Problem 70 - Chemistry: The Central Science in SI Units, Global Edition

Question

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 \(\left(\mathrm{NaNO}_{3}\right)\).

Step-by-Step Solution

Answer

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 1: 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

Step 2: 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

Step 3: Consider dissociation for sodium nitrate

Sodium nitrate is an ionic compound that dissociates in water into sodium ions (Na⁺) and nitrate ions (NO₃⁻). 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

Step 4: 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.

Key Concepts

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.