Applying Models a. You are conducting a freezing-point...

Applying Models a. You are conducting a freezing-point determination in the laboratory by using an aqueous solution of $\mathrm{KNO}_{3}$ . The observed freezing point of the solution is $-1.15^{\circ} \mathrm{C} .$ Using a pure water sample, you recorded the freezing point of the pure solvent on the same thermometer as $0.25^{\circ} \mathrm{C}$ . Determine the molal concentration of KNO $_{3}$ . Assume that there are no forces of attraction between ions. b. You are not satisfied with the result in part (a) because you suspect that you should not ignore the effect of ion interaction. You take a 10.00 $\mathrm{mL}$ sample of the solution. After carefully evaporating the water from the solution, you obtain 0.415 $\mathrm{g}$ KNO. Determine the actual molal concentration of KNO $_{3}$ and the percentage difference between the concentration observed in the freezing-point determination and the actual concentration of $\mathrm{KNO}_{3}$ . Assume that the solution's density is 1.00 $\mathrm{g} / \mathrm{mL}$ .

Applying Models
a. You are conducting a freezing-point determination in the laboratory by using an aqueous solution of $\mathrm{KNO}_{3}$ . The observed freezing point of the solution is $-1.15^{\circ} \mathrm{C} .$ Using a pure water sample, you recorded the freezing point of the pure solvent on the same thermometer as $0.25^{\circ} \mathrm{C}$ . Determine the molal concentration of KNO $_{3}$ . Assume that there are no forces of attraction between ions.
b. You are not satisfied with the result in part (a) because you suspect that you should not ignore
the effect of ion interaction. You take a 10.00 $\mathrm{mL}$ sample of the solution. After carefully evaporating the water from the solution, you obtain 0.415 $\mathrm{g}$ KNO. Determine the actual molal concentration of KNO $_{3}$ and the percentage difference between the concentration observed in the freezing-point determination and the actual concentration of $\mathrm{KNO}_{3}$ . Assume that the solution's density is 1.00 $\mathrm{g} / \mathrm{mL}$ .

Key Concepts

Freezing Point Depression

This is a colligative property where the addition of a non-volatile solute lowers the freezing point of a solvent. The extent of the freezing point depression is directly proportional to the molal concentration of the solute particles in the solution.

Colligative Properties

Colligative properties are characteristics of solutions that depend solely on the number of solute particles present, rather than their specific nature. They include phenomena such as boiling point elevation, vapor pressure lowering, osmotic pressure, and freezing point depression.

Molality

Molality is a concentration unit defined as the number of moles of solute per kilogram of solvent. It is particularly useful in colligative property calculations because it does not change with temperature.

Ion Dissociation

When ionic compounds dissolve in water, they separate into their constituent ions, effectively increasing the number of particles in the solution. This dissociation can affect the magnitude of colligative properties and needs to be taken into account in precise calculations.

van't Hoff Factor

The van't Hoff factor represents the number of particles into which a solute dissociates in solution. It is used to adjust calculations involving colligative properties for the effects of ionization in electrolyte solutions.

Experimental Error Analysis

This concept involves comparing different methods of determining a property (for example, the freezing point depression method versus direct mass determination) to assess the accuracy and potential errors in experimental results. It provides insights into systematic and random errors affecting the reliability of the measurements.

Transcript

00:01 All right.

00:01 So let's look at some experimental data right quick, and we're going to use it to solve for the molality of potassium nitrate solution.

00:11 The van hofye factor for our potassium nitrate solution is we're going to assume it to be two, since we generate a potassium ion and a nitrate ion.

00:20 So it's two particles in total.

00:22 The kf is going to be specific to water, 1 .86, and then the molality is what we're trying to find.

00:30 The delta t or the change in the freezing point is what will be able to determine from the experimental data they give us.

00:37 They tell us that when it was just pure water, it froze at 0 .25 degrees celsius using that thermometer.

00:45 But when they tested the potassium nitrate solution, it froze at negative 1 .15 degrees celsius.

00:53 So in order to get the difference between the two, we'll take the pure temperature, 0 .25, and subtract the new.

01:00 Temperature 1 .15.

01:02 So essentially it'll be like 0 .25 minus negative 1 .15.

01:08 So it really ends up being added for a total difference of 1 .40 degrees celsius.

01:14 So that's what we'll go in for our delta t.

01:18 All right, let's divide and find our molality.

01:30 Okay, so there we go.

01:31 So malality will be equal to the change in temperature divided by the vant -hoff i factor times the freezing point constant for water.

01:42 What that gives us is a malality of 0 .376 malau.

01:48 So this is our experimental answer for part a.

01:51 It now asks us to consider part b, where they do a little bit more of the experiment to try to actually confirm the malality.

01:59 They take 10 mils of the solution and determine that it has 0 .415 grams of k and 03.

02:06 To get molality, we'll need to first convert the grams of solute to moles using the molar mass, where we add up the mass of potassium nitrate or potassium nitrogen, then the three oxygens.

02:20 That will give us a total molar mass of 101 .11 grams per one mole.

02:27 So you do 0 .415 divided by 101 .11, and that gives us a molality that's quite small.

02:35 So let's write it in the sign, or a number of moles that's quite small...

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