A 1.00L flask is filled with 1.30g of argon at 25 degrees Celcius. 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 Pargon in the flask? What is the P.P. of Pethane in flask?

Start with the ideal gas law:

PV = nRT

The following values were provided:

V = 1.00 L

T = 25 C = 298 K

R = 0.0821 liter·atm/mol·K

First we can solve for n for just the argon by just looking up the molecular weight on the periodic table

1.30 g x 1 mol / 39.948 g = 0.0325 mol argon

Now that we know all the variables for the ideal gas law we can solve for pressure which is what we need to know

Pargon = 0.0325 mol x 0.0821 liter·atm/mol·K x 298 K / 1.00 L

Pargon = 0.796 atm

Partial pressure is just the sum of all pressures exerted by all gases present. In this case:

Ptotal = Pargon + Pethane

Therefore:

Pethane = Ptotal - Pargon = 1.450 atm - 0.796 atm = 0.654 atm

The final answers are

Pethane = 0.654 atm

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