ChemistryCBSE Class 12 · Solutions

Elevation of Boiling Point and Depression of Freezing Point

Updated 15 October 2026 · Fact-checked

A non-volatile solute lowers the solvent's vapour pressure, so the solution boils at a higher temperature and freezes at a lower one. For dilute solutions $\Delta T_b = K_b m$ and $\Delta T_f = K_f m$, where $m$ is molality. Measuring either change gives the solute's molar mass: $M_2 = \frac{1000\,K\,w_2}{\Delta T\,w_1}$.

Understand Elevation of Boiling Point and Depression of Freezing Point

Both properties start from one fact you met under Raoult's law: dissolving a non-volatile solute lowers the vapour pressure of the solvent, because solute particles take up part of the surface and fewer solvent molecules escape.

Elevation of boiling point. A liquid boils when its vapour pressure equals the outside pressure (1.013 bar at sea level). Pure water reaches this at 373.15 K. A sugar solution has a lower vapour pressure at every temperature, so it must be heated further before its vapour pressure reaches 1.013 bar. The rise, $Δ T_{b} = T_{b} - T_{b}^{0}$ , is the elevation of boiling point. For dilute solutions it is proportional to the molality: $Δ T_{b} = K_{b} m$ . The constant $K_{b}$ is the molal elevation constant (ebullioscopic constant), the rise produced by a 1 molal solution. It depends only on the solvent; for water it is 0.52 K kg mol⁻¹ and for benzene 2.53 K kg mol⁻¹.

Depression of freezing point. A liquid freezes when the vapour pressure of the liquid equals that of the solid. In a solution, the liquid's vapour pressure is lower, so the two curves meet at a lower temperature. 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 (cryoscopic constant): 1.86 K kg mol⁻¹ for water, 5.12 K kg mol⁻¹ for benzene. NCERT gives the thermodynamic expressions $K_{b} = \frac{R M _{1} T _{b}^{2}}{1000 Δ _{v a p} H}$ and $K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H}$ , which show that the constants come from properties of the solvent alone.

Why molality, not molarity? Molality uses the mass of solvent, which does not change when temperature changes. Molarity uses volume, which expands on heating, so it would change between the freezing and boiling points.

Finding molar mass. Write $m = \frac{w _{2} \times 1000}{M _{2} \times w _{1}}$ with $w_{1}$ in grams of solvent. Substituting, $M_{2} = \frac{1000 K _{b} w _{2}}{Δ T _{b} w _{1}}$ , and the same with $K_{f}$ and $Δ T_{f}$ . This is why these two properties are called colligative: they depend on the number of solute particles, not on what the particles are. Freezing point depression is usually preferred for molar mass because $K_{f}$ is larger than $K_{b}$ for most solvents, so the measured change is bigger and more accurate.

In daily life, ethylene glycol in a car radiator lowers the freezing point of the coolant, and salted or sugared water boils a little above 100 °C. In this topic every solute is a non-electrolyte; salts that split into ions give bigger changes, which the van't Hoff factor handles in the next topic.

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

Elevation of boiling point: $Δ T_{b} = T_{b} - T_{b}^{0} = K_{b} m$; $K_{b}$ (water) = 0.52 K kg mol⁻¹. Valid for dilute solutions of non-volatile non-electrolytes.
Depression of freezing point: $Δ T_{f} = T_{f}^{0} - T_{f} = K_{f} m$; $K_{f}$ (water) = 1.86 K kg mol⁻¹; benzene 5.12 K kg mol⁻¹.
Molality: $m = \frac{n _{2}}{kg of solvent} = \frac{w _{2} \times 1000}{M _{2} \times w _{1}}$; $w_{1}$ is the mass of solvent in grams.
Molar mass of solute: $M_{2} = \frac{1000 \times K \times w _{2}}{Δ T \times w _{1}}$; Use $K_{b}$ with $Δ T_{b}$ , or $K_{f}$ with $Δ T_{f}$ .
Constants from solvent properties: $K_{b} = \frac{R M _{1} T _{b}^{2}}{1000 Δ _{v a p} H}, K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H}$; $M_{1}$ in g mol⁻¹; shows $K_{b}$ and $K_{f}$ depend only on the solvent.
Same solution, both changes: $\frac{Δ T _{f}}{Δ T _{b}} = \frac{K _{f}}{K _{b}}$; Molality cancels: lets you get one change from the other.

