Problem 170 - A Complete Resource Book in Chemistry for JEE Main

Question

Ethylene glycol is used as antifreeze in a cold climate. Mass of ethylene glycol which should be added to \(4 \mathrm{~kg}\) of water to prevent it form freezing at \(-6^{\circ} \mathrm{C}\) will be: \(\left(\mathrm{K}_{f}\right.\) for water \(=1.86 \mathrm{~kg} \mathrm{~mol}^{-1}\), and molar mass of ethylene glycol \(=62 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(204.11 \mathrm{~g}\) (b) \(804.32 \mathrm{~g}\) (c) \(600.20 \mathrm{~g}\) (d) \(302.40 \mathrm{~g}\)

Step-by-Step Solution

Answer

The mass of ethylene glycol needed is 804.32 g (option b).

Step 1: Understand Freezing Point Depression

To determine how much ethylene glycol should be added to 4 kg of water to lower the freezing point to -6°C, we apply the formula for freezing point depression, \ \ \( \Delta T_f = K_f \times m \) \ \ where \( \Delta T_f \) is the lowering in freezing point, \( K_f \) is the cryoscopic constant of water (given as 1.86 kg mol\(^{-1}\)), and \( m \) is the molality of the solution.

Step 2: Calculate Change in Freezing Point

The change in freezing point (\( \Delta T_f \)) is the difference from water's normal freezing point (0°C) to the target freezing point (-6°C). \ \ \( \Delta T_f = 0 - (-6) = 6^{\circ}C \).

Step 3: Set Up Molality Equation

We know \( \Delta T_f = 6^{\circ}C \), and \( K_f = 1.86 \) kg mol\(^{-1}\). Substitute these into the equation \( 6 = 1.86 \times m \) to find the molality \( m \).

Step 4: Solve for Molality

Rearrange the molality equation to solve for \( m \): \ \ \( m = \frac{6}{1.86} = 3.2258 \) mol kg\(^{-1}\). \ \ Molality is defined as moles of solute per kilogram of solvent.

Step 5: Calculate Moles of Ethylene Glycol

With the molality known, calculate the moles of ethylene glycol required for 4 kg of water: \ \ \( 3.2258 \times 4 = 12.9032 \) moles.

Step 6: Convert Moles to Mass

Convert the moles of ethylene glycol to mass using its molar mass: \ \ \( \text{Mass} = \text{Moles} \times \text{Molar Mass} = 12.9032 \times 62 \) g/mol = 799.9984 g.

Step 7: Round and Choose the Closest Option

Round 799.9984 g to the nearest option, which is 804.32 g. Therefore, the correct answer is option (b).

Key Concepts

Colligative Properties

Colligative properties are fascinating because they depend not on the type of particles in a solution, but rather on their number. They are characteristic properties of solutions that change when solutes are added and occur due to disruptions in solvent properties when solutes are introduced.
These properties include:

• Vapor pressure lowering
• Freezing point depression
• Boiling point elevation
• Osmotic pressure

Specifically, freezing point depression is of interest when discussing solutions like antifreeze. When a solute, such as ethylene glycol, is added to a solvent like water, it lowers its freezing point. This is because the solute particles disrupt the formation of the solid phase, making it more difficult for the solvent to freeze at its normal freezing point.
In our specific problem, adding ethylene glycol to water ensures the water remains liquid at lower temperatures, such as in cold climates. The amount of freezing point depression depends directly on the amount of solute particles, a hallmark of colligative properties.

Molality Calculation

Molality is a measure of the concentration of a solution, expressed as moles of solute per kilogram of solvent. It is especially useful in calculating changes in colligative properties, such as freezing point depression, because it does not change with temperature.To determine the molality, we use the equation for freezing point depression: \[ \Delta T_f = K_f \times m \]Where:

• \( \Delta T_f \) is the change in freezing point,
• \( K_f \) is the cryoscopic constant, and
• \( m \) is the molality of the solution.

In the exercise, the change in freezing point \( \Delta T_f \) is calculated as 6°C. With the cryoscopic constant \( K_f \) given as 1.86 kg mol^-1, we rearrange the equation to solve for molality \( m \): \[ m = \frac{6}{1.86} \approx 3.2258 \text{ mol kg}^{-1} \]
This result tells us how concentrated our solution is in terms of solute presence relative to the solvent.

Cryoscopic Constant

The cryoscopic constant \( K_f \) is a critical factor in determining the extent of freezing point depression for a particular solvent. It is defined as the freezing point depression observed per molal concentration of a solute for a given solvent. This constant is crucial because it allows us to predict how much a solute will lower the freezing point of a solvent.For water, the cryoscopic constant is given as 1.86 kg mol^-1. This means that each mole of a solute dissolved in one kilogram of water will decrease the freezing point by 1.86°C. Knowing the cryoscopic constant allows us to calculate the necessary amount of solute needed to achieve a desired freezing point in a solution.
In our example, the ethylene glycol is added specifically to take advantage of water's known cryoscopic constant. By doing so, we can calculate the mass required to lower the freezing point to a target temperature, ensuring that the solution behaves as needed in cold conditions.