The user wants me to solve a problem involving the colligative property of freezing point depression and identify the correct graphical representation.

Step-by-Step Solution:

1. Understand the phenomenon: When a non-volatile solute is added to a pure solvent, the vapor pressure of the solvent decreases. This results in a lowering of the freezing point of the solution. The freezing point is the temperature at which the liquid and solid phases of a substance have the same vapor pressure. On a vapor pressure vs. temperature graph, this is the intersection point of the solid and liquid phase curves.

2. Calculate the moles of solute (ethanol): The molar mass of ethanol (C₂H₅OH) is given as $46$ g/mol. The mass of ethanol added is $34.5$ g. $\text{Moles of ethanol} = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}} = \frac{34.5 \text{ g}}{46 \text{ g/mol}} = 0.75 \text{ mol}$

3. Calculate the molality (m) of the solution: Molality is defined as the number of moles of solute per kilogram of solvent. Mass of solvent (water) = $500$ g = $0.5$ kg. $\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.75 \text{ mol}}{0.5 \text{ kg}} = 1.5 \text{ mol/kg}$

4. Calculate the depression in freezing point (ΔTf): The formula for freezing point depression is $\Delta T_f = K_f \times m$, where $K_f$ is the freezing point depression constant. Given $K_f = 2$ K kg/mol. $\Delta T_f = K_f \times m = (2 \text{ K kg/mol}) \times (1.5 \text{ mol/kg}) = 3 \text{ K}$ The freezing point of the solution will be depressed by $3$ K.

5. Calculate the new freezing point of the solution (Tf): The freezing point of pure water ($T_f^0$) is given as $273$ K. The new freezing point ($T_f$) is given by $T_f = T_f^0 - \Delta T_f$. $T_f = 273 \text{ K} - 3 \text{ K} = 270 \text{ K}$

6. Analyze the graphs: The graphs plot vapor pressure (V.P.) versus temperature (T).

   • The curve for the solid solvent (ice) is shown.

   • The curve for the liquid solvent (pure water) is shown, intersecting the solid's curve at $273$ K, which is the freezing point of pure water.

   • The curve for the solution lies below the curve for the pure liquid solvent because the addition of a solute lowers the vapor pressure.

   • The freezing point of the solution is the temperature where its vapor pressure curve intersects the vapor pressure curve of the solid solvent. We need to find the graph where this intersection occurs at the calculated temperature of $270$ K.

   • Option A: Shows the freezing point of the solution at $271$ K. This is incorrect.

   • Option B: Shows the freezing point of the solution at $275$ K. This represents a freezing point elevation, which is incorrect.

   • Option C: Shows the freezing point of the solution at $270$ K. This matches our calculated value.

   • Option D: Shows the freezing point of the solution at $272$ K. This is incorrect.

Therefore, the graph in option C correctly represents the change in the freezing point.