Problem 1 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A set of solutions is prepared using \(180 \mathrm{~g}\) of water as a solvent and \(10 \mathrm{~g}\) of different non-volatile solutes \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of \(\mathrm{A}=100 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{B}=200 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{C}=10,000 \mathrm{~g}\) \(\left.\mathrm{mol}^{-1}\right]\) (a) \(\mathrm{B}>\mathrm{C}>\mathrm{A}\) (b) \(\mathrm{C}>\mathrm{B}>\mathrm{A}\) (c) \(\mathrm{A}>\mathrm{B}>\mathrm{C}\) (d) \(\mathrm{A}>\mathrm{C}>\mathrm{B}\)

Step-by-Step Solution

Answer

(c) \( \text{A} > \text{B} > \text{C} \).

Step 1: Determine the number of moles of each solute

To find the number of moles, use the formula \( n = \frac{m}{M} \), where \( m \) is the mass and \( M \) is the molar mass of the solute. For solute A, \( n = \frac{10 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ mol} \). For solute B, \( n = \frac{10 \text{ g}}{200 \text{ g/mol}} = 0.05 \text{ mol} \). For solute C, \( n = \frac{10 \text{ g}}{10000 \text{ g/mol}} = 0.001 \text{ mol} \).

Step 2: Calculate the mole fraction of each solute

The mole fraction is given by \( \text{mole fraction} = \frac{n}{n + n_w} \), where \( n_w \) is the moles of water. Moles of water given \( 180 \text{ g} = 180 \div 18 = 10 \text{ mol} \). For A, \( \text{mole fraction} = \frac{0.1}{0.1 + 10} \approx 0.0099 \). For B, \( \text{mole fraction} = \frac{0.05}{0.05 + 10} \approx 0.004975 \). For C, \( \text{mole fraction} = \frac{0.001}{0.001 + 10} \approx 0.0000999 \).

Step 3: Determine the relative lowering of vapor pressure

The relative lowering of vapor pressure is proportional to the mole fraction of the solute in the solution. From Step 2, relative lowering for A is approximately 0.0099, for B it is 0.004975, and for C, it is 0.0000999.

Step 4: Order the solutes based on the relative lowering of vapor pressure

Comparing the values obtained in Step 3, the order from highest to lowest relative lowering is C with approximately 0.0000999, then B at 0.004975, and finally A at 0.0099. Thus, the correct order comparing relative lowering is \( \text{A} > \text{B} > \text{C} \).

Key Concepts

Relative Lowering of Vapor Pressure

The concept of the relative lowering of vapor pressure is a crucial part of understanding colligative properties. These properties depend on the number of solute particles in a solution rather than the type of particles themselves. When a non-volatile solute is added to a solvent, the solution's vapor pressure decreases relative to the pure solvent. This occurs because the solute particles occupy surface space, reducing the number of solvent molecules that can escape into the vapor phase.

The relative lowering of vapor pressure can be mathematically described using Raoult's Law. It states that the lowering of vapor pressure is proportional to the mole fraction of the solute present in the solution:

\[ \frac{\Delta P}{P^0} = x \]

Here, \( \Delta P \) is the decrease in vapor pressure, \( P^0 \) is the vapor pressure of the pure solvent, and \( x \) is the mole fraction of the solute. In simple terms, the more solute molecules present, the greater the reduction in vapor pressure. This key relationship helps us understand the behavior of solutions as more solute is added.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a solution. It is defined as the number of moles of a component divided by the total number of moles in the solution. For a solution with multiple components, it provides a clear picture of the composition without concerning the mass or volume.

The mole fraction can be calculated with the formula:

\[ x_i = \frac{n_i}{n_{total}} \]

In this equation, \( n_i \) represents the moles of the component of interest, and \( n_{total} \) is the total moles of all components in the solution.

In the original exercise, the mole fraction of each solute was calculated to understand their effect on the relative lowering of vapor pressure. Solutions with a higher mole fraction of solute will show a more significant lowering of the vapor pressure, emphasizing the influence of the quantity of solute particles relative to the solvent.

Non-Volatile Solutes

Non-volatile solutes are substances that do not easily vaporize. When added to a solvent, these solute particles remain predominantly in the liquid phase and do not contribute significantly to the vapor above the solution. Their presence leads to a decrease in the solvent's vapor pressure, which is an essential aspect of colligative properties.

In the exercise, non-volatile solutes such as A, B, and C were considered. Their differing molar masses influenced the number of moles present when a fixed mass was used, thus impacting the solution's properties.

Understanding non-volatile solutes helps explain why adding salt to water lowers its freezing point and raises its boiling point—a critical concept in real-world applications like antifreeze selection or culinary practices. This adjustment in vapor pressure by non-volatile solutes is fundamentally linked to changes in solution properties and behavior.