Question

Question #fe55f

Answer 1

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 `"atm"`) is approximately the numerical reciprocal of the version in units of `"M/atm"`.]

I get `"0.048 mols N"_2(g)` in the solution, using `k_H = 90998.8` `"atm"`. You must adjust accordingly using your `100000` `"atm"`.

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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 `"M/atm"`, if I convert my Henry's law constant to what I think is yours...

`k_H ("mine")`

`= 6.1 xx 10^(-4) "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 xx 10^(-5)` `"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_(gas) = chi_(gas(l))k_H`

where:

• `chi_(gas(l)) = (n_(gas))/(n_(gas) + n_(solvent))` 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_(gas)` is the vapor pressure of the gas above the solution. Here, it is in `"atm"`.

RELATING TO PARTIAL PRESSURE IN AIR

Since the air vapor pressure is `"10 atm"`, we know that by the definition of partial pressure, and knowing the percent of `"N"_2` in air is `78%`, the mol fraction of `"N"_2` in air is

`chi_(N_2(v)) = 0.78`.

So, the partial pressure of `"N"_2` in the air above the solution is

`P_(N_2) = chi_(N_2(v))P_(t ot)`

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

GETTING MOLS OF N2

Thus, using the Henry's law constant of `"N"_2` in water at `"298 K"` (not the temperature which was not given in the question!), which I converted above to be `90998.8` `"atm"`, I get a mol fraction of `"N"_2` in solution:

`chi_(N_2(l)) = P_(N_2)/k_H = ("7.8 atm")/("90998.8 atm")`

`=` `0.00008572`

Since we have `"10 L"` of water, assuming its density is `"1000 g/L"`, and the gas hardly changes the volume of the water:

`"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(l)) = 0.00008572 = n_(N_2)/(n_(N_2) + n_(H_2O))`

`= n_(N_2)/(n_(N_2) + 555.09)`

`=> 0.00008572n_(N_2) + 0.00008572(555.09) = n_(N_2)`

`=> color(blue)(n_(N_2)) = (0.04758)/(1 - 0.00008572) = 0.0476`

`~~` `color(blue)("0.048 mols N"_2 " in the solution")`