# Question fe55f

Jul 9, 2017

I have no idea what your Henry's law constant units are...

[It's possible that you are using a mol fraction version of Henry's law, in which ${k}_{H}$ (in $\text{atm}$) is approximately the numerical reciprocal of the version in units of $\text{M/atm}$.]

I get ${\text{0.048 mols N}}_{2} \left(g\right)$ in the solution, using ${k}_{H} = 90998.8$ $\text{atm}$. You must adjust accordingly using your $100000$ $\text{atm}$.

REGARDING THE HENRY'S LAW CONSTANT

I cannot trust the Henry's law constant given here. There are no units given.

Using ${k}_{H}$ in $\text{M/atm}$, if I convert my Henry's law constant to what I think is yours...

${k}_{H} \left(\text{mine}\right)$

$= 6.1 \times {10}^{- 4} \text{mols solute" cdot cancel("L"^(-1) "soln") cdot "atm"^(-1) xx cancel"1 L soln"/(1000 cancel"g water") xx (18.015 cancel"g water")/"1 mol soln}$

$= 1.0989 \times {10}^{- 5}$ ${\text{atm}}^{- 1}$

Then in your case, you may have

1/(k_(H)("mine")) = k_(H) ("yours") = "90998.8 atm", which is somewhat close to $100000$, I suppose...

MOL FRACTION VERSION OF HENRY'S LAW

So, I simply assume the version of Henry's Law shown below:

${P}_{g a s} = {\chi}_{g a s \left(l\right)} {k}_{H}$

where:

• ${\chi}_{g a s \left(l\right)} = \frac{{n}_{g a s}}{{n}_{g a s} + {n}_{s o l v e n t}}$ is the mol fraction of the gas in the solution.
• ${k}_{H}$ is the Henry's law constant (for the PARTICULAR gas at a SPECIFIED temperature in the SPECIFIED solvent!!). This WILL vary depending on your Henry's law equation, and there ARE multiple versions of this equation!! Keep the units straight.
• ${P}_{g a s}$ is the vapor pressure of the gas above the solution. Here, it is in $\text{atm}$.

RELATING TO PARTIAL PRESSURE IN AIR

Since the air vapor pressure is $\text{10 atm}$, we know that by the definition of partial pressure, and knowing the percent of ${\text{N}}_{2}$ in air is 78%, the mol fraction of ${\text{N}}_{2}$ in air is

${\chi}_{{N}_{2} \left(v\right)} = 0.78$.

So, the partial pressure of ${\text{N}}_{2}$ in the air above the solution is

${P}_{{N}_{2}} = {\chi}_{{N}_{2} \left(v\right)} {P}_{t o t}$

= 0.78 ("10 atm") = "7.8 atm"#

GETTING MOLS OF N2

Thus, using the Henry's law constant of ${\text{N}}_{2}$ in water at $\text{298 K}$ (not the temperature which was not given in the question!), which I converted above to be $90998.8$ $\text{atm}$, I get a mol fraction of ${\text{N}}_{2}$ in solution:

${\chi}_{{N}_{2} \left(l\right)} = {P}_{{N}_{2}} / {k}_{H} = \left(\text{7.8 atm")/("90998.8 atm}\right)$

$=$ $0.00008572$

Since we have $\text{10 L}$ of water, assuming its density is $\text{1000 g/L}$, and the gas hardly changes the volume of the water:

$\text{10 L water" ~~ "10000 g water" ~~ "555.09 mols water}$

And so, the mol fraction allows us to solve for the mols of gas in here:

${\chi}_{{N}_{2} \left(l\right)} = 0.00008572 = {n}_{{N}_{2}} / \left({n}_{{N}_{2}} + {n}_{{H}_{2} O}\right)$

$= {n}_{{N}_{2}} / \left({n}_{{N}_{2}} + 555.09\right)$

$\implies 0.00008572 {n}_{{N}_{2}} + 0.00008572 \left(555.09\right) = {n}_{{N}_{2}}$

$\implies \textcolor{b l u e}{{n}_{{N}_{2}}} = \frac{0.04758}{1 - 0.00008572} = 0.0476$

$\approx$ $\textcolor{b l u e}{\text{0.048 mols N"_2 " in the solution}}$