Problem 160 - A Complete Resource Book in Chemistry for JEE Main

Question

\(18 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is added to \(178.2 \mathrm{~g}\) of water. The vapour pressure of water for this aqueous solution at \(100^{\circ} \mathrm{C}\) is (a) \(759.00\) torr (b) \(7.60\) torr (c) \(76.00\) torr (d) \(752.40\) torr

Step-by-Step Solution

Answer

The vapour pressure of the solution is 752.4 torr (option d).

Step 1: Calculate Moles of Glucose

First, calculate the number of moles of glucose. The molar mass of glucose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)) is calculated as: \( 6\times12 + 12\times1 + 6\times16 = 180 \) g/mol. Thus, the moles of glucose are: \( \frac{18}{180} = 0.1 \) mol.

Step 2: Calculate Moles of Water

Next, we calculate the number of moles of water. The molar mass of water (\(\mathrm{H}_{2} \mathrm{O}\)) is \(18\) g/mol. Thus, the moles of water are: \( \frac{178.2}{18} = 9.9 \) mol.

Step 3: Calculate Total Moles in Solution

Add the moles of glucose and moles of water to find the total moles in the solution: \( 0.1 + 9.9 = 10 \) moles.

Step 4: Calculate Mole Fraction of Water

The mole fraction of water is calculated using the formula: \( \text{Mole Fraction} = \frac{\text{Moles of Water}}{\text{Total Moles}} = \frac{9.9}{10} = 0.99 \).

Step 5: Apply Raoult's Law to Find Vapour Pressure

Using Raoult's Law, the vapour pressure of the solution is \( P = \chi \cdot P^0 \), where \( \chi \) is the mole fraction of water and \( P^0 \) is the pure water vapour pressure. At \(100^{\circ} \mathrm{C}\), \( P^0 = 760 \) torr. So, \( P = 0.99 \times 760 = 752.4 \) torr.

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase. In simpler terms, imagine a closed container with a liquid inside it. Some of the liquid molecules will evaporate and form a vapor above the liquid. The pressure created by this vapor is its vapor pressure.

For different liquids, the vapor pressure changes based on temperature and the nature of the liquid itself. For instance, water has a specific vapor pressure at certain temperatures. Raoult's Law helps us understand how this vapor pressure changes when a non-volatile solute, like glucose, is added.

Raoult's Law states that the vapor pressure of a solution is directly affected by the proportion of solvent molecules.

• In this exercise, we calculate the vapor pressure of water in an aqueous solution at 100°C using Raoult's Law, considering that glucose is non-volatile.
• The vapor pressure of the pure solvent (water) at this temperature is 760 torr.

Mole Fraction

The mole fraction is a way to express the concentration of a component (like water or glucose) in a solution. It represents the ratio of the number of moles of one component to the total number of moles of all components in the solution.

To calculate the mole fraction, follow these steps:

• First, find the total number of moles in the solution by adding up the moles of all components.
• Then, determine the mole fraction of a component by dividing the number of moles of that component by the total number of moles.

In the original exercise, we computed the mole fraction for water as follows: \[\text{Mole Fraction of Water} = \frac{\text{Moles of Water}}{\text{Total Moles}} = \frac{9.9}{10} = 0.99\] This value is then used in Raoult's Law to find the new vapor pressure of the solution.

Moles Calculation

Calculating the number of moles is a basic task in chemistry that helps us understand the composition of a mixture or solution. The number of moles of a substance is found using the formula: \[\text{Moles} = \frac{\text{Mass of the Substance}}{\text{Molar Mass}}\] For example, in the original exercise:

• The molar mass of glucose is 180 g/mol. Thus, 18 g of glucose corresponds to \( \frac{18}{180} = 0.1 \) moles.
• Water has a molar mass of 18 g/mol. Hence, 178.2 g of water equals \( \frac{178.2}{18} = 9.9 \) moles.

Summing these gives us the total number of moles in the solution, which is 10 moles in this case. This step is crucial in determining the mole fraction and further in understanding and applying Raoult's Law.

Colligative Properties

Colligative properties are unique because they depend solely on the number of solute particles in a solution and not on their identity. Some common colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

In this exercise, we focus on vapor pressure lowering, which occurs because the addition of a non-volatile solute (such as glucose) reduces the number of solvent molecules at the surface. This decrease leads to a reduction in the vapor pressure of the solvent (water).

• The presence of solute particles disrupts the escape of solvent molecules, thus lowering the vapor pressure compared to the pure solvent.
• This effect is quantitatively described using Raoult's Law, where the new vapor pressure is obtained by multiplying the mole fraction of the solvent by the pure solvent's vapor pressure.

Understanding colligative properties provides insight into how solutes impact the physical properties of solutions, which is crucial for a wide range of chemical applications.