Pure benzene freezes at 5.45ºC at a certain place but a 0.374m solution of tetrachloroethane in benz — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

# Solution: Finding Kf for Benzene **Given Data:** - Freezing point of pure benzene: $T_f = 5.45°C$ - Freezing point of solution: $T_f' = 3.55°C$ - Molality of solution: $m = 0.374$ m - Need to find: $K_f$ for benzene **Step 1: Calculate freezing point depression** $$\Delta T_f = T_f - T_f' = 5.45 - 3.55 = 1.90°C$$ **Step 2: Apply colligative property equation** The freezing point depression is given by: $$\Delta T_f = K_f \times m$$ where $K_f$ is the freezing point depression constant and $m$ is molality. **Step 3: Solve for Kf** $$K_f = \frac{\Delta T_f}{m} = \frac{1.90}{0.374}$$ $$K_f = 5.08 \text{ °C/m (or K·kg/mol)}$$ **Answer: A** is correct because $K_f \approx 5.1°C·kg/mol$ for benzene, which matches the standard literature value and results from straightforward application of the colligative property formula.