An ethylene glycol solution contains 25.2 g of ethylene glycol (C 2 H 6 O 2 ) in 99.5 mL of water. Determine the change in boiling point. Assume a density of 1.00 g/mL for water.

An ethylene glycol solution contains 25.2 grams of ethylene glycol, which is c 28602 in 99.5 milliliters of water. Determine the change in boiling point. Assume a density of 1 gram per milliliter for water. Alright. So we need to find the change in boiling point, so that's delta t b equals I times k b times m. Ethylene glycol is covalent in nature, which means it doesn't break up into ions, so I would just be 1. Here we're placing it in 99.5 ml of water. Water is our solvent, so the boiling point constant of our water solvent is 0.51 degrees Celsius over molality. All we have to do now is we have to determine what our our molality is. Right? So remember molality equals moles of solute divided by kilograms of solvent. Alright. So we're gonna do the easy one part first. So we're gonna change 25.2 grams of ethylene glycol into moles. So we're gonna say here, for every 1 mole, we figure out the grams of ethylene glycol. So it has 2 carbons, 6 hydrogens, and 2 oxygens. Their combined mass is 62.068 grams from the periodic table. So that's gonna give me 0.40601 moles of Ethylene Glycol. Now I need to figure out my kilograms of solvent, which in this case is water. So we have 99.5 ml of water. They give me the density of water is 1, so that's 1 gram of water per 1 milliliter. And then all I have to do is have to change the grams to kilograms. So 1 kilo is 10 to the 3 grams. So that's gonna equal 0.0995 kilograms. So we're gonna take that and plug it in here. So that portion will equal molality, which cancels out with this molality. So when we punch all that in, our change in our boiling point is 2.08 degrees Celsius. So that will represent our delta TB for this particular practice question.

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14. Solutions

Boiling Point Elevation

General Chemistry

14. Solutions

Boiling Point Elevation

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Multiple Choice

An ethylene glycol solution contains 25.2 g of ethylene glycol (C₂H₆O₂) in 99.5 mL of water. Determine the change in boiling point. Assume a density of 1.00 g/mL for water.

18.4°C

9.22°C

2.08°C

0.572°C

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Verified step by step guidance

Calculate the molality of the solution. First, find the moles of ethylene glycol (C2H6O2) using its molar mass. The molar mass of C2H6O2 is calculated as follows: C: 12.01 g/mol * 2 + H: 1.01 g/mol * 6 + O: 16.00 g/mol * 2.

Convert the mass of water from grams to kilograms. Since the density of water is 1.00 g/mL, 99.5 mL of water is equivalent to 99.5 g, which is 0.0995 kg.

Use the formula for molality: molality (m) = moles of solute / kilograms of solvent. Substitute the moles of ethylene glycol and the kilograms of water into this formula to find the molality.

Determine the boiling point elevation using the formula: ΔT_b = i * K_b * m, where ΔT_b is the change in boiling point, i is the van't Hoff factor (which is 1 for ethylene glycol as it does not ionize), K_b is the ebullioscopic constant for water (0.512 °C kg/mol), and m is the molality calculated in the previous step.

Substitute the values into the boiling point elevation formula to find the change in boiling point. This will give you the increase in boiling point due to the presence of ethylene glycol in the solution.