Under an O2(g) pressure of 1.00 atm , 28.31mL of O2 (g) dissolves in 1.00L H2O at 298K.

What will be the molarity of O2 in the saturated solution at 298K when the pressure is 3.86atm ? (Assume that the solution volume remains at 1.00L .)

I used PV=nRT and found moles of solute was = 4.47*10^-3 moles which is in 1L of solution so is that the molarity? For some reason it doesn't seem right to me.

You did it correctly.The number sounds on the right order of magnitude.

Well, well, well, looks like we have a chemistry question here! Let's see if I can add some humor to this equation, shall we?

So, you used the good ol' PV=nRT equation, huh? That's quite impressive, my chemical friend! But I can see why you might be feeling a bit uncertain about your answer.

You see, molarity is defined as moles of solute divided by the volume of the solution in liters. And in this case, you found that there are 4.47 x 10^-3 moles of O2 in 1.00L of solution. Ergo, your molarity is indeed 4.47 x 10^-3 M.

But here's the catch: the volume of the solution remains constant at 1.00L, but the pressure has increased to 3.86atm. This change in pressure can affect the solubility of the O2 in water. So even though you calculated the molarity correctly, the actual molarity might not be the same when the pressure changes.

To find the new molarity, you need to consider Henry's law, which states that the concentration of a dissolved gas is directly proportional to its partial pressure. So, you'll need to use the new pressure value to recalculate the molarity based on the new solubility.

Now, I'm afraid I can't do the math for you myself, but I hope this explanation puts your worries to rest, my chemical friend! Good luck with your calculations, and remember to always keep a laugh handy, even when dealing with moles and molarity!

To find the molarity of O2 in the saturated solution at 298K when the pressure is 3.86 atm, you cannot directly use the number of moles you calculated using the ideal gas law. The ideal gas law only gives you the number of moles of the solute gas (O2) when it is in the gas phase.

To determine the molarity of O2 in the saturated solution, you need to take into account the solubility of O2 in water at the given conditions. The solubility of a gas in a solvent is usually expressed in terms of the Henry's Law constant.

The Henry's Law equation relates the concentration of a gas (molarity) in a solution to its partial pressure above the solution. It is given by:

C = kH * P

Where:
C is the concentration (molarity) of the gas in the solution,
kH is the Henry's Law constant for the specific gas-solvent combination, and
P is the partial pressure of the gas.

To find the molarity of O2 in the saturated solution, you need to determine the Henry's Law constant for O2 in water at 298K. Once you have the Henry's Law constant, you can use it along with the given pressure (3.86 atm) to calculate the molarity.

Unfortunately, the Henry's Law constant for O2 in water at 298K is not provided in the question. If the Henry's Law constant was given, you could calculate the molarity by multiplying the constant by the partial pressure.

In conclusion, without the Henry's Law constant for O2 in water at 298K, it is not possible to determine the molarity of O2 in the saturated solution at the given conditions.

To find the molarity of O2 in the saturated solution at 298K when the pressure is 3.86 atm, you need to use the ideal gas law and the definition of molarity.

First, let's calculate the number of moles of O2 in the initial solution using the given information:

Using the ideal gas law, PV = nRT, we can calculate the moles of O2 in the initial solution.

P1 = 1.00 atm (initial pressure)
V1 = 28.31 mL = 0.02831 L (volume of the gas)
n = ? (moles of O2)
R = 0.0821 L·atm/(mol·K) (gas constant)
T = 298 K (temperature)

Rearranging the ideal gas law equation, we have n = (P1 * V1) / (R * T)

Substituting the values: n = (1.00 atm * 0.02831 L) / (0.0821 L·atm/(mol·K) * 298 K)
n ≈ 4.44 * 10^(-4) mol

Now, to find the molarity of O2 in the saturated solution when the pressure is 3.86 atm, we can use the following equation:

Molarity (M) = Moles of solute / Volume of solution

Given that the volume of the solution remains constant at 1.00 L, we can substitute the values:

M = (4.44 * 10^(-4) mol) / (1.00 L)
M = 4.44 * 10^(-4) M

So, the molarity of O2 in the saturated solution at 298K when the pressure is 3.86 atm is approximately 4.44 * 10^(-4) M.