Asolution is made by dissolving 0.0150 mol of hf in enough water to make 1.00 l of solution. at 26 °c, the osmotic pressure of the solution is 0.449 atm. what is the percent ionization of this acid?

As the osmotic pressure = (molarity)(gas constant)(temp)
 so we can use the given information to find the molarity of the solution so  that would give us that osmotic pressure...
so... 0.449atm = (Molarity)(0.08206 Latm/molK)(299K)
  so the Molarity = 0.0183
 and the total particles in the solution will be concluded by
 this molarity - the given molarity
so... by putting values
0.0183-0.0150 = 0.0030 M
 now we then put this value over the value given molarity which will give us the  percent ionization.
 that is
0.0030M/0.0150M
 = 0.200
 = 20.0%
so the 20.0% is the percent ionozation of this acid
hope it helps

D-no effect on the rate

What do you mean by that? like the contents? the thc level?