Solutions · Lesson 7 of 11
Elevation of Boiling Point
“Connect solvent vapour-pressure lowering to boiling-point elevation and derive equations for concentration and molar mass.”
• Explain why a non-volatile solute raises boiling temperature at fixed external pressure. • Derive and apply the dilute relation ΔT_b=K_bm. • Determine solute molar mass from a measured boiling-point rise. • Use the correct pure-solvent boiling point and state model limitations.
Boiling requires matching the external pressure
Evaporation can occur from a liquid surface below the boiling temperature. Boiling begins when the liquid’s equilibrium vapour pressure can match the external pressure, allowing vapour bubbles to persist inside it. At a fixed external pressure, a pure solvent has a definite boiling temperature T_b°. For water at 1 atm, T_b° is about 373.15 K. At lower pressure, such as at altitude, water boils at a lower temperature.
A dissolved non-volatile solute lowers solvent vapour pressure at a given temperature. At the old pure-solvent boiling temperature, the solution therefore has insufficient vapour pressure to match the same external pressure. Heating raises its vapour pressure until the match occurs. The solution boils at a higher temperature. This argument assumes the solute contributes negligible vapour; it does not assert that every volatile liquid mixture must boil above both pure liquids.
From a pressure shift to a concentration law
For a small composition change, a small temperature rise is sufficient to restore the vapour pressure. Experiments show that this rise is proportional to molality for a dilute solution of independent solute molecules. We write ΔT_b=K_bm, where the proportionality constant is a property of the solvent, not a universal constant for all liquids. Molality is suitable because the solute amount is compared with solvent mass and does not change merely because heating expands the liquid.
Why the solvent constant has this form
A short derivation connects the concentration law to two physical facts: ideal mixing reduces the solvent escaping tendency, and heating increases pure-solvent vapour pressure. At the solution boiling temperature T_b, the ideal relation gives p_ext=x₁p°(T_b); at the pure-solvent boiling temperature, p_ext=p°(T_b°). Dividing eliminates external pressure. Taking a natural logarithm gives ln[p°(T_b)/p°(T_b°)]=−ln x₁.
For a dilute solute, x₁=1−x₂ and −ln(1−x₂)≈x₂≈n₂/n₁. With solvent molar mass M₁ in g mol⁻¹, n₂/n₁=mM₁/1000. Equating these two descriptions of the pressure change and solving for ΔT_b yields the following expression. The 1000 converts grams of solvent per mole into kilograms per mole; it must be removed if M₁ is already in kg mol⁻¹.
A larger enthalpy of vapourisation means vapour pressure changes more strongly with temperature near boiling; a smaller temperature change is then needed for a given composition effect. This explains why K_b depends on both the solvent boiling temperature and its vapourisation enthalpy. The derivation also reveals the approximations: small shifts, dilute composition, ideal mixing and a non-volatile solute.
Calculate a rise first, then the final temperature
Problem
Dissolve 18 g glucose, molar mass 180 g mol⁻¹, in 1.00 kg water. Estimate the boiling temperature at 1 atm using K_b=0.52 K kg mol⁻¹.
- 1.Glucose amount=18/180=0.100 mol and molality=0.100/1.00=0.100 mol kg⁻¹.
- 2.ΔT_b=0.52×0.100=0.052 K.
- 3.Add the rise to the pure-water value at the same pressure: T_b=373.15+0.052=373.202 K, approximately 373.20 K.
- 4.The computed change is small; the recorded precision of the pure boiling temperature limits how many final digits are meaningful.
The equation follows by substituting the mass expression for molality into ΔT_b=K_bm and solving for M₂. A heavier solute gives fewer particles for the same mass, so it produces a smaller rise. Consequently, the inferred molar mass is inversely proportional to the measured rise. A very small difference between two temperatures is experimentally sensitive; uncertainty in that difference can dominate the result.
Problem
Pure benzene boils at 353.23 K. A solution of 1.80 g non-volatile non-electrolyte in 90.0 g benzene boils at 354.11 K. Use K_b=2.53 K kg mol⁻¹ to find molar mass.
