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:
ΔT_b = K_bm
ΔT_b = the amount by which the boiling point is raised
m = molality (moles solute particles per kg of solution)
K_b = molal boiling-point elevation constant (solvent dependent)

Boiling Point of solution = normal boiling point of solvent + ΔT_b

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:
ΔT_f = -K_fm
ΔT_f = the amount by which the freezing point is lowered
m = molality (moles solute particles per kg of solution)
K_f = molal freezing-point depression constant (solvent dependent)

Freezing Point of solution = normal freezing point of solvent + ΔT_f

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Some Boiling-Point Elevation and Freezing-Point Depression Constants

solvent	normal boiling point (^oC)	K_b (^oCm^-1)	normal freezing point (^oC)	K_f (^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
Calculate the freezing point depression: ΔT_f = -K_fm K_f = 1.855 (from table above) m = 0.262 m ΔT_f = -1.855 × 0.262 = -0.486^oC Calculate the freezing point of the solution: T_f (solution) = normal freezing point + ΔT_f T_f (solution) = 0.000 - 0.486 = -0.486^oC	Calculate the boiling point elevation: ΔT_b = K_bm K_b = 0.512 (from table above) m = 0.262 m ΔT_b = 0.512 × 0.262 = 0.134^oC Calculate the boiling point of the solution: T_b (solution) = normal boiling point + ΔT_b T_b (solution) = 100.00 + 0.134 = 100.134^oC

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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
Calculate the freezing point depression: ΔT_f = -K_fm K_f = 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 ΔT_f = -1.855 × 0.30 = -0.557^oC Calculate the freezing point of the solution: T_f (solution) = normal freezing point + ΔT_f T_f (solution) = 0.000 - 0.557 = -0.557^oC	Calculate the boiling point elevation: ΔT_b = K_bm K_b = 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 ΔT_b = 0.512 × 0.30 = 0.154^oC Calculate the boiling point of the solution: T_b (solution) = normal boiling point + ΔT_b T_b (solution) = 100.00 + 0.154 = 100.154^oC

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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 0.275^oC.
Calculate the molar mass of the unknown compound.

Boiling Point Elevation Problem Solution:

Calculate the molality of solute particles:
m = ΔT_b ÷ K_b
ΔT_b = 0.275^oC
K_b = 2.53^oCm^-1 (from table above)
m = 0.275 ÷ 2.53 = 0.109 m

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

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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