JEE Mains Chemistry · Solutions

Elevation of Boiling Point and Depression of Freezing Point

A non-volatile solute raises the boiling point and lowers the freezing point of a solvent by K times the molality, ΔTb = Kb·m and ΔTf = Kf·m.

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Why this matters

Twenty-seven PYQs, fifteen of them numeric, and two from 2026. Sixteen are ΔT = K·m in some direction: a new boiling or freezing point, a molar mass, an antifreeze mass or the ice that separates on cooling. Six are about Kb and Kf themselves, and five read a vapour pressure diagram or test what happens as a solution freezes.

Concept 1 of 3: Elevation and depression, ΔT = K·m

Dissolved particles lower the solvent's vapour pressure, so the solution has to be heated further to boil and cooled further to freeze. Both shifts are proportional to the molality: moles of solute per kilogram of SOLVENT.

Definition

$\Delta T_b = K_b m$ , $\Delta T_f = K_f m$ , with $m = \dfrac{w_2/M_2}{W_1/1000}$ ( $W_1$ in grams of solvent).
Molar mass: $M_2 = \dfrac{1000\,K\,w_2}{\Delta T\,W_1}$ .
Solvent given as a volume: grams $=$ volume $\times$ density.
Several non-electrolytes in one solvent: add their moles.
Equal masses of two solutes in equal solvent: $\Delta T \propto 1/M$ , so the larger molar mass gives the smaller shift.
Two compounds AB and $\mathrm{AB_2}$ : find both molar masses, then solve $A + B$ and $A + 2B$ together.
Ice on cooling: only water freezes. The water still liquid at $-\Delta T_f$ is $n_2 K_f/\Delta T_f$ kg; ice $=$ water at the start minus that. For 0.5 mol sucrose in 1 kg of water cooled to $-1.86$ °C, 0.5 kg stays liquid and 500 g is ice.

Colligative temperature shifts

\Delta T_b = K_b\,m \qquad \Delta T_f = K_f\,m \qquad M_2 = \frac{1000\,K\,w_2}{\Delta T\,W_1}

Worked example

1.8 g of a non-volatile non-electrolyte in 40 g of water lowers the freezing point by 0.465 K. Find its molar mass (

K_f = 1.86

K kg/mol).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q34Moderate

Example 1 · Solutions · Elevation of Boiling Point and Depression of Freezing Point

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and

PQ_{2}

. When 1 g of PQ is dissolved in 50 g of solvent ' A '.

\Delta T_{b}

was 1.176 K while when 1 g of

PQ_{2}

is dissolved in 50 g of solvent ' A ',

\Delta T_{b}

was 0.689 K. (

K_{b}

of ' A '

= 5\text{ }Kkgmol^{- 1}

). The molar masses of elements P and Q (in

gmol^{- 1}

) respectively, are :

The ratio of shifts is the INVERSE ratio of molar masses

For equal masses in equal solvent,

\Delta T \propto 1/M

. If the depressions are in the ratio 1 : 4, the molar masses are in the ratio 4 : 1. Writing 1 : 4 for the masses is the planted option.

Molality is per kilogram of SOLVENT

Divide the solute's moles by the solvent's mass in kg, not by the solution's mass and not by a volume in litres. A solvent given in mL needs its density first.

Answer in the order asked

When a question asks for P and Q 'respectively', the option with the right pair in the wrong order is always present. Match your two values to their letters before choosing.

Ice is pure solvent

On cooling a solution only the solvent freezes. The solute stays behind, so the remaining liquid grows more concentrated. The ice separated is the starting water minus the water still needed to hold the solute at that temperature.

Drill 15 more on elevation and depression, δt = k·m

Concept 2 of 3: Ebullioscopic and cryoscopic constants, Kb and Kf

K_b

and

K_f

are the shifts a 1 mol/kg solution would give. They belong to the solvent: a solvent with a high boiling point and a small heat of vaporisation has a large

K_b

Definition

$K_b = \dfrac{R\,T_b^{2}\,M_1}{1000\,\Delta H_{\text{vap}}}$ , $K_f = \dfrac{R\,T_f^{2}\,M_1}{1000\,\Delta H_{\text{fus}}}$ ( $M_1$ in g/mol, $\Delta H$ in J/mol).
At the freezing point $\Delta S_{\text{fus}} = \Delta H_{\text{fus}}/T_f$ , so $K_f = \dfrac{M_1 R\,T_f}{1000\,\Delta S_{\text{fus}}}$ .
For two solvents with equal $M_1$ : $K_b \propto T_b^{2}/\Delta H_{\text{vap}}$ .
Values: water $K_b = 0.52$ , $K_f = 1.86$ ; benzene $K_b = 2.53$ , $K_f = 5.12$ ; acetic acid $K_f = 3.9$ (all in K kg/mol).
For water $K_f > K_b$ , so the same solution's freezing point falls more than its boiling point rises.
Osmotic pressure, not $\Delta T$ , is used for the molar mass of proteins and polymers: the shifts are too small to read.