- 1.Compute the difference at the same pressure: ΔT_b=354.11−353.23=0.88 K.
- 2.Molality=ΔT_b/K_b=0.88/2.53=0.347826 mol kg⁻¹. In 0.0900 kg benzene, solute amount=0.0313043 mol.
- 3.M₂=1.80/0.0313043=57.5 g mol⁻¹, or about 58 g mol⁻¹ at the precision of the elevation.
- 4.Equivalently, substitute directly into the mass formula. Retaining intermediate digits avoids unnecessary rounding error.
Problem
At 750 mm Hg, pure water boils at 99.63 °C. What mass of sucrose in 500 g water raises its boiling point to 100.00 °C? Use sucrose 342 g mol⁻¹ and K_b=0.52 K kg mol⁻¹.
- 1.The required rise is 100.00−99.63=0.37 K, not zero. The pure-water normal boiling point is not applicable at 750 mm Hg.
- 2.m=0.37/0.52=0.711538 mol kg⁻¹. Required sucrose amount=m×0.500=0.355769 mol.
- 3.Mass=0.355769×342=121.673 g, about 122 g.
- 4.The supplied dilute equation gives an estimate; concentrated solutions may need a more detailed model.
The law describes an instantaneous composition. During prolonged boiling, solvent loss increases molality and the boiling temperature can continue to rise. Specify composition and external pressure when reporting a boiling point.
Quick Quiz
Quiz
Why does a non-volatile solute raise boiling temperature at unchanged pressure?
Explanation: The solution must be heated further to bring its reduced vapour pressure up to the same external pressure.
For K_b=0.52 K kg mol⁻¹ and m=0.20 mol kg⁻¹, what is ΔT_b?
Explanation: Multiply K_b by molality: 0.52×0.20=0.104 K.
Which pure-solvent boiling temperature is needed for a measurement at reduced pressure?
Explanation: The elevation is a difference measured at a fixed external pressure.
What is the unit of K_b when molality is mol kg⁻¹?
Explanation: K kg mol⁻¹ multiplied by mol kg⁻¹ gives K.
At fixed solute and solvent masses, a smaller boiling elevation implies what under the same i=1 model?
Explanation: The mass formula has ΔT_b in the denominator; fewer particles per gram imply a greater molar mass.
Practice Problems
- Calculate the boiling temperature of 0.50 mol urea in 2.0 kg water at 1 atm using K_b=0.52 and pure T_b=373.15 K.
- A 2.00 g solute in 100 g water raises boiling temperature by 0.104 K. Find molar mass with K_b=0.52, assuming i=1.
- Explain why a volatile liquid mixture does not necessarily obey the non-volatile-solute boiling elevation rule.
- Estimate K_b for a solvent with T_b°=350 K, M₁=100 g mol⁻¹ and ΔvapH=35000 J mol⁻¹. Use R=8.314 J mol⁻¹ K⁻¹.
- Two non-electrolytes of molar masses 60 and 180 g mol⁻¹ are separately dissolved in equal masses of the same solvent. Equal solute masses are used. Compare the boiling elevations.
m=0.50/2.0=0.25 mol kg⁻¹. ΔT_b=0.52×0.25=0.13 K. T_b=373.15+0.13=373.28 K. Use solvent mass, not total solution mass. Dividing amount by that mass gives molality, then the predicted rise is added to the pure boiling temperature.
Key Takeaways
• Boiling occurs when equilibrium vapour pressure matches external pressure. • A non-volatile solute lowers vapour pressure at fixed temperature and raises boiling temperature at fixed pressure. • Boiling elevation is ΔT_b=T_b−T_b°. • For dilute ideal solutions with independent solute molecules, ΔT_b=K_bm. • K_b depends on solvent properties and has unit K kg mol⁻¹. • Molar-mass calculations use solvent mass and the measured temperature difference. • Pressure changes and solvent loss must be accounted for when interpreting boiling temperatures.