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 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 formula** The freezing point depression is given by: $$\Delta T_f = K_f \times m \times i$$ For tetrachloroethane in benzene (molecular solute, no ionization): $$i = 1$$ **Step 3: Solve for Kf** $$K_f = \frac{\Delta T_f}{m \times i} = \frac{1.90}{0.374 \times 1}$$ $$K_f = \frac{1.90}{0.374} = 5.08 \text{ K·kg/mol} \approx 5.1 \text{ K·kg/mol}$$ **Why this is correct:** The freezing point depression is directly proportional to molality. Since tetrachloroethane is a non-electrolyte (covalent compound), it doesn't dissociate, so van't Hoff factor remains 1. The calculation straightforwardly yields **Kf ≈ 5.1 K·kg/mol**, which is the literature value for benzene.