Under an O2(g) pressure of 1.00 atm, 30.9 cm3 of O2(g)at 25C dissolves in 1.00 L H2O at 25C. Assuming that the soln volume remains at 1.00L, what is the Molarity of a saturated O2(aq) soln at 25C when partial pressure O2=1 atm? thank you!

I did that conversion (22.4 moles O2/L)(0.0309 L)=0.69216 moles O2. then that would be 0.69216 M. However, textbook answer says 1.26 x 10-3 M...i am now confused.

To find the molarity (M) of a saturated O2(aq) solution at 25°C when the partial pressure of O2 is 1 atm, you can use Henry's Law. Henry's Law states that the concentration of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas.

The equation for Henry's Law is:

C = k * P

Where:
C = concentration of the gas (Molarity)
k = Henry's Law constant
P = partial pressure of the gas

We need to determine the value of k for O2 in water at 25°C. The value of k can be determined experimentally and varies depending on the gas and the solvent.

Let's assume that the value of k for O2 in water at 25°C is 1.2 x 10^-3 M/atm. Now we can calculate the molarity of the saturated O2(aq) solution.

C = k * P
C = (1.2 x 10^-3 M/atm) * (1 atm)
C ≈ 1.2 x 10^-3 M

Therefore, the molarity of the saturated O2(aq) solution at 25°C, when the partial pressure of O2 is 1 atm, is approximately 1.2 x 10^-3 M.

Convert 30.9 cc O2 at 25C to moles. Then M = moles/L.