Given the freezing-point depression ($\Delta T_f$) and the freezing-point depression constant ($K_f$), we can calculate the molality ($m$) of the solute using the formula:

$\Delta T_f = K_f \times m$

Given:

• Freezing-point depression ($\Delta T_f$) = 0.32 K
• Freezing-point depression constant for water ($K_f$) = 1.86 K/m

To Find:

• Molality ($m$)

Solution:

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 this value:

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

Would you like more details or have any questions?

Related Questions:

1. What is the relationship between molality and molarity?
2. How would the freezing-point depression change if a different solvent were used?
3. How do you determine the freezing-point depression constant ($K_f$) for a new solvent?
4. What factors affect the magnitude of the freezing-point depression?
5. How does the presence of electrolytes affect freezing-point depression?

Tip: When working with colligative properties, remember that these properties depend on the number of particles in the solution, not their nature!