How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon inv

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Q. How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.

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Concept understanding — Colligative Properties Association

Colligative Properties: The Intuition First

Imagine you're at a party. The room is full of people dancing — that's your solvent molecules, moving freely. Now, someone brings in a few heavy, slow-moving guests who just stand in one spot — those are your solute particles. They don't dance, they don't interact much, they just take up space.

What happens? The dancers now have less room to move. They bump into the standing guests more often. The whole atmosphere changes — the dancers can't move as freely, they can't escape the room as easily, and the overall "energy" of the party shifts.

That's the core idea of colligative properties. When you add a non-volatile solute (like salt) to a solvent (like water), the solute particles don't do anything special — they just exist in the solution. But their mere presence changes four measurable properties of the solvent:

Vapour pressure decreases
Boiling point increases
Freezing point decreases
Osmotic pressure increases

The key insight: these changes depend only on the number of solute particles, not on what kind of particles they are. One molecule of sugar and one ion of salt (if they don't dissociate) affect these properties identically — provided they're the same number of particles.

Note

This is why "colligative" comes from the Latin colligatus meaning "bound together" — the properties are bound to the quantity of solute, not its identity.

The Precise Statement

Colligative properties are properties of a solution that depend solely on the ratio of the number of solute particles to the number of solvent molecules in a given solution, and not on the chemical nature of the solute.

Mathematically, for a dilute solution of a non-volatile, non-electrolyte solute:

\Delta P = P^0 \cdot x_2

Where:

$\Delta P$ = lowering of vapour pressure
$P^0$ = vapour pressure of pure solvent
$x_2$ = mole fraction of solute
$\Delta T_b$ = elevation in boiling point
$K_b$ = ebullioscopic constant (depends only on solvent)
$m$ = molality of solution
$\Delta T_f$ = depression in freezing point
$K_f$ = cryoscopic constant (depends only on solvent)
$\Pi$ = osmotic pressure
$i$ = van't Hoff factor (accounts for dissociation/association)
$M$ = molarity
$R$ = gas constant
$T$ = absolute temperature

The Crucial Distinction: Association vs. Dissociation

Now, here's where the association part comes in — and it's the twist that catches most students.

The formulas above assume the solute particles remain as individual, independent particles. But in reality:

Dissociation: Some solutes break apart into smaller particles (e.g., NaCl → Na⁺ + Cl⁻). This increases the number of particles, so the colligative effect is larger than expected.
Association: Some solutes clump together into larger particles (e.g., acetic acid in benzene forms dimers: 2 CH₃COOH → (CH₃COOH)₂). This decreases the number of particles, so the colligative effect is smaller than expected.

Watch out

A common mistake: students think "association" means the solute interacts with the solvent. No — association means solute particles bind to each other, reducing the effective particle count. Solvent-solute interactions affect non-colligative properties like solubility.

The van't Hoff Factor $i$

To account for these real-world effects, we introduce the van't Hoff factor:

i = \frac{\text{Actual number of particles in solution}}{\text{Number of formula units dissolved}}

For a non-electrolyte that doesn't associate or dissociate: $i = 1$

For dissociation (e.g., NaCl): $i > 1$ (ideally 2 for NaCl)

For association (e.g., acetic acid dimerizing): $i < 1$

The corrected formulas become:

\Delta T_b = i \cdot K_b \cdot m

\Delta T_f = i \cdot K_f \cdot m

\Pi = i \cdot MRT

A Concrete Example

Consider acetic acid (CH₃COOH) dissolved in benzene. In benzene, acetic acid molecules form hydrogen-bonded dimers:

2 \text{CH}_3\text{COOH} \rightleftharpoons (\text{CH}_3\text{COOH})_2

If you dissolve 1 mole of acetic acid, you might end up with only 0.6 moles of particles (0.4 moles of dimers + 0.2 moles of monomers). So $i = 0.6$ . …

Why this formula?

Colligative Properties & Association: Why the Formula Holds

Let's build this from first principles — understanding why association changes colligative properties, not just memorizing the formula.

The Core Idea: What Are Colligative Properties?

Colligative properties depend only on the number of solute particles in solution, not on their chemical identity. The four key ones are:

Vapor pressure lowering
Boiling point elevation
Freezing point depression
Osmotic pressure

When a solute associates (e.g., two molecules dimerize), the effective number of particles decreases. This is the entire reason the formula changes.

