Question

A G.T.O. has a 22 gal. cooling system. Suppose you fill it with a \(50-50\) solution by volume of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\), ethylene glycol, and water. At what temperature would freezing become a problem? Assume the specific gravity of ethylene glycol is \(1.115\) and the freezing point depression constant of water is \(\left(1.86 \mathrm{C}^{\circ} / \mathrm{mole}\right)\). You might have placed in methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) instead of \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2}\). If the current cost of ethylene glycol is 12 cents \(/ \mathrm{lb}\) and the cost of methanol is 8 cents \(/ \mathrm{lb}\), how much money would you save by using \(\mathrm{CH}_{3} \mathrm{OH}\) ? And yet, ethylene glycol is the more desirable antifreeze. Why? density \(=(.79 \mathrm{~g} / \mathrm{ml})\) for \(\mathrm{CH}_{3} \mathrm{OH}\) and \(3.785\) liter \(=1\) gallon.

Short answer

The freezing point of the 50-50 ethylene glycol and water solution would be -33.45 C°, and using methanol instead of ethylene glycol would save $6.49. However, ethylene glycol is more desirable as an antifreeze due to its better performance, safety, and environmental features.

Step-by-step solution

Calculate the molality of the ethylene glycol and water solution

We have a 22-gallon cooling system filled with a 50-50 solution of ethylene glycol and water. First, we need to find out the specific gravity of the solution and convert it to mass. Specific gravity of ethylene glycol= 1.115 1 gallon = 3.785 liters 22 gallons = 22 gal * 3.78 L/gal = 83.27 L The volume of ethylene glycol in the cooling system is half of the solution as it's a 50-50 mixture: Volume of ethylene glycol = 0.5 * 83.27 L = 41.635 L Now, we will find the mass of ethylene glycol in the solution: Mass of ethylene glycol = Specific gravity * Volume = 1.115 * 41.635 L = 46.43 kg Now, we find the molar mass of ethylene glycol (C2H6O2): Molar mass = 2 * 12.01 (Carbon) + 6 * 1.008 (Hydrogen) + 2 * 16.00 (Oxygen) = 62.07 g/mol Next, we calculate the moles of ethylene glycol in the solution: Moles of ethylene glycol = 46.43 kg * 1000 g/kg / 62.07 g/mol = 748.24 moles Considering the 50-50 mixture, we have an equal volume of water alongside ethylene glycol, which is 41.635 L. To calculate the molality, we need the mass of water in kilograms: Mass of water = Volume of water * Density of water = 41.635 L * 1 kg/L = 41.635 kg Now we can calculate the molality of the solution: Molality = Moles of ethylene glycol / Mass of water = 748.24 moles / 41.635 kg = 17.97 mol/kg

Calculate the freezing point depression temperature

Now that we have the molality of the solution, we can use the freezing point depression constant for water (Kf = 1.86 C°/mol) to calculate the freezing point depression temperature: ΔTf = Kf * molality = 1.86 C°/mol * 17.97 mol/kg = 33.45 C° The freezing point of water is 0 C°, so the freezing point of the 50-50 solution would be 0 C° - 33.45 C° = -33.45 C°. Freezing would become a problem at -33.45 C°.

Calculate the cost difference between ethylene glycol and methanol

The current cost of ethylene glycol is 12 cents/lb and the cost of methanol is 8 cents/lb. Let's first convert the mass of ethylene glycol in the system to lbs: Mass of ethylene glycol = 46.43 kg * 2.205 lb/kg = 102.45 lbs Now we find the cost of ethylene glycol for this solution: Cost of ethylene glycol = 102.45 lbs * \(0.12/lb = \)12.29 In the case of using methanol (CH3OH) instead of ethylene glycol, the density of methanol is 0.79 g/mL. Let's find the mass of methanol for our solution: Mass of methanol = specific gravity * volume = 0.79 * 1000 g/L * 41.635 L = 32.89 kg Now convert mass of methanol to lbs: Mass of methanol = 32.89 kg * 2.205 lb/kg = 72.53 lbs Now we find the cost of methanol for this solution: Cost of methanol = 72.53 lbs * \(0.08/lb = \)5.80 The cost saving from using methanol would be \(12.29 - \)5.80 = $6.49.

Why is ethylene glycol more desirable despite the potential cost savings with methanol?

Ethylene glycol is more desirable as an antifreeze than methanol for several reasons, such as: 1. Ethylene glycol provides a lower freezing point than methanol, which offers better protection in colder weather conditions. 2. Ethylene glycol has a higher boiling point than methanol, which helps to prevent engine overheating. 3. Ethylene glycol is less volatile and flammable than methanol, making it safer to use in a cooling system. 4. Ethylene glycol is less toxic and has lower environmental impact compared to methanol. In conclusion, freezing would become a problem for the ethylene glycol and water solution at -33.45 C°. Using methanol instead of ethylene glycol would save $6.49, but ethylene glycol is still more desirable as an antifreeze due to its better performance, safety, and environmental features.

Key concepts

Molality Calculation

Understanding molality is crucial when dealing with solutions, like antifreeze for a car's cooling system. Molality represents the concentration of a solute in a solvent, calculated as the number of moles of solute per kilogram of solvent.

For example, if you have a solution with ethylene glycol and water, you'll want to determine how much ethylene glycol is present in a certain amount of water to know the solution's antifreeze capability. This is done by calculating the molality using the formula:
\[ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \]
By ensuring proper molality calculation, one can obtain the desired freezing point depression, which is critical for the solution's effectiveness in low temperatures. High molality means more solute particles are present, leading to a lower freezing point, which is exactly what you want for an antifreeze.

Ethylene Glycol as Antifreeze

Ethylene glycol (\(\mathrm{C_2H_6O_2}\)) is a common ingredient in antifreeze formulations, prized for its ability to lower the freezing point of solutions when mixed with water. The physical properties of ethylene glycol, including a high boiling point and low freezing point, make it an ideal substance for maintaining an engine's temperature within an optimal range.

It's important to note that a mixture's capability to protect against freezing is not just about preventing solidification; it is about ensuring that the vehicle's engine operates efficiently under varying temperature conditions. Ethylene glycol meets these requirements and also decreases the risk of overheating due to its boiling point characteristics, making it the preferred choice over other chemicals for a vehicle's cooling system.

Cost Comparison of Antifreeze Chemicals

When selecting an antifreeze chemical, cost is a significant factor. However, it's crucial to consider both the upfront cost and the performance characteristics. As seen in the textbook problem, ethylene glycol and methanol have differing costs per pound. While methanol may be cheaper, it also provides less freezing protection and poses higher risks due to its volatility and toxicity.

A thorough cost comparison includes the following considerations:

• Initial price per unit weight
• Freezing and boiling point properties
• Safety and toxicity levels
• Environmental impact
• Long-term cost efficiency based on performance and required volume

By evaluating these factors, one can understand why a seemingly more expensive option like ethylene glycol might be more cost-effective in the long run for use as an antifreeze agent.