The antifreeze in a car is a 50% by weight solution of ethylene glycol, C2H6O2 in a water solvent. Calculate the boiling point and freezing point of this solution.
(For H2O, Kb=0.52 C/m Kf= 1.86 C/m
I can't even find the formula for this in my book, please help :(
You need to know the mass of the radiator fluid, take 50% for mass ethylene glycol and 50% H2O.
Then mols glycol = grams/molar mass
molality = mols glycol/kg solvent
delta T = Kf*m
Subtract delta T from 0 for new freezing point.
For boiling point, mols is same, m is same, delta T = Kb*m
Add delta T to 100 for new b.p.
To calculate the boiling point and freezing point of the solution, we will use the equations for boiling point elevation and freezing point depression. These equations relate the change in boiling or freezing point to the molal concentration of the solute.
First, let's calculate the molality (m) of the ethylene glycol solution. Molality is defined as the moles of solute (ethylene glycol) per kilogram of solvent (water).
Since we have a weight percent solution, we need to convert the weight percent to molality.
1. Calculate the weight of ethylene glycol:
- Let's assume we have 100 grams of the solution.
- 50% of this weight is ethylene glycol, so the weight of ethylene glycol is:
(50/100) x 100g = 50g
2. Calculate the number of moles of ethylene glycol:
- The molar mass of ethylene glycol (C2H6O2) is:
(12.01 x 2) + (1.01 x 6) + (16.00 x 2) = 62.07 g/mol
- The number of moles of ethylene glycol is:
50g / 62.07 g/mol = 0.805 mol
3. Calculate the mass of water (solvent):
- Since we assumed 100g of the solution and we have 50g of ethylene glycol, the mass of water is:
100g - 50g = 50g
4. Calculate the molality (m):
- The molality is the moles of ethylene glycol divided by the mass of water in kilograms:
m = 0.805 mol / 0.050 kg = 16.1 m
Now, let's use the equations for boiling point and freezing point changes to calculate the boiling point elevation (∆Tb) and freezing point depression (∆Tf).
Boiling Point Elevation:
∆Tb = Kb * m
Freezing Point Depression:
∆Tf = Kf * m
Given:
Kb = 0.52 °C/m (boiling point constant)
Kf = 1.86 °C/m (freezing point constant)
m = 16.1 m (molality)
1. Calculate the boiling point elevation:
∆Tb = 0.52 °C/m * 16.1 m = 8.372 °C
2. Calculate the freezing point depression:
∆Tf = 1.86 °C/m * 16.1 m = 29.946 °C
Finally, we can calculate the boiling point and freezing point of the solution.
Boiling Point:
The boiling point of pure water is 100 °C. The boiling point elevation tells us how much higher the boiling point of the solution is compared to pure water. Therefore, the boiling point of the solution is:
100 °C + ∆Tb = 100 °C + 8.372 °C = 108.372 °C
Freezing Point:
The freezing point of pure water is 0 °C. The freezing point depression tells us how much lower the freezing point of the solution is compared to pure water. Therefore, the freezing point of the solution is:
0 °C - ∆Tf = 0 °C - 29.946 °C = -29.946 °C
Therefore, the boiling point of the solution is 108.372 °C and the freezing point of the solution is -29.946 °C.
To calculate the boiling point and freezing point of the solution, we can use the equations for boiling point elevation and freezing point depression. These equations relate the change in boiling and freezing points to the molality of the solute.
Boiling Point Elevation:
ΔTb = Kb * m
Freezing Point Depression:
ΔTf = Kf * m
In these equations:
- ΔTb is the change in boiling point
- ΔTf is the change in freezing point
- Kb is the boiling point constant for the solvent (water)
- Kf is the freezing point constant for the solvent (water)
- m is the molality of the solute (ethylene glycol)
First, we need to calculate the molality of the solution. To do this, we need to know the mass of ethylene glycol and the mass of water.
Given that the solution is 50% by weight, we can assume that there is an equal mass of water and ethylene glycol. So, for every 100 g of the solution, there would be 50 g of ethylene glycol and 50 g of water.
Next, we need to convert the mass of each component into moles. To do this, we divide the mass by the molar mass.
The molar mass of ethylene glycol (C2H6O2) is:
2 * (12.01 g/mol) + 6 * (1.01 g/mol) + 2 * (16.00 g/mol) = 62.07 g/mol
The number of moles of ethylene glycol is:
moles of ethylene glycol = mass of ethylene glycol / molar mass of ethylene glycol
moles of ethylene glycol = 50 g / 62.07 g/mol
Similarly, for water (H2O), the molar mass is 18.02 g/mol.
The number of moles of water is:
moles of water = mass of water / molar mass of water
moles of water = 50 g / 18.02 g/mol
Now, we have the moles of solute (ethylene glycol) and solvent (water). We can use these values to calculate the molality (m) of the solution.
Molality is defined as the number of moles of solute per kilogram of solvent.
To calculate the molality, we need to convert the mass of water to kilograms:
mass of water in kg = 50 g / 1000 g/kg
Now, we can calculate the molality:
molality = moles of solute / mass of water in kg
Now that we have the molality, we can calculate the change in boiling point (ΔTb) and freezing point (ΔTf) using the equations mentioned previously.
Finally, to calculate the boiling point and freezing point of the solution, we need to add or subtract the changes in boiling point and freezing point from the normal boiling point and freezing point of the solvent (water).
Normal boiling point of water: 100°C
Normal freezing point of water: 0°C
Boiling point of the solution = Normal boiling point of water + ΔTb
Freezing point of the solution = Normal freezing point of water - ΔTf
These calculations will give you the boiling and freezing points of a 50% by weight solution of ethylene glycol in water.