The vapor pressure of acetic acid is 0.033 atm at 25°C. If 5.00g of $\ce{LiF}$ is added to 100g of acetic acid, what would be the vapor pressure of the solution at 25°C?
The lowering in vapor pressure (VP) can be calculated with this equation:
${∆P = X_{solute} * P°_{solvent} * i}$
${∆P}$ is the change (lowering) in VP.
${X_{solute}}$ is the mole fraction of the solute.
${P°_{solvent}}$ is the VP of the pure solvent.
${i}$ is the van't Hoff factor of the solute.
What is the "correct" van't Hoff for this problem, from the perspective of a general chemistry student? I know that $\ce{LiF}$ is not very soluble in water, and I know there are papers on lithium fluoride and acetic acid, but that's all beyond the scope of a general chemistry class.
Therefore, I think it's 2, because $\ce{LiF}$ is expected to be a strong electrolyte. It's ionic, and the solvent is similarly polar (acetic acid has hydrogen bonds). The solvent even resembles water in that both have hydrogen-bonding. So I would expect $\ce{LiF}$ to dissolve fully in acetic acid.
Remember, this is a general chemistry class, and the students aren't expected to know much beyond "like dissolves like." $\ce{LiF}$is polar, and so is acetic acid. So to them, $\ce{LiF}$ should dissolve in acetic acid.
Problem is that I'm in trouble with my supervisor since I told someone that i = 2. My supervisor is telling me that the van't Hoff factor should be 1. I looked over his work to see if he accounted for the dissolving of $\ce{LiF}$when calculating the mole fraction of the solute but he didn't - he just found the mole fraction of $\ce{LiF}$, not the combined mole fractions of $\ce{Li+}$ and $\ce{F-}$.
So, what's the deal with this problem? Is there some weird exception I'm not aware of regarding $\ce{LiF}$ and acetic acid? The van't Hoff factor should be 2, correct?