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

Almost every numerical here is one of three types: find $Δ T$ (and the new boiling or freezing point), find the molar mass, or find how much solute is needed. All use the same formula rearranged.

1List the data with units: $w_{2}$ (g solute), $M_{2}$ , $w_{1}$ (g solvent), $K_{b}$ or $K_{f}$ , and any temperatures.
2Convert the solvent mass to kilograms, or keep it in grams and use the factor 1000 in the combined formula.
3For $Δ T$ : find molality first, $m = \frac{w _{2} / M _{2}}{w _{1} /1000}$ , then multiply by $K$ .
4For the new temperature: add $Δ T_{b}$ to the solvent's boiling point, or subtract $Δ T_{f}$ from its freezing point. Keep K or °C consistently; a difference of 1 K equals 1 °C.
5For molar mass: use $M_{2} = \frac{1000 K w _{2}}{Δ T w _{1}}$ and give the unit g mol⁻¹.
6For 'how much solute': find the required molality $m = Δ T / K$ , multiply by kg of solvent to get moles, then by $M_{2}$ to get grams.
7Check the size: for water, a 1 m solution changes the boiling point by about 0.5 K and the freezing point by about 1.9 K.

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.

Elevation of boiling point in a syrup

Suppose you are making sugar syrup at home and dissolve 342 g of sucrose ( $C_{12} H_{22} O_{11}$ , 342 g mol⁻¹) in 1 kg of water. Your younger brother insists the syrup will boil at exactly 100 °C 'because it is mostly water'.

Using $K_{b}$ for water = 0.52 K kg mol⁻¹, find the boiling point of the syrup at normal atmospheric pressure, treating it as a dilute ideal solution. Explain in one line why it is higher than 100 °C.

Try it first, then show the solutionSolution

Moles of sucrose $= 342/342 = 1$ mol in 1 kg water, so $m = 1$ mol kg⁻¹.
$Δ T_{b} = K_{b} m = 0.52 \times 1 = 0.52$ K.
Boiling point $= 100 + 0.52 = 100.52$ °C.
Sucrose is non-volatile, so it lowers the vapour pressure; the syrup must be heated more before its vapour pressure reaches atmospheric pressure.

Answer: About 100.5 °C (a rise of 0.52 K), because the dissolved sugar lowers the vapour pressure.

Memory hook: Sugar holds water back from escaping, so the pot must get hotter before it boils.

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.

B goes up, F goes down
Boiling point goes Beyond (up), Freezing point Falls (down); both by $K \times m$ .
Use it when: Decides whether to add or subtract $Δ T$ .
One formula, four unknowns
$Δ T \cdot w_{1} \cdot M_{2} = 1000 \cdot K \cdot w_{2}$ : cover the quantity you want and read off the rest.
Use it when: Molar mass, mass of solute or mass of solvent in one line.
Water's numbers: 0.52 and 1.86
Freezing constant is about 3.6 times the boiling constant for water, so freezing point changes more.
Use it when: Quick sense-check of answers and the ratio method.

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 K depends on · 1 mark · MCQBoard level

The molal depression constant $K_{f}$ of a solvent depends on:
(a) the nature of the solvent
(b) the nature of the solute
(c) the molality of the solution
(d) the mass of solute dissolved

Try it first, then show the solutionSolution

$K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H}$ contains only solvent quantities.
The solute and the amounts affect $Δ T_{f}$ , not $K_{f}$ .

Answer: (a) The nature of the solvent.

