Questions: Calculate the freezing point and boiling point of a solution containing 8.00 g of ethylene glycol (C2H6O2) in 928 mL of ethanol. Ethanol has a density of 0.789 g / cm^3. Part A Calculate the freezing point of the solution. (Use Kf=1.99°C / m and normal freezing point =-114.1°C.) Express the answer using four significant figures. TY= C

Solution Steps

First, we need to find the molality of the solution. Molality (\(m\)) is defined as moles of solute per kilogram of solvent.

1. Calculate moles of ethylene glycol (\(\mathrm{C}_2\mathrm{H}_6\mathrm{O}_2\)):

   \[ \text{Molar mass of } \mathrm{C}_2\mathrm{H}_6\mathrm{O}_2 = 2(12.01) + 6(1.008) + 2(16.00) = 62.068 \, \text{g/mol} \]

   \[ \text{Moles of } \mathrm{C}_2\mathrm{H}_6\mathrm{O}_2 = \frac{8.00 \, \text{g}}{62.068 \, \text{g/mol}} = 0.1289 \, \text{mol} \]

2. Calculate mass of ethanol (solvent) in kilograms:

   \[ \text{Density of ethanol} = 0.789 \, \text{g/cm}^3 = 0.789 \, \text{g/mL} \]

   \[ \text{Mass of ethanol} = 928 \, \text{mL} \times 0.789 \, \text{g/mL} = 731.592 \, \text{g} = 0.7316 \, \text{kg} \]

3. Calculate molality:

   \[ m = \frac{0.1289 \, \text{mol}}{0.7316 \, \text{kg}} = 0.1762 \, \text{mol/kg} \]

Use the formula for freezing point depression:

\[ \Delta T_f = K_f \times m \]

Where \(K_f = 1.99^\circ \text{C/m}\).

\[ \Delta T_f = 1.99 \, ^\circ \text{C/m} \times 0.1762 \, \text{mol/kg} = 0.3507 \, ^\circ \text{C} \]

Subtract the freezing point depression from the normal freezing point of ethanol:

\[ T_f = -114.1^\circ \text{C} - 0.3507^\circ \text{C} = -114.4507^\circ \text{C} \]

Final Answer