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Chapter 13: Problem 72

List the following aqueous solutions in order of decreasing freezing point: 0.040 \(\mathrm{m}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{m} \mathrm{KBr}\) 0.030 \(\mathrm{mphenol}\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\)

Short Answer

Expert verified
The aqueous solutions in order of decreasing freezing point are: 1) Glycerin and KBr (both with a freezing point depression of 0.0744 °C) and 2) Phenol (with a freezing point depression of 0.0558 °C).

Step by step solution

01

Determine the van't Hoff factor (i) of each solution.

For glycerin, one molecule of glycerin (C3H8O3) does not dissociate in solution, so i = 1. For KBr, it will dissociate into K+ and Br- ions, so the i = 2. For phenol, one molecule of phenol (C6H5OH) does not dissociate in solution, so i = 1.
02

Calculate the molality (m) of each solute.

The given molality (m) for each solution is: - Glycerin: m = 0.040 m - KBr: m = 0.020 m - Phenol: m = 0.030 m
03

Calculate the freezing point depression of each solution.

The freezing point depression (∆Tf) is given by the equation: ∆Tf = i * Kf * m, where Kf is the molal freezing point depression constant for water, which is 1.86 °C/m. For glycerin: ∆Tf = (1) * (1.86 °C/m) * (0.040 m) = 0.0744 °C For KBr: ∆Tf = (2) * (1.86 °C/m) * (0.020 m) = 0.0744 °C For phenol: ∆Tf = (1) * (1.86 °C/m) * (0.030 m) = 0.0558 °C
04

Order the solutions by decreasing freezing point depression.

From the calculated values: - Glycerin: ∆Tf = 0.0744 °C - KBr: ∆Tf = 0.0744 °C - Phenol: ∆Tf = 0.0558 °C As higher freezing point depression indicates lower freezing point, we can list the solutions in order of decreasing freezing point: 1. Glycerin and KBr (same freezing point depression: 0.0744 °C) 2. Phenol (freezing point depression: 0.0558 °C)

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