Problem 66 - A Complete Resource Book in Chemistry for JEE Main

Question

The freezing point of \(0.1 \mathrm{M}\) solution of glucose is \(-1.86^{\circ} \mathrm{C}\). If an equal volume of \(0.3 \mathrm{M}\) glucose solution is added, the freezing point of the mixture will be (a) \(-5.58^{\circ} \mathrm{C}\) (b) \(-7.44^{\circ} \mathrm{C}\) (c) \(-3.72^{\circ} \mathrm{C}\) (d) \(-2.79^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The freezing point of the mixture is \(-3.72^{\circ} \mathrm{C}\).

Step 1: Understand the Concept

The depression in freezing point is given by the formula: \( \Delta T_f = i \cdot K_f \cdot m \) where \( i \) is the van't Hoff factor (1 for glucose, a non-electrolyte), \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution. For solutions with equal volumes being mixed, the average molality after mixing needs to be calculated and then used to find the new freezing point.

Step 2: Calculate the Initial Depression in Freezing Point

The freezing point depression of the original \(0.1 \text{ M}\) glucose solution is \( -1.86^{\circ} \text{C} \), which means \( \Delta T_f = 1.86 \text{C} \). Since the two solutions have equal volume, we need to find the average concentration after mixing to find the new freezing point depression.

Step 3: Determine the New Average Concentration

When equal volumes of \(0.1 \text{ M}\) and \(0.3 \text{ M}\) glucose solutions are mixed, the average molarity is calculated as \( \frac{0.1 + 0.3}{2} = 0.2 \text{ M}\).

Step 4: Calculate the New Depression in Freezing Point

Using the average molarity of \(0.2 \text{ M}\), the new freezing point depression is \( \Delta T_f = 1.86 \text{C} \times 0.2/0.1 = 3.72 \text{C} \).

Step 5: Find the New Freezing Point of the Solution

Since the pure solvent's (water) freezing point is \(0^{\circ} \text{C}\), subtract the depression from this to get the freezing point of the mixed solution. Thus, the freezing point is \(0 - 3.72 = -3.72^{\circ} \text{C}\).

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating colligative property, which means it relates to the number of solute particles in a solution rather than their specific identities. When solute particles are added to a solvent, like water, the freezing point of the solution becomes lower than that of the pure solvent. This happens because the solute particles interfere with the formation of the solid crystal structure of the solvent.
The formula used to calculate this is:

• \( \Delta T_f = i \cdot K_f \cdot m \)

Here, \( \Delta T_f \) represents the change in the freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant (a property of the solvent), and \( m \) is the molality of the solution.
In the exercise you provided, freezing point depression allows us to understand how mixing two solutions with different concentrations of glucose impacts the overall freezing point of the resulting solution. This approach utilizes the average molality of the solutions which directly influences the total freezing point depression.

Solution Chemistry

Solution chemistry involves understanding how substances dissolve and interact in a solvent, forming a solution. When you dissolve a solute in a solvent, the solute particles become dispersed at the molecular level, creating a homogenous mixture called a solution.
Several factors influence how substances dissolve:

• Temperature: Typically, increasing temperature can increase solubility of solids and liquids.
• Agitation: Stirring helps distribute the solute within the solvent.
• Nature of Solute and Solvent: Like dissolves like, meaning polar solutes dissolve well in polar solvents and non-polar solutes in non-polar solvents.

In the context of the exercise, solution chemistry principles explain how glucose, a non-ionizing solute, dissolves in water, a polar solvent, forming a solution that lowers the freezing point compared to that of pure water. Understanding the concentration of the solution (through molarity and molality) is vital for calculating properties such as freezing point depression.

Van't Hoff Factor

The van't Hoff factor, denoted as \( i \), is essential for accurately determining the colligative properties of a solution. It represents the number of particles into which a solute dissociates in solution. For example, an ionic compound dissociating into multiple ions has a van't Hoff factor greater than one.
In the case of glucose, which is a non-electrolyte, it does not dissociate into ions when dissolved in water. Therefore, its van't Hoff factor equals one: \( i = 1 \). This simplifies calculations for solutions containing glucose since you don't have to account for multiple species.

• The formula for the freezing point depression incorporating the van't Hoff factor is:\[ \Delta T_f = i \cdot K_f \cdot m \]

Understanding the van’t Hoff factor is crucial as it directly impacts calculated values for freezing point depression, boiling point elevation, and osmotic pressure. Its significance is well illustrated in the problem above, where non-dissociating glucose simplifies calculations.