What volume of ethylene glycol (C2H6O2), a nonelectolyte, must be added to 14.0 L water to produce an antifreeze solution with a freezing point of –25.0°C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3) .
The Correct Answer and Explanation is :
To solve this problem, we need to calculate the volume of ethylene glycol (C2H6O2) to be added to 14.0 L of water in order to create a solution with a freezing point of -25.0°C. We’ll use the freezing point depression formula and step through the required calculations.
Step 1: Freezing Point Depression Formula
The freezing point depression is given by: ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m
Where:
- ΔTf\Delta T_f is the freezing point depression (difference between the freezing point of pure water and the solution),
- ii is the van’t Hoff factor (which equals 1 for nonelectrolytes like ethylene glycol),
- KfK_f is the cryoscopic constant (for water, K_f = 1.86 \, ^\circ C \cdot \text{kg/mol}),
- mm is the molality of the solution (moles of solute per kilogram of solvent).
Step 2: Calculate Freezing Point Depression
The freezing point of pure water is 0.0°C. If the solution’s freezing point is -25.0°C, the freezing point depression is: ΔTf=0.0∘C−(−25.0∘C)=25.0∘C\Delta T_f = 0.0^\circ C – (-25.0^\circ C) = 25.0^\circ C
Step 3: Use the Freezing Point Depression Formula
Now we substitute the values into the freezing point depression equation: 25.0^\circ C = 1 \cdot 1.86 \, ^\circ C \cdot \text{kg/mol} \cdot m
Solve for molality: m = \frac{25.0^\circ C}{1.86 \, ^\circ C \cdot \text{kg/mol}} = 13.44 \, \text{mol/kg}
This is the molality of the solution, which means we need 13.44 moles of ethylene glycol per kilogram of water.
Step 4: Calculate Moles of Ethylene Glycol
Since the molality is 13.44 mol/kg, and we have 14.0 L of water (which is approximately 14.0 kg, as the density of water is close to 1.0 g/cm³), we calculate the total moles of ethylene glycol required: moles of ethylene glycol=13.44 mol/kg×14.0 kg=188.16 mol\text{moles of ethylene glycol} = 13.44 \, \text{mol/kg} \times 14.0 \, \text{kg} = 188.16 \, \text{mol}
Step 5: Calculate Mass of Ethylene Glycol
Next, we calculate the mass of ethylene glycol needed. The molar mass of ethylene glycol (C₂H₆O₂) is: Molar mass=2(12.01)+6(1.008)+2(16.00)=62.07 g/mol\text{Molar mass} = 2(12.01) + 6(1.008) + 2(16.00) = 62.07 \, \text{g/mol}
Now calculate the mass: mass of ethylene glycol=188.16 mol×62.07 g/mol=11,687.53 g=11.69 kg\text{mass of ethylene glycol} = 188.16 \, \text{mol} \times 62.07 \, \text{g/mol} = 11,687.53 \, \text{g} = 11.69 \, \text{kg}
Step 6: Convert Mass to Volume
Finally, we use the density of ethylene glycol (1.11 g/cm³) to find the volume of ethylene glycol: volume of ethylene glycol=massdensity=11,687.53 g1.11 g/cm3=10,529.66 cm3=10.53 L\text{volume of ethylene glycol} = \frac{\text{mass}}{\text{density}} = \frac{11,687.53 \, \text{g}}{1.11 \, \text{g/cm}^3} = 10,529.66 \, \text{cm}^3 = 10.53 \, \text{L}
Final Answer:
The volume of ethylene glycol required is 10.53 L.
Explanation:
This calculation uses the freezing point depression principle, which allows us to determine how much solute (ethylene glycol) is required to lower the freezing point of water by a certain amount. Since ethylene glycol is a nonelectrolyte (i.e., it does not dissociate in water), the van’t Hoff factor ii is 1. By determining the molality of the solution that corresponds to a -25.0°C freezing point, we can figure out how many moles of ethylene glycol are necessary, and subsequently how much mass and volume of ethylene glycol are needed to achieve the desired antifreeze effect.