An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass = 62 gmol–1) in 600 g of water. Calculate the freezing point of the solution. (Kf for water = 1·86 Kkg mol–1)

To calculate the freezing point of the solution, we need to use the formula for freezing point depression:

ΔTf = Kf * m

where:

• ΔTf is the change in freezing point,
• Kf is the cryoscopic constant of the solvent (in this case, water), and
• m is the molality of the solution.

First, we need to calculate the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.

The number of moles of ethylene glycol can be calculated as follows:

Moles of ethylene glycol = mass / molar mass = 31 g / 62 g/mol = 0.5 mol

The mass of water needs to be converted from grams to kilograms:

600 g = 0.6 kg

So, the molality of the solution is:

m = moles of solute / mass of solvent in kg = 0.5 mol / 0.6 kg = 0.833 mol/kg

Now we can calculate the change in freezing point:

ΔTf = Kf * m = 1.86 K kg/mol * 0.833 mol/kg = 1.55 K

The freezing point of a solution is lower than that of the pure solvent. The freezing point of pure water is 0°C, so the freezing point of the solution is:

0°C - 1.55°C = -1.55°C

So, the freezing point of the solution is -1.55°C.