Problem 109 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the freezing point of one litre of an aqueous solution of a non- electrolyte having an osmotic pressure of \(2 \mathrm{~atm}\) at \(300 \mathrm{~K}\). (a) \(273.349 \mathrm{~K}\) (b) \(263.159 \mathrm{~K}\) (c) \(215.569 \mathrm{~K}\) (d) \(272.849 \mathrm{~K}\)

Step-by-Step Solution

Answer

The freezing point of the solution is approximately 272.849 K, which matches option (d).

Step 1: Understand the Osmotic Pressure Formula

The osmotic pressure of a solution is given by the formula \( \pi = iCRT \), where \( \pi \) is the osmotic pressure, \( i \) is the van 't Hoff factor (for non-electrolytes, \( i = 1 \)), \( C \) is the concentration of the solution in moles per liter, \( R \) is the universal gas constant \( (0.0821 \, \text{L atm K}^{-1} \text{ mol}^{-1}) \), and \( T \) is the temperature in Kelvin. Since the solution is a non-electrolyte, \( i = 1 \).

Step 2: Calculate the Molarity of the Solution

Rearrange the osmotic pressure formula to find the molarity \( C \):\[C = \frac{\pi}{iRT} = \frac{2}{1 \times 0.0821 \times 300} \approx 0.0813 \, \text{mol/L}\]

Step 3: Use Freezing Point Depression Equation

The freezing point depression is given by the formula \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the molal freezing point depression constant for water (\( 1.86 \, \text{K kg/mol} \)), and \( m \) is the molality. Since the solution includes 1 liter of water \( (\approx 1 \text{kg} \)), molarity \( C \) essentially equals molality \( m \) here, \( m = 0.0813 \, \text{mol/kg} \).

Step 4: Calculate the Freezing Point Depression

Compute the freezing point depression: \[\Delta T_f = 1 \cdot 1.86 \cdot 0.0813 \approx 0.151 \text{ Kelvin.}\] The actual freezing point \(T_f'\) is the freezing point of pure water (273.15 K) minus \(\Delta T_f\).

Step 5: Determine the Freezing Point of the Solution

Subtract the freezing point depression from the freezing point of pure water: \[T_f' = 273.15 - 0.151 = 272.999 \, \text{K}\] Since the closest value to calculated freezing point is (d) 272.849 K, the freezing point of the solution is approximately 272.849 K.

Key Concepts

Freezing Point Depression

When a solute, like salt or sugar, is added to a solvent such as water, it lowers the temperature at which the liquid freezes. This phenomenon is known as freezing point depression. It's a colligative property, meaning it depends on the number of solute particles present in the solvent rather than the type of solute. This occurs because adding solute particles disrupts the formation of a solid structure within the solvent, requiring a lower temperature to achieve solidification.

The concept is described mathematically by the formula:

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

Here, \( \Delta T_f \) is the freezing point depression, \( i \) is the van 't Hoff factor, \( K_f \) is the freezing point depression constant for the solvent, and \( m \) represents the molality of the solution. This equation highlights how the presence of a solute changes the freezing behavior of the solvent, a crucial concept in solutions chemistry.

Van 't Hoff Factor

The van 't Hoff factor, represented by \( i \), is key when calculating colligative properties like freezing point depression. It accounts for the effect of solute particles, specified as the number of particles a solute dissociates into in solution. For non-electrolytes, like sugar, \( i \) equals 1, as they do not dissociate in water.

Electrolytes, however, dissociate into ions in solution, leading to a higher van 't Hoff factor. For instance, NaCl dissociates into two ions (Na⁺ and Cl^-), resulting in \( i = 2 \). This factor plays a crucial role in modifying the degree of freezing point depression or any other colligative property.

• \( i = 1 \) for non-electrolytes
• \( i > 1 \) for electrolytes

Understanding this factor enables predictions on how different solutes affect the properties of solutions.

Freezing Point of Solution

To determine the freezing point of a solution, you start with the freezing point of the pure solvent and subtract the calculated freezing point depression. This tells you how much the presence of solute has lowered the freezing point of the solvent.

Mathematically, this is expressed as:

• \( T_f' = T_f - \Delta T_f \)

Where \( T_f \) represents the initial freezing point of the pure solvent, and \( T_f' \) indicates the new freezing point of the solution.
For instance, in the context of this exercise, with pure water freezing at 273.15 K and a freezing point depression of 0.151 K, the solution's freezing point becomes 272.999 K. This demonstrates how solute particles influence the freezing characteristics of solutions, an important concept when working with solutions in chemistry.