Questions: Experiment 2: Suppose a student performed a similar experiment with a different unknown. Given the data in the table, what is the van't Hoff factor of the unknown substance? Kf of water 1.86 C / m freezing point of pure water 0.0 C molality of solution 2.075 mol / kg freezing point of solution -3.86 C i=

Solution Steps

The problem requires us to find the van't Hoff factor (\(i\)) of an unknown substance using the given data. The van't Hoff factor is a measure of the effect of solute particles on the colligative properties of a solution. We will use the formula for freezing point depression to find \(i\).

The formula for freezing point depression is:

\[ \Delta T_f = i \cdot K_f \cdot m \]

where:

• \(\Delta T_f\) is the change in freezing point,
• \(i\) is the van't Hoff factor,
• \(K_f\) is the cryoscopic constant (freezing point depression constant) of the solvent,
• \(m\) is the molality of the solution.

The change in freezing point (\(\Delta T_f\)) is the difference between the freezing point of pure water and the freezing point of the solution:

\[ \Delta T_f = 0.0^\circ \text{C} - (-3.86^\circ \text{C}) = 3.86^\circ \text{C} \]

Substitute the known values into the freezing point depression formula and solve for \(i\):

\[ 3.86 = i \cdot 1.86 \cdot 2.075 \]

Rearrange to solve for \(i\):

\[ i = \frac{3.86}{1.86 \cdot 2.075} \]

Calculate \(i\):

\[ i = \frac{3.86}{3.8595} \approx 1.0001 \]

Final Answer