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 Information:** - Freezing point of pure benzene: $T_f = 5.45°C$ - Freezing point of solution: $T_f' = 3.55°C$ - Molality of $CCl_4$ (tetrachloroethane): $m = 0.374$ mol/kg **Step 1: Calculate freezing point depression** $$\Delta T_f = T_f - T_f' = 5.45 - 3.55 = 1.90°C$$ **Step 2: Apply the freezing point depression formula** $$\Delta T_f = K_f \times m \times i$$ where $K_f$ is the freezing point depression constant and $i$ is the van 't Hoff factor. **Step 3: Determine the van 't Hoff factor** Tetrachloroethane ($CCl_4$) is a non-electrolyte and doesn't dissociate in benzene, so: $$i = 1$$ **Step 4: Solve for Kf** $$K_f = \frac{\Delta T_f}{m \times i} = \frac{1.90}{0.374 \times 1}$$ $$K_f = \frac{1.90}{0.374} = 5.08 \text{ K·kg/mol} \approx 5.1 \text{ K·kg/mol}$$ **Answer: Option A** (which is $K_f \approx 5.1$ K·kg/mol) This matches the standard literature value for benzene's freezing point depression constant.