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

Question

\(\mathrm{K}_{i}\) 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

(b) 93 g

Step 1: Understand the Freezing Point Depression Formula

To solve the problem, we need to apply the formula for freezing point depression: \[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the change in freezing point, \(K_f\) is the freezing point depression constant, \(m\) is the molality, and \(i\) is the van't Hoff factor. For ethylene glycol, \(i=1\) because it is a nonelectrolyte and doesn't dissociate into ions.

Step 2: Calculate the Change in Freezing Point

We are given that the water's freezing point should be lowered to \(-2.8^{\circ} C\). The normal freezing point of water is \(0^{\circ} C\). Therefore, the change in freezing point \(\Delta T_f\) is:\[\Delta T_f = 0^{\circ} C - (-2.8^{\circ} C) = 2.8^{\circ} C\]

Step 3: Solve for Molality

Plug \(\Delta T_f\) and \(K_f\) into the freezing point depression formula to solve for molality \(m\):\[2.8 = 1 \cdot 1.86 \cdot m\]Solving for \(m\), we get:\[m = \frac{2.8}{1.86} \approx 1.505\ \text{mol/kg}\]

Step 4: Determine the Moles of Solute Required

Molality is defined as moles of solute per kilogram of solvent. Since there is 1.0 kg of water, the moles of ethylene glycol \(n\) required are:\[n = 1.505\ \text{mol/kg} \times 1\ \text{kg} = 1.505\ \text{mol}\]

Step 5: Convert Moles to Grams

To find the grams of ethylene glycol needed, multiply the moles by the molar mass of ethylene glycol (\(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\)), which is \(62.07\ \text{g/mol}\):\[\text{mass} = 1.505\ \text{mol} \times 62.07\ \text{g/mol} \approx 93.4\ \text{g}\]Rounded to the nearest gram, this is roughly 93 g.

Key Concepts

Colligative Properties

Colligative properties are special characteristics of solutions affected by the number of solute particles, not by the type of particles. This makes these properties distinctive because they don't depend on the chemical identity of the solute. For instance, when a solute is added to a solvent, it can impact various physical properties.

Some key colligative properties include:

• Freezing Point Depression
• Boiling Point Elevation
• Vapor Pressure Lowering
• Osmotic Pressure

The decrease in the freezing point of water when ethylene glycol is added is a typical example of freezing point depression. This happens because the solute particles interfere with the formation of a solid structure, hence requiring a lower temperature to freeze.

In the context of an automobile radiator, using a solute like ethylene glycol prevents the water from freezing at the regular 0°C. As a utility, ethylene glycol's presence allows engines to function in cooler temperatures without damage from ice formation. It's a practical application of colligative properties in daily life.

Molality

Molality, often symbolized as "m," is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent, and the units are mol/kg. While similar to molarity, which measures concentration as moles per liter of solution, molality uses mass instead and does not change with temperature.

Molality comes in handy for calculations involving colligative properties because it accounts for variations in volume due to temperature changes. When solutes alter the physical properties of a solution, precise calculations are essential, making molality a preferred measure.

To calculate molality, follow these basic steps:

• Determine the number of moles of the solute.
• Measure the mass of the solvent in kilograms.
• Divide the moles of solute by the mass of solvent to determine molality.

In our specific problem with ethylene glycol, calculating its molality helped determine how much needs to be added to the water to achieve a desired freezing point.

Van't Hoff Factor

The Van't Hoff factor, denoted as "i," plays a crucial role in adjusting for the effect of solute particles in solution chemistry. It represents the number of particles a solute dissociates into in a solution. For nonelectrolytes, such as ethylene glycol, the Van't Hoff factor is 1 since they do not dissociate into ions.

Understanding the Van't Hoff factor is vital when dealing with electrolytes, as these compounds split into multiple ions, impacting the colligative properties. For instance, sodium chloride ( salt) has a Van't Hoff factor of 2 because each molecule separates into two ions: Na⁺ and Cl^-.

In calculations involving freezing point depression and other similar properties, the Van't Hoff factor helps correct the formula. By multiplying the factor with other variables, we gain more accurate results regarding physical changes in the solution.

In our automotive example, using a Van't Hoff factor of 1 was straightforward since ethylene glycol doesn't split into multiple ions. That simplicity made the calculations easier, providing a clear demonstration of how these factors are used effectively in practical scenarios.