ChemistryISC Class 12 · Solutions

Elevation of Boiling Point and Depression of Freezing Point (ISC Class 12)

Updated 15 October 2026 · Fact-checked

A non-volatile solute lowers the vapour pressure of a solvent, so the solution boils at a higher temperature and freezes at a lower one. For dilute solutions ΔTb = Kb m and ΔTf = Kf m, where m is molality. Rearranged, M = 1000 K w₂ / (ΔT w₁) gives the molar mass of the solute.

Understand Elevation of Boiling Point and Depression of Freezing Point

Both effects begin with lowering of vapour pressure. When a non-volatile solute is dissolved, some surface positions are taken by solute particles, so fewer solvent molecules escape and the vapour pressure of the solution is lower than that of the pure solvent at every temperature.

Elevation of boiling point. A liquid boils when its vapour pressure equals the external pressure. The solution's vapour pressure curve lies below the solvent's, so the solution must be heated to a higher temperature before its vapour pressure reaches 1 atm. The rise $Δ T_{b} = T_{b} - T_{b}^{0}$ is the elevation of boiling point. For dilute solutions it is proportional to molality: $Δ T_{b} = K_{b} m$ . The constant $K_{b}$ is the molal elevation constant or ebullioscopic constant: the elevation produced by a 1 molal solution of a non-volatile non-electrolyte (an extrapolated value, as the law holds only for dilute solutions). Its unit is K kg mol⁻¹; for water it is 0.52, for benzene 2.53.

Depression of freezing point. At the freezing point, the vapour pressure of the liquid equals the vapour pressure of the solid. Only pure solvent freezes out, so the solid's vapour pressure curve is unchanged. The solution's lower curve meets the solid's curve at a lower temperature, so the solution freezes below the pure solvent. The fall $Δ T_{f} = T_{f}^{0} - T_{f}$ is the depression of freezing point, and $Δ T_{f} = K_{f} m$ . $K_{f}$ is the molal depression constant or cryoscopic constant (water 1.86, benzene 5.12, camphor much larger).

Deriving the molar mass formula. Molality is moles of solute per kilogram of solvent: if $w_{2}$ g of solute of molar mass $M_{2}$ is dissolved in $w_{1}$ g of solvent, $m = \frac{w _{2} / M _{2}}{w _{1} /1000} = \frac{1000 w _{2}}{M _{2} w _{1}}$ . Substituting in $Δ T = K m$ gives $Δ T = \frac{1000 K w _{2}}{M _{2} w _{1}}$ , so

$M_{2} = \frac{1000 K _{b} w _{2}}{Δ T _{b} w _{1}}$ and $M_{2} = \frac{1000 K _{f} w _{2}}{Δ T _{f} w _{1}}$ .

Where the constants come from. Thermodynamics gives $K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H}$ and $K_{b} = \frac{R M _{1} T _{b}^{2}}{1000 Δ _{v a p} H}$ , with $M_{1}$ the molar mass of the solvent in g mol⁻¹, $T$ in kelvin and the enthalpies in J mol⁻¹. Writing $l_{f} = Δ_{f u s} H / M_{1}$ (latent heat of fusion per gram) gives $K_{f} = \frac{R T _{f}^{2}}{1000 l _{f}}$ . So $K_{f}$ and $K_{b}$ depend only on the solvent, never on the solute. Water's $K_{f}$ (1.86) is bigger than its $K_{b}$ (0.52) mainly because its enthalpy of fusion is far smaller than its enthalpy of vaporisation.

Colligative. Both effects depend on the number of solute particles. For electrolytes multiply by the van't Hoff factor: $Δ T_{b} = i K_{b} m$ , $Δ T_{f} = i K_{f} m$ . Freezing-point depression is the more common method in practice because $K_{f}$ is larger and the reading is not affected by changes in atmospheric pressure. Neither method suits proteins: their solutions give changes too small to measure, which is why osmotic pressure is used for them.

