Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

The elevation in boiling point for \(13.44 \mathrm{~g}\) of \(\mathrm{CuCl}_{2}\) dissolved in \(1 \mathrm{~kg}\) of water as solvent will be \(\left(\mathrm{K}_{\mathrm{b}}=0.52\right.\) \(\mathrm{kg} / \mathrm{J}\), molar mass of \(\left.\mathrm{CuCl}_{2}=134.4 \mathrm{~g} / \mathrm{mol}\right)\) (a) \(0.05\) (b) \(0.10\) (c) \(0.16\) (d) \(0.20\).

Step-by-Step Solution

Answer

The boiling point elevation is approximately 0.16 K. Option (c) is correct.

Step 1: Determine Moles of Solute

First, calculate the number of moles of the solute, \( \mathrm{CuCl}_{2} \). Use the formula: \[\text{moles of } \mathrm{CuCl}_2 = \frac{\text{mass of } \mathrm{CuCl}_2}{\text{molar mass of } \mathrm{CuCl}_2}\]Substitute the given values:\[\text{moles of } \mathrm{CuCl}_2 = \frac{13.44 \text{ g}}{134.4 \text{ g/mol}} = 0.1 \text{ mol}\]

Step 2: Determine Van’t Hoff Factor

For \( \mathrm{CuCl}_2 \), dissociation occurs as follows:\[\mathrm{CuCl}_2 \rightarrow \mathrm{Cu}^{2+} + 2\mathrm{Cl}^{-}\]This means the Van’t Hoff factor, \( i \), is 3 because 3 ions are formed from one formula unit of \( \mathrm{CuCl}_2 \).

Step 3: Apply Boiling Point Elevation Formula

The formula for boiling point elevation is:\[\Delta T_b = i \cdot K_b \cdot m\]Where \( \Delta T_b \) is the elevation in boiling point, \( i \) is the Van’t Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.

Step 4: Calculate Molality

The molality \( m \) is calculated as:\[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\]Given that the mass of water is \( 1 \text{ kg} \):\[m = \frac{0.1 \text{ mol}}{1 \text{ kg}} = 0.1 \text{ mol/kg}\]

Step 5: Calculate Boiling Point Elevation

Substitute the values into the boiling point elevation formula:\[\Delta T_b = 3 \cdot 0.52 \text{ kg/J} \cdot 0.1 \text{ mol/kg}\]\[\Delta T_b = 0.156 \text{ K}\]Thus, the elevation in boiling point is approximately 0.16 K.

Key Concepts

Van’t Hoff factor

The Van’t Hoff factor, often represented as \(i\), is a crucial concept when dealing with colligative properties like boiling point elevation. It measures the number of particles a solute produces in solution. In simple terms, it reflects how many pieces a solute "breaks" into when it dissolves in a solvent.

• For non-electrolytes: Substances that do not ionize in solution have an \(i\) value of 1. For example, sugar dissolves in water without breaking into ions, so its Van’t Hoff factor is 1.
• For electrolytes: These substances, like \(\text{CuCl}_2\), break into ions when dissolved. For \(\text{CuCl}_2\), each unit dissociates into one \(\text{Cu}^{2+}\) ion and two \(\text{Cl}^-\) ions, making a total of three particles. Therefore, the Van’t Hoff factor is 3.

Understanding the Van’t Hoff factor is essential for calculating properties related to solutions, as having more particles impacts properties like boiling point and freezing point. It's a measure that connects the chemical composition with physical changes in solutions.

molality

Molality is a measure of the concentration of a solute in a solution. Unlike molarity, which depends on the volume of the solution, molality is solely based on the mass of the solvent. This makes it particularly useful for studying colligative properties, which depend on the number of solute particles rather than their volume. Molality, denoted by \(m\), is calculated using the formula:\[m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\]Since molality uses mass instead of volume, it remains constant with temperature changes, thus providing a stable measure for calculations. In the given exercise, the molality is calculated as follows:

• Moles of \(\text{CuCl}_2\): 0.1 mol (from the original data)
• Mass of solvent: 1 kg of water
• Resulting molality: \(0.1 \text{ mol/kg}\)

This concept is pivotal when applying formulas like the boiling point elevation, as it directly links the concentration of solute to the change in boiling point.

CuCl2 dissociation

Dissociation of \(\text{CuCl}_2\) is an exemplary demonstration of how ionic compounds behave in aqueous solutions. When \(\text{CuCl}_2\) is added to water, it separates into its constituent ions. This is a critical process for many applications in chemistry, as it effectively increases the number of solute particles present in the solution.The dissociation of \(\text{CuCl}_2\) can be expressed by the equation:\[\text{CuCl}_2 \rightarrow \text{Cu}^{2+} + 2\text{Cl}^-\]This equation tells us:

• Each formula unit of \(\text{CuCl}_2\) yields three ions.
• The copper ion, \(\text{Cu}^{2+}\), carries a 2+ charge.
• Each chloride ion, \(\text{Cl}^-\), carries a negative charge.

It's this breakdown into three ions that affects the solution's properties. More particles usually mean more significant changes in boiling and freezing points, making the knowledge of how \(\text{CuCl}_2\) dissociates essential for predicting and calculating these colligative properties.