An ethylene glycol solution contains 30.2 g of ethylene glycol (C2H6O2) in 87.0 mL of water. (Assume a density of 1.00 g/mL for water.) FInd the freezing point and boiling point of the solution
dT = kf*m
mols ethylene glycol = grams/molar mass, then,
molality = mols ethylene glycol/kg H2O
Plug in above and solve for dT, subtract from zero C to obtain new freezing point.
Same for boiling point except, dT = Kb*molality. Find dT and add to 100 C to find new boiling point for H2O.
To find the freezing and boiling points of the solution, we need to use the concept of colligative properties.
First, let's calculate the molality (moles of solute per kilogram of solvent) of the ethylene glycol in the solution.
Step 1: Calculate the mass of water in the solution:
Mass of water = volume of water x density of water
= 87.0 mL x 1.00 g/mL
= 87.0 g
Step 2: Calculate the molality (m) of the ethylene glycol:
m = moles of solute / mass of solvent (in kg)
moles of solute = mass of ethylene glycol / molar mass of ethylene glycol
= 30.2 g / 62.07 g/mol (molar mass of ethylene glycol)
We have grams of ethylene glycol, and we need to convert it to moles.
30.2 g / 62.07 g/mol ≈ 0.486 mol
mass of solvent (in kg) = mass of water / 1000 (since 1 kg = 1000 g)
mass of solvent = 87.0 g / 1000 = 0.0870 kg
Now, let's calculate the molality (m):
m = 0.486 mol / 0.0870 kg ≈ 5.586 m
Now, we can use the formulas for freezing point depression and boiling point elevation to find the freezing and boiling points of the solution.
Freezing Point Depression:
∆Tf = Kf x m
where ∆Tf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality.
The cryoscopic constant for water is approximately 1.86 °C/m.
∆Tf = 1.86 °C/m x 5.586 m
≈ 10.37 °C
The freezing point of pure water is 0 °C. Therefore, the freezing point of the solution is:
Freezing point = 0 °C - 10.37 °C
≈ -10.37 °C
Boiling Point Elevation:
∆Tb = Kb x m
where ∆Tb is the change in boiling point, Kb is the ebullioscopic constant, and m is the molality.
The ebullioscopic constant for water is approximately 0.512 °C/m.
∆Tb = 0.512 °C/m x 5.586 m
≈ 2.86 °C
The boiling point of pure water is 100 °C. Therefore, the boiling point of the solution is:
Boiling point = 100 °C + 2.86 °C
≈ 102.86 °C
So, the freezing point of the solution is approximately -10.37 °C, and the boiling point is approximately 102.86 °C.
To find the freezing point and boiling point of the solution, we need to know the molality of the ethylene glycol (C2H6O2) in the solution. Molality (m) is defined as the number of moles of solute divided by the mass of the solvent in kilograms.
First, let's calculate the molality (m) of the solution:
1. Calculate the moles of ethylene glycol (C2H6O2):
- Given mass of ethylene glycol = 30.2 g
- Determine the molar mass of ethylene glycol (C2H6O2):
- Atomic mass of C = 12.01 g/mol (carbon)
- Atomic mass of H = 1.008 g/mol (hydrogen)
- Atomic mass of O = 16.00 g/mol (oxygen)
- Molar mass of C2H6O2 = (2 * 12.01 g/mol) + (6 * 1.008 g/mol) + (2 * 16.00 g/mol) = 62.07 g/mol
- Calculate the number of moles of ethylene glycol:
- Moles of C2H6O2 = Mass of C2H6O2 / Molar mass of C2H6O2
- Moles of C2H6O2 = 30.2 g / 62.07 g/mol = 0.487 mol
2. Calculate the mass of the water:
- Given volume of water = 87.0 mL
- Given density of water = 1.00 g/mL
- Mass of water = Volume of water * Density of water
- Mass of water = 87.0 mL * 1.00 g/mL = 87.0 g
3. Convert mass of water to kilograms:
- Mass of water in kg = Mass of water / 1000
- Mass of water in kg = 87.0 g / 1000 = 0.087 kg
4. Calculate the molality of the solution:
- Molality (m) = Moles of solute / Mass of solvent (in kg)
- Molality (m) = 0.487 mol / 0.087 kg = 5.60 mol/kg
Now that we have the molality (m) of the solution, we can use the freezing point depression (ΔTf) and boiling point elevation (ΔTb) formulas to calculate the freezing and boiling points:
1. Freezing Point Depression (ΔTf):
- ΔTf = Kf * m
- The Kf (cryoscopic constant) for water is approximately 1.86 °C/m (degrees Celsius per molality)
- ΔTf = 1.86 °C/m * 5.60 mol/kg = 10.4 °C
2. Boiling Point Elevation (ΔTb):
- ΔTb = Kb * m
- The Kb (ebullioscopic constant) for water is approximately 0.512 °C/m (degrees Celsius per molality)
- ΔTb = 0.512 °C/m * 5.60 mol/kg = 2.8672 °C
Now, we can calculate the freezing point and boiling point of the solution:
1. Freezing Point:
- The freezing point of pure water is 0 °C
- Freezing point of solution = Freezing point of pure water - ΔTf
- Freezing point of solution = 0 °C - 10.4 °C = -10.4 °C
2. Boiling Point:
- The boiling point of pure water is 100 °C
- Boiling point of solution = Boiling point of pure water + ΔTb
- Boiling point of solution = 100 °C + 2.8672 °C = 102.8672 °C
Therefore, the freezing point of the solution is -10.4 °C, and the boiling point of the solution is 102.8672 °C.