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?

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