After performing a freezing point depression experiment I had to find the molecular weight of glycol. Now I have to answer the question what value did I assume "i" to equal and why did I make this assumption.
My teacher told us the value of i to use for this was 1. I am not sure why though. Could someome please explain. Thank you for your help!!
Ethylene glycol does not ionize; remember i is the number of particles. NaCl has two when dissolved in water so i = 2. But ethylene glycol doesn't ionize. You get 6.02E23 particles when you dissolve a mole of ethylene glycol. When you dissolve a mole of NaCl you get 2*6.02E23 particles.
oh ok I understand thank you so much for all of your help!!!
In freezing point depression experiments, we use the equation ΔT = Kf * i * molality, where ΔT is the change in freezing point, Kf is the cryoscopic constant, i is the van't Hoff factor, and molality is the concentration of a solute in terms of moles of solute per kilogram of solvent.
The van't Hoff factor (i) represents the number of particles into which a compound dissociates or associates in a solution. It takes into account the effect of solute-solvent interactions on the colligative properties of the solution. For non-electrolytes, i is typically equal to 1 since they do not dissociate or associate in solution.
In the case of glycol, which is a non-electrolyte, we assume an i value of 1 because it does not dissociate into ions or associate with the solvent molecules. This assumption simplifies the equation and allows us to correctly calculate the molecular weight of the solute based on the observed freezing point depression.
Therefore, in your freezing point depression experiment with glycol, you assumed the value of i to be 1 because glycol does not dissociate into ions or associate with the solvent molecules, making it a non-electrolyte.