Problem 19 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The elevation in boiling point of a solution of \(13.44 \mathrm{~g}\) of \(\mathrm{CuCl}_{2}\) in \(1 \mathrm{~kg}\) of water using the following information will be (Molecular weight of \(\mathrm{CuCl}_{2}=134.4\) and \(K_{b}=0.52 \mathrm{~K}\) molal \(\left.^{-1}\right)\) (a) \(0.16\) (b) \(0.05\) (c) \(0.1\) (d) \(0.2\)

Step-by-Step Solution

Answer

(b) 0.05

Step 1: Calculate Moles of Solute

First, we need to calculate how many moles of \( \mathrm{CuCl}_{2} \) are present in the solution. Use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \] Given the mass of \( \mathrm{CuCl}_{2} \) is \( 13.44 \text{ g} \) and its molecular weight is \( 134.4 \text{ g/mol} \), the moles of \( \mathrm{CuCl}_{2} \) are: \[ \text{moles} = \frac{13.44}{134.4} = 0.1 \text{ mol} \]

Step 2: Calculate Molality of Solution

Molality (\( m \)) is defined as moles of solute per kilogram of solvent. Since we have \( 0.1 \text{ mol} \) of \( \mathrm{CuCl}_{2} \) and \( 1 \text{ kg} \) of water, the molality of the solution is: \[ m = 0.1 \text{ mol/kg} \]

Step 3: Calculate Elevation in Boiling Point

The elevation in boiling point (\( \Delta T_b \)) can be calculated using the formula: \[ \Delta T_b = K_b \times m \] where \( K_b = 0.52 \text{ K molal}^{-1} \) and molality \( m = 0.1 \text{ mol/kg} \). Substituting these values gives: \[ \Delta T_b = 0.52 \times 0.1 = 0.052 \text{ K} \]

Step 4: Choose Closest Answer Option

Compare the calculated elevation \( \Delta T_b = 0.052 \text{ K} \) with the given choices. The answer closest to this value is option (b) \( 0.05 \).

Key Concepts

Molality

Molality is a concentration term in chemistry that measures the moles of solute per kilogram of solvent. It is represented by the symbol \( m \) and provides a way to express the concentration of a solution without being affected by the temperature changes since it relates to mass, not volume. Calculating molality involves dividing the moles of dissolved substance (solute) by the mass of the solvent in kilograms. In this exercise, the solution contains copper(II) chloride, \( \mathrm{CuCl}_2 \), in water. For example, with 0.1 moles of \( \mathrm{CuCl}_2 \) dissolved in 1 kg of water, the molality is straightforwardly computed as \( m = 0.1 \text{ mol/kg} \). By using molality, one can accurately predict changes in physical properties of solutions, like boiling point elevation, irrespective of external factors.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the concentration of solute particles but not their identity. These include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. These properties arise because adding solute particles affects how the solvent's molecules interact.

• Boiling Point Elevation: Adding a non-volatile solute like \( \mathrm{CuCl}_2 \) to water raises the boiling point of the solvent. This is because solute particles disrupt the formation of vapor, requiring more energy (heat) to bring the solvent to its boiling point.
• Formula: The boiling point elevation \( \Delta T_b \) is quantitatively determined using \( \Delta T_b = K_b \times m \), where \( K_b \) is the ebullioscopic constant.

Solutes that dissolve fully, such as electrolytes like \( \mathrm{CuCl}_2 \), contribute to a more significant boiling point elevation because they break into more particles. Understanding these properties helps predict how substances will alter the behavior of solutions.

Molecular Weight Calculation

The molecular weight (or molar mass) is the mass of one mole of a chemical compound. It is essential for converting between grams of a substance and moles, an integral step in many chemical calculations. In the problem, knowing that the molecular weight of \( \mathrm{CuCl}_2 \) is 134.4 g/mol helps calculate how many moles are present in a given sample.
The mole is a fundamental unit in chemistry representing a specific number of particles (Avogadro's number), and knowing this allows for:

• Converting mass to moles: Divide the mass of your sample by the compound's molecular weight. For instance, 13.44 g of \( \mathrm{CuCl}_2 \) results in \( 0.1 \) moles (\( 13.44/134.4 \)).
• Determining suitable quantities for reactions: Ensures that stoichiometric relationships in reactions are maintained so that reactants fully convert to products.

Understanding how to work with molecular weights lays the groundwork for assessing reaction costs and amounts of substances required in various chemical processes.