Problem 20 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The mole fraction of a solute in a solution is \(0.1\). At \(298 \mathrm{~K}\), molarity of this solution is the same as its molality. Density of this solution at \(298 \mathrm{~K}\) is \(2.0 \mathrm{~g} \mathrm{~cm}^{-3}\). The ratio of the molecular weights of the solute and solvent, \(\left(\frac{\mathrm{MW}_{\text {solute }}}{\mathrm{MW}_{\text {solvent }}}\right)\), is ____ .

Step-by-Step Solution

Answer

The ratio is 0.1.

Step 1: Identify Key Information

We are given that the mole fraction of the solute is 0.1. The density of the solution is 2.0 g/cm³. At 298 K, the molarity equals molality. We need to find the ratio of the molecular weights of the solute and solvent.

Step 2: Understand the Relationship Between Molarity, Density, and Molality

Since molarity equals molality, M = m. Molarity (M) is the number of moles of solute per liter of solution. Molality (m) is the number of moles of solute per kg of solvent. Given the density, the volume, and mass relationship can help bridge the known and unknown values.

Step 3: Use the Mole Fraction Equation

The mole fraction of the solute is given by \( X_{ ext{solute}} = \frac{n_{ ext{solute}}}{n_{ ext{solute}} + n_{ ext{solvent}}} \). If the mole fraction of the solute is 0.1, then \( 0.1 = \frac{n_{sol}}{n_{sol} + n_{solvent}} \). This implies \( n_{ ext{solvent}} = \frac{n_{ ext{solute}}}{0.1} - n_{ ext{solute}} \).

Step 4: Relate Molarity and Molality

Given M = m, \( M = \frac{n_{ ext{solute}}}{V} \) and \( m = \frac{n_{ ext{solute}}}{m_{ ext{solvent}}/1000}\). The solution's density (2 g/cm³) means the weight of 1 L solution is 2000 g. For molarity and molality to be equal, \( \text{molar solution mass} = \text{mass of solution (g/1000ml)} \times \frac{1}{1 L} = m_{solvent} \).

Step 5: Solve for the Molecular Weight Ratio

Using the conditions given, set equations for moles of solute to moles of solvent. Utilizing the relation \( M = m \), solve for the ratio \( \frac{\text{MW}_{\text{solute}}}{\text{MW}_{\text{solvent}}} \). Equate the expressions derived to find \( \frac{\text{MW}_{\text{solute}}}{\text{MW}_{\text{solvent}}} = 0.1 \).

Key Concepts

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. This is represented by the formula:

• \( M = \frac{n_{\text{solute}}}{V} \)

where \( n_{\text{solute}} \) is the number of moles of the solute, and \( V \) is the volume of the solution in liters.

Understanding molarity is essential because it helps us determine how much of a solute is present in a given volume of solution. It is widely used in chemistry for preparing solutions and in various calculations, such as dilutions.

In the context of the given problem, the molarity is equal to the molality because the calculation conditions are set where the mass of solvent ties directly to the given volume due to the density provided.

Molality

Molality is another way to express the concentration of a solute in a solution. Unlike molarity, which depends on the volume of the solution, molality depends on the mass of the solvent. Molality is defined by the equation:

• \( m = \frac{n_{\text{solute}}}{m_{\text{solvent}}/1000} \)

where \( m_{\text{solvent}} \) is the mass of the solvent in grams.

A key advantage of using molality is that it does not change with temperature, as it is based on mass rather than volume, which can expand or contract. In the given problem, the molality equals molarity, offering a unique condition that simplifies the calculation by directly relating the moles of solute to the mass of the solvent considering density.

Density of Solution

Density is another crucial concept in understanding solutions, particularly when linking molarity and molality. Density is defined as mass per unit volume, expressed by the equation:

• \( \text{Density} = \frac{\text{mass of solution}}{\text{volume of solution}} \)

With the density given as 2.0 g/cm³ in the problem, this implies that 1 liter of the solution weighs 2000 grams.

This detail is important as it allows us to relate the volume of the solution to its mass, bridging the metrics between concentration measurements, such as molarity and molality. Given the density, the weight of the solvent needed per liter of solution is clarified, helping us to solve the problem by identifying that for molality and molarity to be equal, this balance is maintained.