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) \(27 \mathrm{~g}\)

Short answer

Add 93 g of ethylene glycol to the water.

Step-by-step solution

Identify Given Information

We are given: - Freezing point depression constant for water, \(K_t = 1.86 \, \text{K kg mol}^{-1}\).- Mass of water in the radiator, \(m = 1.0 \, \text{kg}\).- Desired freezing point of the solution, \(-2.8^{\circ} \text{C}\). - Chemical formula for ethylene glycol, \(\text{C}_2 \text{H}_6 \text{O}_2\).

Calculate Freezing Point Depression

The freezing point depression \(\Delta T_f\) can be calculated as follows:\[\Delta T_f = \text{freezing point of pure solvent} - \text{freezing point of solution} = 0^{\circ} \text{C} - (-2.8^{\circ} \text{C}) = 2.8\, \text{C}\].

Use Freezing Point Depression Formula

The formula for freezing point depression is given by:\[\Delta T_f = K_t \cdot m \cdot i\]where \(m\) is the molality of the solution, and \(i\) is the van 't Hoff factor (which is 1 for nonelectrolytes like ethylene glycol). Therefore:\[2.8 = 1.86 \cdot m\]Solve for \(m\):\[m = \frac{2.8}{1.86}\approx 1.505\, \text{mol kg}^{-1}\]

Calculate Moles of Solute Needed

Molality is defined as moles of solute per kilogram of solvent:\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} = 1.505\, \text{mol kg}^{-1}\].Given that the mass of water (solvent) is 1 kg, the moles of ethylene glycol needed is 1.505 mol.

Convert Moles to Grams

To find grams of ethylene glycol, use its molar mass. The molar mass of \(\text{C}_2 \text{H}_6 \text{O}_2\) is:\[2(12.01) + 6(1.01) + 2(16.00) = 62.08\, \text{g/mol}\].Now, calculate grams:\[\text{Grams of } \text{C}_2 \text{H}_6 \text{O}_2 = 1.505 \, \text{mol} \times 62.08\, \text{g/mol} \approx 93.4\, \text{g}\].

Round and Choose Answer

Round 93.4 g to the nearest whole number to match the given options. The closest answer is 93 g. Thus, the correct answer is option (b) 93 g.

Key concepts

Ethylene Glycol as an Antifreeze Agent

Ethylene Glycol, with the chemical formula \(\text{C}_2 \text{H}_6 \text{O}_2\), is a colorless, odorless liquid primarily used as an antifreeze agent in automotive engines. It lowers the freezing point of the solution when mixed with water, which helps protect the engine from freezing in cold weather conditions.

Its effectiveness arises from its ability to disrupt the hydrogen bonding between water molecules, thereby reducing the temperature at which the solution freezes. Furthermore, Ethylene Glycol is miscible with water, meaning it can be mixed in any proportion to achieve the desired freezing point depression, a key characteristic that makes it ideal for this purpose.

However, it is essential to handle Ethylene Glycol with care as it is toxic if ingested. Thus, make sure to store it safely and clean any spills immediately. Besides its use in antifreeze, it also acts as a precursor in the manufacture of polymers and serves in industrial applications like hydraulic fluids and plasticizers.

Understanding Molality

Molality is a concentration term used to describe the amount of solute in a solution. Specifically, it is defined as the number of moles of solute per kilogram of solvent.

Compared to molarity, which considers the volume of the solution, molality offers an advantage for freezing point depression calculations because it remains unchanged with temperature, providing more accurate results in varying conditions.

Mathematically, molality \(m\) is expressed as:

• \(m = \frac{\text{moles of solute}}{\text{kg of solvent}}\)

In our scenario, the moles of ethylene glycol required to lower the water's freezing point to \(-2.8^{\circ}\text{C}\) is calculated using the given freezing point depression constant for water, \(K_t\).

Because we have 1 kg of water, the simplification of the molality calculation becomes straightforward, allowing us to approach the process with ease.

Van 't Hoff Factor in Non-electrolytes

The Van 't Hoff Factor \(i\) is a figure that indicates the effect of a solute on the colligative properties of solutions. For non-electrolytes like ethylene glycol, \(i\) equals 1, as they do not dissociate into ions in solution.

Colligative properties, such as freezing point depression, depend on the number of particles in a solution rather than their identity. In the case of ethylene glycol, since it does not break apart into smaller particles, the factor remains as 1.

When calculating the effect of a solute on the freezing point of a solvent, the Van 't Hoff Factor is crucial. If the solute were an electrolyte, it would split into multiple ions, and the factor would reflect that dissociation.

Here's what to remember about the Van 't Hoff Factor:

• Non-electrolytes like ethylene glycol have \(i = 1\).
• Electrolytes have a higher \(i\) depending on the number of ions formed.
• It plays a key role in accurately determining colligative properties.

Knowing this aspect helps in preparing precise calculations for solutions, ensuring that the desired physical properties are achieved effectively.