Question

Antifreeze solutions are aqueous solutions of ethylene glycol, C2H6O2 ( d = 1.12 g/ml) . In Connecticut, cars are “winterized” by filling radiators with an antifreeze solution that will protect the engine for as low as -20 oF.


a. What is the minimum molality of antifreeze solution required?

b. How many milliliters of ethylene glycol need to be added to 250 ml of water (dw = 1.00 g/ml) to
prepare the solution in (a)

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Answer 1

Answer:

check ur text book

Explanation:

Answer 2

To determine the minimum molality of an antifreeze solution required to protect an engine at -20 oF and the amount of ethylene glycol needed to prepare the solution, use the following method:

a. We need to calculate the molality of the antifreeze solution required to protect the engine at -20 oF. For this, we can use the equation:

ΔTf = Kf * molality

• where ΔTf is the freezing point depression
• Kf is the freezing point depression constant (1.86 oC/m for water), and molality is the molality of the antifreeze solution.

• At -20 oF, the freezing point depression is 52 oC, which is equivalent to 52/1.86 = 27.96
• Therefore, the minimum molality of antifreeze solution required to protect the engine at -20 oF is 27.96 m.

b. To determine the amount of ethylene glycol needed to prepare the solution, we can use the formula:

• mass of solute = molality * molar mass * mass of solvent
• In this case, the solvent is water and the mass of solvent is 250 ml, which is equivalent to 250 g (since the density of water is 1 g/ml). The molar mass of ethylene glycol is 62 g/mol.
• Using the minimum molality of 27.96 m from part (a), we can calculate the mass of ethylene glycol needed as follows:
• mass of ethylene glycol = 27.96 * 62 * 0.25
• mass of ethylene glycol = 432.6 g

Therefore, we need to add 432.6 g (or approximately 433 ml) of ethylene glycol to 250 ml of water to prepare an antifreeze solution with a minimum molality of 27.96 m.

To learn more about molality from the given link.

https://brainly.in/question/51743564

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