Problem 77 - A Complete Resource Book in Chemistry for JEE Main

Question

\(1.00 \mathrm{~g}\) of a non-electrolyte solute (molar mass \(250 \mathrm{~g} \mathrm{~mol}^{-1}\) ) was dissolved in \(51.2 \mathrm{~g}\) of benzene. If the freezing point depression constant, \(\mathrm{K}_{i}\) of benzene is \(5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\), the freezing point of benzene will be lowered by (a) \(0.5 \mathrm{~K}\) (b) \(0.2 \mathrm{~K}\) (c) \(0.4 \mathrm{~K}\) (d) \(0.3 \mathrm{~K}\)

Step-by-Step Solution

Answer

The freezing point lowering is 0.4 K, corresponding to option (c).

Step 1: Identify Given Values

We are given that the mass of the solute is 1.00 g, the molar mass of the solute is 250 g/mol, and the mass of benzene is 51.2 g. The freezing point depression constant \( \mathrm{K}_{\mathrm{i}} \) of benzene is 5.12 K kg/mol.

Step 2: Convert Solvent Mass

Convert the mass of benzene from grams to kilograms for consistency in units. \[ 51.2 \text{ g} = 0.0512 \text{ kg} \]

Step 3: Calculate Moles of Solute

To find moles of the solute, use the formula: \[ \text{moles} = \frac{\text{mass of solute}}{\text{molar mass of solute}} \]Substituting the given values:\[ \text{moles} = \frac{1.00 \text{ g}}{250 \text{ g/mol}} = 0.004 \text{ mol} \]

Step 4: Calculate Molality

Use the definition of molality:\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]Plug in the values:\[ m = \frac{0.004 \text{ mol}}{0.0512 \text{ kg}} = 0.0781 \text{ mol/kg} \]

Step 5: Calculate Freezing Point Depression

Use the formula for freezing point depression:\[ \Delta T_f = K_f \cdot m \]Where \( K_f = 5.12 \text{ K kg/mol} \) and \( m = 0.0781 \text{ mol/kg} \)Thus,\[ \Delta T_f = 5.12 \times 0.0781 = 0.4 \text{ K} \]

Step 6: Select the Correct Answer

From the given options, the freezing point of benzene will be lowered by \(0.4 \text{ K}\). Therefore, the correct answer is option (c).

Key Concepts

Colligative Properties

Colligative properties are unique features of solutions that depend solely on the number of solute particles in a given amount of solvent, not on the type of solute. Key colligative properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. These properties arise because adding solute particles disrupts solvent properties. As more solute particles are introduced, effects like freezing point depression become evident. The higher the number of solute particles, the greater the depression of the freezing point, making it a crucial aspect of understanding solution behavior.
Colligative properties allow chemists to determine characteristics of solvents and solutes in addition to accessing important quantitative data. In this exercise, freezing point depression, which measures how the freezing point of a solvent is lowered when a solute is added, is vital. The formula used is closely tied to these properties and calculates how much the freezing point drops because of the presence of solute particles.

Molality Calculation

Molality is a concentration unit which helps in determining colligative properties, like freezing point depression. It is represented by the symbol \( m \) and defined as moles of solute per kilogram of solvent. This measurement differs from molarity, which is moles per liter of solution. Molality is more advantageous when temperature changes occur, as it is not affected by temperature and pressure variations unlike volume-based measurements.

To calculate molality:

• First, calculate the moles of solute by dividing the mass of the solute by its molar mass.
• Then, determine the mass of the solvent in kilograms.
• Finally, divide the moles of solute by the mass of the solvent to obtain molality.

In this exercise:

• Moles of solute are calculated as \( \frac{1.00 \text{ g}}{250 \text{ g/mol}} = 0.004 \text{ mol} \).
• The mass of benzene (solvent) in kilograms is 0.0512 kg.
• Molality becomes \( \frac{0.004 \text{ mol}}{0.0512 \text{ kg}} = 0.0781 \text{ mol/kg} \).

This accurate measure of concentration ensures proper calculation of how colligative effects, such as freezing point depression, are quantified.

Non-electrolyte Solutions

Non-electrolyte solutions are composed of solutes that do not dissociate into ions when dissolved in a solvent. Unlike electrolytes, non-electrolytes retain their molecular identity in solution. Due to this nature, they do not conduct electricity. Common examples include sugar and most organic compounds, such as benzene and alcohol.
In the context of freezing point depression, non-electrolytes are particularly useful for studying the colligative properties of solutions. Since they do not break into ions, the number of particles added per solute molecule is consistent and predictable. This allows for straightforward calculations using colligative formulas.

In this exercise, a non-electrolyte solute is used, meaning the calculations are straightforward, without needing adjustments for ion separation. The molar mass is critical as it enables the conversion from mass to moles, a key step for calculating molality and ultimately for determining the extent of freezing point depression in the solution.