$$\Delta T_f = K_f \times m$$

Where:

- \(\Delta T_f\) is the freezing point depression

- \(K_f\) is the cryoscopic constant of the solvent

- \(m\) is the molality of the solution

In this case, the solvent is ether, which has a \(K_f\) value of 2.25 °C/m. The molality of the solution is:

$$m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}$$

We have 0.500 mol of solute and 500.0 g of solvent. To convert grams to kilograms, we divide by 1000:

$$m = \frac{0.500 \text{ mol}}{0.500 \text{ kg}} = 1.00 \text{ m}$$

Now we can substitute the values of \(K_f\) and \(m\) into the equation for \(\Delta T_f\):

$$\Delta T_f = 2.25 \text{ °C/m} \times 1.00 \text{ m} = 2.25 \text{ °C}$$

The freezing point of the solution is therefore 2.25 °C lower than the freezing point of pure ether.