[Rutherford]

Calculate the freezing point depression of an acetic acid solution (c=0.1M, ρ=1.006g/cm³) if K_a=1.75*10^-5.

I suppose that it is a water solution, so K=1.86KKg/mol.
For a non-electrolyte the formula is Δt=K*b (b-molality).
For a strong electrolyte: Δt=i*K*b (i is Hoff's factor).
What is the formula for a weak electrolyte?

[Arkcon]

And which do you have when you dissolve acetic acid in water?  Do you have an ionized compound, like HCl?  Or do you have a dissolved molecular compound, like sucrose.  Why did they give you K_a?

[Rutherford]

I have both the non-dissociated acetic molecules, and the CH3COO^- and H⁺ ions. How to combine these into one equation, and why did they give the density?

[Borek]

What is the meaning of the Van 't Hoff factor?

How are you going to calculate molality?

[Rutherford]

I was on the wrong track. If I assume that I have 1dm³ of the solution there is 0.1mol of acetic acid and 1000g of water in it so the molality is 0.1mol/kg.
i=1+α(k-1) where k is the number of ions produced when one molecule dissociates. Therefore i=1+(k-1)*sqrt(Ka/c)=1.013.
Δt=1.013*1.86*0.1=0.188. The right answer is 0.191. I believe its because of the approximation I used for α. Thanks so much.