Elevation of Boiling Point and Depression of Freezing Point: formulas and key facts

Elevation of boiling point: $Δ T_{b} = T_{b} - T_{b}^{0} = K_{b} m$; $m$ = molality in mol kg⁻¹; $K_{b}$ in K kg mol⁻¹ (water 0.52).
Depression of freezing point: $Δ T_{f} = T_{f}^{0} - T_{f} = K_{f} m$; $K_{f}$ for water 1.86, benzene 5.12 K kg mol⁻¹.
Molality from masses: $m = \frac{1000 w _{2}}{M _{2} w _{1}}$; $w_{1}$ is the mass of solvent in grams, not the mass of solution.
Molar mass of solute: $M_{2} = \frac{1000 K w _{2}}{Δ T w _{1}}$; Use $K_{b}$ with $Δ T_{b}$ or $K_{f}$ with $Δ T_{f}$ .
Constants from thermodynamic data: $K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H}, K_{b} = \frac{R M _{1} T _{b}^{2}}{1000 Δ _{v a p} H}$; $R = 8.314$ J K⁻¹ mol⁻¹, $M_{1}$ in g mol⁻¹, enthalpy in J mol⁻¹; equivalently $K_{f} = \frac{R T _{f}^{2}}{1000 l _{f}}$ .
Electrolytes: $Δ T_{b} = i K_{b} m, Δ T_{f} = i K_{f} m$; $i$ = van't Hoff factor.

How to solve Elevation of Boiling Point and Depression of Freezing Point questions

ISC numericals here are of three kinds: find the boiling or freezing point, find a molar mass, or link the two effects through molality. Work in this order.

1Read which effect is given and which is asked. Find $Δ T$ as a positive difference (in K or °C; a difference is the same in both).
2Write the formula you will use, $Δ T = K m$ or $M_{2} = \frac{1000 K w _{2}}{Δ T w _{1}}$ , before substituting.
3Use the mass of solvent $w_{1}$ in grams in the 1000 form (or in kg in $m = n / k g$ ). Never use the mass of the solution.
4If the solute is an electrolyte, include $i$ ; otherwise take $i = 1$ .
5To go from boiling point to freezing point (or back), find $m$ from one equation and put it in the other: $\frac{Δ T _{f}}{Δ T _{b}} = \frac{K _{f}}{K _{b}}$ .
6For the final temperature, add $Δ T_{b}$ to the solvent's boiling point or subtract $Δ T_{f}$ from its freezing point, and give the unit.
7For molecular formula questions, divide the molar mass by the atomic mass to find atoms per molecule (for example $S_{8}$ ).

Elevation of Boiling Point and Depression of Freezing Point in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Salt in cooking water barely changes its boiling point

Suppose a cook claims that adding a spoonful of salt (10 g of NaCl) to 2.0 kg of water makes the pasta cook much faster because the water boils far hotter. Assume NaCl dissociates completely ( $i = 2$ ) and $K_{b}$ for water is 0.52 K kg mol⁻¹.

Calculate the elevation of boiling point and judge the claim. How much salt would be needed in 1.0 kg of water to raise the boiling point by about 1 K?

Try it first, then show the solutionSolution

$m = \frac{10/58.5}{2.0} = 0.0855$ mol kg⁻¹.
$Δ T_{b} = i K_{b} m = 2 \times 0.52 \times 0.0855 = 0.089$ K.
A rise of less than 0.1 K cannot change cooking time noticeably, so the claim is wrong; the salt is for taste.
For $Δ T_{b} = 1.04$ K in 1 kg we need $m = 1$ , i.e. 58.5 g of NaCl per kg, far saltier than anyone would eat.

Answer: $Δ T_{b} \approx 0.09$ K, too small to matter. About 58.5 g of NaCl per kg of water would be needed for a 1 K rise.

Memory hook: Colligative effects scale with particles per kilogram of solvent: a pinch of salt makes only a tiny change.

Tricks to remember Elevation of Boiling Point and Depression of Freezing Point

Short memory aids. Say each one aloud once, then use it on a question the same day.

Boiling goes Up, Freezing goes Down
Solute is a 'blanket': it keeps the solvent liquid over a wider range, boiling point up and freezing point down.
Use it when: Sign of $Δ T$ and which way to adjust the final temperature.
Thousand K w over delta-T W
$M = \frac{1000 \cdot K \cdot w}{Δ T \cdot W}$ : small $w$ (solute) on top with $K$ ; big $W$ (solvent) below with $Δ T$ .
Use it when: Recalling the molar mass formula without mixing up the masses.
Kf over Kb for water: about 3.6
1.86 / 0.52 ≈ 3.6, so in water the freezing-point drop is about 3.6 times the boiling-point rise for the same solution.
Use it when: Quick checks when converting one effect into the other.

Elevation of Boiling Point and Depression of Freezing Point: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: What the constant depends on · 1 mark · MCQBoard level

The molal elevation constant of a liquid depends on:
(a) the nature of the solute
(b) the nature of the solvent
(c) the molality of the solution
(d) the volume of the solution

Try it first, then show the solutionSolution

$K_{b} = \frac{R M _{1} T _{b}^{2}}{1000 Δ _{v a p} H}$ contains only properties of the solvent.
It is a constant for a given solvent, whatever the solute or concentration.

