Calculation of the Freezing Point of 5% Glucose Solution

To calculate the freezing point of a 5% glucose solution in water, we will first determine the molality of the solution using the freezing point depression formula. Then, we will use this molality value to calculate the freezing point of the solution.

Step 1: Determine the Molality of the Solution
The molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is glucose and the solvent is water.

The given concentration is 5% (w/w) glucose solution, which means that 5 grams of glucose is present in 100 grams of solution.

To calculate the molality, we need to convert the mass of glucose to moles using the molar mass of glucose (M = 180 g/mol). Then, we divide the moles by the mass of water (solvent) in kilograms.

Moles of Glucose:
Mass of glucose = 5% of 100g = 5g
Moles of glucose = Mass of glucose / Molar mass of glucose
= 5g / 180 g/mol
= 0.0278 mol

Moles of Water:
Mass of water = Total mass of solution - Mass of glucose
= 100g - 5g
= 95g

To convert grams to kilograms, we divide the mass by 1000.
Mass of water = 95g / 1000
= 0.095 kg

Now, we can calculate the molality of the solution.
Molality (m) = Moles of solute / Mass of solvent in kg
= 0.0278 mol / 0.095 kg
= 0.292 mol/kg

Step 2: Calculate the Freezing Point of the Solution
The freezing point depression (ΔTf) is a colligative property that depends on the molality of the solution. It is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The freezing point depression is given by the equation:
ΔTf = Kf * m

Where:
ΔTf = Freezing point depression
Kf = Cryoscopic constant (molal freezing point depression constant)
m = Molality of the solution

Given that the freezing point of the 4% sucrose solution is 271.15 K, we can use this information to find the cryoscopic constant, Kf.

Calculating Kf:
ΔTf = Freezing point of pure solvent - Freezing point of solution
ΔTf = 0 K (pure solvent) - 271.15 K (4% sucrose solution)
ΔTf = -271.15 K

Now, we can rearrange the freezing point depression equation to solve for Kf:
Kf = ΔTf / m

Kf = (-271.15 K) / (0.292 mol/kg)
Kf ≈ -929.38 K·kg/mol

Calculating the Freezing Point of 5% Glucose Solution:
Using the calculated cryoscopic constant and the molality of the