5.00kg glycol, C2H4(OH)2, [this is anti-freeze!] is added to your radiator. If your radiator contains 12.0kg of water, what would be the freezing point of the water-antifreeze mixture? The molal freezing point depression constant for water is 1.86 degrees celsius x kg/mole.
I need help solving it
To find the freezing point of the water-antifreeze mixture, we need to calculate the molality (m) of the antifreeze in the solution and then use the freezing point depression formula.
Step 1: Find the number of moles of glycol (C2H4(OH)2) in 5.00 kg.
To do this, we need to calculate the molar mass of glycol.
The molar mass of C2H4(OH)2 = (2*12.01) + (4*1.01) + (2*16.00) + (2*1.01) = 62.07 g/mol.
Now, convert the mass of glycol to moles:
Moles of glycol = Mass of glycol / Molar mass of glycol = 5000 g / 62.07 g/mol = 80.57 mol.
Step 2: Calculate the molality (m) of the antifreeze in the solution.
Molality (m) is defined as the moles of solute per kilogram of solvent.
Here, the solvent is water, and the mass of water is 12.0 kg.
Molality (m) = Moles of glycol / Mass of water (in kg) = 80.57 mol / 12.0 kg = 6.711 mol/kg.
Step 3: Use the freezing point depression formula to calculate the freezing point depression (∆Tf).
The freezing point depression formula is: ∆Tf = Kf * m,
where Kf is the molal freezing point depression constant for water.
Given: Kf = 1.86 degrees Celsius x kg/mole.
∆Tf = 1.86 degrees Celsius x kg/mole * 6.711 mol/kg = 12.477 degrees Celsius.
Step 4: Calculate the freezing point of the water-antifreeze mixture.
The freezing point depression (∆Tf) is calculated as the difference between the freezing point of pure water (0 degrees Celsius) and the freezing point of the water-antifreeze mixture (Tf mixture).
∆Tf = Tf pure water - Tf mixture.
Rearranging the formula, we get:
Tf mixture = Tf pure water - ∆Tf = 0 degrees Celsius - 12.477 degrees Celsius = -12.477 degrees Celsius.
Therefore, the freezing point of the water-antifreeze mixture is approximately -12.477 degrees Celsius.
To determine the freezing point of the water-antifreeze mixture, we need to calculate the molality of the solution and then use the freezing point depression equation.
Step 1: Calculate the number of moles of glycol (C2H4(OH)2) and water (H2O):
- Calculate the number of moles of glycol:
- Given mass of glycol = 5.00 kg
- Molar mass of glycol (C2H4(OH)2) = 2(12.01 g/mol) + 4(1.01 g/mol) + 2(16.00 g/mol) = 62.07 g/mol
- Number of moles of glycol = mass of glycol / molar mass of glycol = 5.00 kg * (1000 g/kg) / 62.07 g/mol
- Calculate the number of moles of water:
- Given mass of water = 12.0 kg
- Molar mass of water (H2O) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
- Number of moles of water = mass of water / molar mass of water = 12.0 kg * (1000 g/kg) / 18.02 g/mol
Step 2: Calculate the molality of the solution:
- Given that the mass of water remains unchanged:
- Mass of water = 12.0 kg
- Molality (m) is defined as the number of moles of solute (glycol) divided by the mass of the solvent (water) in kg:
- Molality (m) = moles of glycol / mass of water
Step 3: Use the freezing point depression equation to find the freezing point:
The freezing point depression (ΔTf) is given by the equation:
ΔTf = Kf * m
Given:
- ΔTf: freezing point depression
- Kf: molal freezing point depression constant for water = 1.86°C x kg/mole (as given)
- m: molality of the solution (calculated in Step 2)
Rearranging the equation, we can solve for the freezing point depression:
ΔTf = Kf * m
ΔTf / Kf = m
Finally, we can substitute the values into the equation and solve for the freezing point depression (ΔTf).
Step 4: Calculate the freezing point:
The freezing point of a solution is obtained by subtracting the freezing point depression from the freezing point of the pure solvent (water). The freezing point of pure water is 0°C.
Freezing point of the water-antifreeze mixture = 0°C - ΔTf
By following these steps and plugging in the given values, you should be able to calculate the freezing point of the water-antifreeze mixture.