Chemistry · Solutions · NEET
A pure liquid boils when its vapour pressure equals the outside air pressure (1 atm). When you dissolve a non-volatile solute, the vapour pressure of the solution goes DOWN. So the solution must be heated to a HIGHER temperature to reach 1 atm again. That extra temperature is the elevation of boiling point, ΔTb = Tb − Tb°, where Tb is the solution's boiling point and Tb° is the pure solvent's boiling point. Example from NCERT: 1 mol sucrose in 1000 g water boils at 373.52 K instead of 373.15 K.
For dilute solutions, ΔTb is directly proportional to molality: ΔTb = Kb·m. Here m is molality (moles of solute per kg of solvent). Kb is the Molal Elevation Constant, also called the Ebullioscopic Constant or Boiling Point Elevation Constant. Kb is a fixed value for each SOLVENT (it does not depend on the solute). For water Kb = 0.52 K kg mol⁻¹. So Kb tells you how much the boiling point rises for a 1 molal solution.
The units of Kb are K kg mol⁻¹ (kelvin kilogram per mole). Since ΔTb = Kb·m and molality m has units mol kg⁻¹, when you multiply Kb (K kg mol⁻¹) by m (mol kg⁻¹) the kg and mol cancel and you are left with K, a temperature change. A common NEET trap is writing Kb units as K mol⁻¹ — that is wrong; the kg must be there.
A non-volatile solute takes up space at the liquid surface, so fewer solvent molecules can escape into the vapour. This lowers the vapour pressure of the solution. Because the vapour pressure is now lower, you need a higher temperature to make it equal to 1 atm and boil. That is why sea water boils above 100°C. The effect depends only on HOW MANY solute particles are present, not their chemical nature — that is why it is a colligative property.
Yes, for solutes that split into ions (electrolytes). The full formula is ΔTb = i·Kb·m, where i is the van't Hoff factor (number of particles one formula unit gives). NaCl → Na⁺ + Cl⁻ so i = 2. Na₂SO₄ → 2Na⁺ + SO₄²⁻ so i = 3. Glucose and urea do not split, so i = 1. To compare which solution boils highest, compare the value i × m (effective particle concentration) — the largest wins.
Step 1: Find molality m = (moles of solute)/(kg of solvent). Step 2: Multiply ΔTb = i·Kb·m (use i = 1 if the solute is a non-electrolyte like sugar). Step 3: Add ΔTb to the pure solvent boiling point. For water, boiling point of solution = 100°C + ΔTb. Watch the units — Kb is per kg, so convert grams of solvent to kg.
At 100 °C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be:
Which of the following aqueous solutions will exhibit highest boiling point?
Which amongst the following aqueous solutions of electrolytes will have minimum elevation in boiling point?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Kb is a property of the SOLVENT only. It does not depend on which solute you add. For water Kb = 0.52 K kg mol⁻¹, for benzene it is about 2.53 K kg mol⁻¹. This is why the same solute causes different boiling point rises in different solvents.
Because it depends only on the NUMBER of solute particles dissolved, not on their chemical identity. One mole of glucose particles and one mole of urea particles raise the boiling point by the same amount, since both give i = 1.
Tb is the actual boiling point of the solution (for example 101°C). ΔTb is only the RISE compared to pure solvent: ΔTb = Tb − Tb°. In NEET numericals you usually calculate ΔTb first, then add it to the pure solvent's boiling point to get Tb.
Yes, both come from the same cause — the solute lowers vapour pressure. Adding a solute raises the boiling point (ΔTb = i·Kb·m) and lowers the freezing point (ΔTf = i·Kf·m). Both are colligative properties and both use molality and the van't Hoff factor i.
By combining ΔTb = Kb·m with m = (w₂/M₂)/(w₁ in kg), you can rearrange to find the molar mass M₂ of an unknown solute: M₂ = (Kb × w₂ × 1000)/(ΔTb × w₁ in g). This is a common NEET numerical.
