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.
Using the ideal gas law, we can solve for the moles of O2(g) at the given conditions of 1.00 atm and 28.31 mL:
PV = nRT
n = PV / RT
n = (1.00 atm * 28.31 mL) / (0.08206 L atm/mol K * 298 K)
n = 4.47 x 10^-3 moles
Since the volume of the solution remains constant at 1.00 L, the molarity of the solution will also remain constant at 4.47 x 10^-3 M. This is because the number of moles of solute is proportional to the concentration of the solution. Therefore, even if the pressure changes, the concentration of the solution will remain the same.
Well, if it doesn't seem right to you, then maybe we should trust your gut feeling and explore other possibilities!
Let's see, you used the ideal gas law to calculate the moles of O2, which is a good start. However, molarity is defined as moles of solute divided by the volume of the solution in liters. In this case, the volume of the solution is always 1.00L.
So, you already have the moles of O2, but to find the molarity, you need to divide that by the volume of the solution. Since the volume remains constant at 1.00L, your answer of 4.47 x 10^-3 moles is indeed the molarity of O2 in the saturated solution at 298K when the pressure is 3.86atm.
Yep, sometimes the right answer can be surprisingly simple!
To calculate the molarity of O2 in the saturated solution at 298K when the pressure is 3.86 atm, you first need to use the ideal gas law equation, PV = nRT, to find the number of moles of O2 present in the solution.
Given:
Pressure (P1) = 1.00 atm
Volume (V1) = 28.31 mL = 0.02831 L
Temperature (T) = 298 K
Molar mass of O2 = 32 g/mol (approximately)
Step 1: Convert the given pressure to moles of O2.
n1 = (P1 * V1) / (R * T)
n1 = (1.00 atm * 0.02831 L) / (0.0821 L*atm/mol*K * 298K)
n1 = 0.001184 mol
So, there are approximately 0.001184 moles of O2 in the saturated solution at P1.
Step 2: Calculate the molarity of O2.
Molarity (M) is defined as moles of solute per liter of solution.
Molarity (M1) = n1 / V
M1 = 0.001184 mol / 1.00 L
M1 = 0.001184 M
Therefore, the molarity of O2 in the saturated solution at P1 (1.00 atm) and 298K is 0.001184 M.
To calculate the molarity of O2 in the saturated solution at a different pressure (P2), you can use the same approach.
To find the molarity of O2 in the saturated solution, you need to first calculate the moles of O2 in the solution at the given pressure of 3.86 atm.
You correctly used the ideal gas law, PV = nRT, to find the moles of O2 at the initial pressure of 1.00 atm. However, since the solution volume remains constant at 1.00L, you can use the same value of moles obtained initially to calculate the molarity at the new pressure.
Here's how you can calculate the molarity:
Step 1: Calculate the moles of O2 at the initial pressure:
Use the ideal gas law equation to find the moles of O2 in the 1.00L solution at 1.00 atm:
PV = nRT
(1.00 atm) * (1.00 L) = (n) * (0.0821 L*atm/mol*K) * (298 K)
n = 0.0408 moles
Step 2: Calculate the molarity at the new pressure:
Since the solution volume remains constant at 1.00L, the number of moles of O2 remains the same. Therefore, the moles of O2 in the saturated solution at the new pressure of 3.86 atm is still 0.0408 moles.
Molarity (M) is defined as the moles of solute divided by the volume of the solution in liters. In this case, the volume of the solution is 1.00L.
Molarity = Moles of solute / Volume of solution
M = 0.0408 moles / 1.00 L
M = 0.0408 M
So, the molarity of O2 in the saturated solution at 298K, when the pressure is 3.86 atm, is 0.0408 M.