Under an O2 pressure of 1.00 atm, 28.31 mL of O2 dissolves in 1.00 L H2O at 25 degrees.
Determine the molarity of O2 in an aqueous solution at equilibrium with air at normal atmospheric pressure. The volume percent of O2 in air is 20.80 \%.
(Have no idea how to go through with this problem.)
4 answers
See my response below.
Okay, so this is how I think I should set it up? can you confirm it at least. Anyways I would try to find moles of O2 using ideal gas law (1.00 atm)(0.283)/(0.0821)(298) = n moles O2.
then I would multiply 0.208 by the 1 L of water??
Then divide the moles of O2 / by L of water to get molarity?
then I would multiply 0.208 by the 1 L of water??
Then divide the moles of O2 / by L of water to get molarity?
sorry would i multiply the moles of O2 by the 1(0.208) to get the concentration which is proportional to the molarity?
The first part, if you wish to work in liter-atm, appears to be ok. But I think that first part is to allow you to calculate k
partial pressure O2 = kc
So you have partial pressure O2 in atm and c in moles/L, that gets k.
Go from there. Using k from above and the new pressure, calculate moles under the new conditions with the same formula of p = kc.
partial pressure O2 = kc
So you have partial pressure O2 in atm and c in moles/L, that gets k.
Go from there. Using k from above and the new pressure, calculate moles under the new conditions with the same formula of p = kc.
5 answers