Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.22 m glycerol (C 3 H 8 O 3 ) in ethanol, (b) 0.240 mol of naphthalene (C 10 H 8 ) in 2.45 mol of chloroform, (c) 1.50 g NaCl in 0.250 kg of water, (d) 2.04 g KBr and 4.82 g of glucose (C 6 H 12 O 6 ) in 188 g of water.

Hello everyone today, we are being asked to determine the freezing and boiling points Of a solution composed of 10 g of sucrose dissolved in 250 g of water. So the first thing you want to do is we want to determine the morality that we're working with. And morality can be expressed as morality is equal to our We have 10 g and we are going to have to Multiply that by our Mueller mass of our sucrose or a compound which in this case is 342.3 g per mole. And they were going to multiply by our weight or a mass which is 250 g. But when we convert it to kg by using our conversion factor of We can say that one g is 10 to the negative third kg to give us 0.25 kg. Once we do that, we get a morality value of 0.117, we're gonna know that with em for units and now we can determine the freezing point. So we can say the freezing point noted by F. P is going to be our total change in our temperature. So you go to our KF or our constant times our morality. And so that would look like The 1. degrees of Celsius over em multiplied by the morality that we just found, which is 0.117 morality. And that would give us a temperature of 0.22 degrees Celsius. And so in finding our T F RR temperature, we're going to take the freezing point of water. And we're going to say that zero degrees Celsius and subtract that from our delta T. 0.22 degrees Celsius and get -0.22°C for our freezing point. We're gonna go and highlight that and then boiling point. We're going to go ahead and a note by BP it's gonna be a very similar process. We're going to say our total boiling point elevation is going to be equal to our KB. Or our constant times our morality and that's going to look very similar. But instead the boiling point constant is going to be 0.513 degrees Celsius over m times the morality that we sold 4. modalities that gives us 0.6 degrees Celsius. And then solving for T. B. We have to take our boiling point of water which is 100 degrees Celsius. And we have our our boiling point of sucrose. Excuse me. And we add our Delta T. B. Which is .06°C. and we get an answer of 400.06°C. That's our final answer. I hope this helped. And until next time

Ch.13 - Properties of Solutions

Brown - Chemistry: The Central Science 14th Edition

Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook

All textbooks

Brown 14th Edition

Ch.13 - Properties of Solutions

Problem 73

Chapter 13, Problem 73

Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.22 m glycerol (C₃H₈O₃) in ethanol, (b) 0.240 mol of naphthalene (C₁₀H₈) in 2.45 mol of chloroform, (c) 1.50 g NaCl in 0.250 kg of water, (d) 2.04 g KBr and 4.82 g of glucose (C₆H₁₂O₆) in 188 g of water.

Verified step by step guidance

Identify the colligative properties involved: freezing point depression and boiling point elevation.

Use the formula for freezing point depression: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant for chloroform, and \( m \) is the molality of the solution.

Calculate the molality \( m \) of the solution using the formula \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). Here, the solute is naphthalene and the solvent is chloroform.

Use the formula for boiling point elevation: \( \Delta T_b = i \cdot K_b \cdot m \), where \( K_b \) is the boiling point elevation constant for chloroform.

Determine the new freezing and boiling points by subtracting \( \Delta T_f \) from the normal freezing point and adding \( \Delta T_b \) to the normal boiling point of chloroform.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the number of solute particles in a given amount of solvent, rather than the identity of the solute. These properties include boiling point elevation and freezing point depression, which are crucial for calculating the changes in freezing and boiling points when a solute is added to a solvent.

Freezing Point Depression

Freezing point depression is a colligative property that describes the decrease in the freezing point of a solvent when a solute is dissolved in it. The extent of freezing point depression can be calculated using the formula ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van 't Hoff factor, Kf is the freezing point depression constant of the solvent, and m is the molality of the solution.

Boiling Point Elevation

Boiling point elevation is another colligative property that refers to the increase in the boiling point of a solvent when a solute is added. This can be calculated using the formula ΔTb = i * Kb * m, where ΔTb is the change in boiling point, i is the van 't Hoff factor, Kb is the boiling point elevation constant of the solvent, and m is the molality of the solution. Understanding this concept is essential for determining the new boiling point of the solution.