Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

By dissolving \(10 \mathrm{~g}\) of a non-volatile solute in \(100 \mathrm{~g}\) of benzene, the boiling point rises by \(1^{\circ} \mathrm{C}\). The molecular mass of solute is \(\left[\mathrm{K}_{\mathrm{b}}\right.\) for benzene \(\left.=2.53 \mathrm{~K} \mathrm{~m}^{-1}\right]\) (a) \(235 \mathrm{~g}\) (b) \(352 \mathrm{~g}\) (c) \(250 \mathrm{~g}\) (d) \(253 \mathrm{~g}\)

Step-by-Step Solution

Answer

The molecular mass of the solute is 253 g/mol, option (d).

Step 1: Understand the Concept

This problem involves using the concept of Boiling Point Elevation. The formula for boiling point elevation is \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the change in boiling point, \( i \) is the van't Hoff factor (which is 1 for non-electrolytes), \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.

Step 2: Rearrange the Formula

Since the van't Hoff factor \( i = 1 \) (because the solute is non-volatile and non-electrolyte), the formula simplifies to \( \Delta T_b = K_b \cdot m \). Rearranging for molality \( m \), we have \( m = \frac{\Delta T_b}{K_b} \). Substitute the given values: \( \Delta T_b = 1^{\circ}C \) and \( K_b = 2.53 \; \text{K} \; \text{m}^{-1} \).

Step 3: Calculate Molality (m)

Substitute the values in the formula obtained: \( m = \frac{1}{2.53} \approx 0.395 \; \text{mol/kg} \). This represents the number of moles of solute per kilogram of solvent.

Step 4: Use Molality to Find Moles of Solute

Molality is moles of solute per kilogram of solvent. We have \( m = 0.395 \; \text{mol/kg} \) and the mass of benzene as the solvent is \(100 \; \text{g} = 0.1 \; \text{kg}\). Thus, \( n = m \times \text{mass of solvent in kg} = 0.395 \times 0.1 = 0.0395 \; \text{mol} \).

Step 5: Determine Molecular Mass

We know the mass of solute is \( 10 \; \text{g} \) and we have calculated \( n = 0.0395 \; \text{mol} \). Molecular mass \( M \) is \( \frac{10 \; \text{g}}{0.0395 \; \text{mol}} \approx 253 \; \text{g/mol} \).

Step 6: Choose the Correct Answer

From the calculations, the molecular mass of the solute is \( 253 \; \text{g/mol} \), which corresponds to option (d).

Key Concepts

Molecular Mass Calculation

Molecular mass, often referred to as molar mass, is a fundamental concept in chemistry that quantifies the mass of a given substance in terms of grams per mole. It's the measure of how much one mole of a chemical compound weighs. This concept is crucial when we are tasked with solving boiling point elevation problems because it allows us to connect the macroscopic properties (like mass) to the microscopic number of particles (moles).
To calculate molecular mass, we use the formula: \[ M = \frac{\text{mass of solute in grams}}{\text{number of moles of solute}} \]
The molecular mass helps us determine the precise weight of individual molecules in a sample. This value is especially important in quantitative chemistry tasks or when involved in reactions where stoichiometry plays a vital role. In this exercise, through the boiling point elevation data and molality, we determined the moles of solute, which was then used to compute the molecular mass of the unknown solute.

Molality

Molality is a measure of the concentration of a solute in a solution. Specifically, it is defined as the number of moles of solute per kilogram of solvent used in the solution.
The mathematical representation of molality is given by:\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Unlike molarity, which uses liters of solution, molality relies on the mass of the solvent. This property offers several advantages particularly in situations where temperature changes are considered, because molality remains unaffected by temperature fluctuations.
In the context of boiling point elevation, knowing the molality helps us find the boiling point change caused by the solute. In our given problem, we used the change in boiling point and the ebullioscopic constant to calculate molality, providing insight into how concentrated the solution is with respect to the substance dissolved in it.

Ebullioscopic Constant

The ebullioscopic constant is a crucial factor in determining how a solute affects the boiling point of a solvent. It provides a measure for the extent of boiling point elevation when a solute is dissolved. The ebullioscopic constant (\(K_b\)) is specific to each solvent and given in units of Kelvin per molal (\(K \, m^{-1}\)).
When a non-volatile solute is introduced into a solvent, the boiling point of that solvent increases. This change is described by the formula: \[ \Delta T_b = K_b \cdot m \]where \(\Delta T_b\) is the boiling point elevation and \(m\) is the molality of the solution. Using this relationship, we can assess how solutes with different molalities alter the boiling point, which is invaluable in calculating properties like molecular mass.
In the exercise, with a change in boiling point given and a known \(K_b\) for benzene, we could efficiently understand how concentrated the solute was through molality, and ultimately calculate the molecular mass of the unknown solute. This reliance on the ebullioscopic constant highlights its importance in practical chemical calculations.