Example 2: Boiling point of a glucose solution · 2 marks · Very short answerBoard level

18 g of glucose ( $C_{6} H_{12} O_{6}$ , 180 g mol⁻¹) is dissolved in 500 g of water. Calculate the boiling point of the solution at 1 atm. ( $K_{b}$ for water = 0.52 K kg mol⁻¹; boiling point of water = 373.15 K)

Try it first, then show the solutionSolution

Moles of glucose $= 18/180 = 0.10$ mol; $m = \frac{0.10}{0.500} = 0.20$ mol kg⁻¹ (½ mark).
$Δ T_{b} = 0.52 \times 0.20 = 0.104$ K (1 mark).
$T_{b} = 373.15 + 0.104 = 373.25$ K (½ mark).

Answer: 373.25 K (a rise of 0.104 K).

Example 3: Molar mass from freezing point depression · 3 marks · Short answerHarder board level

1.25 g of a non-electrolyte solute dissolved in 40 g of benzene lowers its freezing point by 0.50 K. $K_{f}$ for benzene is 5.12 K kg mol⁻¹. Find the molar mass of the solute.

Try it first, then show the solutionSolution

Formula: $M_{2} = \frac{1000 \times K _{f} \times w _{2}}{Δ T _{f} \times w _{1}}$ (1 mark).
Substitute: $M_{2} = \frac{1000 \times 5.12 \times 1.25}{0.50 \times 40}$ (1 mark).
$M_{2} = \frac{6400}{20} = 320$ g mol⁻¹ (1 mark).

Answer: 320 g mol⁻¹.

Example 4: How much antifreeze? · 3 marks · Short answerHarder board level

How much ethylene glycol ( $C_{2} H_{6} O_{2}$ , 62 g mol⁻¹) must be added to 5.0 kg of water so that it does not freeze above −6.0 °C? What will be the boiling point of this coolant at 1 atm? ( $K_{f}$ = 1.86 K kg mol⁻¹, $K_{b}$ = 0.52 K kg mol⁻¹ for water)

Try it first, then show the solutionSolution

Required $Δ T_{f} = 6.0$ K, so $m = \frac{6.0}{1.86} = 3.226$ mol kg⁻¹ (1 mark).
Moles for 5.0 kg $= 3.226 \times 5.0 = 16.13$ mol; mass $= 16.13 \times 62 = 1000$ g $= 1.0$ kg (1 mark).
$Δ T_{b} = 0.52 \times 3.226 = 1.68$ K, so the coolant boils at about 101.7 °C (1 mark).

Answer: About 1.0 kg of ethylene glycol; boiling point about 101.7 °C.

Example 5: One change from the other · 3 marks · Short answerHOTS / competency

An aqueous solution of a non-volatile non-electrolyte boils at 100.26 °C at 1 atm. (a) Without calculating the molar mass, find its freezing point. (b) Why is freezing point depression usually preferred over boiling point elevation for finding molar mass? ( $K_{b}$ = 0.52, $K_{f}$ = 1.86 K kg mol⁻¹)

Try it first, then show the solutionSolution

$Δ T_{b} = 0.26$ K, so $m = \frac{0.26}{0.52} = 0.50$ mol kg⁻¹ (1 mark).
$Δ T_{f} = 1.86 \times 0.50 = 0.93$ K, so the freezing point is $- 0.93$ °C. Equivalently $Δ T_{f} = Δ T_{b} \times \frac{K _{f}}{K _{b}}$ (1 mark).
(b) $K_{f}$ is larger than $K_{b}$ for water (and most solvents), so the same solution gives a larger, more precisely measurable temperature change; the experiment also avoids heating, which could decompose some solutes (1 mark).

Answer: (a) −0.93 °C. (b) $K_{f} > K_{b}$ gives a larger, more accurate reading, and no heating is needed.