Answer: (b) the nature of the solvent.

Example 2: Boiling point of a urea solution · 2 marks · Short answerBoard level

Calculate the boiling point of a solution containing 6.0 g of urea ( $M = 60$ g mol⁻¹) in 250 g of water. Water boils at 373.15 K and $K_{b} = 0.52$ K kg mol⁻¹.

Try it first, then show the solutionSolution

$m = \frac{6.0/60}{0.250} = 0.40$ mol kg⁻¹ (½ mark).
$Δ T_{b} = K_{b} m = 0.52 \times 0.40 = 0.208$ K (1 mark).
$T_{b} = 373.15 + 0.208 = 373.36$ K (½ mark).

Answer: About 373.36 K (100.21 °C).

Example 3: Molecular formula of sulphur · 3 marks · Short answerHarder board level

1.28 g of sulphur dissolved in 50.0 g of carbon disulphide lowers the freezing point of the solvent by 0.383 K. $K_{f}$ for carbon disulphide is 3.83 K kg mol⁻¹. Calculate the molar mass of sulphur in this solution and deduce its molecular formula. (S = 32)

Try it first, then show the solutionSolution

$M_{2} = \frac{1000 K _{f} w _{2}}{Δ T _{f} w _{1}}$ (1 mark).
$M_{2} = \frac{1000 \times 3.83 \times 1.28}{0.383 \times 50.0} = 256$ g mol⁻¹ (1 mark).
Atoms per molecule $= \frac{256}{32} = 8$ , so the molecular formula is $S_{8}$ (1 mark).

Answer: $M = 256$ g mol⁻¹; sulphur exists as $S_{8}$ molecules.

Example 4: Derive and apply the boiling-point formula · 3 marks · Short answerHarder board level

(a) Starting from $Δ T_{b} = K_{b} m$ , derive an expression for the molar mass of a non-volatile solute in terms of $w_{2}$ g of solute and $w_{1}$ g of solvent. (b) 3.0 g of a non-volatile non-electrolyte dissolved in 60 g of ethanol raises its boiling point by 0.50 K. Find its molar mass. ( $K_{b}$ for ethanol $= 1.20$ K kg mol⁻¹)

Try it first, then show the solutionSolution

(a) Moles of solute $= w_{2} / M_{2}$ ; kilograms of solvent $= w_{1} /1000$ ; so $m = \frac{1000 w _{2}}{M _{2} w _{1}}$ (½ mark).
$Δ T_{b} = \frac{1000 K _{b} w _{2}}{M _{2} w _{1}}$ , hence $M_{2} = \frac{1000 K _{b} w _{2}}{Δ T _{b} w _{1}}$ (1 mark).
(b) $M_{2} = \frac{1000 \times 1.20 \times 3.0}{0.50 \times 60} = 120$ g mol⁻¹ (1½ marks).

Answer: (a) $M_{2} = \frac{1000 K _{b} w _{2}}{Δ T _{b} w _{1}}$ . (b) 120 g mol⁻¹.

Example 5: Kf from enthalpy, then from one effect to the other · 5 marks · Long answerHOTS / competency

(a) The enthalpy of fusion of ice is 6.01 kJ mol⁻¹ at 273.15 K. Calculate $K_{f}$ for water. ( $R = 8.314$ J K⁻¹ mol⁻¹, $M_{H_{2} O} = 18$ g mol⁻¹) [2]
(b) An aqueous solution of glucose freezes at −0.93 °C. At what temperature will it boil at 1 atm? ( $K_{f} = 1.86$ , $K_{b} = 0.52$ K kg mol⁻¹) [2]
(c) Why is $Δ T_{f}$ larger than $Δ T_{b}$ for the same aqueous solution? [1]

Try it first, then show the solutionSolution

(a) $K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H} = \frac{8.314 \times 18 \times ( 273.15 ) ^{2}}{1000 \times 6010}$ (1 mark) $= 1.86$ K kg mol⁻¹ (1 mark).
(b) $m = \frac{Δ T _{f}}{K _{f}} = \frac{0.93}{1.86} = 0.50$ mol kg⁻¹ (1 mark).
$Δ T_{b} = 0.52 \times 0.50 = 0.26$ K, so the solution boils at 100.26 °C (373.41 K) (1 mark).
(c) $K_{f} > K_{b}$ for water: the enthalpy of fusion (6.01 kJ mol⁻¹) is much smaller than the enthalpy of vaporisation, and $K$ varies inversely with that enthalpy (1 mark).

