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 closest to \(804.32 \mathrm{~g}\).

Step 1: Understand the Formula for Freezing Point Depression

To determine the mass of ethylene glycol needed, we use the formula for freezing point depression: \[ \Delta T_f = i \cdot K_f \cdot m \] where \( \Delta T_f \) is the depression in freezing point, \( i \) is the van't Hoff factor (for ethylene glycol, \( i = 1 \)), \( K_f \) is the freezing point depression constant for water, and \( m \) is the molality of the solution.

Step 2: Calculate the Required Depression in Freezing Point

The freezing point of pure water is \(0^{\circ}C\). We want the solution to freeze at \(-6^{\circ}C\). Thus, the change in freezing point is: \[ \Delta T_f = 0 - (-6) = 6^{\circ}C \]

Step 3: Rearrange the Formula for Molality

Using the formula from Step 1, solve for molality \( m \):\[ m = \frac{\Delta T_f}{i \cdot K_f} = \frac{6}{1 \times 1.86} = 3.2258 \text{ mol/kg} \]

Step 4: Calculate the Moles of Solute (Ethylene Glycol)

Molality is defined as moles of solute per kilogram of solvent. For 4 kg of water:\[ \text{Moles of ethylene glycol} = 3.2258 \times 4 = 12.9032 \text{ moles} \]

Step 5: Convert Moles of Ethylene Glycol to Mass

Convert the moles calculated in Step 4 to mass using the molar mass of ethylene glycol:\[ \text{Mass of ethylene glycol} = 12.9032 \text{ moles} \times 62 \text{ g/mol} = 800 \text{ g} \]

Step 6: Match the Calculation to the Nearest Option

Compare the calculated mass with the provided options. The closest answer is option (b) \(804.32 \text{ g}\).

Key Concepts

Ethylene Glycol

Ethylene glycol, a simple, colorless liquid, is a key ingredient in antifreeze solutions. This organic compound helps lower the freezing point of water, preventing it from becoming ice in cold temperatures. It is especially useful in regions where temperatures dip below freezing. Ethylene glycol forms a solution with water that disrupts the regular crystal structure of ice. This disruption requires a lower temperature to solidify the mixture, hence preventing freezing. The molecule itself has a chemical formula of \(C_2H_6O_2\) and a molar mass of 62 g/mol.

When ethylene glycol is mixed with water, the mixture exhibits what is known as freezing point depression—a colligative property. This means that the property depends on the number of solute particles in the solvent, rather than their specific type. It's important to note that in its liquid form, ethylene glycol is quite toxic, so careful handling is crucial.

Molality

Molality is a measure of the concentration of a solution that specifically describes the number of moles of solute per kilogram of solvent. This concentration measure is used in calculating colligative properties because it does not change with temperature, unlike molarity which involves volumes that can expand or contract with temperature variations.

In the scenario of the antifreeze solution, molality helps us determine how many moles of ethylene glycol are necessary per kilogram of water to achieve the desired freezing point depression. The formula used is: \[ m = \frac{\Delta T_f}{i \cdot K_f} \] where \(\Delta T_f\) is the change in freezing point, \(i\) is the van't Hoff factor, and \(K_f\) is the cryoscopic constant. By calculating this value, we can determine how concentrated the solution must be to prevent water from freezing.

Van't Hoff Factor

The van't Hoff factor \(i\) accounts for the effect a solute has on the solution's colligative properties and can vary depending on the nature of the solute. It represents the number of particles the solute forms in solution. For non-electrolytes like ethylene glycol, the van't Hoff factor is 1 because it doesn't dissociate into more particles in the solution.

In electrolytes, this factor would be greater than 1, reflecting how many ions the solute forms upon dissolving. For ethylene glycol, since it remains as single entities in the solution, the factor is straightforward. The simplicity of its use in calculating freezing point depression makes it a particularly useful concept in chemistry, especially for students comparing electrolytes to nonelectrolytes.

KF value

The \(K_f\) value, also known as the cryoscopic constant, is a property of the solvent that indicates how much the freezing point of the solvent is lowered per molal concentration of the solute. For water, \(K_f = 1.86 \text{ kg mol}^{-1}\). This constant is unique to each solvent and crucial when calculating freezing point depressions.

It's important because it quantifies how sensitive a solvent is to the addition of a solute. In this exercise, the \(K_f\) value for water helps us determine how much ethylene glycol we need to add to achieve the desired change in freezing point. This constant is used alongside the molality and the van't Hoff factor in the freezing point depression equation, ensuring our solution calculation is accurate.