Problem 37 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A solution of a non-volatile solute in water freezes at \(-0.30^{\circ} \mathrm{C}\). The vapour pressure of pure water at \(298 \mathrm{~K}\) is \(23.51 \mathrm{~mm} \mathrm{Hg}\) and \(K_{f}\) for water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} .\) Calculate the vapour pressure of this solution at \(298 \mathrm{~K}\).

Step-by-Step Solution

Answer

The vapour pressure of the solution at 298 K is approximately 23.44 mm Hg.

Step 1: Calculating the Molality (m)

We start by finding the molality of the solution using the formula for freezing point depression: \\[\Delta T_f = K_f \times m\] where \(\Delta T_f = 0.30^{\circ}C\) and \(K_f = 1.86 \, \text{K}\, \text{kg}\, \text{mol}^{-1}\). Thus: \\[m = \frac{\Delta T_f}{K_f} = \frac{0.30}{1.86}\approx 0.1613 \, \text{mol} \cdot \text{kg}^{-1}\]

Step 2: Determining Mole Fraction of Solute

Assuming one kg of water, the moles of solute (from the molality) would be 0.1613 mol. The moles of water can be calculated as it has a molar mass of 18 g/mol: \\[\text{Moles of water} = \frac{1000}{18} \approx 55.56 \, \text{mol}\] \[\text{Total moles} = 55.56 + 0.1613 = 55.7213\] \[\text{Mole fraction of water} = \frac{55.56}{55.7213}\approx 0.9971\]

Step 3: Calculating Vapour Pressure of Solution

Using Raoult's Law: \(P_{\text{solution}} = \text{Mole fraction of water} \times P^0_{\text{water}}\), where \(P^0_{\text{water}} = 23.51 \, \text{mm Hg}\). Thus: \\[P_{\text{solution}} = 0.9971 \times 23.51 \approx 23.44 \, \text{mm Hg}\]

Key Concepts

Molality

Molality is a measure of the concentration of a solute in a solution. Unlike molarity, which is sensitive to the volume changes with temperature, molality is defined in terms of mass, making it a more stable measurement for various conditions. Here's a closer look at what molality actually entails:

• Definition: Molality ( \( m \) ) is the number of moles of solute dissolved in one kilogram of solvent.
• Formula: To calculate molality, the formula is: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
• Application: Molality is particularly useful for freezing point depression, boiling point elevation, and various other colligative properties. It gives an accurate measure as the solvent's mass does not change with temperature.

In the given exercise, the freezing point depression data were used to calculate molality, revealing the direct relationship between the solute concentration and the extent to which it lowers the freezing point. Such calculations are crucial in chemistry to prepare solutions with specific properties while considering changes due to environmental factors.

Raoult's Law

Raoult's Law is fundamental in understanding how the vapour pressure of a solvent is affected by adding a solute. It helps predict how a solution behaves differently from the pure solvent. Here's why Raoult's Law is significant in this context:

• Basic Principle: The vapour pressure of a solution is directly proportional to the mole fraction of the solvent in the solution.
• Mathematical Representation: Expressed through the formula: \[ P_{\text{solution}} = \text{Mole fraction of solvent} \times P^0_{\text{solvent}} \]
• Importance: It shows that the presence of a non-volatile solute lowers the vapour pressure of the solution compared to the pure solvent.

For this exercise, Raoult's Law is used to determine the vapour pressure of the given solution of water and non-volatile solute. After finding the mole fraction of the solvent, which is water in this case, the law helps calculate the reduced vapour pressure, capturing the effect of dissolved solutes in practical terms.

Vapour Pressure

The concept of vapour pressure is crucial to understand when studying solutions, especially in relation to colligative properties like freezing point depression and boiling point elevation. Let's dive deeper into what vapour pressure means:

• Definition: Vapour pressure is the pressure exerted by the vapour when a liquid is in dynamic equilibrium with its vapour. It is essentially a measure of a liquid's tendency to evaporate.
• Factors Affecting Vapour Pressure:
  • Temperature: As temperature increases, vapour pressure increases because more molecules have adequate energy to escape into the gas phase.
  • Nature of the Liquid: Volatile liquids have higher vapour pressures than less volatile ones.
• Impact of Solutes: When a non-volatile solute is added to a solvent like water, the vapour pressure decreases due to fewer solvent molecules at the surface able to escape into the vapour phase.

In the exercise, the goal was to calculate the vapour pressure of a solution at a specific temperature. This involves understanding how the addition of a solute modifies the physical properties of the solvent, directly impacting its vapour pressure and the overall behaviour of the solution.