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, \(K_{f}\) 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 of benzene is lowered by 0.4 K (option c).

Step 1: Identify the formula for freezing point depression

The formula to calculate the change in freezing point (freezing point depression) is \( \Delta T_f = K_f \cdot m \), where \( \Delta T_f \) is the depression in freezing point, \( K_f \) is the freezing point depression constant, and \( m \) is the molality of the solution.

Step 2: Calculate the molality of the solution

Molality \( m \) is defined as moles of solute per kilogram of solvent. First, calculate the moles of solute: \( \text{moles of solute} = \frac{1.00 \text{ g}}{250 \text{ g/mol}} = 0.004 \text{ mol} \). The mass of benzene is \(51.2 \text{ g}\), or \(0.0512 \text{ kg}\). Therefore, molality \( m = \frac{0.004 \text{ mol}}{0.0512 \text{ kg}} \approx 0.0781 \text{ mol/kg} \).

Step 3: Calculate the freezing point depression

Use the formula from Step 1 with the values from Step 2. \( \Delta T_f = 5.12 \text{ K kg/mol} \times 0.0781 \text{ mol/kg} \approx 0.4 \text{ K} \).

Step 4: Determine the correct answer choice

The calculated depression in the freezing point \( \Delta T_f = 0.4 \text{ K} \) matches option (c). Therefore, the freezing point of benzene will be lowered by \(0.4 \text{ K}\).

Key Concepts

Molality

Molality is an important concept in chemistry when it comes to solutions. Unlike molarity, which depends on volume, molality is a concentration measure that focuses on the mass of the solvent. It is defined as the number of moles of solute per kilogram of solvent. This means that, regardless of temperature changes, molality remains constant because it's independent of volume expansion or contraction.

• To calculate molality, divide the moles of solute by the kilograms of solvent.
• For example, in our exercise, the moles of solute are found by taking the mass of the solute and dividing it by its molar mass, which gives us 0.004 mol.
• The mass of the solvent, benzene in this case, is then converted to kilograms (0.0512 kg).
• Plugging these into the formula, the molality of this benzene solution is approximately 0.0781 mol/kg.

This method offers an accurate measure of how concentrated a solution is in terms of the solute being used.

Non-electrolyte Solute

A non-electrolyte solute is a substance that, when dissolved in a solvent, does not dissociate into ions. This means the solution does not conduct electricity. Such substances typically dissolve as molecules rather than as charged particles.

• For instance, in our exercise, the given solute is a non-electrolyte with a molar mass of 250 g/mol.
• When 1.00 g of this non-electrolyte is dissolved in benzene, it only contributes to the number of moles present without creating ions.
• This is significant because when calculating the freezing point depression, the presence of ions does not influence the outcome, meaning more predictable behavior in solutions is expected.

Understanding this helps us realize that the freezing point depression in the context of a non-electrolyte is governed mainly by the number of particles dissolved, not by electrical conductivity.

Freezing Point Depression Constant (K_f)

The freezing point depression constant, denoted as \( K_f \), is a specific value for each solvent that measures the degree to which the freezing point is lowered for a 1 molal solution of a solute.

• This constant is an inherent property of the solvent and depends on its nature.
• In our scenario with benzene, \( K_f \) is given as 5.12 K kg/mol. This means for every mole of solute added per kilogram of benzene, the freezing point will decrease by 5.12 K.

To use \( K_f \) effectively, it is applied in the freezing point depression formula: \( \Delta T_f = K_f \cdot m \). This equation shows how \( K_f \) works together with the solution's molality to calculate the change in freezing point, demonstrating the relationship between solvent properties and solution behavior.

Molar Mass Calculation

Calculating molar mass is a crucial step in many chemical processes, especially when determining the concentration of solutions.

• Molar mass is defined as the mass of one mole of a substance and is typically expressed in g/mol.
• In our exercise, the molar mass of the non-electrolyte is provided as 250 g/mol.
• To find the number of moles of solute: take the mass of the solute and divide it by its molar mass, yielding 0.004 mol in our example.
• It’s essential to ensure accuracy in molar mass calculations since it directly affects the calculated molality and, subsequently, the freezing point depression, ensuring our results are reliable and precise.

Understanding how to accurately calculate molar mass is key to solving many types of problems related to solutions and their physical properties.