Why is the freezing point depression of potassium ethanoate greater than that of camphor or urea (in benzoic acid), with equal moles of each?

Wouldn't the Kethanoate break up into two ions per molecule?

To understand why the freezing point depression of potassium ethanoate (KCH3COO) is greater than that of camphor or urea (in benzoic acid), we need to consider the concept of colligative properties.

Colligative properties depend on the concentration of solute particles in a solution, rather than the specific nature of the solute. The key factor in determining the extent of freezing point depression is the number of solute particles present in the solution.

In the case of potassium ethanoate, each molecule dissociates into two ions when it dissolves in water - one potassium ion (K+) and one ethanoate ion (CH3COO-). So, for each mole of potassium ethanoate, we have two moles of solute particles (one mole of potassium ions and one mole of ethanoate ions).

On the other hand, camphor and urea do not dissociate into ions when they dissolve in benzoic acid. Instead, they remain as individual molecules in the solution.

Since freezing point depression is directly proportional to the concentration of solute particles, potassium ethanoate (with two ions per molecule) will have a greater effect on the freezing point depression compared to camphor or urea (with no dissociation into ions).

In summary, the freezing point depression of potassium ethanoate is greater than that of camphor or urea because potassium ethanoate dissociates into two ions per molecule, whereas camphor and urea remain as individual molecules.