Problem 51 - A Complete Resource Book in Chemistry for JEE Main

Question

The vapours pressure of water at \(23^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~mm}\). of \(\mathrm{Hg} .0 .1\) mole of glucose is dissolved in \(178.2 \mathrm{~g}\) of water. What is the vapour pressure (in \(\mathrm{mm}\) of \(\mathrm{Hg}\) ) of the resultant solution? (a) \(19.0\) (b) \(19.602\) (c) \(19.402\) (d) \(19.202\)

Step-by-Step Solution

Answer

The vapour pressure of the solution is 19.602 mmHg.

Step 1: Understanding the Formula for Vapour Pressure Lowering

We will use Raoult's Law, which states that the vapour pressure of the solution is the vapour pressure of the pure solvent times the mole fraction of the solvent in the solution. The formula can be expressed as \( P_{solution} = P_{solvent} \times \chi_{solvent} \), where \( P_{solution} \) is the vapour pressure of the solution, \( P_{solvent} \) is the vapour pressure of the pure solvent, and \( \chi_{solvent} \) is the mole fraction of the solvent.

Step 2: Calculate the Moles of Water

Given that the mass of water is \(178.2 \text{ g}\) and the molar mass of water is \(18.02 \text{ g/mol}\), we can calculate the moles of water. \[ \text{Moles of water} = \frac{178.2}{18.02} \approx 9.89 \text{ moles} \]

Step 3: Determine the Mole Fraction of the Solvent

To find the mole fraction of water, use the formula \( \chi_{solvent} = \frac{n_{water}}{n_{water} + n_{glucose}} \), where \( n_{water} \) is the moles of water and \( n_{glucose} \) is the moles of glucose, which is given as 0.1 mole. \[ \chi_{solvent} = \frac{9.89}{9.89 + 0.1} \approx 0.99 \]

Step 4: Apply Raoult's Law to Find Vapour Pressure

Using the calculated mole fraction of water in the solution and the initial vapour pressure of pure water \( P_{solvent} = 19.8 \text{ mmHg} \), apply the formula: \[ P_{solution} = 19.8 \times 0.99 \approx 19.602 \text{ mmHg} \]

Step 5: Conclusion

The vapour pressure of the resultant solution is determined to be approximately \(19.602 \text{ mmHg}\). Thus, the correct answer is option (b).

Key Concepts

Vapour Pressure

Vapour pressure is a key concept in solution chemistry. It refers to the pressure exerted by the vapour of a liquid in equilibrium with its liquid phase at a given temperature. This equilibrium signifies that the rate of evaporation of the liquid equals the rate of condensation of the vapour.
In simpler terms, vapour pressure is the tendency of molecules to escape into the gas phase. For pure liquids, the vapour pressure is a characteristic value at each particular temperature. However, when a non-volatile solute is dissolved in a liquid, the vapour pressure of the resulting solution is typically lower than that of the pure solvent.

• The decrease in vapour pressure is due to solute molecules occupying surface space, preventing some solvent molecules from escaping into the vapour phase.
• Raoult's Law helps us quantify this change in vapour pressure for ideal solutions.

Mole Fraction

The mole fraction is a measure that helps us understand the composition of a mixture or a solution. It is defined as the ratio of the number of moles of a particular component to the total number of moles in the mixture.
In a solution, mole fraction provides insight into the concentration of the components without involving units like molarity, which depend on volume. We calculate the mole fraction as follows:

• Formula: \[ \chi_{i} = \frac{n_{i}}{n_{total}} \]where \( \chi_{i} \) represents the mole fraction of component \( i \), \( n_{i} \) is the number of moles of component \( i \), and \( n_{total} \) is the total number of moles of all components in the solution.
• In our problem, the mole fraction is calculated for water (the solvent), considering both water and glucose (the solute).

The mole fraction closely ties into Raoult's Law, as it is used to determine how the presence of solute affects the solvent's vapour pressure.

Solution Chemistry

Solution chemistry is the study of the properties and behavior of solutions, focusing significantly on how solutes and solvents interact. A solution consists of two major parts: the solute (the substance being dissolved) and the solvent (the substance doing the dissolving).
In this scenario, we are dealing with a solution made by dissolving glucose in water. Understanding solution chemistry involves several important concepts:

• Solubility: The ability of a solute to dissolve in a solvent at a specific temperature and pressure.
• Concentration: Indicates how much solute is present in a certain quantity of solvent or solution, expressible in units like molarity, molality, or as in this problem, mole fraction.
• Colligative Properties: Properties that depend on the number of solute particles in a solution, not the identity of solute particles. Vapour pressure lowering is one such property.

These concepts help us understand why the presence of glucose in the water affects the vapour pressure, illustrating how solute-solvent interactions lead to real-world phenomena like boiling point elevation and freezing point depression.