To calculate the molality of solutes in the solution, we can use the provided freezing point depression equation:

$\Delta T_f = K_f \cdot m$

Where:

• $\Delta T_f$ is the freezing point depression,
• $K_f$ is the freezing-point depression constant,
• $m$ is the molality of the solution.

Given values:

• $\Delta T_f = 0.32 \, \text{K}$,
• $K_f = 1.86 \, \text{K/m}$.

Now, we can rearrange the equation to solve for molality ($m$):

$m = \frac{\Delta T_f}{K_f}$

Substituting the given values:

$m = \frac{0.32 \, \text{K}}{1.86 \, \text{K/m}}$

Let's calculate the molality:

$m = \frac{0.32}{1.86}$The molality of the solutes in the solution is approximately $0.172 \, \text{mol/kg}$.

Would you like further details or have any questions? Here are some related topics:

1. How does molality differ from molarity?
2. What factors affect freezing-point depression in a solution?
3. Can you explain why freezing-point depression occurs at the molecular level?
4. How would you calculate the mass of a solute needed to achieve a specific molality?
5. How does the presence of multiple solutes affect the freezing point of a solution?

Tip: Molality is temperature-independent because it's based on the mass of the solvent, not the volume, which makes it particularly useful in freezing-point depression and boiling-point elevation calculations.