Partial Pressure of Kr in a N2/Kr Mixture (0.505 atm)

AI Thread Summary
The discussion revolves around calculating the partial pressure of Krypton (Kr) in a mixture of gases. Given a total pressure of 0.505 atm for a mixture of 5.40 g of Nitrogen (N2) and 10.2 g of Kr, the calculation involves determining the mole fraction of Kr. Participants clarify that the problem assumes constant volume and temperature, allowing the use of the ideal gas law (PV=nRT) if needed. The correct approach is to find the mole fraction of Kr and then multiply it by the total pressure to obtain the partial pressure. The focus remains on understanding the definition of partial pressure and applying it correctly to solve the problem.
parwana
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A mixture of 5.40 g of N2 and 10.2 g of Kr exerts a pressure of 0.505 atm.
What is the partial pressure, in atm, of Kr?
 
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parwana, please show what you've tried, when you need help with a problem.
 
I just don't get the problem at all, i mean does this mean volume and temperature are constant and if so do i make up any numbers for the temp and volume and substitute in the equation pv=nrt??

pleASE help
 
i got it, i got the mole fraction of Ke and then times thta by total pressure.
 
parwana said:
i got it, i got the mole fraction of Ke and then times thta by total pressure.

If you meant "Kr", then yes, that's correct. With any such problem, you want to start from the definitions. In this case, the definition of 'partial pressure' is all you need.
 
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