.0727 mol
Now we can find the partial pressure of argon:
Par = (moles of argon)(RT/V) = (0.0727 mol)(0.0821 L.atm/mol.K)(298 K) / 1.00 L ≈ 2.18 atm
Therefore, the partial pressure of argon, Par, in the flask is approximately 2.18 atm.