# Arrange the following solutions in order of decreasing freezing points: 0.10 m Na3PO4,0.35 m NaCl,0.20 m MgCl2,0.15 m C6H1206,0.15 m CH3COOH (:Madisen:)

The solution with the lowest molality will have the highest freezing point (smallest freezing point depression).

Assume that all the salts are completely dissociated. Calculate the molality of the ions in the solutions.

#### (e) 0.15 m CH₃COOH------------------- weak electrolyte, slightly greater than 0.15 m

Thus, the above solutions will be arranged as follows: (d)> (e)> (a)> (c)> (b) (decreasing freezing points).

The freezing point will be depressed most by the solution that contains the most solute particles. You should try to classify each solute as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. All three solutions have the same concentration, so comparing the solutions is straightforward.

HCl is a strong electrolyte, so under ideal conditions it will completely dissociate into two particles per molecule. The concentration of particles will be 1.00 m.

Acetic acid is a weak electrolyte, so it will only dissociate to a small extent. The concentration of particles will be greater than 0.50 m, but less than 1.00 m.

Glucose is a nonelectrolyte, so glucose molecules remain as glucose molecules in solution. The concentration of particles will be 0.50 m. For these solutions, the order in which the freezing points become lower is: 0.50 m glucose > 0.50 m acetic acid > 0.50 m HCl.

In other words, the HCl solution will have the lowest freezing point (greatest freezing point depression).

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