Calculate the depression in freezing point produced by dissolving $1.00\ ext{g}$ of glucose ($M = 180\ ext{g mol}^{-1}$) in $50\ ext{g}$ of water. ($K_f$ of water $= 1.86\ ext{K kg mol}^{-1}$.)

Question

Calculate the depression in freezing point produced by dissolving $1.00\ 	ext{g}$ of glucose ($M = 180\ 	ext{g mol}^{-1}$) in $50\ 	ext{g}$ of water. ($K_f$ of water $= 1.86\ 	ext{K kg mol}^{-1}$.)

Accepted Answer

Concept understanding — Depression of Freezing Point

Depression of Freezing Point

Imagine a cold winter morning. You see water on the road turning to ice at 0°C. But if you sprinkle salt on that ice, it melts — even though the temperature is still below zero. That’s the same phenomenon that keeps roads safe in snowy countries and makes ice cream freeze in a churn. The salt lowers the freezing point of water.

That is the core idea: when you dissolve a non-volatile solute (like salt, sugar, or urea) in a solvent (like water), the freezing point of the solution becomes lower than that of the pure solvent. This drop is called the depression of freezing point, denoted by $\Delta T_f$ .

Why does this happen? The intuition

In a pure liquid, molecules at the surface escape into the solid (freeze) when the temperature is low enough — the solid and liquid are in equilibrium at the freezing point. Now add a solute. The solute particles sit between solvent molecules, getting in the way. For the solvent to freeze, its molecules must arrange themselves into an orderly crystal lattice. The solute particles disrupt this order — they make it harder for the solvent to solidify.

Think of it like trying to pack a suitcase full of neatly stacked blocks. If you throw in a few marbles, the blocks can’t settle as tightly. You’d need to cool the system further (lower the temperature) to force the blocks into place. That extra cooling is the depression.

Note

The solute must be non-volatile (it doesn’t evaporate) and non-electrolyte (it doesn’t break into ions) for the simplest formula to work. If the solute dissociates (like NaCl → Na⁺ + Cl⁻), the effect is larger — but that’s a refinement you’ll meet later.

The precise statement

For a dilute solution, the depression in freezing point is directly proportional to the molality of the solution (moles of solute per kilogram of solvent).

\Delta T_f = K_f \cdot m

Where:

$\Delta T_f = T_f^\circ - T_f$ (pure solvent freezing point minus solution freezing point)
$K_f$ = cryoscopic constant or molal freezing point depression constant — a property of the solvent alone (units: K kg mol⁻¹)
$m$ = molality of the solution

\Delta T_f = K_f \cdot m

Each solvent has its own $K_f$ . For water, $K_f = 1.86 \text{ K kg mol}^{-1}$ . That means: a 1 molal aqueous solution freezes at $-1.86^\circ\text{C}$ (instead of $0^\circ\text{C}$ ).

How it helps find molar mass

If you dissolve a known mass $w_2$ of an unknown solute in a known mass $w_1$ of solvent, measure the freezing point depression $\Delta T_f$ , you can calculate the molar mass $M_2$ of the solute.

Start from the definition of molality:

m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{w_2 / M_2}{w_1 / 1000}

Substitute into $\Delta T_f = K_f \cdot m$ :

\Delta T_f = K_f \cdot \frac{w_2 \times 1000}{M_2 \times w_1}

Rearrange for $M_2$ :

M_2 = \frac{K_f \cdot w_2 \cdot 1000}{\Delta T_f \cdot w_1}

Tip

This is the most common exam formula. Remember: $w_1$ is in grams, $w_2$ in grams, and the factor 1000 converts grams of solvent to kilograms.

A quick example

You dissolve 5.00 g of a non-electrolyte in 100 g of water. The freezing point drops to $-0.93^\circ\text{C}$ . Find the molar mass.

Given: $K_f = 1.86$ , $\Delta T_f = 0.93$ , $w_2 = 5.00$ , $w_1 = 100$ . …

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Moles of glucose $= 1.00/180 \approx 0.005556\ \text{mol}$ . Molality $m = 0.005556\ \text{mol} / 0.050\ \text{kg} \approx 0.1111\ \text{mol kg}^{-1}$ . By $\Delta T_f = K_f m$ , $\Delta T_f = 1.86 \times 0.1111 \approx 0.2067\ \text …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. In cold countries ethylene glycol is added to water in radiators of cars during winter. This results in:$
(a) Lowering of freezing point
(b) Reducing viscosity
(c) Reducing specific heat
(d) Making water better conductor of electricity
›Reveal solutionSolution

Ethylene glycol is added as an antifreeze; it lowers the freezing point of water so the coolant doesn't freeze in cold winters.