Answer: (a) $K_{f} \approx 1.86$ K kg mol⁻¹. (b) 100.26 °C. (c) Because $K_{f}$ (1.86) is larger than $K_{b}$ (0.52), which follows from water's small enthalpy of fusion compared with its enthalpy of vaporisation.

Where students lose marks in Elevation of Boiling Point and Depression of Freezing Point

The slips examiners see most often, and what to write instead.

Mistake: Wrong formula or wrong $Δ T$
ISC examiners noted wrong boiling-point-elevation formulas and miscalculated $Δ T_{b}$ in a 2025 question.
Wrong working
$Δ T_{b} = 100.70$ (the boiling point itself)
Right working
$Δ T_{b} = 100.70 - 100.00 = 0.70$ K
Do this: Always subtract to get $Δ T$ first, then substitute in $M_{2} = \frac{1000 K _{b} w _{2}}{Δ T _{b} w _{1}}$ .
Usually costs: Up to 2 marks.
Mistake: Using the mass of the solution instead of the solvent
Students add solute and solvent masses for $w_{1}$ .
Wrong working
$w_{1} = 1.5 + 75 = 76.5$ g
Right working
$w_{1} = 75$ g (solvent only)
Do this: Molality is per kilogram of solvent.
Usually costs: The answer mark.
Mistake: Subtracting $Δ T_{b}$ or adding $Δ T_{f}$
The direction of change is reversed in the final step.
Do this: Boiling point rises; freezing point falls. Check the answer against common sense.
Usually costs: ½ to 1 mark.
Mistake: Saying $K_{b}$ depends on the solute
Students confuse $K_{b}$ with $Δ T_{b}$ .
Do this: $K_{b}$ and $K_{f}$ are properties of the solvent; $Δ T$ depends on the amount of solute.
Usually costs: 1 mark in objective questions.

Exam tips

ISC asks for the derivation of the molar mass expression; learn the three lines: definition of molality, substitution, rearrangement.
Keep $Δ T$ in kelvin; a temperature difference in °C has the same number.
When both $K_{f}$ and $K_{b}$ are given, the molality is the bridge between the two effects.
For ionic solutes, write $i$ explicitly in the formula line, even if it is 1 for the solute in question.
Give molar masses with the unit g mol⁻¹ and, where asked, deduce the molecular formula from them.

Practice questions on Elevation of Boiling Point and Depression of Freezing Point

