Question

A 1.00 L flask is filled with 1.25 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.450 atm.

A) What is the partial pressure of argon, PAr, in the flask?
B) What is the partial pressure of ethane, Pethane, in the flask?

Solution

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Answered 2 years ago
Answered 2 years ago

Part A

We first need to calculate the number of moles of Argon

n=1.2539.95=0.031 molesn=\dfrac{1.25}{39.95}=0.031\text{ moles}

We will use the ideal gas law to determine the partial pressure of Argon from

PV=nRTPV=nRT

P(1)=0.031(0.0821)(25+273)P(1)=0.031(0.0821)(25+273)

P=0.75 atm\boxed{P=0.75\text{ atm}}

Part B

The partial pressure of ethane can be found by by subtracting the pressure found in Part A for Argon from the total Pressure

Pe=PtotalPaP_{e}=P_{total}-P_{a}

Pe=1.450.75P_{e}=1.45-0.75

Pe=0.7 atm\boxed{P_e=0.7\text{ atm}}

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