# Freezing Point Depression Calculation Let's solve the problem step by step: ## Given Data - Mass of glucose (\( m_{solute} \)) = **7.2 g** - Mass of water (\( m_{solvent} \)) = **100 g** = **.100 kg** - Molar mass of glucose (\( M_{solute} \)) = **180 g/mol** - Freezing point depression constant (\( K_f \)) for water = **1.86 K kg/mol** - Freezing point of pure water (\( T_f^ \)) = **°C** --- ## Step 1: Find Moles of Glucose \[ \text{Moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{7.2\ \text{g}}{180\ \text{g/mol}} = .04\ \text{mol} \] --- ## Step 2: Calculate Molality of the Solution Molality (\( m \)) is defined as: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] \[ m = \frac{.04\ \text{mol}}{.100\ \text{kg}} = .4\ \text{mol/kg} \] --- ## Step 3: Calculate Freezing Point Depression (\( \Delta T_f \)) The formula for freezing point depression: \[ \Delta T_f = K_f \times m \] \[ \Delta T_f = 1.86\ \text{K kg/mol} \times .4\ \text{mol/kg} = .744\ \text{K} \] --- ## Step 4: Calculate the New Freezing Point \[ \text{Freezing point of solution} = T_f^ - \Delta T_f \] \[ = ^\circ\text{C} - .744\ \text{K} = -.744^\circ\text{C} \] --- ## **Final Answer** **The freezing point of the solution is \(-.744^\circ \text{C}\).**