An ethylene glycol solution contains 28.3 g of ethylene glycol, C 2 H 6 O 2 in 97.2 mL of water. Calculate the freezing point of the solution. The density of water 1.00 g/mL.

An ethylene glycol solution contains 28.3 grams of ethylene glycol, which is c c 2 h six zero two, in 97.2 ml of water. Calculate the freezing point of the solution, we're told the density of water is 1 gram per milliliter. Alright. So here we need to calculate the freezing point. So we need to figure out change in freezing point. So delta t f equals I times k f times m. Ethylene glycol is a covalent solute, so it doesn't break up into ions. Therefore, the van't Hoff factor is 1. The solvent here is water, so its k f is 1.86 degrees Celsius over molality. Then what we need to do is we need to figure out the moles of ethylene glycol. Okay. So here we're gonna figure out the moles of ethylene glycol. For every 1 mole, the mass is 62.068 grams. So plug that here. And then we need to figure out the kilograms of water, so we have 97.2 ml's. We're told that the density of water is 1 gram per 1 ml. And then remember 1 kilogram is 10 to the 3 grams. Okay. So this comes out to 0.0972 kilograms. So we plug that here. So when we do that we get 8.72 degrees Celsius, and freezing point of our new solution equals freezing point of our pure solvent water minus delta t f. Remember, pure water freezes at 0 degrees Celsius minus 8.72 degrees Celsius, so our answer here is negative 8.72 degrees Celsius. So this will be the new freezing point of our created solution.

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9. Solutions

Freezing Point Depression

GOB Chemistry

9. Solutions

Freezing Point Depression

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Multiple Choice

An ethylene glycol solution contains 28.3 g of ethylene glycol, C₂H₆O₂ in 97.2 mL of water. Calculate the freezing point of the solution. The density of water 1.00 g/mL.

–8.72°C

–0.848°C

–0.541°C

–17.4°C

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Verified step by step guidance

First, calculate the molality of the solution. Molality (m) is defined as the moles of solute per kilogram of solvent. Start by finding the moles of ethylene glycol (C2H6O2) using its molar mass. The molar mass of C2H6O2 is approximately 62.07 g/mol. Use the formula: moles = mass (g) / molar mass (g/mol).

Next, determine the mass of the water in kilograms. Since the density of water is 1.00 g/mL, the mass of 97.2 mL of water is 97.2 g. Convert this mass to kilograms by dividing by 1000.

Calculate the molality of the solution using the formula: molality (m) = moles of solute / kilograms of solvent. Use the moles of ethylene glycol calculated in step 1 and the mass of water in kilograms from step 2.

Use the freezing point depression formula to find the change in freezing point: ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van't Hoff factor (which is 1 for ethylene glycol as it does not ionize), Kf is the freezing point depression constant for water (1.86 °C/m), and m is the molality calculated in step 3.

Finally, subtract the change in freezing point (ΔTf) from the normal freezing point of water (0°C) to find the freezing point of the solution. This will give you the final answer.