Step 1: Identify the properties.
Boiling point elevation and freezing point depression are both colligative properties. They depend on the molality of the solution, not on the kind of solute.
Step 2: Find the boiling point rise.
Pure water boils at 100 degrees. The solution boils at 100.5, so the rise is \[ \Delta T_b = 100.5 - 100 = 0.5\,^{\circ}\text{C} \]
Step 3: Write both formulas.
The two equations are $\Delta T_b = K_b\,m$ and $\Delta T_f = K_f\,m$. The molality $m$ is the same in both because it is the same solution.
Step 4: Divide to remove molality.
Dividing one by the other cancels $m$. \[ \frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b} \]
Step 5: Put in the numbers.
So $\dfrac{\Delta T_f}{0.5} = \dfrac{1.86}{0.512}$, which gives \[ \Delta T_f = \frac{1.86 \times 0.5}{0.512} \approx 1.816\,^{\circ}\text{C} \]
Step 6: Find the freezing point.
Pure water freezes at 0 degrees, so the freezing point drops below zero by 1.816. \[ \boxed{T_f = -1.816\,^{\circ}\text{C}} \]