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: Using the mass of solution instead of solvent
Students divide by the total mass when the question gives a percentage.
Wrong working
4% sucrose: $m = \frac{4/342}{0.100}$
Right working
$m = \frac{4/342}{0.096}$ (100 g solution has 96 g water)
Do this: Molality is per kilogram of solvent; subtract the solute mass first.
Usually costs: ½ to 1 mark.
Mistake: Reporting $Δ T$ as the new boiling or freezing point
The question asks for the freezing point but the student stops at the depression.
Wrong working
Freezing point = 0.372 °C
Right working
$Δ T_{f} = 0.372$ K, so freezing point $= 0 - 0.372 = - 0.372$ °C
Do this: Finish with one more line: add for boiling, subtract for freezing.
Usually costs: ½ mark.
Mistake: Forgetting the factor 1000 or the units of $K$
Mixing grams and kilograms in the combined formula.
Wrong working
$M_{2} = \frac{5.12 \times 1.25}{0.50 \times 40} = 0.32$
Right working
$M_{2} = \frac{1000 \times 5.12 \times 1.25}{0.50 \times 40} = 320$ g mol⁻¹
Do this: If $w_{1}$ is in grams, the 1000 must be there. Check that the answer is a sensible molar mass.
Usually costs: 1 mark (final answer).
Mistake: Applying $Δ T = K m$ to salts without $i$
Students treat NaCl or $C a C l_{2}$ like glucose.
Do this: This formula is for non-electrolytes. For electrolytes use $Δ T = i K m$ (van't Hoff factor).
Usually costs: 1-2 marks in a 3-mark numerical.

Exam tips

Write the formula first, then substitute with units; the formula line carries a mark even if arithmetic slips.
State $K_{b}$ and $K_{f}$ with units K kg mol⁻¹ when you quote them.
If the question gives temperatures in °C, a change of 1 °C equals 1 K, so $Δ T$ needs no conversion.
For 'explain' questions, always start from lowering of vapour pressure; for freezing, say the solid and liquid phases have equal vapour pressure at the freezing point.
Round only at the end; carry three or four significant figures through the working.