Adding a non-volatile solute like ethylene glycol to water is a classic application of the colligative property of depression in freezing point, $\Delta T_f = K_f \times m$ . This lowers the freezing point of the radiator coolant well below $0°C$ , preventing it from freezing and expanding (which could crack the engine block) in cold climates. The s …

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CBSE 2026Set SEM31 markMCQ
$Q. Which of the following solutions has the lowest freezing point?$
(a) 1 (m) sucrose
(b) 1 (m) KCl
(c) 1 (m) MgCl2
(d) 1 (m) glucose
›Reveal solutionSolution
Freezing-point depression is a colligative property, proportional to i x m. MgCl2 has the highest van't Hoff factor (i = 3), so lowest freezing point. Correct option (c).

Depression in freezing point: dTf = i x Kf x m. At equal molality (1 m) and same Kf, the solution giving the most dissolved particles depresses the freezing point most.

Sucrose (i = 1) -> 1 particle

Glucose (i = 1) -> 1 particle

KCl -> K+ + Cl- (i = 2) -> 2 particles …
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CBSE 2025Set ANNUAL1 mark
$Q. Explain the following — Molal Depression Constant$
›Reveal solutionSolution

Kf (cryoscopic constant) links depression in freezing point directly to molality.

The molal depression constant (freezing point depression constant, or cryoscopic constant, $K_f$ ) is defined as the depression in freezing point produced when 1 mole of a solute is dissolved in 1 kg (1000 g) of a solvent.

It relates the observed depression in freezing point ( $\Delta T_f$ ) to molality ( $m$ ) via:

$\Delta T_f = K_f \times m$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Equimolar aqueous solution with the lowest freezing point is:$
(a) Potassium sulphate
(b) Sodium Chloride
(c) Glucose
(d) Urea
›Reveal solutionSolution
For equimolar solutions, freezing-point depression ΔTf = i·Kf·m increases with the van't Hoff factor i (number of particles produced per formula unit); K2SO4 gives i≈3, the highest of the four, so it has the lowest (most depressed) freezing point.

ΔTf = i × Kf × m, where m is molality (same for all four since they are equimolar) and Kf is the same solvent constant. So the compound with the largest i shows the largest ΔTf, i.e. the lowest freezing point.

K2SO4 → 2K+ + SO4^2-: i ≈ 3

NaCl → Na+ + Cl-: i ≈ 2

Glucose: a non-electrolyte, does not dissociate: i = 1 …
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CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: The unit of freezing point depression constant (Kf) is ________.$
›Reveal solutionSolution

The freezing point depression constant Kf has SI unit K kg mol^-1, obtained from the defining relation ΔTf = Kf x m.

The depression in freezing point is related to molality by:

ΔTf = Kf x m

where ΔTf is in kelvin (K) and molality m is in mol/kg (mol kg^-1). Rearranging:

Kf = ΔTf / m

Units: Kf = K / (mol kg^-1) = K kg mol^-1

…

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CBSE 2024Set 56/3/11 markMCQ
$Q. For the following question, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Addition of ethylene glycol to water lowers its freezing point. Reason (R) : Ethylene glycol is insoluble in water due to lack of its ability to form hydrogen bonds with water molecules.$
›Reveal solutionSolution

Ethylene glycol does lower water's freezing point (colligative property), but the reason is wrong: ethylene glycol is highly soluble in water because it forms hydrogen bonds. Assertion true, Reason false → (C).

The question tests two ideas: colligative properties (freezing-point depression) and the solubility behaviour of ethylene glycol in water. Let's unpack each statement.

Understanding the Assertion

Ethylene glycol, $\text{HOCH}_2\text{CH}_2\text{OH}$ , is a common antifreeze. When you dissolve a non-volatile solute in a solvent, the freezing point of the solution drops below that of the pure solvent. This is a colligative property — it depends on the number of solute particles, not their identity.

The freezing-point depression is given by
$\Delta T_f = K_f \cdot m$
where $K_f$ is the cryoscopic constant of the solvent (water) and $m$ is the molality of the solution. Adding ethylene glycol introduces solute particles that disrupt the orderly arrangement of water molecules trying to form ice, so the solution must be cooled further to freeze.

Assertion (A) is true: ethylene glycol lowers the freezing point of water.

Examining the Reason

Now the Reason claims ethylene glycol is insoluble in water because it cannot form hydrogen bonds with water.

Look at the structure of ethylene glycol: it has two hydroxyl groups ( $-\text{OH}$ ). Each $-\text{OH}$ is polar and can act as both a hydrogen-bond donor (the H) and acceptor (the O lone pairs). Water molecules, $\text{H}_2\text{O}$ , are also excellent hydrogen-bond donors and acceptors.

When ethylene glycol dissolves in water, extensive hydrogen bonding occurs between the $-\text{OH}$ groups of ethylene glycol and water molecules. This is why ethylene glycol is miscible with water in all proportions — it is highly soluble, not insoluble. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following is a colligative property?$
(a) Relative lowering of fluid pressure
(b) Decrease in boiling point
(c) Decrease in freezing point
(d) Change in volume after mixing
›Reveal solutionSolution
Colligative properties depend only on the number of solute particles. Judging the options exactly as printed, (c) 'Decrease in freezing point' (depression of freezing point) is the genuine colligative property; option (a) as printed reads 'fluid pressure', which is not a standard colligative term.

