Problem 164 - A Complete Resource Book in Chemistry for JEE Main

Question

The vapour pressure of water at \(20^{\circ} \mathrm{C}\) is \(17.5 \mathrm{~mm} \mathrm{Hg}\). If \(18 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is added to \(178.2 \mathrm{~g}\) of water at \(20^{\circ} \mathrm{C}\), the vapour pressure of the resulting solution will be (a) \(17.675 \mathrm{~mm} \mathrm{Hg}\) (b) \(15.750 \mathrm{~mm} \mathrm{Hg}\) (c) \(16.500 \mathrm{~mm} \mathrm{Hg}\) (d) \(17.325 \mathrm{~mm} \mathrm{Hg}\).

Step-by-Step Solution

Answer

The vapor pressure of the solution is 17.325 mm Hg, option (d).

Step 1: Define Raoult's Law

Raoult's law states that the vapor pressure of a solution is the sum of the vapor pressures of each volatile component, multiplied by the mole fraction of each component. For an ideal solution, the vapor pressure of the solvent can be expressed as \( P_{ ext{solution}} = P_{ ext{solvent}}^0 \times X_{ ext{solvent}} \), where \( P_{\text{solvent}}^0 \) represents the vapor pressure of the pure solvent, and \( X_{\text{solvent}} \) is the mole fraction of the solvent.

Step 2: Calculate Moles of Solvent and Solute

First, we need to calculate the moles of water (solvent) and glucose (solute). The molecular weight of water is 18 g/mol and that of glucose is 180 g/mol. Thus, the moles of water are \( \frac{178.2 \, \text{g}}{18 \, \text{g/mol}} = 9.9 \, \text{mol} \), and the moles of glucose are \( \frac{18 \, \text{g}}{180 \, \text{g/mol}} = 0.1 \, \text{mol} \).

Step 3: Calculate Mole Fraction of the Solvent

The mole fraction of water (solvent) is calculated using the formula \( X_{\text{solvent}} = \frac{\text{moles of solvent}}{\text{moles of solvent} + \text{moles of solute}} \). Therefore, \( X_{\text{water}} = \frac{9.9 \, \text{mol}}{9.9 \, \text{mol} + 0.1 \, \text{mol}} = \frac{9.9}{10} = 0.99 \).

Step 4: Calculate Vapor Pressure of the Solution

Now apply Raoult's Law: \( P_{\text{solution}} = P_{\text{water}}^0 \times X_{\text{water}} \). Given that \( P_{\text{water}}^0 = 17.5 \, \text{mm Hg} \), we calculate \( P_{\text{solution}} = 17.5 \, \text{mm Hg} \times 0.99 = 17.325 \, \text{mm Hg} \).

Step 5: Choose the Correct Answer

Based on our calculations, the vapor pressure of the solution is \( 17.325 \, \text{mm Hg} \), which corresponds to option (d).

Key Concepts

Understanding Vapor Pressure

Vapor pressure is a crucial concept in chemistry, especially when dealing with solutions. It refers to the pressure exerted by the vapor that is in equilibrium with its liquid phase. This equilibrium occurs when the rate of evaporation equals the rate of condensation. In simple terms, vapor pressure indicates how readily molecules escape from a liquid to the vapor phase.
For pure substances, the vapor pressure depends solely on temperature – as temperature increases, the molecules have more energy to escape into the vapor phase, increasing the vapor pressure. However, when a solute is introduced into a solvent, like glucose in water, the vapor pressure of the solvent typically decreases. This is because the addition of a solute results in fewer solvent molecules at the surface, leading to a reduction in the number of molecules escaping into the vapor phase.

Mole Fraction Explained

The mole fraction is a way of expressing the concentration of a component in a mixture. It plays a significant role in applying Raoult's Law. The mole fraction of a substance is the ratio of the number of moles of that substance to the total number of moles in the mixture.
Mathematically, it's expressed as:

• The mole fraction of the solvent (\(X_{ ext{solvent}}\)),

is given by:\[X_{ ext{solvent}} = \frac{n_{ ext{solvent}}}{n_{ ext{solute}} + n_{ ext{solvent}}} \]
For example, if we have 9.9 moles of water and 0.1 mole of glucose, the mole fraction of water\(X_{ ext{water}}\)can be calculated as:\[X_{ ext{water}} = \frac{9.9}{9.9 + 0.1} = 0.99\]
Thus, the mole fraction helps us understand the proportion of solvent in the mixture and directly influences the vapor pressure calculation according to Raoult's Law.

What is an Ideal Solution?

An ideal solution is a solution that perfectly follows Raoult's Law. In such solutions, the interactions between molecules of different components are similar to interactions between molecules of the same component. This means that neither the evaporation nor the mixing process alters the energy distribution within the liquid significantly.
Ideal solutions are often theoretical, but they provide a baseline for understanding real solutions. In practice, solutions may deviate from ideal behavior due to differences in molecular forces between solute and solvent, but when a solution behaves ideally, its vapor pressure can be accurately predicted using the mole fraction and the vapor pressure of the pure components.
The concept of an ideal solution is essential for studying the colligative properties of solutions, which depend on the number of particles dissolved in the solution rather than their identity.

The Calculation of Moles

Calculating the number of moles in a substance is a basic yet crucial task in chemistry. Moles provide a bridge between the mass of a substance and the number of atoms or molecules it contains, enabling quantitative analysis across chemical reactions and processes.
The formula used is:

• \( ext{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \)

For example, when calculating the moles of water, given its mass and molar mass (18 g/mol), we use:\[\text{moles of water} = \frac{178.2 \, \text{g}}{18 \, \text{g/mol}} = 9.9 \, \text{mol} \]
Similarly, for glucose with a molar mass of 180 g/mol, we find:\[\text{moles of glucose} = \frac{18 \, \text{g}}{180 \, \text{g/mol}} = 0.1 \, \text{mol}\]
Understanding and accurately performing these calculations are fundamental, as they underpin the determination of concentrations, reaction stoichiometry, and the subsequent calculations of properties such as vapor pressure in solutions.