The van't Hoff Factor: The Bridge

We define the van't Hoff factor $i$ as:

i = \frac{\text{actual number of particles in solution}}{\text{number of formula units dissolved}}

For a non-electrolyte that does not associate, $i = 1$ .

For association, $i < 1$ .

Example: Dimerization of Benzoic Acid in Benzene

Benzoic acid ( $C_6H_5COOH$ ) forms dimers in benzene:

2 \, C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2

If we dissolve $n$ moles of monomer, but only $n/2$ moles of dimer exist, then:

i = \frac{n/2}{n} = 0.5

Deriving the Modified Formula

Step 1: Start with the Normal Colligative Formula

For freezing point depression (the most common exam case):

\Delta T_f = K_f \cdot m

where $m$ is the molality of the solute (moles per kg solvent).

Step 2: Replace $m$ with Effective Molality

Because only the number of particles matters, we replace $m$ with $i \cdot m$ :

\Delta T_f = K_f \cdot (i \cdot m)

This is the general formula for any colligative property when association or dissociation occurs.

Step 3: Express $i$ in Terms of Degree of Association

Let:

$\alpha$ = degree of association (fraction of molecules that associate)
$n$ = number of molecules that combine to form one associated particle (e.g., $n=2$ for dimerization)

For a dimerization ( $n=2$ ):

Initially: 1 mole of monomer
After association: $(1 - \alpha)$ moles remain as monomer, and $\alpha/2$ moles of dimer form
Total particles = $(1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$

Thus:

i = \frac{1 - \frac{\alpha}{2}}{1} = 1 - \frac{\alpha}{2}

General formula for association of $n$ molecules:

i = 1 - \alpha + \frac{\alpha}{n} = 1 - \alpha\left(1 - \frac{1}{n}\right)

Why This Makes Physical Sense …

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Sprinkling salt on snow-covered roads lowers the freezing point of water through a phenomenon called freezing point depression, causing the ice to melt even at temperatures below $0^\circ\text{C}$ .

When salt is sprinkled on snow or ice, it helps clear the roads by exploiting a fundamental property of solutions known as freezing point depression, which is a type of colligative property. Colligative properties are those properties of solutions that depend solely on the number of solute particles in a given amount of solvent, and not on the identity of the solute particles themselves.

Here's how the phenomenon works:

The Nature of Ice and Water: Pure water freezes at $0^\circ\text{C}$ ( $32^\circ\text{F}$ ) at standard atmospheric pressure. Below this temperature, water exists as solid ice. However, even at temperatures slightly below $0^\circ\text{C}$ , there's often a very thin, quasi-liquid layer of water molecules on the surface of ice.
Introducing the Solute (Salt): When salt (commonly sodium chloride, $\text{NaCl}$ , or calcium chloride, $\text{CaCl}_2$ ) is sprinkled on snow or ice, it dissolves in this thin layer of liquid water. This forms an aqueous salt solution.
Disruption of Crystal Formation: For water to freeze and form ice crystals, its molecules must arrange themselves into a highly ordered, crystalline lattice structure. When salt dissolves in water, it dissociates into ions (e.g., $\text{NaCl}$ dissociates into $\text{Na}^+$ and $\text{Cl}^-$ ions). These dissolved salt ions act as impurities within the water.
Lowering the Freezing Point: The presence of these solute ions interferes with the ability of water molecules to come together and form the stable, ordered crystal structure required for freezing. The water molecules are more attracted to the dissolved ions, and their movement is hindered, making it more difficult for them to settle into a solid lattice. Consequently, a lower temperature is required to slow down the water molecules sufficiently for them to overcome the disruptive effect of the solute particles and form ice.
Melting the Ice: If the ambient temperature is, for example, $-5^\circ\text{C}$ , pure water would normally be frozen solid. However, a salt solution might have a freezing point of, say, $-10^\circ\text{C}$ . Since the ambient temperature ( $-5^\circ\text{C}$ ) is above the new freezing point of the salt solution ( $-10^\circ\text{C}$ ), the ice will melt and remain in its liquid state. This process continues as long as the ambient temperature is above the freezing point of the salt solution formed.
The Role of Concentration: The extent of freezing point depression is directly proportional to the concentration of solute particles in the solution. More salt (up to its solubility limit) means more dissolved ions, leading to a greater depression of the freezing point. …

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Here are the common mistakes students make on this question, along with how to avoid each one.

