Freezing Point Depression

The freezing point depression is a colligative property of a solution, which depends on the number of solute particles present in the solution. It is given by the equation:

ΔTf = Kf * m

Where ΔTf is the freezing point depression, Kf is the cryoscopic constant, and m is the molality of the solution.

Given Information

In this case, we are given that a 5% solution of cane sugar (sucrose) in water has a freezing point of 271 K. We are also given that the freezing point of pure water is 273.15 K.

Calculating Molality

To determine the freezing point depression of a 5% solution of glucose in water, we first need to calculate the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.

Molality (m) = moles of solute / mass of solvent (in kg)

Since we are given the percentage concentration of the solution, we can assume that 5% refers to 5 grams of glucose in 100 grams of water. The molar mass of glucose is 180.16 g/mol.

Number of moles of glucose = (mass of glucose / molar mass of glucose)
= (5 g / 180.16 g/mol)
≈ 0.0277 mol

Mass of water = (100 g - 5 g) = 95 g

Molality (m) = (0.0277 mol / 0.095 kg)
≈ 0.2916 mol/kg

Calculating Freezing Point Depression

Now that we have the molality of the glucose solution, we can use the freezing point depression equation to calculate the change in freezing point.

ΔTf = Kf * m

We are given that the freezing point of pure water is 273.15 K. Therefore, the change in freezing point (ΔTf) can be calculated as:

ΔTf = Kf * m
= Kf * 0.2916 mol/kg

We don't have the value of the cryoscopic constant (Kf) for glucose in water, so we cannot calculate the exact freezing point depression without that information. However, we can conclude that the freezing point of a 5% glucose solution in water will be lower than the freezing point of pure water (273.15 K) due to the presence of the solute particles.