Freezing Point Depression: Videos & Practice Problems

The phenomenon of freezing point depression occurs when a solute is added to a pure solvent, resulting in a decrease in the solvent's freezing point. The normal freezing point (abbreviated as fp) refers to the freezing point of the pure solvent before any solute is introduced, while the freezing point of the solution (denoted as fp solution) is the freezing point after the solute has been added. This transformation from solvent to solution is crucial in understanding how solutes affect freezing behavior.

The freezing point depression can be quantified using the formula:

$\Delta t_f = i \cdot k_f \cdot m$

In this equation, $\Delta t_f$ represents the change in freezing point, i is the van't Hoff factor (which accounts for the number of particles the solute dissociates into), k_f is the freezing point constant of the solvent (expressed in degrees Celsius per molality), and m is the molality of the solution, defined as moles of solute per kilogram of solvent.

To determine the freezing point of a solution, the relationship can be expressed as:

$fp solution = fp - \Delta t_f$

Common solvents such as water, benzene, chloroform, and ethanol each have distinct normal freezing points and unique freezing point constants. While it is not necessary to memorize these values, they serve as useful references for understanding the effects of solutes on freezing points. Importantly, the more solute that is added to the solvent, the greater the decrease in the freezing point, illustrating the significant impact of solute concentration on the physical properties of solutions.

To calculate the freezing point of a solution containing glucose (C₆H₁₂O₆) dissolved in water, we start with the formula for the freezing point depression:

ΔT_f = I × K_f × m

In this equation:

• ΔT_f is the change in freezing point.
• I is the van 't Hoff factor, which is 1 for glucose since it is a covalent compound and does not dissociate into ions.
• K_f is the freezing point depression constant for water, which is 1.86 °C kg/mol.
• m is the molality of the solution.

First, we need to determine the number of moles of glucose. The molar mass of glucose is calculated as follows:

Molar mass of glucose = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.156 g/mol

Next, we convert the mass of glucose into moles:

Number of moles of glucose = 110.7 g × (1 mol / 180.156 g) ≈ 0.614 moles

Now, we convert the mass of water from grams to kilograms:

Mass of water = 302.6 g = 0.3026 kg

Next, we calculate the molality (m) of the solution:

Molality (m) = moles of solute / kg of solvent = 0.614 moles / 0.3026 kg ≈ 2.03 mol/kg

Now we can calculate ΔT_f:

ΔT_f = I × K_f × m = 1 × 1.86 °C kg/mol × 2.03 mol/kg ≈ 3.78 °C

Finally, we find the new freezing point of the solution:

Freezing point of the solution = Freezing point of pure solvent - ΔT_f = 0 °C - 3.78 °C = -3.78 °C

Thus, the freezing point of the glucose solution is approximately -3.78 °C.