Checking each option as printed:

(a) Relative lowering of fluid pressure - the standard colligative property is relative lowering of vapour pressure. As actually printed (both the English and Hindi versions use 'fluid pressure' / 'taral daab') this is not a recognised colligative property, so it cannot be taken as the intended correct term.

(b) Decrease in boiling point - wrong direction: a non-volatile solute raises (elevates) the boiling point, it does not decrease it. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Assertion [A] : When NaCl is added to water, a depression in freezing point is observed. Reason [R] : The lowering of vapour pressure of a solution causes depression in the freezing point.$
(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.
›Reveal solutionSolution

Adding NaCl lowers water's vapour pressure, and this lowering of vapour pressure is exactly what causes the solution's freezing point to fall below that of pure water.

[A] Adding a non-volatile solute like NaCl to water does depress its freezing point — this is a well-established colligative property, and is TRUE.

…

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CBSE 2022Set E1 markMCQ
$Q. Which of the following will have maximum depression in freezing point ?$
(a) K2SO4
(b) NaCl
(c) Urea
(d) Glucose
›Reveal solutionSolution
Depression in freezing point (a colligative property) increases with the number of particles: van't Hoff factor i. K2SO4 (i = 3) gives the most.

Depression in freezing point: delta Tf = i x Kf x m.

For equal molality, the electrolyte producing the most particles wins:

K2SO4 -> 2K+ + SO4^2- , i = 3

NaCl -> Na+ + Cl- , i = 2

Urea, non-electrolyte, i = 1 …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. Equimolal aqueous solutions of NaCl and KCl are prepared. If the freezing point of NaCl is -2 degree C, the freezing point of KCl solution is expected to be:$
(a) -1 degree C
(b) -2 degree C
(c) 0 degree C
(d) -4 degree C
›Reveal solutionSolution

Freezing point depression depends on the number of particles a solute produces in solution (colligative property). NaCl and KCl both dissociate into 2 ions per formula unit, so equimolal solutions of the two give the same depression.

Freezing point depression is given by ΔTf = i × Kf × m, where i is the van't Hoff factor (number of particles produced per formula unit), Kf is the molal freezing point depression constant of water, and m is the molality.

NaCl → Na+ + Cl- gives i = 2.

KCl → K+ + Cl- gives i = 2.

…

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CBSE 2021Set A1 markMCQ
$Q. In comparison to 0.01 M solution of glucose, the depression in freezing point of 0.01 M MgCl2 solution is$
(a) same
(b) about twice
(c) about three times
(d) about six times
›Reveal solutionSolution
Depression in freezing point is a colligative property proportional to the van't Hoff factor i; MgCl2 gives i = 3 versus i = 1 for glucose.

Depression in freezing point is given by delta Tf = i x Kf x m, where i is the van't Hoff factor (number of particles produced per formula unit) and m is the molality (here comparable at equal molarity for dilute aqueous solutions).

Glucose is a non-electrolyte; it does not dissociate, so i = 1. …
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CBSE 2020Set ANNUAL1 markMCQ
$Q. Which of the following 0.1 M aqueous solutions is likely to have the highest depression in freezing point?$
(a) Na2SO4
(b) NaCl
(c) Glucose
(d) Na3PO4
›Reveal solutionSolution
Depression in freezing point, $\Delta T_f = i K_f m$ , so at the same molal concentration the electrolyte that dissociates into the most ions (highest $i$ ) gives the largest depression.

For a fixed concentration (0.1 M) and the same solvent (same $K_f$ ), $\Delta T_f \propto i$ , the van't Hoff factor, which equals the number of particles (ions) each formula unit dissociates into (assuming complete dissociation):

Glucose: non-electrolyte, $i = 1$

NaCl $\to$ Na $^+$ + Cl $^-$ : $i = 2$

Na $_2$ SO $_4$ $\to$ 2Na $^+$ + SO $_4^{2-}$ : $i = 3$

Na $_3$ PO $_4$ $\to$ 3Na $^+$ + PO $_4^{3-}$ : $i = 4$ …
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Q.Calculate the depression in freezing point produced by dissolving 1.00 g1.00\ \text{g}1.00 g of glucose (M=180 g mol−1M = 180\ \text{g mol}^{-1}M=180 g mol−1) in 50 g50\ \text{g}50 g of water. (KfK_fKf​ of water =1.86 K kg mol−1= 1.86\ \text{K kg mol}^{-1}=1.86 K kg mol−1.)

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Q.Calculate the depression in freezing point produced by dissolving $1.00\ \text{g}$ of glucose ( $M = 180\ \text{g mol}^{-1}$ ) in $50\ \text{g}$ of water. ( $K_f$ of water $= 1.86\ \text{K kg mol}^{-1}$ .)

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