Mistake 1: Confusing the Phenomenon with "Boiling Point Elevation"

The Mistake:

Students often write that salt increases the melting point of ice, or they incorrectly state that the phenomenon is "boiling point elevation" (which applies to liquids turning into gas).

Why it happens:

Students memorize "colligative properties" as a list (boiling point elevation, freezing point depression, etc.) and pick the wrong one because they see "salt + water" and think of cooking (where salt raises the boiling point of water).

How to Avoid:

Anchor the concept to the phase change. The road has solid ice (snow). We want it to become liquid water. This is melting.
Memorize the specific rule: Adding a non-volatile solute (salt) to a solvent (water) lowers the freezing point. This is called Freezing Point Depression.
Use a simple memory trick: "Salt depresses the ice — it makes it sad and weak, so it melts at a lower temperature."

Mistake 2: Forgetting the "Why" — The Colligative Nature

The Mistake:

Students simply state "salt lowers the freezing point" without explaining why it happens. They treat it as a magic fact.

Why it happens:

They focus on the application (clearing roads) and skip the mechanism (the physical chemistry).

How to Avoid:

Explain the particle interference. Pure water freezes at $0^\circ\text{C}$ because water molecules arrange into a crystal lattice. Salt (NaCl) dissolves into $\text{Na}^+$ and $\text{Cl}^-$ ions. These ions get in the way of the water molecules trying to form the ice lattice.
Use the "party guest" analogy: Imagine water molecules trying to hold hands to form a solid (ice). Salt ions are like clumsy guests who bump into them, making it harder for them to hold hands. You need to make it colder (lower temperature) to force them to hold hands despite the interference.
Key formula (conceptual): $\Delta T_f = i \cdot K_f \cdot m$ . The "i" (van't Hoff factor) is 2 for NaCl (one Na⁺, one Cl⁻), which is why it's effective.

Mistake 3: Ignoring the "Hilly Areas" Context

The Mistake:

Students give a generic answer about salt melting ice, but fail to connect it to the specific conditions of hilly areas (sub-zero temperatures).

Why it happens:

They treat the question as a standard "salt on ice" problem without reading the location-specific detail.

How to Avoid:

State the temperature condition explicitly. In hilly areas, the ambient temperature is often below $0^\circ\text{C}$ (e.g., $-5^\circ\text{C}$ or $-10^\circ\text{C}$ ). Pure ice will not melt at these temperatures.
Explain the shift: Salt lowers the freezing point of water to, say, $-10^\circ\text{C}$ or lower (depending on concentration). Now, at the existing ambient temperature of $-5^\circ\text{C}$ , the ice is above its new freezing point, so it melts.
Write it clearly: "At $-5^\circ\text{C}$ , pure ice is stable. But a salt solution freezes at, for example, $-10^\circ\text{C}$ . Since $-5^\circ\text{C} > -10^\circ\text{C}$ , the ice melts."

Mistake 4: Using the Wrong Salt or Wrong Mechanism

The Mistake:

Students say "common salt (NaCl) is sprinkled" but then describe a chemical reaction (like an exothermic dissolution) as the reason for melting.

Why it happens:

They confuse the physical colligative effect with a chemical heat release.

How to Avoid: …

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

COMEDK 2026Set 2026-A1 markMCQ
$5 \%$
›Reveal solutionSolution
Freezing-point depression depends on molality, not mass percent — since glucose has a smaller molar mass than sucrose, a 5% glucose solution has more solute particles per kg of water, giving a larger depression. Using the given sugar-solution data as a reference (without needing water's real $K_f$ ), the glucose solution's freezing point works out to about 269 K, option (C).

Concept and Intuition

For a non-electrolyte, $\Delta T_f = K_f \cdot m$ , where $m$ is molality. Both solutions here are 5% by mass (5 g solute per 100 g solution, so 95 g water), but sucrose ( $\mathrm{C_{12}H_{22}O_{11}}$ , $M\approx342$ ) is much heavier per mole than glucose ( $\mathrm{C_6H_{12}O_6}$ , $M\approx180$ ), so the same mass of glucose gives more moles — and a bigger depression. Since $K_f$ is the same for both solutions (same solvent), we can find the glucose depression as a ratio of the sugar depression, without needing water's textbook $K_f$ value.

Step-by-step reasoning

Depression for the sugar solution.

