Choose the aqueous solution below with the lowest freezing point. These are all solutions of nonvolitile solutes and you should assume ideal van't Hoff factors where applicable.

a. 0.075 m NaI
b. 0.075 m (NH4)3PO4
c. 0.075 m NaBrO4
d. 0.075 m LiCN
e. 0.075 m KNO2

I can't find enough information on this subject to make it clear?

A little reasoning is what you need.

The freezing point is
delta T = i*kf*m
You know Kf is the same for all of them. You know, from the problem, that m is the same (0.075m) for all of them, so what's the difference? The only difference is i. A small i gives a smaller delta T than a large i, right?
Note: You COULD calculate the answers for all given then you would KNOW the freezing point of each and you could choose that way but there is no need to do that. For example, A is
i*Kf*m = 2*1.86*0.075 = 0.279 so freezing point is -0.279 degrees C.
The easy way is to use i. Largest i gives largest delta T and that gives smallest (lowest) freezing point.

To determine the aqueous solution with the lowest freezing point, we need to compare the van't Hoff factors (i) and the concentration (molarity) of each solution. The van't Hoff factor represents the number of particles an individual molecule of a solute dissociates into in solution. For example, NaI dissociates into two particles, Na+ and I-, so its van't Hoff factor is 2.

The freezing point depression is directly proportional to the van't Hoff factor and the molality (molarity) of the solution. The formula for freezing point depression is:

ΔTf = Kf × i × m

Where:
- ΔTf is the change in freezing point
- Kf is the cryoscopic constant (a characteristic of the solvent)
- i is the van't Hoff factor
- m is the molality (molarity) of the solution

In this case, we can assume ideal van't Hoff factors for all the solutes.

Now, by comparing the solutions listed:
a. 0.075 m NaI
The van't Hoff factor for NaI is 2.

b. 0.075 m (NH4)3PO4
(NH4)3PO4 dissociates into 4 particles: 3NH4+ and PO4^3-. The van't Hoff factor for (NH4)3PO4 is 4.

c. 0.075 m NaBrO4
NaBrO4 dissociates into 2 particles: Na+ and BrO4-. The van't Hoff factor for NaBrO4 is 2.

d. 0.075 m LiCN
LiCN dissociates into 2 particles: Li+ and CN-. The van't Hoff factor for LiCN is 2.

e. 0.075 m KNO2
KNO2 dissociates into 2 particles: K+ and NO2-. The van't Hoff factor for KNO2 is 2.

Now, we need to compare the solutions based on their van't Hoff factors and molality.
Since all the solutions have the same molality (0.075 m), the solution with the lowest freezing point will have the highest van't Hoff factor.

From the given options, (b) 0.075 m (NH4)3PO4 has the highest van't Hoff factor (4) compared to the other solutions with a van't Hoff factor of 2. Therefore, solution (b) has the lowest freezing point.

To determine the aqueous solution with the lowest freezing point, we need to consider the concept of freezing point depression. The freezing point of a solution is lower than that of the pure solvent due to the presence of a solute.

The freezing point depression is directly proportional to the molality of the solution. The molality (m) is defined as the number of moles of solute dissolved in 1 kg of solvent.

In this given scenario, all solutions have the same concentration of 0.075 m, so we can compare them based on the nature of the solute.

The freezing point depression depends on the dissociation or ionization of solutes in water. Higher dissociation will contribute to a greater freezing point depression.

We need to consider the van't Hoff factor (i) to determine the extent of dissociation. The van't Hoff factor represents the number of particles into which one formula unit dissociates in the solution.

Let's calculate the van't Hoff factors for each solute:

a. NaI dissociates into Na+ and I- ions. Therefore, the van't Hoff factor is 2.
b. (NH4)3PO4 dissociates into 3NH4+ and PO4(3-) ions. The van't Hoff factor is 4.
c. NaBrO4 dissociates into Na+, BrO4(-) ions. The van't Hoff factor is 2.
d. LiCN dissociates into Li+ and CN- ions. The van't Hoff factor is 2.
e. KNO2 dissociates into K+ and NO2- ions. The van't Hoff factor is 2.

Now, we can determine which solution will have the lowest freezing point by comparing the van't Hoff factors.

The solution with the highest van't Hoff factor will have the greatest freezing point depression, indicating the lowest freezing point. Therefore, the solution with the highest van't Hoff factor is (b) 0.075 m (NH4)3PO4 since it has a van't Hoff factor of 4.

Therefore, the aqueous solution with the lowest freezing point is 0.075 m (NH4)3PO4.