Problem 62 - A Complete Resource Book in Chemistry for JEE Main

Question

The rise in the boiling point of a solution containing \(1.8 \mathrm{~g}\) of glucose in \(100 \mathrm{~g}\) of solvent is \(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is (a) \(1 \mathrm{~K} / \mathrm{m}\) (b) \(0.1 \mathrm{~K} / \mathrm{m}\) (c) \(0.01 \mathrm{~K} / \mathrm{m}\) (d) \(10 \mathrm{~K} / \mathrm{m}\)

Step-by-Step Solution

Answer

The molal elevation constant is \( 1 \text{ K/m} \) (option a).

Step 1: Understanding the problem

We need to calculate the molal elevation constant, represented as \( K_b \), using the given data about the solution: its boiling point elevation, the mass of glucose, and the mass of the solvent. The boiling point elevation formula is \( \Delta T_b = i \times K_b \times m \), where \( \Delta T_b \) is the boiling point increase, \( i \) is the van't Hoff factor (which is 1 for glucose since it doesn't dissociate), and \( m \) is the molality.

Step 2: Calculating the moles of solute

Calculate the number of moles of glucose (\( C_6H_{12}O_6 \)) using its molar mass, which is 180 g/mol. Apply the equation: \( \text{Moles of glucose} = \frac{1.8 \text{ g}}{180 \text{ g/mol}} = 0.01 \text{ moles} \).

Step 3: Finding the molality of the solution

Molality (\( m \)) is calculated by dividing the moles of solute by the mass of the solvent in kilograms. Convert 100 g of solvent to kilograms, which is 0.1 kg. Then, \( m = \frac{0.01 \text{ moles}}{0.1 \text{ kg}} = 0.1 \text{ mol/kg} \).

Step 4: Applying the boiling point elevation formula

Use the formula for boiling point elevation: \( \Delta T_b = K_b \times m \). Given \( \Delta T_b = 0.1 \degree C \) and \( m = 0.1 \text{ mol/kg} \), substitute them into the equation: \( 0.1 = K_b \times 0.1 \).

Step 5: Solving for the molal elevation constant

Isolate \( K_b \) from the equation: \( K_b = \frac{0.1}{0.1} = 1 \). This means the molal elevation constant \( K_b \) is \( 1 \text{ K/m} \).

Key Concepts

Molal Elevation Constant

The molal elevation constant, symbolized as \( K_b \), is a crucial parameter in boiling point elevation problems in chemistry. It represents how much the boiling point of a solvent increases per molal concentration of solute added. The unit of \( K_b \) is Kelvin per molal (\( \text{K/m} \)), indicating the boiling point rise in Kelvin per molal (mol/kg) concentration of solute.

This constant is specific to each solvent and depends on its properties. In the context of making solutions, understanding \( K_b \) helps us predict how a solute will alter a solvent's boiling point, which is essential for applications like antifreeze formulations. This property arises because solute particles disrupt solvent-solvent interactions, requiring more heat (energy) for the solvent to transition to a gaseous state.

• Higher \( K_b \) means the solvent's boiling point is greatly affected by dissolved solutes.
• The boiling point elevation can be calculated using \( \Delta T_b = i \times K_b \times m \), where \( \Delta T_b \) is the change in boiling point, \( i \) is the van't Hoff factor, and \( m \) is the molality.

Molality

Molality is a measure of the concentration of a solute in a solution and is denoted by the symbol \( m \). It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which depends on the volume of solution, molality is related to the mass of the solvent, making it a more stable concentration measure over a range of temperatures.

To calculate molality, you need to know the mass of the solute, its molar mass, and the mass of the solvent in kilograms. Simply divide the moles of solute by the mass of the solvent in kilograms:

• For example, with 0.01 moles of glucose dissolved in 0.1 kg of solvent, the molality \( m \) is \( \frac{0.01}{0.1} = 0.1 \text{ mol/kg} \).

Molality is particularly useful in colligative property calculations, such as those related to freezing-point depression or boiling-point elevation, since these properties depend on the number of solute particles rather than their volume.

Van't Hoff Factor

The van't Hoff factor, \( i \), quantifies the effect of solute particles on colligative properties of solutions. It represents the number of particles into which a solute dissociates in solution. For glucose, \( i = 1 \) because it doesn't dissociate into smaller particles in solution.

Understanding the van't Hoff factor is essential when calculating colligative properties because it directly affects properties such as boiling-point elevation and freezing-point depression.

• For example, electrolytes like \( ext{NaCl} \) have \( i = 2 \) because they dissociate into sodium and chloride ions.
• Non-electrolytes like glucose maintain their molecular structure, resulting in \( i = 1 \).

Correctly applying the van't Hoff factor will ensure precise calculations in chemical solutions by accounting for the actual number of solute particles affecting the solvent's properties.

Glucose Solution

A glucose solution is simply water with dissolved glucose, a type of sugar with the molecular formula \( C_6H_{12}O_6 \). When glucose dissolves, it does not dissociate into ions, making it a non-electrolyte.

This property of glucose significantly simplifies calculating the boiling point elevation because the van't Hoff factor, \( i \), is 1. In practical applications, glucose solutions are often used in biochemistry and medical fields due to glucose's role in cellular processes.

• The molality of a glucose solution is determined by the amount of glucose and the mass of the solvent, making it crucial for colligative property calculations.
• In exercises, understanding the behavior of glucose solutions helps predict how they alter the boiling or freezing point of water.

Being a familiar substance in both food and the body, glucose provides an excellent model for studying solution properties without the complexity of ionic dissociation.