Question

Arrange \(0.10 \mathrm{~m}\) aqueous solutions of the following solutes in order of decreasing freezing point and boiling point. (a) \(\mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{MgSO}_{4}\)

Short answer

Question: Arrange the following aqueous solutions of solutes in order of decreasing freezing point and boiling point: (a) \(\mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3}\), (b) \(\mathrm{CH}_{3} \mathrm{OH}\), (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), (d) \(\mathrm{MgSO}_{4}\). Answer: The order of decreasing freezing point and boiling point for the given solutes is: \(\mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3} \to\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \to\) \(\mathrm{MgSO}_{4} \to\) \(\mathrm{CH}_{3} \mathrm{OH}\).

Step-by-step solution

Identify the number of ions produced by each solute in solution

(a) \(\mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3}\): In water, this solute will dissociate into 1 Al³⁺ ion and 3 ClO₃⁻ ions, totaling in 4 ions. (b) \(\mathrm{CH}_{3} \mathrm{OH}\): This solute will not dissociate in water. It will remain as a single molecule, producing only 1 particle. (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\): In water, this solute will dissociate into 2 NH₄⁺ ions and 1 Cr₂O₇²⁻ ion, totaling in 3 ions. (d) \(\mathrm{MgSO}_{4}\): In water, this solute will dissociate into 1 Mg²⁺ ion and 1 SO₄²⁻ ion, totaling in 2 ions.

Arrange the solutes in order of decreasing freezing point and boiling point

The solutes can be arranged in the order of decreasing freezing point and boiling point based on the number of ions they produce in water. More ions lead to a greater decrease in freezing point (more negative) and a greater increase in boiling point (higher values). The order will be: 1. \(\mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3}\) (4 ions) 2. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (3 ions) 3. \(\mathrm{MgSO}_{4}\) (2 ions) 4. \(\mathrm{CH}_{3} \mathrm{OH}\) (1 ion) So, the order of decreasing freezing point and boiling point for the given solutes is: \(\mathrm{Al}\left(\mathrm{ClO}_{3}\right)_{3} \to\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \to\) \(\mathrm{MgSO}_{4} \to\) \(\mathrm{CH}_{3} \mathrm{OH}\).