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.
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.
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