Calculate the freezing point and boiling point of an antifreeze solution that is \(50.0 \%\) by mass of ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) in water. Ethylene glycol is a nonelectrolyte.

The formulas for freezing point depression and boiling point elevation are: \(ΔT_f = K_f × m × i\) \(ΔT_b = K_b × m × i\) where: - \(ΔT_f\) is the freezing point depression - \(ΔT_b\) is the boiling point elevation - \(K_f\) is the freezing point depression constant of the solvent (water in our case) - \(K_b\) is the boiling point elevation constant of the solvent (water in our case) - \(m\) is the molality of the solute (ethylene glycol) in the solution - \(i\) is the van't Hoff factor of the solute (equal to 1 for a nonelectrolyte) The values for \(K_f\) and \(K_b\) for water are 1.86 °C/m and 0.512 °C/m, respectively.

Since we know the mass percentage of ethylene glycol in the antifreeze solution, we can use it to calculate the molality of ethylene glycol in the solution: Let's assume we have 100 g of the antifreeze solution (so we can easily calculate the mass of ethylene glycol and water in the solution). Mass of ethylene glycol = 50.0 g (as it's 50% by mass) Mass of water = 50.0 g (since the remaining 50% is water) Now we need to convert the mass of ethylene glycol to moles: Molar mass of ethylene glycol (\(HOCH_2CH_2OH\)) = (2 × 1) + (6 × 12) + (6 × 1) + (1 × 16) = 62 g/mol Moles of ethylene glycol = mass / molar mass = 50.0 g / 62 g/mol = 0.806 mol Now we can calculate molality: \(m = \cfrac{\text{moles of solute}}{\text{mass of solvent (kg)} }\) \(m = \cfrac{0.806\,\text{mol}}{0.050\,\text{kg}} = 16.13\,\text{mol/kg}\)

Now that we have the molality of ethylene glycol in the solution, we can calculate the freezing point depression and boiling point elevation using the formulas: \(ΔT_f = K_f × m × i\) \(ΔT_f = 1.86\,\text{°C/m} × 16.13\,\text{mol/kg} × 1\) \(ΔT_f = 30.0\,\text{°C}\) \(ΔT_b = K_b × m × i\) \(ΔT_b = 0.512\,\text{°C/m} × 16.13\,\text{mol/kg} × 1\) \(ΔT_b = 8.26\,\text{°C}\)

Finally, we can calculate the freezing point and boiling point of the antifreeze solution by subtracting the freezing point depression and adding the boiling point elevation to the normal freezing and boiling points of water, respectively: Freezing point of the antifreeze solution = Normal freezing point of water - \(ΔT_f\) Freezing point of the antifreeze solution = 0°C - 30.0°C = -30.0 °C Boiling point of the antifreeze solution = Normal boiling point of water + \(ΔT_b\) Boiling point of the antifreeze solution = 100°C + 8.26°C = 108.26 °C Now we have the freezing point and boiling point of the antifreeze solution containing 50.0% by mass ethylene glycol in water. The freezing point is -30.0 °C, and the boiling point is 108.26 °C.