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.

• A. 18.4°C
• B. 9.22°C
• C. 2.08°C
• D. 0.572°C

Answer: (C)

Verified step by step guidance

1. 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.
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.
3. 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.
4. 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.
5. 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.