# How to determine which aqueous solution has the largest freezing point depression

Question

This is more of a conceptual question. Let's say that we are given different aqueous solutions and we are asked to choose the one with the highest freezing point depression, which do we choose? Is there a pattern? do we choose the one with salts inside of it?

Example question

Which aqueous solution exhibits the largest freezing point depression? \begin{align}(\mathrm{A})&\ 1.0\ \mathrm{M}\ \ce{KBr} & (\mathrm{C})&\ 0.5\ \mathrm{M}\ \ce{MgCl2} \\ (\mathrm{B})&\ 0.75\ \mathrm{M}\ \ce{C6H12O6}& (\mathrm{D}) &\ 0.25\ \mathrm{M}\ \ce{Ga2(SO4)3}\end{align}

now this question is a bit different because it has concentrations, so do we choose the one with the highest concentrations or the one with salts?

BTW

But my hypothesis is that we only look at the ones that contains salts/ions inside of them and then the one with the highest concentration. Please let me know if this is correct or not

Freezing point depression is a function of the amount of dissolved substance — but substance must be treated as distinct particles rather than compound. This is an important distinction because ionic compounds, when they dissolve in water, typically separate into hydrated anions and hydrated cations — two distinct species as far as freezing point depression is concerned.

Thus, for ionic compounds you first need to determine how many individual ions are generated and then multiply that by the overall concentration. In your examples:

• $1~\mathrm{M}\ \ce{KCl}$. One cation, one anion. Thus, there are two moles of dissolved particles per mole of compound; giving $2~\mathrm{M}$ effective particle concentration.

• $0.75~\mathrm{M}\ \ce{C6H21O6}$. This compound is not ionic. Thus, it does not dissociate and one mole dissolved gives one mole of particles. $0.75~\mathrm{M}$ effective particle concentration.

• $0.5~\mathrm{M}\ \ce{MgCl2}$. This again is ionic but we get two moles of anions per mol of compound in addition to one mole of cations. Thus three moles of ions per mole of compound. Effective particle concentration $1.5~\mathrm{M}$.

• $0.25~\mathrm{M}\ \ce{Ga2(SO4)3}$. Here, we have a total of two moles of cations and three of anions per mole of compound. Thus, the effective particle concentration is five times the original or $1.25~\mathrm{M}$.

The actual formula for freezing point depression is based on a constant factor of degrees per dissolved mole; and this analysis was used to confirm that ionic compounds separate into ions and to determine molecular weight in the past.

Depression of freezing point of a solvent(here water) is a colligative property of a solution i.e., it depends on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. Therefore, more the number of solute particles, more is the depression of freezing point of the solvent. Also ionic compounds when dissolve in water, dissociate into anions and cations which depends on dissociation constant, so the van't Hoff factor, i must also be taken into consideration. So, applying above stated logic, option (A) is the most appropriate answer for the given question.