Automotive antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile electrolyte. Calculate the boiling point and freezing point of a 25.0 mass percent of ethylene glycol in solution.
25% by mass means 25 g ethylene glycol + 75 g water (assuming the density of ethylene glycol is 1.0 g/mL which I doubt). If you have a density, you can change those numbers.
Then delta T = Kf*molality
molality = moles/kg solvent.
moles = g/molar mass.
Freezing point part is done the same way but use Kf instead of Kb.
Kf is 1.86 for water.
Kb is 0.51 for water.
Anti
To calculate the boiling point and freezing point of a solution consisting of 25.0 mass percent of ethylene glycol (C2H6O2), we need to apply the concept of colligative properties. The boiling point and freezing point of a solution are affected by the presence of solute particles.
First, let's calculate the molality of the solution, which is the moles of solute per kilogram of solvent.
Mass of ethylene glycol = 25.0 mass percent of solution
Mass of solution = 100 g (assume 100 g to make calculations easier)
Mass of ethylene glycol = (25.0 / 100) * 100 g = 25 g
Molecular weight of ethylene glycol (C2H6O2) = 2(12.01) + 6(1.01) + 2(16.00) = 62.07 g/mol
Number of moles of ethylene glycol = Mass of ethylene glycol / Molecular weight of ethylene glycol
Number of moles of ethylene glycol = 25 g / 62.07 g/mol = 0.4028 mol
Since the ethylene glycol is a nonvolatile solute, it does not significantly contribute to the vapor pressure of the solution. Therefore, the boiling point elevation is negligible, and we can assume the boiling point of the solution is the same as that of the pure solvent.
The freezing point depression can be calculated using the equation:
ΔTf = Kf * molality
where:
ΔTf = freezing point depression
Kf = Cryoscopic constant (specific to solvent)
molality = Moles of solute / Mass of solvent (in kg)
For ethylene glycol, Kf = 12.0 °C kg/mol.
Assuming water as the solvent (which is common for antifreeze solutions), the molality can be calculated as follows:
Mass of water = Mass of solution - Mass of solute = 100 g - 25 g = 75 g
molality = Moles of solute / Mass of solvent (in kg)
= 0.4028 mol / (75 g / 1000) kg
= 5.37 mol/kg
Now we can calculate the freezing point depression:
ΔTf = Kf * molality
= 12.0 °C kg/mol * 5.37 mol/kg
= 64.44 °C
Therefore, the freezing point of the solution will be lowered by approximately 64.44 °C.
As mentioned earlier, the boiling point elevation is negligible for nonvolatile solutes, so the boiling point of the solution would be close to that of the pure solvent, which is 100 °C for water at standard atmospheric pressure.
To calculate the boiling and freezing points of a solution with ethylene glycol, we can use the concept of colligative properties. Colligative properties depend only on the number of solute particles in the solution, not on their chemical nature. One of the most important colligative properties is the change in boiling and freezing points.
The equation used to calculate the boiling point elevation (∆Tb) and freezing point depression (∆Tf) is:
∆Tb = Kb * m
∆Tf = Kf * m
Where:
∆Tb: Change in boiling point
∆Tf: Change in freezing point
Kb: Boiling point elevation constant (molal boiling point constant)
Kf: Freezing point depression constant (molal freezing point constant)
m: Molality of the solution (moles of solute/kg of solvent)
For ethylene glycol, the values are:
Kb = 0.512 °C/m
Kf = -1.86 °C/m
Step 1: Convert the mass percentage of ethylene glycol to the molality of the solution.
Let's assume we have 100 grams of the solution, therefore we would have 25 grams of ethylene glycol (mass percent = 25.0%). The molar mass of C2H6O2 is 62.08 g/mol, so:
mol of ethylene glycol = (grams of ethylene glycol) / (molar mass of ethylene glycol)
mol of ethylene glycol = 25 g / 62.08 g/mol
Step 2: Calculate the molality of the solution.
Molality (m) is calculated by dividing the number of moles of solute by the mass of the solvent in kg:
molality of the solution = (mol of solute) / (kg of solvent)
Step 3: Calculate the boiling point elevation and freezing point depression.
Using the formulas mentioned earlier:
∆Tb = Kb * m
∆Tf = Kf * m
Substitute the known values:
∆Tb = 0.512 °C/m * (molality of the solution)
∆Tf = -1.86 °C/m * (molality of the solution)
Step 4: Calculate the boiling point and freezing point of the solution.
To obtain the final boiling and freezing points, add or subtract the change in temperature (∆Tb or ∆Tf) to the boiling point and freezing point of the pure solvent, respectively.
Boiling point of the solution = boiling point of the pure solvent + ∆Tb
Freezing point of the solution = freezing point of the pure solvent + ∆Tf
For example, if the boiling point of the pure solvent is 100 °C and the change in boiling point (∆Tb) is 3 °C, then:
Boiling point of the solution = 100 °C + 3 °C = 103 °C
Similarly, if the freezing point of the pure solvent is 0 °C and the change in freezing point (∆Tf) is -4 °C, then:
Freezing point of the solution = 0 °C - 4 °C = -4 °C
So, based on the provided values, you can apply this process to calculate the boiling point and freezing point of a 25.0 mass percent of ethylene glycol in solution.