A 1-L flask is filled with 1.45 g of argon at 25 degrees C. A sample of ethane vapor is added to the same flask until the total pressure is 1.35 atm. What is the partial pressure of argon in the flask?

figure the moles of argon in 1.45g.

PartialPressure*Volume=nRT solve for partial pressure of Argon

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0.888

To determine the partial pressure of argon in the flask, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, let's find the number of moles of argon present in the flask:

1. Calculate the molar mass of argon (Ar):
The molar mass of argon (Ar) is approximately 39.95 g/mol.

2. Convert the mass of argon to moles:
Number of moles = Mass / Molar mass
Number of moles = 1.45 g / 39.95 g/mol
Number of moles = 0.0363 mol (rounded to four decimal places)

Next, using the ideal gas law equation, we can calculate the initial pressure of argon (P1) in the flask when it only contains argon:

P1 * V = n1 * R * T

We need to rearrange the equation to solve for P1:

P1 = (n1 * R * T) / V

Given:
- V = 1 L (volume of the flask)
- n1 = 0.0363 mol (number of moles of argon)
- R is the ideal gas constant which is approximately 0.0821 L·atm/(mol·K)
- T = 25 °C = 298 K (temperature in Kelvin)

Plugging in the given values:

P1 = (0.0363 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1 L
P1 = 0.897 atm (rounded to three decimal places)

The initial pressure of argon in the flask, when it only contains argon, is 0.897 atm.

Lastly, we need to calculate the partial pressure of argon (P2) in the flask when the total pressure is 1.35 atm:

P2 = Ptotal - Pethane

Given:
- Ptotal = 1.35 atm (total pressure in the flask)
- Pethane = 1.35 atm (since the ethane vapor is added to the same flask)

Plugging in the given values:

P2 = 1.35 atm - 1.35 atm
P2 = 0 atm

Therefore, the partial pressure of argon in the flask is 0 atm (or zero).