$\Delta T_f(\text{sugar}) = 273.15 - 271 = 2.15\ \text{K}$

Ratio of molalities. For the same mass of solute (5 g) in the same mass of water (95 g), molality is inversely proportional to molar mass: …
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COMEDK 2026Set 2026-M1 markMCQ
$Q. Identify the correct statement (A) A hypotonic solution is more concentrated with respect to the other solution separated from it by a semi permeable membrane (B) At the freezing point of a solution containing a non-volatile solute, the vapour pressure of the liquid solvent becomes less than that of the solid solvent (C) In case of dissociation of solute particles in a solution, the van't Hoff factor is more than unity because the observed molar mass has a lesser value (D) The molal boiling point elevation constant for a solvent is directly proportional to its enthalpy of vaporisation$
›Reveal solutionSolution
The key idea is to test each statement against fundamental solution chemistry: osmosis, freezing-point depression, van’t Hoff factor, and boiling-point elevation. Only statement (C) is correct.

Concept and Intuition

This question checks your understanding of four distinct topics in physical chemistry:

Hypotonic vs. hypertonic solutions (osmosis)

Freezing point depression (vapour pressure of solid vs. liquid solvent)

Van’t Hoff factor (effect of dissociation on colligative properties)

Molal boiling point elevation constant (its relation to enthalpy of vaporisation)

Each statement must be evaluated carefully — common pitfalls arise from mixing up definitions or misremembering proportionalities.

Step-by-step reasoning

Statement (A):

“A hypotonic solution is more concentrated with respect to the other solution separated from it by a semipermeable membrane.”

A hypotonic solution has a lower solute concentration than the solution on the other side of the membrane.

Water moves from the hypotonic (less concentrated) side to the hypertonic (more concentrated) side.

Therefore, a hypotonic solution is less concentrated, not more. ➜ False.

Statement (B):

“At the freezing point of a solution containing a non-volatile solute, the vapour pressure of the liquid solvent becomes less than that of the solid solvent.”

At the freezing point of a pure solvent, the vapour pressures of solid and liquid are equal.

For a solution, the vapour pressure of the liquid solvent is lowered by the solute (Raoult’s law).

At the new freezing point, the vapour pressure of the liquid solvent equals that of the solid solvent (both are lower than the pure solvent’s freezing point).

So the vapour pressure of the liquid is not less than that of the solid — they are equal at equilibrium. ➜ False.

Statement (C):

“In case of dissociation of solute particles in a solution, the van't Hoff factor is more than unity because the observed molar mass has a lesser value.”

Van’t Hoff factor $i = \frac{\text{actual number of particles in solution}}{\text{number of formula units dissolved}}$ .

For dissociation, $i > 1$ (e.g., NaCl → 2 ions, so $i \approx 2$ ).

Colligative properties depend on particle count. Observed molar mass $M_{\text{obs}} = \frac{M_{\text{theoretical}}}{i}$ . …
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COMEDK 2025Set 2025-A1 markMCQ
$0.4 \mathrm{~g} \mathrm{CH}_3 \mathrm{COOH}$
›Reveal solutionSolution
The problem uses freezing-point depression to find the van’t Hoff factor for acetic acid in benzene, then relates that factor to the degree of dimerisation. The percentage association is about 94.54%, so the correct option is (C).

Concept & Intuition

Freezing-point depression is a colligative property: it depends on the number of solute particles in solution, not their identity. When acetic acid dimerises in benzene, two molecules join to form one dimer, reducing the total particle count. The observed depression is smaller than expected for monomeric acid. By comparing the experimental van’t Hoff factor $i$ (actual particles per formula unit) with the ideal value of 1, we can calculate the fraction of acid that has dimerised.

Step-by-step solution

Calculate the expected freezing-point depression if no dimerisation occurred

Mass of acetic acid = $0.4\ \mathrm{g}$

Molar mass of $\mathrm{CH_3COOH}$ = $60\ \mathrm{g/mol}$

Moles of acid = $\frac{0.4}{60} = 0.006667\ \mathrm{mol}$

Mass of benzene = $40\ \mathrm{g} = 0.040\ \mathrm{kg}$

Molality (if monomeric) = $\frac{0.006667}{0.040} = 0.1667\ \mathrm{mol/kg}$

Using $\Delta T_f = K_f \cdot m$ :

$\Delta T_f(\text{ideal}) = 5.12 \times 0.1667 = 0.8533\ ^\circ\mathrm{C}$

Find the van’t Hoff factor $i$ from the observed depression

Observed $\Delta T_f = 0.45\ ^\circ\mathrm{C}$

The van’t Hoff factor is defined as:

$i = \frac{\text{observed } \Delta T_f}{\text{ideal } \Delta T_f} = \frac{0.45}{0.8533} = 0.5274$
This $i < 1$ confirms association (dimerisation).