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which 0.10 mol kg⁻¹ aqueous solution has the highest boiling point (assume complete dissociation)?
(a) KCl
(b) urea
(c) BaCl₂
(d) glucose
Hint
Count particles: $i \times m$ .
Answer
(c) BaCl₂, which gives three ions ( $i = 3$ ).
Q2 · 2 marks · Short answerBoard level
Define molal elevation constant and give its unit. Explain why a solution of a non-volatile solute boils at a higher temperature than the pure solvent.
Hint
Link boiling to vapour pressure equal to external pressure.
Answer
$K_{b}$ is the elevation of boiling point produced when 1 mol of a non-volatile non-electrolyte is dissolved in 1 kg of solvent; unit K kg mol⁻¹. The solute lowers the vapour pressure, so the solution must be heated to a higher temperature before its vapour pressure equals atmospheric pressure.
Q3 · 2 marks · Short answerHarder board level
Calculate the freezing point of a solution of 3.42 g of sucrose ( $M = 342$ g mol⁻¹) in 50.0 g of water. ( $K_{f} = 1.86$ K kg mol⁻¹)
Hint
$m = 0.20$ mol kg⁻¹.
Answer
$m = \frac{0.0100}{0.0500} = 0.20$ ; $Δ T_{f} = 1.86 \times 0.20 = 0.372$ K; freezing point −0.372 °C (272.78 K).
Q4 · 3 marks · Short answerHarder board level
1.5 g of a non-electrolyte X in 75 g of water freezes at −0.62 °C. (a) Find the molar mass of X. (b) At what temperature does the same solution boil at 1 atm? ( $K_{f} = 1.86$ , $K_{b} = 0.52$ K kg mol⁻¹)
Hint
Get $M$ first, then the molality.
Answer
(a) $M = \frac{1000 \times 1.86 \times 1.5}{0.62 \times 75} = 60$ g mol⁻¹. (b) $m = \frac{1.5/60}{0.075} = 0.333$ ; $Δ T_{b} = 0.52 \times 0.333 = 0.173$ K; boils at about 100.17 °C.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An aqueous sugar solution boils above 100 °C at 1 atm.
Reason (R): Dissolved sugar lowers the vapour pressure of water, so a higher temperature is needed for the vapour pressure to reach atmospheric pressure.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Boiling point is where vapour pressure equals external pressure.
Answer
(a).
Q6 · 2 marks · Short answerHarder board level
An aqueous solution of a non-electrolyte boils at 100.13 °C at 1 atm. Without finding its molar mass, calculate its freezing point. ( $K_{b} = 0.52$ , $K_{f} = 1.86$ K kg mol⁻¹)
Hint
The molality is common to both equations.
Answer
$m = \frac{0.13}{0.52} = 0.25$ mol kg⁻¹; $Δ T_{f} = 1.86 \times 0.25 = 0.465$ K; freezing point −0.465 °C.
Q7 · 3 marks · Short answerHOTS / competency
Benzene ( $M = 78$ g mol⁻¹) boils at 353.3 K and its enthalpy of vaporisation there is 30.8 kJ mol⁻¹. Calculate its molal elevation constant and comment on how it compares with the tabulated value of 2.53 K kg mol⁻¹. ( $R = 8.314$ J K⁻¹ mol⁻¹)
Hint
$K_{b} = \frac{R M _{1} T _{b}^{2}}{1000 Δ _{v a p} H}$ .
Answer
$K_{b} = \frac{8.314 \times 78 \times ( 353.3 ) ^{2}}{1000 \times 30 800} = 2.63$ K kg mol⁻¹, close to the tabulated 2.53. The small difference comes from the approximations in the derivation and the enthalpy value used.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a polar research team treats sea water as a 0.55 mol kg⁻¹ solution of NaCl that dissociates completely. In winter, ice forms on the sea surface, and the ice is found to be almost pure water.
(i) Estimate the freezing point of this sea water. ( $K_{f} = 1.86$ K kg mol⁻¹) [1]
(ii) Why is the ice almost free of salt? [1]
(iii) If half the water freezes out, what happens to the freezing point of the brine left behind? Give the new value. [1]
Hint
Only pure solvent freezes; the salt stays in the liquid.
Answer
(i) $Δ T_{f} = 2 \times 1.86 \times 0.55 = 2.05$ K, so about −2.0 °C. (ii) When a solution freezes, pure solvent separates as the solid; the solute stays in the liquid. (iii) The same salt is now in half the water, so molality doubles to 1.10 mol kg⁻¹ and the freezing point falls to about −4.1 °C.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define cryoscopic constant.Show answer
The depression of freezing point produced by a 1 molal solution of a non-volatile non-electrolyte; unit K kg mol⁻¹.
2.Write $K_{f}$ in terms of thermodynamic data.Show answer
$K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H}$ (or $\frac{R T _{f}^{2}}{1000 l _{f}}$ ).
3. $K_{b}$ and $K_{f}$ of water?Show answer
0.52 and 1.86 K kg mol⁻¹.
4.Why does a solution freeze below the pure solvent?Show answer
Its vapour pressure is lower, so it becomes equal to the vapour pressure of the solid solvent only at a lower temperature.
5.Formula for molar mass from freezing-point depression?Show answer
$M_{2} = \frac{1000 K _{f} w _{2}}{Δ T _{f} w _{1}}$ .
6.Why are these methods not used for proteins?Show answer
The changes are too small to measure for dilute solutions of very large molecules; osmotic pressure is used instead.

Elevation of Boiling Point and Depression of Freezing Point: frequently asked questions

What is the difference between Kb and ΔTb?

K_b is a constant for a solvent (the elevation for a 1 molal solution). \Delta T_b is the actual rise for a particular solution and equals K_b m.

Why is depression of freezing point a colligative property?

For a dilute solution \Delta T_f = iK_f m, so it depends only on the number of solute particles per kilogram of solvent, not on their identity.

Is the derivation of Kf in the ISC syllabus?

The CISCE syllabus includes the definition and mathematical expression of the cryoscopic and ebullioscopic constants with derivation, so learn how \Delta T = Km leads to the molar mass formula and the relation of K to enthalpy.

Why does salt melt ice on roads?

The salt dissolves in the thin layer of water on the ice, forming a solution whose freezing point is below 0 °C, so the ice melts at temperatures where pure water would stay frozen.

Why is Kf larger than Kb for water?

K is inversely proportional to the enthalpy change of the phase transition, and water's enthalpy of fusion is much smaller than its enthalpy of vaporisation.