Kf for water is 1.86K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8°C ?

To determine how many grams of ethylene glycol (C2H6O2) you need to add to lower the freezing point of a solution to -2.8°C, we can use the concept of freezing point depression. This principle states that the freezing point of a solvent decreases when a solute is added. The formula to calculate the freezing point depression is:

Understanding the Freezing Point Depression Formula

The equation is:

ΔTf = Kf * m

Where:

• ΔTf is the change in freezing point (°C)
• Kf is the freezing point depression constant of the solvent (in this case, water, which is 1.86 K kg mol^-1)
• m is the molality of the solution (in moles of solute per kg of solvent)

Step-by-Step Calculation

First, let’s clarify what we know:

• The freezing point of pure water is 0°C.
• We want to lower it to -2.8°C.
• The change in freezing point (ΔTf) is therefore:

ΔTf = 0°C - (-2.8°C) = 2.8°C

Next, we can rearrange the freezing point depression formula to find the molality (m):

m = ΔTf / Kf

Substituting the values:

m = 2.8°C / 1.86 K kg mol^-1 = 1.505 m

Finding the Number of Moles of Ethylene Glycol

Now that we have the molality, we can determine how many moles of ethylene glycol we need. Since molality is defined as moles of solute per kilogram of solvent, we can use the mass of the water in the radiator:

Mass of water = 1.0 kg

Using the formula:

Moles of solute = m * mass of solvent (in kg)

Moles of ethylene glycol = 1.505 mol/kg * 1.0 kg = 1.505 moles

Calculating the Mass of Ethylene Glycol

Next, we need to convert moles of ethylene glycol to grams. First, we need its molar mass:

• Carbon (C): 12.01 g/mol * 2 = 24.02 g/mol
• Hydrogen (H): 1.008 g/mol * 6 = 6.048 g/mol
• Oxygen (O): 16.00 g/mol * 2 = 32.00 g/mol

Total molar mass of C2H6O2 = 24.02 g/mol + 6.048 g/mol + 32.00 g/mol = 62.068 g/mol

Now, you can calculate the mass of ethylene glycol needed:

Mass = moles * molar mass

Mass = 1.505 moles * 62.068 g/mol ≈ 93.41 grams

The Final Answer

To achieve a freezing point of -2.8°C, you will need to add approximately 93.41 grams of ethylene glycol to the 1.0 kg of water in the automobile radiator. This process shows how colligative properties, such as freezing point depression, rely on the number of solute particles and not their identity, which is quite fascinating in the study of solutions!