Question
Asked Nov 7, 2019
If a gaseous mixture is made by combining 3.97 g Ar and 3.90 g Kr in an evacuated 2.50 L container at 25.0 ∘C, what are the partial pressures of each gas, ?Ar and ?Kr, and what is the total pressure, ?total, exerted by the gaseous mixture?
?Ar=
 
atm
?Kr=
 
atm
?total=
 
atm
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Expert Answer

Step 1

Calculation of number of moles:

3.97 g
mass
0.0993 mol
Moles of Ar =
Molar mass
39.95g /mol
3.90g
mass
=0.04653mol
Moles of Kr
Molarmass 83.80g /mol
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3.97 g mass 0.0993 mol Moles of Ar = Molar mass 39.95g /mol 3.90g mass =0.04653mol Moles of Kr Molarmass 83.80g /mol

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Step 2

According to the Ideal gas...

For Ar
(0.0993mol)(273 +25) K(0.0821 atm.L.mol1 K1)
nRT
Р.
Ar
V
2.50L
0.9717 atm
For Kr
nRT (0.04653mol) (273+25) K (0.0821atm.L.molK1
Р
Kr
2.50L
0.4553 atm.
help_outline

Image Transcriptionclose

For Ar (0.0993mol)(273 +25) K(0.0821 atm.L.mol1 K1) nRT Р. Ar V 2.50L 0.9717 atm For Kr nRT (0.04653mol) (273+25) K (0.0821atm.L.molK1 Р Kr 2.50L 0.4553 atm.

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