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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: bpeck8 on January 23, 2010, 03:44:01 PM

Title: Need help with two problems.
Post by: bpeck8 on January 23, 2010, 03:44:01 PM
First problem:
As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.58 atm and a temperature of 25.0 C, what is the solubility of N2 in a diver's blood? [Use the value of the Henry's law constant k calculated in Part A,  6.26 * 10^-4 mol/Latm

Assume that the composition of the air in the tank is the same as on land and that all of the dissolved nitrogen remains in the blood.

I am having trouble figuring out the partial pressure. Any help?

part A had a partial pressure of .78atm. THis is part a:
Air is a mixture of gases that is about 78.0% N2 by volume. When air is at standard pressure and 25.0 C, the N2 component will dissolve in water with a solubility of 4.88* 10^-4 M. What is the value of Henry's law constant for N2 under these conditions?


Second problem:
At what temperature would a 2.00 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 C/m.

I am just lost as to how to approach this.

Thanks,
Brent
Title: Re: Need help with two problems.
Post by: UG on January 23, 2010, 05:17:21 PM
part A had a partial pressure of .78atm. THis is part a:
Air is a mixture of gases that is about 78.0% N2 by volume. When air is at standard pressure and 25.0 C, the N2 component will dissolve in water with a solubility of 4.88* 10^-4 M.
Well, if the partial pressure of nitrogen at standard pressure, i.e 1 atm (well, I assume you are using 1atm as the STP) is 0.78 atm then what is the partial pressure of nitrogen at 2.58 atm?

Second problem:
At what temperature would a 2.00 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86 C/m.

Well the freezing point depression formula is simply,  :delta:T = Kfmsolute
And if you throw in the van't Hoff factor, i it becomes  :delta:T= i Kfmsolute

Title: Re: Need help with two problems.
Post by: bpeck8 on January 24, 2010, 01:31:04 AM
Thanks for you help UG.

I got for problem 1: 1.26*10^-3 mol/L

and I got for problem 2: -7.07 C, just a little correction on your formula, there should be a - sign in front of the i. took me a little while to figure out why i was getting the answer wrong until i looked at it closer and realized it.