int depression

$$i$$ is the van 't Hoff factor (since propane is a non-electrolyte, $$i=1$$)

$$K_f$$ is the freezing point depression constant ($$5.12^{\circ} \mathrm{C}/\mathrm{m}$$ for benzene)

$$m$$ is the molality of the solution

The molality (m) is given by the formula:

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

First, we need to calculate the moles of propane. The molar mass of propane (C₃H₈) is:

$$3 \times 12.01 \, \text{g/mol} + 8 \times 1.01 \, \text{g/mol} = 44.09 \, \text{g/mol}$$

Thus, the moles of propane are:

$$\frac{20 \, \text{g}}{44.09 \, \text{g/mol}} = 0.4536 \, \text{moles}$$

Next, we need to convert the mass of benzene from grams to kilograms:

$$400 \, \text{g} = 0.4 \, \text{kg}$$

Now, we can calculate the molality (m) of the solution:

$$m = \frac{0.4536 \, \text{moles}}{0.4 \, \text{kg}} = 1.134 \, \text{mol/kg}$$

Now, we can calculate the freezing point depression:

$$\Delta T_f = 1 \times 5.12 \, ^{\circ} \mathrm{C}/\mathrm{m} \times 1.134 \, \text{mol/kg} = 5.804 \, ^{\circ} \mathrm{C}$$

Finally, we can determine the new freezing point of the solution by subtracting the freezing point depression from the normal freezing point of benzene:

$$\text{Freezing point of solution} = 5.48^{\circ} \mathrm{C} - 5.804^{\circ} \mathrm{C} = -0.324^{\circ} \mathrm{C}$$

Thus, the closest answer to our calculated value is:

Option A: $$-0.34^{\circ} \mathrm{C}$$

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