The solvent constants

K_b = \frac{R\,T_b^{2}\,M_1}{1000\,\Delta H_{\text{vap}}} \qquad K_f = \frac{R\,T_f^{2}\,M_1}{1000\,\Delta H_{\text{fus}}} = \frac{M_1 R\,T_f}{1000\,\Delta S_{\text{fus}}}

Worked example

Solvents X and Y have the same molar mass. X boils at 300 K with

\Delta H_{\text{vap}} = 30

kJ/mol; Y boils at 450 K with

\Delta H_{\text{vap}} = 45

kJ/mol. Find

K_b(\text{Y})/K_b(\text{X})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q107Moderate

Example 2 · Solutions · Elevation of Boiling Point and Depression of Freezing Point

Given below are two statements: Statement (I) : Molal depression constant

K_{f}

is given by

\frac{M_{1}RT_{f}}{\Delta S_{fus}}

, where symbols have their usual meaning. Statement (II) :

K_{f}

for benzene is less than the

K_{f}

for water. In the light of the above statements, choose the most appropriate answer from the options given below :

For water, Kf is larger than Kb

Water's

K_f = 1.86

is more than three times its

K_b = 0.52

. A statement that the boiling point of water rises more than its freezing point falls, for the same solution, is false.

Benzene's Kf is larger than water's

Benzene has

K_f = 5.12

K kg/mol against water's 1.86. The same molality freezes benzene almost three times as far below its normal freezing point.

Drill 5 more on ebullioscopic and cryoscopic constants, kb and kf

Concept 3 of 3: Vapour pressure diagrams and what freezes out

Draw vapour pressure against temperature. The solution's curve sits below the solvent's at every temperature, so it reaches 1 atm at a higher temperature (it boils later) and meets the solid solvent's curve at a lower one (it freezes later on cooling).

Definition

Boiling point: where the liquid's curve reaches the external pressure.
Freezing point: where the liquid's curve crosses the curve of the SOLID solvent.
$\Delta T_b$ is the gap between the two liquids' boiling points; $\Delta T_f$ the gap between their freezing points.
Only the solvent solidifies; the solute stays dissolved.

Feature	What happens	Why
Solution's vapour pressure curve	Lies below the pure solvent's curve at every temperature	The non-volatile solute lowers the vapour pressure
Boiling point	The solution reaches 1 atm (760 mmHg) at a higher temperature	Its vapour pressure starts lower, so it must be heated further
Freezing point	The solution meets the solid solvent's curve at a lower temperature	Its lower vapour pressure matches the solid's only at a lower temperature
What freezes out	Pure solid solvent	The solute stays in the liquid 'Only solute molecules solidify' is the planted false statement.
Solution as ice forms	Grows more concentrated and its freezing point keeps falling	Water leaves as ice while the solute stays
Salt on ice at 0 °C	The ice melts and the mixture cools below 0 °C	Brine freezes below 0 °C, a freezing mixture that keeps ice cream frozen

The solution's curve below the solvent's explains both shifts: a higher boiling point and a lower freezing point.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q47Moderate

Example 3 · Solutions · Elevation of Boiling Point and Depression of Freezing Point

In the depression of freezing point experiment A. Vapour pressure of the solution is less than that of pure solvent B. Vapour pressure of the solution is more than that of pure solvent C. Only solute molecules solidify at the freezing point D. Only solvent molecules solidify at the freezing point Choose the most appropriate answer from the options given below:

Only the solvent freezes

At the freezing point of a solution, pure solvent crystallises and the solute is left in the liquid. A statement that the solute solidifies, or that both do, is false.

The solution's vapour pressure is lower, not higher

Every colligative effect starts from the lowered vapour pressure. A statement that the solution's vapour pressure is more than the solvent's contradicts Raoult's law.

Drill 4 more on vapour pressure diagrams and what freezes out

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Elevation and depression, ΔT = K·m
Colligative temperature shifts
$\Delta T_b = K_b\,m \qquad \Delta T_f = K_f\,m \qquad M_2 = \frac{1000\,K\,w_2}{\Delta T\,W_1}$
Ebullioscopic and cryoscopic constants, Kb and Kf
The solvent constants
$K_b = \frac{R\,T_b^{2}\,M_1}{1000\,\Delta H_{\text{vap}}} \qquad K_f = \frac{R\,T_f^{2}\,M_1}{1000\,\Delta H_{\text{fus}}} = \frac{M_1 R\,T_f}{1000\,\Delta S_{\text{fus}}}$

Reference tables (1)

Vapour pressure diagrams and what freezes out6 rows

Feature	What happens	Why
Solution's vapour pressure curve	Lies below the pure solvent's curve at every temperature	The non-volatile solute lowers the vapour pressure
Boiling point	The solution reaches 1 atm (760 mmHg) at a higher temperature	Its vapour pressure starts lower, so it must be heated further
Freezing point	The solution meets the solid solvent's curve at a lower temperature	Its lower vapour pressure matches the solid's only at a lower temperature
What freezes out	Pure solid solvent	The solute stays in the liquid 'Only solute molecules solidify' is the planted false statement.
Solution as ice forms	Grows more concentrated and its freezing point keeps falling	Water leaves as ice while the solute stays
Salt on ice at 0 °C	The ice melts and the mixture cools below 0 °C	Brine freezes below 0 °C, a freezing mixture that keeps ice cream frozen

The solution's curve below the solvent's explains both shifts: a higher boiling point and a lower freezing point.

Watch out for (8)

The ratio of shifts is the INVERSE ratio of molar masses
→ Elevation and depression, ΔT = K·m
Molality is per kilogram of SOLVENT
→ Elevation and depression, ΔT = K·m
Answer in the order asked
→ Elevation and depression, ΔT = K·m
Ice is pure solvent
→ Elevation and depression, ΔT = K·m
For water, Kf is larger than Kb
→ Ebullioscopic and cryoscopic constants, Kb and Kf
Benzene's Kf is larger than water's
→ Ebullioscopic and cryoscopic constants, Kb and Kf
Only the solvent freezes
→ Vapour pressure diagrams and what freezes out
The solution's vapour pressure is lower, not higher
→ Vapour pressure diagrams and what freezes out

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