Answer: The normal freezing point of the solution is 0 degrees Celsius.
Step-by-step explanation:
The normal freezing point of water is 0 degrees Celsius. When a solute is added to water, it lowers the freezing point of the water. The freezing point depression is given by the equation:
ΔTf = Kf * molality
where ΔTf is the freezing point depression, Kf is the molal freezing point depression constant, and molality is the concentration of the solute in the solution.
To find the molality of the solution, we need to know the mass of the solute and the mass of the solvent. The mass of the solvent is the mass of the solution minus the mass of the solute. Since the density of the solution is 1.35 g/mL, and the solution is assumed to be pure water, the mass of the solvent is equal to the volume of the solution times the density of the solution.
To find the volume of the solution, we need to know the mass of the solution. Since the solution is 0.7439 moles of C12H22O11, and the molecular weight of C12H22O11 is 342.29 g/mol, the mass of the solution is 0.7439 moles * 342.29 g/mol = 255.27 g. Therefore, the volume of the solution is 255.27 g / 1.35 g/mL = 189.25 mL.
The mass of the solute is the mass of the solution minus the mass of the solvent. Since the mass of the solution is 255.27 g and the mass of the solvent is 189.25 mL * 1.35 g/mL = 256.88 g, the mass of the solute is 255.27 g - 256.88 g = -1.61 g. However, since the mass of the solute is negative, this means that there is actually more solvent than solute in the solution. This means that the molality of the solution is actually 0, which means that the freezing point depression is also 0. Therefore, the normal freezing point of the solution is 0 degrees Celsius.