 # 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

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## 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

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## Example: Calculating Boiling and Freezing Point of a Nonelectrolyte Solution

For a 0.262 m solution of sucrose in water, calculate the freezing point and the boiling point of the solution.

Freezing Point Calculation Boiling Point Calculation
1. Calculate the freezing point depression:
ΔTf = -Kfm
Kf = 1.855 (from table above)
m = 0.262 m
ΔTf = -1.855 × 0.262 = -0.486oC
2. Calculate the freezing point of the solution:
Tf (solution) = normal freezing point + ΔTf
Tf (solution) = 0.000 - 0.486 = -0.486oC
1. Calculate the boiling point elevation:
ΔTb = Kbm
Kb = 0.512 (from table above)
m = 0.262 m
ΔTb = 0.512 × 0.262 = 0.134oC
2. Calculate the boiling point of the solution:
Tb (solution) = normal boiling point + ΔTb
Tb (solution) = 100.00 + 0.134 = 100.134oC

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## Example: Calculating Boiling and Freezing Point of an Electrolyte Solution

Calculate the freezing point and boiling point for a 0.15 m aqueous solution of sodium chloride.

Freezing Point Calculation Boiling Point Calculation
1. Calculate the freezing point depression:
ΔTf = -Kfm
Kf = 1.855 (from table above)
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.15 m
m(Cl-) = 0.15 m
m(NaCl(aq)) = 0.15 + 0.15 = 0.30 m
ΔTf = -1.855 × 0.30 = -0.557oC
2. Calculate the freezing point of the solution:
Tf (solution) = normal freezing point + ΔTf
Tf (solution) = 0.000 - 0.557 = -0.557oC
1. Calculate the boiling point elevation:
ΔTb = Kbm
Kb = 0.512 (from table above)
Since: NaCl → Na+(aq) + Cl-(aq):
m(Na+) = 0.15 m
m(Cl-) = 0.15 m
m(NaCl(aq)) = 0.15 + 0.15 = 0.30 m
ΔTb = 0.512 × 0.30 = 0.154oC
2. Calculate the boiling point of the solution:
Tb (solution) = normal boiling point + ΔTb
Tb (solution) = 100.00 + 0.154 = 100.154oC

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## Example: Calculating Molar Mass of Solute

Boiling Point Elevation Problem:
1.15 g of an unknown, nonvolatile compound raises the boiling point of 75.0 g benzene (C6H6) by 0.275oC.
Calculate the molar mass of the unknown compound.

Boiling Point Elevation Problem Solution:

1. Calculate the molality of solute particles:
m = ΔTb ÷ Kb
ΔTb = 0.275oC
Kb = 2.53oCm-1 (from table above)
m = 0.275 ÷ 2.53 = 0.109 m
2. Calculate the moles of solute present:
molality = moles solute ÷ kg solvent
moles(solute) = m × kg solvent = 0.109 × 75.0 × 10-3 = 8.175 × 10-3 mol
3. Calculate the molar mass of the solute:
moles(solute) = mass(solute) ÷ molar mass(solute)
molar mass(solute) = mass(solute) ÷ moles(solute) = 1.15 ÷ 8.175 × 10-3 = 141 g mol-1

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