Relate $i$ to the degree of dimerisation Let $\alpha$ be the fraction of acetic acid molecules that dimerise. For every 2 monomers that react, we get 1 dimer. Start with 1 mole of monomer:

Moles that dimerise = $\alpha$ → these form $\alpha/2$ moles of dimer

Moles that remain monomer = $1 - \alpha$ …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{X}$
›Reveal solutionSolution

$\Delta T_b$ gives van't Hoff factor $i=0.5$ ; for dimerisation $i=1-\alpha/2$ , so $\alpha=1=100\%$ .

$\Delta T_b = 101.04 - 100 = 1.04\ ^\circ\mathrm{C}$
Using $\Delta T_b = i\,K_b\,m$ :
$i = \frac{\Delta T_b}{K_b\,m} = \frac{1.04}{0.52 \times 4} = 0.5$
For association of $n$ monomers into one particle, $i = 1 - \alpha\left(1 - \tfrac{1}{n}\right)$ . For a dimer $n=2$ : …
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COMEDK 2024Set 2024-M1 markMCQ
$60 \mathrm{~g}$
›Reveal solutionSolution

A freezing-point depression of $9.3\,^\circ\text{C}$ needs a $5\ m$ solution, which requires only $193.55\ \text{g}$ of water to stay liquid, so $250 - 193.55 = \mathbf{56.45\ g}$ of ice separates out — option (B).

Concept

As the solution is cooled, pure ice crystallises out. Ice keeps forming until the remaining liquid water is concentrated enough that its freezing point equals the bath temperature. So we find how much liquid water is needed to give a freezing-point depression of $9.3\,^\circ\text{C}$ ; the rest of the original water has frozen.

Step 1 — Molality required for $\Delta T_f = 9.3\ \text{K}$
$\Delta T_f = K_f \cdot m \;\Rightarrow\; m = \frac{\Delta T_f}{K_f} = \frac{9.3}{1.86} = 5\ \text{mol kg}^{-1}$
Step 2 — Moles of ethylene glycol (non-volatile solute, does not freeze) …
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KCET 2023Set D-21 markMCQ
$5.85\% \left(\frac{w}{w}\right)$
›Reveal solutionSolution
Work with 100 g of solution, get the moles of each solute, apply the van't Hoff factor for partial ionisation, compute the molality in the water only, and use $\Delta T_f = i\,K_f\,m$ .

1. Take a 100 g basis of the solution

Because the strengths are given as % (w/w):

mass of AB $= 5.85$ g

mass of $\mathrm{XY_2} = 9.50$ g

mass of water (solvent) $= 100 - 5.85 - 9.50 = 84.65$ g $= 0.08465$ kg

2. Moles of each solute
$n_{AB} = \frac{5.85}{58.5} = 0.1\ \text{mol}, \qquad n_{XY_2} = \frac{9.50}{95} = 0.1\ \text{mol}$
3. van't Hoff factors (partial ionisation)

For a solute giving $n$ particles with degree of ionisation $\alpha$ :
$i = 1 + (n-1)\alpha$
AB $\rightarrow \mathrm{A^+ + B^-}$ , so $n = 2$ , $\alpha = 0.80$ :
$i_{AB} = 1 + (2-1)(0.80) = 1.8$
$\mathrm{XY_2}$ $\rightarrow \mathrm{X^{2+} + 2Y^-}$ (Y is monovalent, so X must be divalent), $n = 3$ , $\alpha = 0.60$ :
$i_{XY_2} = 1 + (3-1)(0.60) = 1 + 1.2 = 2.2$
4. Total effective (particle) moles
$n_{\text{eff}} = i_{AB}\,n_{AB} + i_{XY_2}\,n_{XY_2} = (1.8)(0.1) + (2.2)(0.1) = 0.18 + 0.22 = 0.40\ \text{mol}$
5. Effective molality
$m_{\text{eff}} = \frac{0.40\ \text{mol}}{0.08465\ \text{kg}} = 4.725\ \text{mol kg}^{-1}$
6. Depression of freezing point …
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COMEDK 2023Set 2023-M1 markMCQ
$Q. Abnormal colligative properties are observed only when the dissolved non-volatile solute in a given dilute solution (A) is a non-electrolyte (B) offers an intense colour (C) associates and dissociates (D) offers no colour$
›Reveal solutionSolution

Abnormal colligative properties occur because association or dissociation changes the actual number of solute particles relative to the number of formula units dissolved.

