Problem 171 - A Complete Resource Book in Chemistry for JEE Main

Question

\(K_{t}\) for water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} .\) If your automobile radiator holds \(1.0 \mathrm{~kg}\) of water, how many grams of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) must you add to get the freezing point of the solution lowered to \(-2.8^{\circ} \mathrm{C}\) ? (a) \(39 \mathrm{~g}\) (b) \(93 \mathrm{~g}\) (c) \(72 \mathrm{~g}\) (d) \(27 \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct answer is (b) 93 g.

Step 1: Formula for Freezing Point Depression

The formula for freezing point depression is \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution. For ethylene glycol, \( i = 1 \) because it does not ionize in solution.

Step 2: Calculate Change in Freezing Point

The initial freezing point of pure water is \( 0^{\circ} \mathrm{C} \). We need the solution to have a freezing point of \( -2.8^{\circ} \mathrm{C} \). Therefore, the change in freezing point, \( \Delta T_f = 0^{\circ} \mathrm{C} - (-2.8^{\circ} \mathrm{C}) = 2.8^{\circ} \mathrm{C} \).

Step 3: Calculate Molality

Rearrange the freezing point depression formula to solve for molality: \( m = \frac{\Delta T_f}{i \cdot K_f} \). Substitute the known values: \( m = \frac{2.8}{1 \cdot 1.86} = 1.5054 \frac{\mathrm{mol}}{\mathrm{kg}} \).

Step 4: Convert Molality to Moles of Solute

Molality \( (m) \) is given in moles of solute per kilogram of solvent. Since we have \( 1.0 \mathrm{~kg} \) of water, the moles of ethylene glycol needed is the same as the molality: \( 1.5054 \mathrm{~mol} \).

Step 5: Calculate Mass of Ethylene Glycol

The molar mass of ethylene glycol \( \left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right) \) is \( 62.07 \mathrm{~g/mol} \). Therefore, the mass of ethylene glycol required is \( 1.5054 \mathrm{~mol} \times 62.07 \mathrm{~g/mol} = 93.41 \mathrm{~g} \).

Key Concepts

Cryoscopic Constant

The cryoscopic constant, denoted as \( K_f \), is a measure of a solvent's ability to lower its freezing point when a solute is dissolved in it. It's a special property intrinsic to each solvent. For instance, water has a well-known cryoscopic constant of \( 1.86 \, \mathrm{K} \, \mathrm{kg} \, \mathrm{mol}^{-1} \).
This means that for every mole of solute particles added per kilogram of water, the freezing point is expected to depress by \( 1.86 \) degrees Celsius. The value of this constant takes into account the inherent resistance of a liquid to freeze when external molecules interfere with the solid structure during crystallization.
Understanding the cryoscopic constant is key to calculating the extent of freezing point depression in solutions.

Molality

Molality (\( m \)) is a concentration measure describing the number of moles of solute per kilogram of solvent. It is an important factor used in colligative properties like freezing point depression. Unlike molarity, which changes with temperature due to volume changes, molality remains constant.
For example, in a freezing point depression scenario like ours, it's important to calculate the molality to understand how many moles of a solute such as ethylene glycol are present in water. The formula to find molality 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.

• Molality is particularly useful when dealing with temperature variations, as it does not change unlike molarity, making it more reliable for such applications.

Van't Hoff Factor

The van't Hoff factor (\( i \)) is a dimensionless quantity used to account for the effect of a solute in a solution based on the number of particles upon dissolution.
For molecules like ethylene glycol, which do not ionize in solution, this factor is \( i = 1 \). In other words, each molecule of ethylene glycol results in one particle in solution, simplifying calculations since no ionization occurs.
Understanding the van't Hoff factor can significantly aid in calculating colligative properties because it reflects how many pieces a solute splits into in a solvent. This, in turn, affects the extent to which the properties change based on particle concentration, not merely the solute's mass.