Boiling Point Elevation and Freezing Point Depression |
Key Concepts
Boiling Point Elevation:
- A liquid boils at the temperature at which its vapor pressure equals atmospheric pressure.
- The presence of a solute lowers the vapor pressure of the solution at each temperature, making it necessary to heat the solution to a higher temperature to boil the solution.
- In dilute solutions with a nonvolatile solute, the boiling point elevation is proportional to the molality of the solute particles:
ΔTb = Kbm
ΔTb = the amount by which the boiling point is raised
m = molality (moles solute particles per kg of solution)
Kb = molal boiling-point elevation constant (solvent dependent)
- Boiling Point of solution = normal boiling point of solvent + ΔTb
Freezing Point Depression:
- A solute lowers the freezing point of a solvent.
- In dilute solutions, the freezing point depression is proportional to the molality of the solute particles:
ΔTf = -Kfm
ΔTf = the amount by which the freezing point is lowered
m = molality (moles solute particles per kg of solution)
Kf = molal freezing-point depression constant (solvent dependent)
- Freezing Point of solution = normal freezing point of solvent + ΔTf
Some Boiling-Point Elevation and Freezing-Point Depression Constants
| solvent |
normal boiling
point (oC) |
Kb (oCm-1) |
normal freezing
point (oC) |
Kf (oCm-1) |
|---|
| benzene |
80.2 |
2.53 |
5.5 |
5.12 |
|
| water |
100.0 |
0.512 |
0.000 |
1.855 |
|
acetic acid
(ethanoic acid) |
118.5 |
3.07 |
16.6 |
3.90 |
|
| camphor |
208.3 |
5.95 |
178.4 |
40.0 |
|
| naphthalene |
218.0 |
5.65 |
80.2 |
6.9 |
Example: Calculating Boiling and Freezing Point of a Nonelectrolyte Solution
For a 0.262m solution of sucrose in water, calculate the freezing point and the boiling point of the solution.
| Freezing Point Calculation |
Boiling Point Calculation |
|---|
- Calculate the freezing point depression:
ΔTf = -Kfm
Kf = 1.855 (from table above)
m = 0.262m
ΔTf = -1.855 x 0.262 = -0.486oC
- Calculate the freezing point of the solution:
Tf (solution) = normal freezing point + ΔTf
Tf (solution) = 0.000 - 0.486
= -0.486oC
|
- Calculate the boiling point elevation:
ΔTb = Kbm
Kb = 0.512 (from table above)
m = 0.262m
ΔTb = 0.512 x 0.262 = 0.134oC
- Calculate the boiling point of the solution:
Tb (solution) = normal boiling point + ΔTb
Tb (solution) = 100.00 + 0.134
= 100.134oC
|
Example: Calculating Boiling and Freezing Point of an Electrolyte Solution
Calculate the freezing point and boiling point for a 0.15m aqueous solution of sodium chloride.
| Freezing Point Calculation |
Boiling Point Calculation |
|---|
- Calculate the freezing point depression:
ΔTf = -Kfm
Kf = 1.855 (from table above)
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.15m
m(Cl-) = 0.15m
m(NaCl(aq)) = 0.15 + 0.15 = 0.30m
ΔTf = -1.855 x 0.30 = -0.557oC
- Calculate the freezing point of the solution:
Tf (solution) = normal freezing point + ΔTf
Tf (solution) = 0.000 - 0.557
= -0.557oC
|
- Calculate the boiling point elevation:
ΔTb = Kbm
Kb = 0.512 (from table above)
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.15m
m(Cl-) = 0.15m
m(NaCl(aq)) = 0.15 + 0.15 = 0.30m
ΔTb = 0.512 x 0.30 = 0.154oC
- Calculate the boiling point of the solution:
Tb (solution) = normal boiling point + ΔTb
Tb (solution) = 100.00 + 0.154
= 100.154oC
|
Example: Calculating Molecular Mass (Formula Weight) of Solute
1.15g of an unknown, nonvolatile compound raises the boiling point of 75.0g benzene (C6H6) by 0.275oC.
Calculate the molecular mass (formula weight) of the unknown compound.
- Calculate the molality of solute particles:
m = ΔTb ÷ Kb
ΔTb = 0.275oC
Kb = 2.53oCm-1 (from table above)
m = 0.275 ÷ 2.53 = 0.109m
- Calculate the moles of solute present:
molality = moles solute ÷ kg solvent
n(solute) = m x kg solvent = 0.109 x 75.0 x 10-3 = 8.175 x 10-3 mol
- Calculate the molecular mass (formula weight) of the solute:
n(solute) = mass(solute) ÷ MM(solute)
MM(solute) = mass(solute) ÷ n(solute) = 1.15 ÷ 8.175 x 10-3 = 141 g/mol
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| Raoult's Law
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Colligative Properties of Solutions
Molecular Mass (Formula Weight)
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