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:** - Pure benzene freezing point: $T_f° = 5.45°C$ - Solution freezing point: $T_f = 3.55°C$ - Molality of $C_2Cl_4$: $m = 0.374$ m **Step 1: Calculate freezing point depression** $$\Delta T_f = T_f° - T_f = 5.45 - 3.55 = 1.90°C$$ **Step 2: Apply freezing point depression equation** The colligative property formula is: $$\Delta T_f = K_f \times m$$ **Step 3: Solve for Kf** $$K_f = \frac{\Delta T_f}{m} = \frac{1.90}{0.374}$$ $$K_f = 5.08 \text{ K·kg/mol} \approx 5.1 \text{ K·kg/mol}$$ **Why this is correct:** - Tetrachloroethane is a non-electrolyte, so $i = 1$ (no ionization) - The freezing point depression depends only on solute molality and the solvent's cryoscopic constant - The calculated value (~5.1) matches the known $K_f$ for benzene