Problem 75 - Chemistry: The Central Science

Question

How many grams of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) must be added to \(1.00 \mathrm{~kg}\) of water to produce a solution that freezes at \(-5.00^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Approximately 166.96 grams of ethylene glycol (C2H6O2) must be added to 1.00 kg of water to produce a solution that freezes at -5.00°C.

Step 1: Calculate the freezing-point depression

The freezing-point depression, ΔTf, is the difference between the freezing point of the pure solvent and the freezing point of the solution. In this case, ΔTf is the difference between the freezing point of pure water, 0°C, and the freezing point of the solution, -5.00°C. ΔTf = 0 - (-5.00) ΔTf = 5.00°C

Step 2: Use the freezing-point depression formula to find the molality of the solution

The freezing-point depression can be calculated using the formula: ΔTf = Kf ⋅ molality Where Kf is the cryoscopic constant of a solvent (in the case of water, Kf = 1.86 °C/molal), and molality is the number of moles of solute per kilogram of solvent. Rearrange the formula to find the molality: molality = ΔTf / Kf Now substitute the known values to find the molality: molality = 5.00°C / 1.86 °C/molal molality ≈ 2.69 molal

Step 3: Determine the moles of ethylene glycol needed to achieve the desired molality

We know that molality is the number of moles of solute per kilogram of solvent. In this case, we have 1.00 kg of solvent (water) and want to achieve a molality of 2.69 molal. moles of ethylene glycol = molality × mass of solvent moles of ethylene glycol = 2.69 molal × 1.00 kg moles of ethylene glycol ≈ 2.69 mol

Step 4: Convert the moles of ethylene glycol to grams

To find the mass of ethylene glycol needed, we need to multiply the moles of ethylene glycol by its molar mass. The molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol. mass of ethylene glycol = moles of ethylene glycol × molar mass of ethylene glycol mass of ethylene glycol = 2.69 mol × 62.07 g/mol mass of ethylene glycol ≈ 166.96 g Therefore, approximately 166.96 grams of ethylene glycol must be added to 1.00 kg of water to produce a solution that freezes at -5.00°C.