Question

A 1.00 l flask is filled with 1.10 g of argon at 25 ∘c. a sample of ethane vapor is added to the same flask until the total pressure is 1.200 atm .what is the partial pressure of argon, par, in the flask?

Answer 1

Assuming ideal gas, we can solve the total number of moles in the system as:

PV=nRT
Solving for n,
n = PV/RT = (1.2 atm)(1 L)/(0.0821 L·atm/mol·K)(25 + 273 K)
n = 0.049

Now, compute for he moles of Argon knowing that its molar mass is 39.95 g/mol.

Mol Ar: 1.10 g/39.95 g/nol = 0.0275 mol
Mol fraction of Ar = 0.0275/0.049 = 0.562

Thus,
Partial Pressure = Total Pressure * Mol Fraction
Partial Pressure = (1.2 atm)(0.562) = 0.674 atm