A 5 mL sample of an unknown liquid is placed into a 250 mL flask that...
A 5 mL sample of an unknown liquid is placed into a 250 mL flask that has had all of the air removed. The pressure measured in the flask is found to be 39.4 kPa. In a seperate trial the student puts 214 mL of the same liquid into the same flask at the same conditions. What will the pressure measure (in kPa) in the second trial?
I have tried P1V1 = P2V2 and got the answer P2=268 kPa, but it is saying that it is the wrong answer.
Homework Answers
Vapor presssure is the equilibrium pressure exerted where the rate of evaporation of liquid equals the rate of condenssation. This vapor pressure of liquid is a function of temperature and it being internsive property does not depends upon the mass of the liquid. Hence, the pressure even when 214ml of liquid is taken at the same temperature remains at 39.4 Kpa.
Since this is the pressure exerted by liquid, gas law cannot be used.
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i dont understand why the molecular weight i calculated
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made in my lab that would account for the difference.
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help with identifiying each unknown in extraction and washing
experiment. the ir spectra are given below, along with data, and
the lab sheet is attached.
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5. When all of the sample to be collected has left the
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be acidic, basic or neutral?
6. How many H+ ions would be displaced from the ion
exchange column by 1 Cu2+ ion?
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with which ion before the sample solution is added?
8. Ions are displaced from the resin binding sites by
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