Calculate the expected boiling and freezing point for the...

Calculate the expected boiling and freezing point for the solution in the previous problem. If you had to bring this syrup to the boiling point for a recipe, would you expect it to take much more time than it takes to boil the same amount of pure water? Why or why not? Would the syrup freeze in a typical freezer $\left(-18^{\circ} \mathrm{C}\right) ?$ Why or why not?

Key Concepts

Freezing Point Depression

Freezing point depression is another colligative property where the freezing temperature of a solvent is lowered when a solute is present. The addition of solute molecules disrupts the crystallization process, meaning more energy must be removed (i.e., a lower temperature achieved) to freeze the solution than would be necessary for the pure solvent.

Colligative Properties

Colligative properties are characteristics of solutions that depend on the number of solute particles in a solvent rather than their specific identity. These properties include boiling point elevation and freezing point depression, and they are key to understanding how adding a solute like sugar to water can alter its physical behavior compared to the pure solvent.

Boiling Point Elevation

Boiling point elevation is a colligative property whereby the boiling point of a solvent is raised when a solute is dissolved in it. This happens because the solute particles interfere with the formation of vapor bubbles by reducing the solvent's vapor pressure, meaning the solution must be heated to a higher temperature to achieve boiling compared to the pure solvent.

Practical Implications in Cooking

In cooking or food preparation, understanding the impact of solutes on boiling and freezing points is critical. A solution with an elevated boiling point may require more energy or time to reach the boiling condition compared to pure water, although the actual time difference can be influenced by various factors like heat transfer efficiency. Similarly, a lowered freezing point indicates that such a solution might not freeze at the temperatures typical of a household freezer, which is crucial when considering texture and processing of culinary products.

Transcript

00:01 For this next question, you're asked to calculate the expected boiling and freezing points of the solution prepared in problem 135.

00:15 To calculate the boiling point and the freezing point, you need to use the boiling point and freezing point equations with their corresponding k values, kf, the freezing point constant of water, or kb, the boiling point constant of water.

00:36 When using these equations, the only thing we don't do.

00:39 Have in order to calculate the change in temperature is molality.

00:42 So we will calculate molality from the solution prepared in problem 135.

00:49 You add 13 .62 grams of sucrose.

00:54 Molality is moles of the sucrose, most of the solute per kilogram of solvent.

01:00 So to get moles of the solute sucrose, we'll divide by the molar mass of sucrose.

01:05 Now we have moles sucrose.

01:08 Now we need to divide by the kilograms of the solvent water.

01:11 We have 241 .5 milliliters of water used to prepare the solution.

01:18 So we can convert the milliliters of water into grams of water with density.

01:25 This density was provided at 0 .997 grams per mil liter.

01:29 Then when we have grams, we simply divide by a thousand to get kilograms.

01:33 So moles solute per kilogram solvent gives us molality .165 molal sucrose.

01:45 Now we can calculate the boiling point elevation, molality multiplied by .512, gives us an increase from 100 degrees celsius of .0846, so it now boils at 100 .0846 or .08545...

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