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 vapor pressure of the solution is 752.40 torr, option (d).

Step 1: Calculate Moles of Glucose

First, calculate the number of moles of glucose using the formula: \[ \text{Moles of Glucose} = \frac{\text{Mass of Glucose}}{\text{Molar Mass of Glucose}} \]Given that glucose \((\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6)\) has a molar mass of \(180 \mathrm{~g/mol}\), the calculation is: \[ \frac{18 \mathrm{~g}}{180 \mathrm{~g/mol}} = 0.1 \text{ moles} \]

Step 2: Calculate Moles of Water

Calculate the number of moles of water using its molar mass \((18 \mathrm{~g/mol})\): \[ \text{Moles of Water} = \frac{178.2 \mathrm{~g}}{18 \mathrm{~g/mol}} = 9.9 \text{ moles} \]

Step 3: Determine Mole Fraction of Water

Calculate the mole fraction of water in the solution. The mole fraction of water is given by: \[ X_{\text{water}} = \frac{\text{Moles of Water}}{\text{Total Moles}} = \frac{9.9}{9.9 + 0.1} = \frac{9.9}{10} = 0.99 \]

Step 4: Calculate Vapor Pressure of the Solution

Use Raoult's law to calculate the vapor pressure of the aqueous solution. Raoult's law states that: \[ P_{\text{solution}} = X_{\text{water}} \times P^{\circ}_{\text{water}} \] Where \(P^{\circ}_{\text{water}} = 760 \text{ torr}\) at \(100^{\circ} \mathrm{C}\).Thus, \[ P_{\text{solution}} = 0.99 \times 760 \text{ torr} = 752.4 \text{ torr} \]

Step 5: Choose the Correct Answer

From the options given:(a) 759.00 torr (b) 7.60 torr (c) 76.00 torr (d) 752.40 torr The correct answer is: \((d) 752.40 \text{ torr}\).

Key Concepts

Vapor Pressure Lowering

Vapor pressure lowering is a phenomenon observed when a non-volatile solute is added to a solvent, such as water. The presence of the solute decreases the number of solvent molecules at the surface, thus reducing the pressure exerted by the vapor above the liquid. This effect is classified as a colligative property, meaning it depends on the number of solute particles rather than their identity.
To understand how this works, imagine the solvent particles freely moving and able to escape as vapor in a pure solvent. When solute particles are present, they occupy space at the surface, blocking solvent molecules from escaping as easily. This leads to a lowered vapor pressure compared to the pure solvent.
The extent of vapor pressure lowering can be calculated using Raoult's Law, which connects the vapor pressure of a solution to its composition.

Raoult's Law

Raoult's Law is a fundamental principle used to explain how the presence of a solute affects the vapor pressure of a solvent. It states that the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent present. In formula terms:
\[ P_{\text{solution}} = X_{\text{solvent}} \times P^{\circ}_{\text{solvent}} \]
Here, \(P_{\text{solution}}\) is the vapor pressure of the solution, \(X_{\text{solvent}}\) is the mole fraction of the solvent, and \(P^{\circ}_{\text{solvent}}\) is the vapor pressure of the pure solvent.
This law applies best to ideal solutions where interactions between molecules of different components are similar to those among molecules of the same component.
In the exercise example, using Raoult's Law and the mole fraction of water calculated, the vapor pressure of the solution was found to be lower, at 752.4 torr, than that of pure water at 760 torr.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture. It is the ratio of the number of moles of one component to the total number of moles in the solution. Mathematically, it is expressed as:
\[X_{\text{component}} = \frac{\text{Moles of Component}}{\text{Total Moles in Solution}}\]
In the exercise, to find the mole fraction of water, first calculate the moles of water and glucose. Adding these gives the total moles in the solution.
For water, the mole fraction is calculated as 0.99, indicating that water makes up most of the solution. Mole fractions require no units since they're ratios. These values are crucial in applying Raoult's Law to find changes in vapor pressure.

Molar Mass Calculation

Calculating molar mass is essential for determining the number of moles, a critical step in solving many chemistry problems. The molar mass of a compound is the mass of one mole of its molecules, typically expressed in grams per mole (g/mol).
To find the molar mass, sum the atomic masses of all the atoms in the molecule. For example, the molar mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) is calculated by adding the masses of 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms, totaling 180 g/mol.
Using this, you can convert a known mass of a substance to moles through division. This was done in the original exercise for both glucose and water, allowing subsequent calculations of mole fractions and applications of Raoult's Law.