estimate the freezing point of an aqueous of 10.0 g of glucose dissolved in 500.0 g of water

delta T=iKm

K=1.853C/m

find delta T. (hint: m=mol glucose/kg H2O)Then subtract that value from 0 degrees Celsius.

http://library.thinkquest.org/C006669/data/Chem/colligative/colligative.html

To estimate the freezing point of a solution, we can use the formula for freezing point depression:

ΔT = Kf * molality

where ΔT is the change in temperature (in degrees Celsius), Kf is the cryoscopic constant for water (1.86 °C·kg/mol), and molality is the molality of the solution (moles of solute per kilogram of solvent).

To calculate the molality, we first need to find the moles of solute (glucose) and the mass of the solvent (water).

Given:
Mass of glucose (solute) = 10.0 g
Mass of water (solvent) = 500.0 g

First, let's calculate the moles of solute (glucose):
Molar mass of glucose (C6H12O6) = 180.16 g/mol
Moles of glucose = mass of glucose / molar mass of glucose
= 10.0 g / 180.16 g/mol

Next, calculate the mass of water in kilograms:
Mass of water (solvent) = 500.0 g
= 500.0 g / 1000 (to convert to kg)

Now, we can calculate the molality:
Molality = moles of solute / mass of water (in kg)

With the molality calculated, we can calculate the freezing point depression using the equation ΔT = Kf * molality.

Finally, we can estimate the freezing point of the solution by subtracting the freezing point depression from the normal freezing point of water (0°C).

Please note that this is an estimation as it assumes ideal behavior and does not consider any other factors that may affect the freezing point.