Which of the following solutes in aqueous solution would be expected to exhibit the smallest freezing-point lowering (assuming ideal behavior?

0.1 m NaCl
.2 m CH3COOH
.05 m Al2(SO4)3
.1 m MgCl2
.25 m NH3

What would be the steps to picking the right option? Because just picking .05 because it appears to be the smallest amount is likely wrong.

delta T = i*Kf*m

Since Kf is constant we can forget that. dT depends upon i and m so
i*m for each.
NaCl is 0.1*2 = ? for i = 2
CH3COOH is 0.2*1 for i = 1
Al2(SO4)3 is 0.05*5 for i = 5
MgCl2 i = 3
NH3 i = 1
The greatest i*m causes the greatest dT (or the smallest i*m causes the least dT).

It’s nacl

0.1 m MgCl2

To determine the solute that would exhibit the smallest freezing-point lowering, you need to compare the number of particles produced by each solute in the solution. According to the Colligative Properties, the extent of freezing-point depression depends on the number of solute particles, rather than their individual mass or size. The more particles there are, the greater the freezing-point depression.

Here are the steps to determine the solution with the smallest freezing-point lowering:

1. Identify the solutes in the given options: NaCl, CH3COOH, Al2(SO4)3, MgCl2, and NH3.
2. Determine the number of particles produced by each solute when dissolved in water.
3. Compare the number of particles for each solute.
4. Select the solute with the fewest particles as the one that would exhibit the smallest freezing-point lowering.

So, let's calculate the number of particles produced for each solute:

1. NaCl: NaCl dissociates into Na+ and Cl- ions. Therefore, it produces two particles.
2. CH3COOH: CH3COOH does not fully dissociate in water but rather partially ionizes. It produces one particle.
3. Al2(SO4)3: Al2(SO4)3 dissociates into Al3+ and (SO4)2- ions. Therefore, it produces four particles.
4. MgCl2: MgCl2 dissociates into Mg2+ and 2 Cl- ions. Therefore, it produces three particles.
5. NH3: NH3 does not dissociate or ionize in water; it remains as NH3 molecules. Therefore, it produces one particle.

Now, comparing the number of particles:

- NaCl: 2 particles
- CH3COOH: 1 particle
- Al2(SO4)3: 4 particles
- MgCl2: 3 particles
- NH3: 1 particle

Therefore, the solute with the smallest freezing-point lowering would be CH3COOH (0.2 m) because it produces the fewest particles.

It is worth noting that this calculation assumes ideal behavior, where all solutes and solvents behave as expected. In reality, some deviations may occur, but the principle of comparing particle numbers still holds.