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


A. 0.1 m NaCl
B. 0.2 m CH3COOH
C. 0.1 m MgCl2
D. 0.05 m Al2(SO4)3
E. 0.25 m NH3

delta T = i*K*m

Since K is a constant it can be ignored and you are interested in the largest number for i*m where i = # particles (which you get from the number of ions) and m = molality.

To determine which of these solutes would exhibit the largest freezing-point lowering, we need to consider the concept of colligative properties. The freezing-point lowering is a colligative property, which means it depends only on the number of solute particles, not their identity.

The degree of freezing-point lowering is directly proportional to the number of solute particles in solution. In other words, the more solute particles there are, the greater the freezing-point lowering.

To determine the number of solute particles each option generates, we need to consider the dissociation of solutes in water. When an ionic compound dissolves in water, it typically dissociates into ions, whereas molecular compounds like CH3COOH or NH3 do not dissociate.

Now, let's calculate the number of solute particles for each option:

A. 0.1 m NaCl:
NaCl dissociates into Na+ and Cl- ions, so we have 1 Na+ ion and 1 Cl- ion, giving a total of 2 solute particles.

B. 0.2 m CH3COOH:
CH3COOH does not dissociate, so we have only 1 solute particle.

C. 0.1 m MgCl2:
MgCl2 dissociates into 1 Mg2+ ion and 2 Cl- ions, giving a total of 3 solute particles.

D. 0.05 m Al2(SO4)3:
Al2(SO4)3 dissociates into 2 Al3+ ions and 3 SO42- ions, giving a total of 5 solute particles.

E. 0.25 m NH3:
NH3 does not dissociate, so we have only 1 solute particle.

Comparing the number of solute particles for each option, we find that D. 0.05 m Al2(SO4)3 generates the largest number of solute particles (5 solute particles). Therefore, option D would be expected to exhibit the largest freezing-point lowering.