Answer:
= -49.76 °C
Step-by-step explanation:
To calculate the freezing point depression, we can use the following equation:
ΔTf = Kf × m
where ΔTf is the freezing point depression, Kf is the freezing point depression constant (also known as the cryoscopic constant), and m is the molality of the solute.
In this case, the molality of the glucose solution is:
m = (moles of solute) / (mass of solvent in kg)
We are given that the concentration of glucose is 4.40 m, which means that there are 4.40 moles of glucose in 1 kg of solvent. The molar mass of glucose is 180.16 g/mol, so the mass of 4.40 moles of glucose is:
mass = 4.40 mol × 180.16 g/mol = 792.7 g
Therefore, the mass of solvent is:
mass of solvent = 1000 g - 792.7 g = 207.3 g
Converting the mass of solvent to kg, we get:
mass of solvent = 207.3 g / 1000 g/kg = 0.2073 kg
Now we can calculate the molality of the solution:
m = (4.40 mol) / (0.2073 kg) = 21.23 m/kg
Finally, we can use the freezing point depression constant and molality to calculate the freezing point depression:
ΔTf = Kf × m = 2.53 °C/m × 21.23 m/kg = 53.76 °C
To find the freezing point of the solution, we need to subtract the freezing point depression from the pure freezing point of the solvent:
Freezing point = 4.00 °C - 53.76 °C = -49.76 °C
Therefore, the freezing point of the glucose solution is approximately -49.76 °C. This means that the solution will freeze at a lower temperature than the pure solvent, due to the presence of the solute (glucose). The greater the concentration of solute, the greater the freezing point depression and the lower the freezing point of the solution.