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Chapter 13: Problem 70

Arrange the following aqueous solutions, each \(10 \%\) by mass in solute, in order of increasing boiling point: glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right),\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) sodium nitrate \(\left(\mathrm{NaNO}_{3}\right)\).

Short Answer

Expert verified
The order of increasing boiling point for the given aqueous solutions with 10% by mass solute is sucrose (\(0.324 \, \text{mol/kg}\)) < glucose (\(0.617 \, \text{mol/kg}\)) < sodium nitrate (\(2.614 \, \text{mol/kg}\)).

Step by step solution

01

Calculate the moles of solute in each solution

The mass percentage of each solute in the solutions is 10% by mass. That means, for every 100 g of solution, there are 10g of solute and 90g of solvent (water). We will now calculate the moles of solute in each solution using their respective molar masses: - For glucose (\(C_{6}H_{12}O_{6}\)), Molar mass = 180 g/mol Moles of glucose = \(\frac{10 \, \text{g}}{180 \, \text{g/mol}}\)= 0.0556 moles - For sucrose (\(C_{12}H_{22}O_{11}\)), Molar mass = 342 g/mol Moles of sucrose = \(\frac{10 \, \text{g}}{342 \, \text{g/mol}}\)= 0.0292 moles - For sodium nitrate (\(NaNO_{3}\)), Molar mass = 85 g/mol Moles of sodium nitrate = \(\frac{10 \, \text{g}}{85 \, \text{g/mol}}\)= 0.1176 moles
02

Calculate the molality of each solution

Now, we will calculate the molality of each solution using the previously calculated moles of solute and the mass of solvent (water) in kilograms. - Molality of glucose = \(\frac{0.0556 \, \text{moles}}{0.090 \, \text{kg}}\)= 0.617 mol/kg - Molality of sucrose = \(\frac{0.0292 \, \text{moles}}{0.090 \, \text{kg}}\)= 0.324 mol/kg - Molality of sodium nitrate = \(\frac{0.1176 \, \text{moles}}{0.090 \, \text{kg}}\)= 1.307 mol/kg
03

Consider dissociation for sodium nitrate

Sodium nitrate is an ionic compound that dissociates in water into sodium ions (Na鈦�) and nitrate ions (NO鈧冣伝). Thus, instead of one particle, two particles (ions) are present for every mole of sodium nitrate in solution. Therefore, we must double the molality of sodium nitrate to account for the dissociation: - Adjusted molality of sodium nitrate = \(2 \times 1.307 \, \text{mol/kg}\) = 2.614 mol/kg
04

Arrange the solutions in order of increasing boiling point

We can now arrange the solutions in order of increasing molality, as a greater number of solute particles leads to a higher boiling point: 1. Sucrose (\(0.324 \, \text{mol/kg}\)) 2. Glucose (\(0.617 \, \text{mol/kg}\)) 3. Sodium nitrate (\(2.614 \, \text{mol/kg}\)) Thus, the order of increasing boiling point is: sucrose < glucose < sodium nitrate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Molar Mass
Molar mass is a fundamental concept in chemistry. It allows us to find the mass of one mole of a substance, measured in grams per mole (g/mol). By knowing the molar mass, we can calculate how many moles are in a given sample of a substance.
For example:
  • The molar mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) is 180 g/mol.
  • The molar mass of sucrose (\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\)) is 342 g/mol.
  • The molar mass of sodium nitrate (\(\text{NaNO}_3\)) is 85 g/mol.

When solving chemical problems, we often need to convert from grams to moles using these molar masses. This conversion helps in determining how a substance behaves in chemical reactions and solutions.
The Role of Molality in Solutions
Molality is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality does not change with temperature, making it a reliable concentration measure.
To find molality:
  • Calculate the moles of solute.
  • Divide by the mass of the solvent in kilograms.

For example, let's say you have sucrose, glucose, and sodium nitrate:
  • The molality of glucose is calculated as 0.617 mol/kg.
  • The molality of sucrose is 0.324 mol/kg.
  • For sodium nitrate, it is initially 1.307 mol/kg; however, ionic dissociation alters this value.

Molality plays a critical role in determining properties like boiling point elevation.
Ionic Dissociation in Solution
Ionic dissociation refers to the process where ionic compounds dissolve in water, breaking down into individual ions. This process is crucial for understanding how ionic compounds behave in solutions and affects properties like boiling point.
For example:
  • Sodium nitrate (\(\text{NaNO}_3\)) dissociates into sodium ions (\(\text{Na}^+\)) and nitrate ions (\(\text{NO}_3^-\)) when dissolved in water.

This dissociation means that a single formula unit of sodium nitrate creates two particles in solution. Consequently, for sodium nitrate, you must double the molality when considering its effect on boiling point.
Understanding ionic dissociation is essential for predicting and controlling the behavior of solutions in various chemical applications.
Aqueous Solutions and Their Properties
An aqueous solution is a solution where water acts as the solvent. These solutions are vital in chemistry because many substances dissolve in water, enabling various chemical reactions.
Key properties of aqueous solutions include:
  • Solvent polarity, which impacts solute solubility.
  • Ability to conduct electricity if ions are present.
  • Changes in boiling and freezing points based on solute concentration.

In the context of boiling point elevation:
  • Sucrose and glucose are non-electrolytes, so they do not produce ions. Their effect on boiling point is based only on their molality.
  • Sodium nitrate is an electrolyte, resulting in more particles in solution due to dissociation.

Aqueous solutions are used in various fields, from biological systems to industrial processes, highlighting the importance of understanding their nature and behavior.

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Most popular questions from this chapter

At \(20^{\circ} \mathrm{C}\), the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(10 \mathrm{kPa}\), and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is \(2.9 \mathrm{kPa}\). Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fraction of a solution that has a vapor pressure of \(4.7 \mathrm{kPa}\) at \(20^{\circ} \mathrm{C} ?\) (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of \(5.00 \mathrm{~g}\) of lauryl alcohol in \(0.100 \mathrm{~kg}\) of benzene freezes at \(4.1^{\circ} \mathrm{C}\). What is the molar mass of lauryl alcohol from this data?

When ammonium chloride dissolves in water, the solution becomes colder. (a) Is the solution process exothermic or endothermic? (b) Why does the solution form?

You take a sample of water that is at room temperature and in contact with air and put it under a vacuum. Right away, you see bubbles leave the water, but after a little while, the bubbles stop. As you keep applying the vacuum, more bubbles appear. A friend tells you that the first bubbles were water vapor, and the low pressure had reduced the boiling point of water, causing the water to boil. Another friend tells you that the first bubbles were gas molecules from the air (oxygen, nitrogen, and so forth) that were dissolved in the water. Which friend is mostly likely to be correct? What, then, is responsible for the second batch of bubbles? [Section 13.4]

An ionic compound has a very negative \(\Delta H_{\text {soln }}\) in water. (a) Would you expect it to be very soluble or nearly insoluble in water? (b) Which term would you expect to be the largest negative number: \(\Delta H_{\text {solvent }}, \Delta H_{\text {solute }}\), or \(\Delta H_{\text {mix }} ?\)
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