Question

A solution 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?

Answer 1

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