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: \(\left(K_{f}\right.\) 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

Add 804.32 g of ethylene glycol to prevent freezing at -6°C.

Step 1: Understanding the Problem

To find the mass of ethylene glycol needed to lower the freezing point of water to -6°C, we use the formula 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.

Step 2: Calculate Freezing Point Depression

The initial freezing point of water is 0°C, and it needs to be lowered to -6°C. The change in freezing point \( \Delta T_f = 0°C - (-6°C) = 6°C \).

Step 3: Determine Molality (m)

Rearrange the freezing point depression formula: \( m = \frac{\Delta T_f}{K_f} = \frac{6}{1.86} \approx 3.23 \text{ mol/kg} \). This is the molality of the ethylene glycol in water.

Step 4: Calculate Moles Needed

Molality (m) is defined as the moles of solute per kilogram of solvent. Therefore, the moles of ethylene glycol needed is \( m \times \text{mass of water in kg} = 3.23 \times 4 = 12.92 \text{ moles} \).

Step 5: Find the Mass of Ethylene Glycol

The molar mass of ethylene glycol is \(62 \text{ g/mol}\). The mass needed is \(12.92 \text{ moles} \times 62 \text{ g/mol} = 801.04 \text{ grams}\).

Step 6: Select the Closest Answer

Among the provided options, option (b) \(804.32\) grams is the closest to our calculated mass of \(801.04\) grams.

Key Concepts

Cryoscopic Constant

The cryoscopic constant, often represented as \( K_f \), is a critical factor in understanding freezing point depression. This constant is specific to each solvent and describes how much the freezing point is lowered per unit of molality of a solute. In other words, it tells us how effectively a substance can lower the freezing temperature of a liquid when dissolved in it.

For water, the cryoscopic constant is \( 1.86 \text{ kg mol}^{-1} \). This means, each molal of a dissolved solute will reduce the freezing point of water by \( 1.86^{ ext{o}} ext{C}\). This property is essential when substances like ethylene glycol are added to water to prevent freezing in cold climates.

Key points about cryoscopic constant:

• It's solvent-specific, so different solvents have different \( K_f \) values.
• Helps calculate the freezing point depression, which is the change from the pure solvent's freezing temperature.
• Essential in applications like antifreeze, where maintaining a liquid state is critical under low temperatures.

Molality Calculation

The calculation of molality is a fundamental step in determining how much a solute will affect a solvent's properties, such as its freezing point.
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, it does not change with temperature, because it's based on mass, not volume. This makes it particularly useful in colligative properties like freezing point depression.

To find molality, use this formula:

• \( m = \frac{\Delta T_f}{K_f} \)

Where:

• \( \Delta T_f \) is the difference in freezing points (given as 6°C in the original exercise).
• \( K_f \) is the cryoscopic constant. For water, it is \( 1.86 \, \text{kg mol}^{-1} \).

In the exercise, we calculated:

• \( m \approx 3.23 \, \text{mol/kg} \).

This means a solution with 3.23 moles of ethylene glycol per kilogram of water will lower the freezing point by 6°C.

Ethylene Glycol as Antifreeze

Ethylene glycol is a common choice for antifreeze because of its properties and effectiveness. Its ability to lower the freezing point of water allows it to prevent the ice formation in vehicle engines and other systems exposed to cold temperatures.

Here's why ethylene glycol is preferred:

• **Freezing Point Depression**: When ethylene glycol is mixed with water, it disrupts the formation of the ice lattice, thus lowering the freezing point.
• **Boiling Point Elevation**: It also helps in elevating the boiling point of water, making it ideal for use in cooling systems of automobiles, which need to function efficiently at a wide range of temperatures.
• **Cost-Effective and Readily Available**: Ethylene glycol is relatively cheap and easy to produce.
• **Moderate Toxicity**: While toxic if ingested, it is less harmful when handled properly compared to some alternatives.

The process of using ethylene glycol involves calculating the required amount to achieve a desired freezing point. This is crucial in places with extremely cold winters, ensuring that water in engines doesn't freeze, which could otherwise lead to significant damage.