A solution is prepared by dissolving 25.0 g of glucose, C6H12O6, in 125 g of water at 25oC. (show work please)

Determine the freezing point of the solution.

mols sucrose = grams/molar mass

solve for mols.

molality = m = mols/kg solvent
solve for m

dT = Kf*m
solve for dT and subtract from 0.

A solution is made by dissolving 50.0 g of glucose (C6H12O6) with 600.0 g of water. What is the vapor pressure of this solution at 80.0℃? (nonvolatile)

To determine the freezing point of the solution, we need to use the formula:

ΔT = Kf * m

where:
ΔT = change in freezing point
Kf = freezing point depression constant for the solvent (water)
m = molality of the solution

Step 1: Calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
Given:
Mass of glucose (C6H12O6) = 25.0 g
Mass of water = 125 g

Molar mass of glucose (C6H12O6) = 180.16 g/mol
Number of moles of glucose = mass / molar mass = 25.0 g / 180.16 g/mol = 0.1387 mol

Mass of water converted to kg = 125 g / 1000 g/kg = 0.125 kg

Molality (m) = number of moles of solute / mass of solvent in kg
m = 0.1387 mol / 0.125 kg = 1.11 mol/kg

Step 2: Find the freezing point depression constant for water (Kf).
The freezing point depression constant for water is 1.86 °C/m.

Step 3: Calculate the change in freezing point (ΔT).
ΔT = Kf * m
ΔT = 1.86 °C/m * 1.11 mol/kg
ΔT = 2.0626 °C

Step 4: Determine the freezing point of the solution.
The freezing point of pure water is 0 °C.
Freezing point of the solution = Freezing point of pure water - ΔT
Freezing point of the solution = 0 °C - 2.0626 °C
Freezing point of the solution = -2.0626 °C

Therefore, the freezing point of the solution is approximately -2.06 °C.

To determine the freezing point of the solution, you need to use the concept of freezing point depression, which states that the presence of solute particles in a solvent lowers its freezing point.

First, you need to calculate the molality (m) of the solution, which is defined as the number of moles of solute (glucose) per kilogram of solvent (water).

1. Calculate the moles of glucose:
The molar mass of glucose (C6H12O6) is:
6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

m = mass / molar mass
m = 25.0 g / 180.18 g/mol
m ≈ 0.1387 mol

2. Calculate the mass of water in kilograms:
125 g = 0.125 kg

3. Calculate the molality (m):
m = moles of solute / mass of solvent (kg)
m = 0.1387 mol / 0.125 kg
m ≈ 1.11 mol/kg

Next, you need to use the formula for freezing point depression:

ΔTf = Kf * m

where ΔTf is the freezing point depression, Kf is the cryoscopic constant (constant for the solvent), and m is the molality of the solution.

The cryoscopic constant for water is 1.86 oC·kg/mol.

4. Calculate the freezing point depression (ΔTf):
ΔTf = 1.86 oC·kg/mol * 1.11 mol/kg
ΔTf ≈ 2.06 oC

Finally, you need to subtract the freezing point depression from the normal freezing point of pure water (0 oC) to find the freezing point of the solution.

5. Calculate the freezing point of the solution:
Freezing point = 0 oC - 2.06 oC
Freezing point ≈ -2.06 oC

Therefore, the freezing point of the solution is approximately -2.06 oC.