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? Show all of the work needed to solve this problem.
mols glucose = grams/molar mass.
Sole for mols.
molality = m = mols/kg solvent.
Solve for m.
delta T = Kf*m
Solve for delta T. That's the freezing point depression.
To find the freezing-point depression of the solvent, you can use the formula:
ΔT = i * Kf * m
Where:
ΔT is the freezing-point depression
i is the van't Hoff factor (the number of particles into which a compound dissociates)
Kf is the freezing point constant
m is the molality of the solution (the number of moles of solute per kilogram of solvent)
First, let's calculate the molality (m) of the solution.
Step 1: Calculate the number of moles of glucose (C6H12O6)
To calculate the number of moles, you need to divide the mass of glucose by its molar mass.
Molar mass of glucose (C6H12O6) = (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
Number of moles of glucose = mass of glucose / molar mass of glucose
= 10.20 g / 180.18 g/mol
≈ 0.0566 mol
Step 2: Calculate the molality (m)
Molality (m) = moles of solute / mass of solvent in kg
= 0.0566 mol / (355 g / 1000)
= 0.0566 mol / 0.355 kg
≈ 0.159 mol/kg
Now, let's calculate the freezing-point depression (ΔT).
Step 3: Plug the values into the formula
ΔT = i * Kf * m
In this case, glucose does not dissociate into multiple particles, so the van't Hoff factor (i) is 1.
Given that the freezing point constant (Kf) is -1.86 °C/m, and the molality (m) is 0.159 mol/kg, we can substitute these values into the formula.
ΔT = 1 * (-1.86 °C/m) * 0.159 mol/kg
Step 4: Calculate ΔT
ΔT = -1.86 °C/m * 0.159 mol/kg
≈ -0.29574 °C
Therefore, the freezing-point depression of the solvent is approximately -0.29574 °C.