A glucose solution contains 55.0g of glucose C6H12O6 in 475g of water.

Compute the freezing point and boiling point of the solution. Assume a density of 1.00g/mL for water.

mols glucose = grams/molar mass. Solve for mols.

m = mols/Kg solvent
m = mols/0.475 = ?

Substitute m into
delta T = Kf*m and subtract from 0 C for the new freezing point

and delta T = Kb*m and add to 100 for the new boiling point.
Kf = 1.86 and Kb = 0.51

To calculate the freezing and boiling points of the solution, we need to use the concept of molality.

1. Calculate the moles of glucose (C6H12O6):
Molar mass of glucose (C6H12O6) = 180.16 g/mol
Moles of glucose = Mass of glucose / Molar mass of glucose
Moles of glucose = 55.0 g / 180.16 g/mol

2. Calculate the moles of water:
Moles of water = Mass of water / Molar mass of water
Molar mass of water = 18.02 g/mol
Moles of water = 475 g / 18.02 g/mol

3. Calculate the molality of the glucose solution:
Molality (m) = Moles of solute / Mass of solvent (in kg)
Mass of solvent = Mass of water / Density of water
Density of water = 1.00 g/mL = 1.00 g/cm³
Mass of solvent = 475 g / 1.00 g/cm³ = 475 mL = 0.475 kg
Molality (m) = Moles of glucose / Mass of water (in kg)
Molality (m) = (55.0 g / 180.16 g/mol) / 0.475 kg

4. Calculate the freezing point depression:
The freezing point depression (ΔTF) is given by the equation:
ΔTF = Kf × m
where Kf is the cryoscopic constant for water (1.86 °C/m)

Freezing point depression (ΔTF) = (1.86 °C/m) × (moles of glucose / mass of water (in kg))

5. Calculate the boiling point elevation:
The boiling point elevation (ΔTB) is given by the equation:
ΔTB = Kb × m
where Kb is the ebullioscopic constant for water (0.512 °C/m)

Boiling point elevation (ΔTB) = (0.512 °C/m) × (moles of glucose / mass of water (in kg))

6. Calculate the freezing point:
Freezing point = Freezing point of pure water - Freezing point depression (ΔTF)

7. Calculate the boiling point:
Boiling point = Boiling point of pure water + Boiling point elevation (ΔTB)

Using the given values, we can now calculate the freezing and boiling points of the solution.

To compute the freezing point and boiling point of the glucose solution, we need to use the concept of molality and the cryoscopic and ebullioscopic constants.

1. Calculate the molality of the glucose solution:
Molality (m) is defined as the number of moles of solute (glucose) per kilogram of solvent (water).
First, we need to determine the number of moles of glucose:
Molar mass of glucose (C6H12O6): 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol
Number of moles of glucose = Mass of glucose (55.0 g) / Molar mass of glucose (180.18 g/mol) = 0.305 mol

Next, calculate the mass of water in kilograms:
Mass of water (475 g) / 1000 = 0.475 kg

Molality (m) = Number of moles of solute / Mass of solvent in kilograms
Molality (m) = 0.305 mol / 0.475 kg ≈ 0.642 m

2. Calculate the freezing point depression (∆Tf):
∆Tf = m * Kf
where Kf is the cryoscopic constant, which represents the change in the freezing point per molal concentration.

For water, Kf = 1.86 °C/m

∆Tf = 0.642 m * 1.86 °C/m = 1.194 °C

The freezing point depression (∆Tf) represents the amount by which the freezing point of the solution is lowered compared to the pure solvent.

3. Calculate the freezing point of the solution:
The freezing point of pure water is 0 °C.
Freezing point of solution = Freezing point of pure solvent - ∆Tf
Freezing point of solution = 0 °C - 1.194 °C = -1.194 °C

So, the freezing point of the glucose solution is approximately -1.194 °C.

4. Calculate the boiling point elevation (∆Tb):
∆Tb = m * Kb
where Kb is the ebullioscopic constant, which represents the change in the boiling point per molal concentration.

For water, Kb = 0.512 °C/m

∆Tb = 0.642 m * 0.512 °C/m = 0.329 °C

The boiling point elevation (∆Tb) represents the amount by which the boiling point of the solution is raised compared to the pure solvent.

5. Calculate the boiling point of the solution:
The boiling point of pure water is 100 °C.
Boiling point of solution = Boiling point of pure solvent + ∆Tb
Boiling point of solution = 100 °C + 0.329 °C = 100.329 °C

So, the boiling point of the glucose solution is approximately 100.329 °C.