Eugenol, the active ingredient in cloves, has a formula of C10H12O2. What is the boiling point of a solution containing 1.14g of eugenol dissolved in 10.0g of benzene? The boiling point of benzene is 80.1C and the boiling point elevation constant for benzene is 2.53C/m.
I got .0678mol for eugenol, is that correct?Do I use Delta T=(2.53C/m)(.0678mol)? I got a little confused.
I don't get that.
moles = grams/molar mass
moles eugenol = 1.14/164.2 = 0.00694.
molality = mols/kg solvent = 0.00694/0.010 = ?? (use m in the following).
delta T = Kb*m
delta T = 2.53*m
To find the boiling point of a solution, you can use the boiling point elevation formula:
ΔTb = Kbp * b * i
where:
- ΔTb is the boiling point elevation (in Celsius)
- Kbp is the boiling point elevation constant for the solvent (in this case, benzene) which is 2.53 °C/m
- b is the molality of the solute (eugenol), which is the number of moles of solute per kilogram of solvent
- i is the van 't Hoff factor, which represents the number of particles the solute molecule dissociates into in the solution
To find the number of moles of eugenol (C10H12O2), you correctly calculated it as 0.0678 mol.
Next, we need to find the molality (b) of the eugenol solution. Molality is defined as moles of solute per kilogram of solvent. In this case, the solute is eugenol (C10H12O2) and the solvent is benzene (C6H6).
Molar mass of eugenol (C10H12O2) = (12.01 * 10) + (1.01 * 12) + (16.00 * 2) = 164.2 g/mol
Using the mass of eugenol (1.14 g) and mass of benzene (10.0 g), we can calculate the molality:
b = (0.0678 mol) / [(1.14 g / 164.2 g/mol) + (10.0 g / 78.11 g/mol)]
b = 0.0038 mol/kg
Now that we have the molality (b) and the boiling point elevation constant (Kbp), we can substitute these values into the boiling point elevation formula:
ΔTb = (2.53 °C/m) * (0.0038 mol/kg) * i
Since eugenol is a non-electrolyte and does not dissociate into ions, the van 't Hoff factor (i) is 1.
ΔTb = (2.53 °C/m) * (0.0038 mol/kg) * 1
ΔTb = 0.0096 °C
Finally, to find the boiling point of the solution, add the boiling point elevation (ΔTb) to the boiling point of the pure solvent (benzene):
Boiling point of solution = Boiling point of benzene + ΔTb
Boiling point of solution = 80.1 °C + 0.0096 °C
Thus, the boiling point of the solution containing 1.14g of eugenol dissolved in 10.0g of benzene is approximately 80.11 °C.
To find the number of moles of eugenol (C10H12O2), you can use the given mass and the molar mass of eugenol.
1. Calculate the molar mass of eugenol:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol
Total molar mass = (12.01 * 10) + (1.01 * 12) + (16.00 * 2) = 164.22 g/mol
2. Calculate the number of moles of eugenol:
Moles = Mass / Molar mass
Moles = 1.14 g / 164.22 g/mol
Moles ≈ 0.00695 mol (rounded to five decimal places)
Therefore, you are correct that the number of moles of eugenol is approximately 0.00695 mol.
Now, to find the boiling point elevation, you can use the boiling point elevation formula:
ΔT = K * m
where ΔT is the boiling point elevation, K is the boiling point elevation constant, and m is the molality of the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent.
3. Calculate the molality of the solution:
Mass of solvent = 10.0 g = 0.01 kg
Molality (m) = Moles of solute / Mass of solvent
Molality = 0.00695 mol / 0.01 kg
Molality = 0.695 mol/kg
4. Calculate the boiling point elevation:
ΔT = (2.53 °C/m) * (0.695 mol/kg)
ΔT ≈ 1.76 °C (rounded to two decimal places)
Finally, to find the boiling point of the solution, add the boiling point elevation to the boiling point of benzene:
Boiling point of solution = Boiling point of benzene + ΔT
Boiling point of solution = 80.1 °C + 1.76 °C
Boiling point of solution ≈ 81.86 °C (rounded to two decimal places)