1
$\begingroup$

I have tried for several hours to answer this question with no luck. I believe I may be missing a couple of steps or I am not utilizing the given information properly. Nonetheless, some guidance will be greatly appreciated.

The concentration of $\ce{N2}$ in the ocean at 25C is 445 $\ce{\mu}$M. The Henry's Law constant for $\ce{N2}$ is 0.61 x $\ce 10^{-3}$ mol $\ce L^{-1}$ atm $\ce L^{-1}$.

Part A:

Calculate the mass of $\ce{N2}$ in a liter of ocean water.

4.45 x $\ce 10^{-4}$ mol/kg = molality

$\ce{N2}$ in moles = 28.014 g/mol

Using Molality Equation

Molality = $\frac{amount of solute (mols)}{mass of solvent(kg)}$

I want kilograms, thus:

4.45 x $\ce 10^{-4}$ $\frac{mol}{kg}$ = $\frac{28.014 g}{mol}$

Where do I go from here?

Part B:

Calculate the partial pressure of $\ce{N2}$ in the atmosphere

Use C = K x $P_(gas)$

4.45 x $\ce 10^{-4}$ = (0.61 x $\ce 10^{-3}$ mol $\ce L^{-1}$ atm $\ce L^{-1}$)x

x = 0.73 atm

$\endgroup$
  • 2
    $\begingroup$ No one can help me with this question? $\endgroup$ – Cetshwayo Feb 14 '15 at 16:58
  • 1
    $\begingroup$ I formatted your question using MathJax meta.chemistry.stackexchange.com/questions/86/… This makes your question computer readable and typing out your whole is generally the way to go, so it can potentially help others (and in increases the chance of a response.) If you hit edit on your question, you can see some of the functions I used to format it. I kept your image in case I made an error. Best of luck finding the answer to your question, I'm studying something fairly similar right now. $\endgroup$ – Melanie Shebel Feb 14 '15 at 22:18
3
$\begingroup$

Part A: Why do you want to use the molality?! You have the concentration i.e the number of moles in one liter. So the mass of nitrogen in one liter is simply:

Number of moles in one liter $\times$ molar mass.

Part B: It's correct.

$\endgroup$
  • $\begingroup$ Thnaks. I do not understand the definition well enough. $\endgroup$ – Cetshwayo Feb 14 '15 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.