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

Verified

Part A

We first need to calculate the number of moles of Argon

$n=\dfrac{1.25}{39.95}=0.031\text{ moles}$

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

$PV=nRT$

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

$\boxed{P=0.75\text{ atm}}$

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

$P_{e}=P_{total}-P_{a}$

$P_{e}=1.45-0.75$

$\boxed{P_e=0.7\text{ atm}}$

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