± Dalton\'s Law of Partial Pressure A 1.00 L flask is filled with 1.45 g of argo
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Question
± Dalton's Law of Partial Pressure A 1.00 L flask is filled with 1.45 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 . Part A What is the partial pressure of argon, PAr, in the flask? Express your answer to three significant figures and include the appropriate units. PAr = SubmitHintsMy AnswersGive UpReview Part Incorrect; Try Again Part B What is the partial pressure of ethane, Pethane, in the flask? Express your answer to three significant figures and include the appropriate units. Pethane = SubmitHintsMy AnswersGive UpReview Part
Explanation / Answer
Part A)
Argon partial pressure is 0.734 atm.
So, you know that you start with a certain mass of argon in the flask. The first thing you need to do is figure out exactly how many moles of argon you have
1.45 g Ar1 mole Ar40.0 g=0.03625 moles Ar
Now you add an unknown number of moles of ethane vapor to the flask. Since you know volume, total pressure, and temperature, we can use the ideal gas equation to figure out how many moles of both the gases can be found in the flask.
PV=nRT ntotal=Ptotal V / RT
Therefore, ntotal=1.450 atm*1.00 L / 0.082atmLmolK*(273+ 25)K=0.05934 moles
So, your mixture contains a total of 0.05934 moles of gas, argon and ethane. Now, you can express the Argon partial pressure by using its mole fraction and the total pressure in the flask
Pargon=argonPtotal, where
Pargon - the Argon partial pressure;
argon - its mole fraction, i.e. the ratio between the number of moles of argon and the total number of moles present in the mixture;
Ptotal - the total pressure in the flask;
Since you know how many moles of argon you have, its partial pressure will be
Pargon=0.03625 moles Ar *1.450 atm / 0.05934 moles gas
Pargon=0.8857 atm
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