A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

To calculate the freezing-point depression, we need to use the formula:

ΔTf = Kf * m

Where:
ΔTf = freezing-point depression (in °C)
Kf = freezing point depression constant (in °C/m)
m = molality of the solution (in mol/kg)

First, we need to calculate the molality of the solution.

Step 1: Calculate the moles of glucose.
To convert grams to moles, use the formula:
moles = mass / molar mass

The molar mass of glucose (C6H12O6) is:
(6 * atomic mass of carbon) + (12 * atomic mass of hydrogen) + (6 * atomic mass of oxygen)
= (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol)
= 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
≈ 180.18 g/mol

moles of glucose = 15.5 g / 180.18 g/mol
≈ 0.086 moles

Step 2: Calculate the moles of water.
The molar mass of water (H2O) is 2 * atomic mass of hydrogen + atomic mass of oxygen:
= (2 * 1.01 g/mol) + 16.00 g/mol
= 18.02 g/mol

moles of water = 245 g / 18.02 g/mol
≈ 13.59 moles

Step 3: Calculate the total moles of the solution.
Total moles of the solution = moles of glucose + moles of water
= 0.086 moles + 13.59 moles
≈ 13.68 moles

Step 4: Calculate the mass of the solvent (water).
mass of water = 245 g

Step 5: Calculate the molality of the solution.
Molality (m) = moles of solute / mass of solvent (in kg)
mass of solvent = 245 g = 0.245 kg

m = 0.086 moles / 0.245 kg
≈ 0.351 mol/kg

Now, we can calculate the freezing-point depression using the formula mentioned earlier.

ΔTf = Kf * m
Given that the freezing point constant (Kf) is -1.86 °C/m, and molality (m) is 0.351 mol/kg:

ΔTf = -1.86 °C/m * 0.351 mol/kg
≈ -0.65286 °C

Therefore, the freezing-point depression of the solvent is approximately -0.65286 °C.

To find the freezing-point depression of the solvent, we need to use the formula:

ΔTf = Kf * m

Where:
- ΔTf is the freezing-point depression
- Kf is the freezing point constant
- m is the molality of the solution

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

1. Calculate the 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

moles of glucose = mass of glucose / molar mass of glucose
= 15.5 g / 180.18 g/mol

2. Calculate the mass of the solvent (water):
mass of water = 245 g

3. Calculate the molality (moles of solute per kilogram of solvent):
mass of solvent (in kg) = mass of water / 1000
= 245 g / 1000
= 0.245 kg

molality (m) = moles of solute / mass of solvent (in kg)
= moles of glucose / 0.245 kg

4. Calculate the freezing-point depression:
ΔTf = Kf * m
= -1.86 °C/m * molality

Now, substitute the value of molality in the formula and calculate ΔTf.