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
The molal elevation constant is the elevation in boiling point produced when the solution has a molality of:
(a) 0.1 m (b) 1 m (c) 10 m (d) 1 M
Hint
Put $m = 1$ in $Δ T_{b} = K_{b} m$ .
Answer
(b) 1 molal (1 mol of solute in 1 kg of solvent).
Q2 · 2 marks · Very short answerBoard level
Why does a solution of a non-volatile solute boil at a higher temperature than the pure solvent? Why is molality used instead of molarity in this formula?
Hint
Link boiling to vapour pressure, and molarity to volume.
Answer
The solute lowers the vapour pressure, so a higher temperature is needed to make it equal to atmospheric pressure. Molality depends on masses, which do not change with temperature; molarity depends on volume, which does.
Q3 · 2 marks · Very short answerBoard level
Calculate the freezing point of a solution of 3.0 g of urea ( $N H_{2} C O N H_{2}$ , 60 g mol⁻¹) in 250 g of water. ( $K_{f}$ = 1.86 K kg mol⁻¹)
Hint
$m = 0.05/0.25$ .
Answer
$m = 0.20$ mol kg⁻¹; $Δ T_{f} = 0.372$ K; freezing point $= - 0.372$ °C (272.78 K).
Q4 · 2 marks · Very short answerHarder board level
When 2.40 g of a non-volatile solute is dissolved in 75 g of benzene, its boiling point rises by 0.81 K. Calculate the molar mass of the solute. ( $K_{b}$ for benzene = 2.53 K kg mol⁻¹)
Hint
Use $M_{2} = \frac{1000 K _{b} w _{2}}{Δ T _{b} w _{1}}$ .
Answer
$M_{2} = \frac{1000 \times 2.53 \times 2.40}{0.81 \times 75} = 99.95 \approx 100$ g mol⁻¹.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The value of $K_{f}$ for water is the same whether glucose or urea is dissolved in it.
Reason (R): The molal depression constant depends only on the properties of the solvent.
(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
Look at what appears in the expression for $K_{f}$ .
Answer
(a). $K_{f} = \frac{R M _{1} T _{f}^{2}}{1000 Δ _{f u s} H}$ has only solvent terms.
Q6 · 3 marks · Short answerHarder board level
31 g of ethylene glycol (62 g mol⁻¹) is mixed with 400 g of water. Calculate (a) the freezing point and (b) the boiling point of the solution at 1 atm. ( $K_{f}$ = 1.86, $K_{b}$ = 0.52 K kg mol⁻¹)
Hint
Moles = 0.5; molality = 1.25 mol kg⁻¹.
Answer
$m = 1.25$ mol kg⁻¹. (a) $Δ T_{f} = 2.325$ K, freezing point $\approx - 2.33$ °C (270.83 K). (b) $Δ T_{b} = 0.65$ K, boiling point $\approx 100.65$ °C (373.80 K).
Q7 · 3 marks · Short answerHOTS / competency
A 4% (by mass) aqueous solution of sucrose (342 g mol⁻¹) freezes at −0.227 °C. Without using the value of $K_{f}$ , calculate the freezing point of a 4% (by mass) aqueous solution of glucose (180 g mol⁻¹).
Hint
Both have 4 g solute in 96 g water, so $Δ T_{f}$ is inversely proportional to molar mass.
Answer
$\frac{Δ T _{f} ( glucose )}{Δ T _{f} ( sucrose )} = \frac{342}{180}$ , so $Δ T_{f} = 0.227 \times 1.9 = 0.431$ K. Freezing point $\approx - 0.43$ °C.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a biochemistry student wants to find the molar mass of a newly isolated protein, expected to be about 50 000 g mol⁻¹. She dissolves 1.0 g of it in 100 g of water and plans to measure the freezing point depression with a thermometer that reads to 0.01 K.
(a) Calculate the expected $Δ T_{f}$ . ( $K_{f}$ = 1.86 K kg mol⁻¹) (2 marks)
(b) Can her thermometer detect it? (1 mark)
(c) Which colligative property should she use instead, and why? (1 mark)
Hint
Moles of protein are tiny: $2 \times 1 0^{- 5}$ mol.
Answer
(a) $m = \frac{2.0 \times 1 0 ^{- 5}}{0.100} = 2.0 \times 1 0^{- 4}$ mol kg⁻¹; $Δ T_{f} = 1.86 \times 2.0 \times 1 0^{- 4} = 3.7 \times 1 0^{- 4}$ K. (b) No; it is far below 0.01 K. (c) Osmotic pressure: it gives a measurable value even for very dilute solutions and is measured at room temperature, so the protein is not damaged.

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.Formula for elevation of boiling point?Show answer
$Δ T_{b} = K_{b} m$ .
2. $K_{f}$ of water?Show answer
1.86 K kg mol⁻¹.
3.At the freezing point, which two vapour pressures are equal?Show answer
Those of the solid solvent and the liquid (solution or solvent).
4.Molar mass formula from $Δ T_{f}$ ?Show answer
$M_{2} = \frac{1000 K _{f} w _{2}}{Δ T _{f} w _{1}}$ .
5.Another name for $K_{b}$ ?Show answer
Ebullioscopic constant (molal elevation constant).
6.Why add ethylene glycol to a radiator in cold places?Show answer
It depresses the freezing point of the water so the coolant does not freeze.

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

What is elevation of boiling point in Class 12 Chemistry?

It is the rise in boiling point of a solvent when a non-volatile solute is dissolved in it. For a dilute solution \Delta T_b = K_b m, where K_b is the molal elevation constant and m the molality.

Why is freezing point lowered by adding a solute?

The solute lowers the vapour pressure of the liquid, so the liquid and solid reach equal vapour pressure only at a lower temperature. That lower temperature is the new freezing point.

What are the units of $K_b$ and $K_f$?

K kg mol⁻¹ (kelvin kilogram per mole).

Why are these called colligative properties?

Because, for dilute solutions, they depend only on the number of solute particles per kilogram of solvent, not on the identity of the solute.

Do I need to convert °C to K for $\Delta T$?

No. A temperature difference of 1 °C equals 1 K. Convert only when the question asks for the final temperature in kelvin.