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.)
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?
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.
To determine the molarity of O2 in an aqueous solution at equilibrium with air at normal atmospheric pressure, we need to follow a few steps.
Step 1: Calculate the moles of O2 dissolved in 1.00 L of water.
To do this, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.00 atm)
V = volume (28.31 mL converted to L, which is 0.02831 L)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (25 ºC converted to Kelvin, which is 298 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
n = (1.00 atm)(0.02831 L) / (0.0821 L·atm/mol·K)(298 K)
n = 0.001216 mol
Therefore, 0.001216 moles of O2 are dissolved in 1.00 L of water.
Step 2: Calculate the total volume of the solution.
To calculate the total volume, we need to take into account the volume of water and the volume of air. Since we know the volume percent of O2 in air is 20.80% and the total volume is 1.00 L, we can calculate the volume of air:
Volume of O2 in air = (Volume percent of O2 in air / 100) * total volume
Volume of O2 in air = (20.80 / 100) * 1.00 L
Volume of O2 in air = 0.208 L
Step 3: Calculate the molarity of O2.
Molarity is defined as moles of solute divided by liters of solution. In this case, the solute is O2 and the solution is the combination of water and air.
Molarity of O2 = moles of O2 / total volume of solution
Total volume of solution = volume of water + volume of air
Total volume of solution = 1.00 L + 0.208 L
Total volume of solution = 1.208 L
Molarity of O2 = 0.001216 mol / 1.208 L
Molarity of O2 = 0.001006 M
Therefore, the molarity of O2 in the aqueous solution at equilibrium with air at normal atmospheric pressure is 0.001006 M.