Problem 70 - Chemistry The Central Science In Si Units

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 is: sucrose < glucose < sodium nitrate.

Step 1: Identify Boiling Point Elevation Concept

The boiling point of a solution is higher than that of the pure solvent due to the presence of a solute. This phenomenon is known as boiling point elevation, which is given by the formula: \[ \Delta T_b = i imes K_b imes m \]where \( \Delta T_b \) is the boiling point elevation, \( i \) is the van't Hoff factor (number of particles the solute dissociates into), \( K_b \) is the ebullioscopic constant (depends on the solvent, which is water here), and \( m \) is the molality of the solution.

Step 2: Determine van't Hoff Factor

The van't Hoff factor \( i \) reflects the number of particles a solute produces in solution:- For glucose \((\text{C}_6\text{H}_{12}\text{O}_6)\), \( i = 1 \) because it does not dissociate.- For sucrose \((\text{C}_{12}\text{H}_{22}\text{O}_{11})\), \( i = 1 \) as it also does not dissociate.- For sodium nitrate \((\text{NaNO}_3)\), \( i = 2 \) because it dissociates into \( \text{Na}^+ \) and \( \text{NO}_3^- \).

Step 3: Calculate Molality of Solutions

Assuming the density of water is approximately 1 g/mL, for a 10% solution by mass, there are 10 g of solute in 100 g of solution:- For glucose, molar mass is approximately 180 g/mol, molality \( m \approx \frac{10/180}{0.1} \approx 0.556 \text{ mol/kg} \).- For sucrose, molar mass is approximately 342 g/mol, molality \( m \approx \frac{10/342}{0.1} \approx 0.292 \text{ mol/kg} \).- For sodium nitrate, molar mass is approximately 85 g/mol, molality \( m \approx \frac{10/85}{0.1} \approx 1.176 \text{ mol/kg} \).

Step 4: Rank Solutions Based on Boiling Point

Calculate the boiling point elevation for each:- Glucose: \( \Delta T_b \approx 1 \times K_b \times 0.556 \)- Sucrose: \( \Delta T_b \approx 1 \times K_b \times 0.292 \)- Sodium nitrate: \( \Delta T_b \approx 2 \times K_b \times 1.176 \)Since \( K_b \) is a constant for water, only the product of \( i \) and \( m \) matters. Sodium nitrate has the highest product due to both dissociation and molality, leading to the highest boiling point. Next is glucose, then sucrose. Therefore, the order is sucrose \( < \) glucose \( < \) sodium nitrate.

Key Concepts

Van't Hoff Factor

The van't Hoff factor, denoted as \( i \), is a crucial concept in understanding boiling point elevation and other colligative properties of solutions. It represents the number of particles a solute dissociates into upon dissolving in a solvent. This factor directly influences the magnitude of boiling point elevation in a solution.

Let's break it down further:

• For non-electrolytes like glucose \((\text{C}_6\text{H}_{12}\text{O}_6)\) and sucrose \((\text{C}_{12}\text{H}_{22}\text{O}_{11})\), the van't Hoff factor \( i = 1 \), since they do not dissociate into multiple particles in aqueous solutions.
• However, for electrolytes such as sodium nitrate \((\text{NaNO}_3)\), the factor is higher. Sodium nitrate dissociates into two ions: \( \text{Na}^+ \) and \( \text{NO}_3^- \), thus \( i = 2 \).

This distinction is essential because the increase in the number of particles does not just affect the boiling point, but also the freezing point depression and osmotic pressure of solutions. Always consider the van't Hoff factor when dealing with solutions, especially when calculating colligative properties.

Molality

Molality is another key part of the puzzle when it comes to boiling point elevation. Represented by \( m \), molality is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of the solution, molality depends solely on the mass, making it independent of temperature changes.

Here's how molality is applied in our example:

• In a 10% by mass solution of glucose, where glucose's molar mass is approximately 180 g/mol, calculate molality as \( m = \frac{10/180}{0.1} = 0.556 \ \text{mol/kg} \).
• For sucrose, with a molar mass of 342 g/mol, the calculation yields \( m = \frac{10/342}{0.1} = 0.292 \ \text{mol/kg} \).
• Sodium nitrate, with a molar mass of 85 g/mol, gives \( m = \frac{10/85}{0.1} = 1.176 \ \text{mol/kg} \).

Understanding and calculating molality accurately is crucial because it directly influences the boiling point elevation when combined with the van't Hoff factor and the ebullioscopic constant.

Aqueous Solution

An aqueous solution is one where the solvent is water. Water's ability to dissolve various substances makes it a universal solvent, ideal for studying colligative properties such as boiling point elevation.

In the context of our exercise, each solution is a 10% by mass aqueous solution. Understanding this term is simple yet profound:

• The term "aqueous" indicates that water is the solvent in the solution.
• When dealing with aqueous solutions, the ebullioscopic constant \( K_b \) is specific to water, streamlining calculations related to boiling point elevation.
• Our exercise focuses on aqueous solutions of glucose, sucrose, and sodium nitrate, showcasing different behaviors due to the diverse van't Hoff factors.

Recognizing the nature of aqueous solutions helps better comprehend how solutes affect properties like boiling point and why water is typically used in chemical exercises and laboratory settings due to its versatility and predictive constants.