A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m?

moles = grams/molar mass

Solve for moles.

molality = moles/kg solvent
Solve for molality

delta T = Kf*molality
Solve for delta T, then for the freezing point knowing the normal freezing point is zero C.

To determine the freezing-point depression of the solvent, we need to calculate the moles of solute (glucose) and then use that to find the change in freezing point.

Step 1: Calculate the moles of glucose (C6H12O6):
To do this, we need to know the molar mass of glucose, which is the sum of the atomic masses of its constituent atoms.

Molar mass of C6H12O6 = (6 * atomic mass of C) + (12 * atomic mass of H) + (6 * atomic mass of O)
= (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

Now, we can calculate the moles of glucose by dividing the mass of glucose by its molar mass.

Moles of glucose = Mass of glucose / Molar mass of glucose
= 10.20 g / 180.18 g/mol
= 0.0566 mol

Step 2: Calculate the change in freezing point:
The change in freezing point (∆Tf) is given by the formula:

∆Tf = Kf * m

where Kf is the freezing point constant and m is the molality of the solution.

Molality (m) is defined as the ratio of moles of solute to the mass of the solvent in kilograms.

Mass of water = 355 g
Mass of water in kg = 355 g / 1000 = 0.355 kg

Molality (m) = Moles of solute / Mass of solvent in kg
= 0.0566 mol / 0.355 kg
= 0.1594 mol/kg

Now, we can calculate the change in freezing point:

∆Tf = -1.86 °C/m * 0.1594 mol/kg

∆Tf = -0.296 °C

Therefore, the freezing-point depression of the solvent is -0.296 °C.