Colligative properties (relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, osmotic pressure) depend only on the number of solute particles, not their nature.

When a solute dissociates (e.g. $\mathrm{NaCl \to Na^+ + Cl^-}$ ), the particle count increases, so the observed colligative property is larger than expected (van't Hoff factor $i>1$ ). …
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COMEDK 2021Set 2021-B1 markMCQ
$Q.The mole fraction of the solute in a dilute aqueous solution of glucose with a vapour pressure of 750 mm at 373 K is : (A) 1/35 (B) 1/76 (C) 1/25 (D) 1/80$
›Reveal solutionSolution

By Raoult's law, $x_{solute} = (P^0 - P)/P^0 = 10/760 = 1/76$ .

At 373 K (100 °C) the vapour pressure of pure water $P^0 = 760$ mm.

Relative lowering of vapour pressure = mole fraction of solute: …
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KCET 2020Set A-11 markMCQ
$M_{0.96}O$
›Reveal solutionSolution
Impose electrical neutrality on the non-stoichiometric oxide and solve for the number of $M^{3+}$ ions.

Step 1 — The concept: metal-deficiency defect. In an oxide written $M_{0.96}O$ the metal sites are deficient (a cation vacancy defect, as in FeO / wüstite). The crystal must still be electrically neutral, so some cations are oxidised from $M^{2+}$ to $M^{3+}$ to make up the missing positive charge.

Step 2 — Set up the balance. Take $1$ mol of the oxide: it contains $0.96$ mol of metal and $1$ mol of $O^{2-}$ (total negative charge $= 2$ ).

Let

$a$ = mol of $M^{2+}$ ,

$b$ = mol of $M^{3+}$ .

Then
$a + b = 0.96 \qquad \text{(total metal)}$
$2a + 3b = 2 \qquad \text{(charge neutrality: total } + \text{ charge} = 2)$ …
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KCET 2020Set A-11 markMCQ
$[K_b = 0.52$
›Reveal solutionSolution

Molar mass of $X = 26\ \text{g mol}^{-1}$ — option (A).

$X$ dimerises: $2X \rightleftharpoons X_2$ , degree of association $\alpha = 0.80$ . The van't Hoff factor is

$i = 1 - \dfrac{\alpha}{2} = 1 - \dfrac{0.80}{2} = 0.60.$

Elevation of boiling point: $\Delta T_b = i\,K_b\,m$ , with molality $m = \dfrac{2.5/M}{0.100} = \dfrac{25}{M}\ \text{mol kg}^{-1}$ . …
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KCET 2019Set A-11 markMCQ
$L^{-1}$
›Reveal solutionSolution
The solute tetramerises, so the van’t Hoff factor is less than 1. Using the observed freezing point depression and the formula $\Delta T_f = i \, K_f \, m$ , we find the apparent molar mass, then correct for association to get the true molar mass. The answer is 62 g/mol.

The key idea: when a solute associates (forms larger molecules), the number of particles in solution is less than the number of formula units dissolved. The freezing point depression depends on the actual number of particles, so we first find the apparent molar mass from the given $\Delta T_f$ , then use the degree of association to find the true molar mass.

For tetramerisation, 4 molecules of A combine to form one tetramer. If the degree of association is $\alpha$ , the van’t Hoff factor $i$ is given by:
$i = 1 - \alpha + \frac{\alpha}{n}$
where $n = 4$ here. With $\alpha = 0.80$ (80% association), we can compute $i$ directly.

Calculate the van’t Hoff factor $i$

For tetramerisation: $n = 4$ , $\alpha = 0.80$ .

$i = 1 - \alpha + \frac{\alpha}{4} = 1 - 0.80 + 0.20 = 0.40$
So only 40% of the original number of particles remain, on average.

Find the apparent molar mass from the observed depression

The freezing point depression is:

$\Delta T_f = i \, K_f \, m$
where $m$ is the molality of the solution if no association occurred (i.e., using the formula mass of A). Let the true molar mass of A be $M$ g/mol.

Molality (if no association) = $\frac{2.5 / M}{0.100} = \frac{25}{M}$ mol/kg.

Plug into the equation:
$0.3 = 0.40 \times 1.86 \times \frac{25}{M}$

Solve for $M$

$0.3 = 0.40 \times 1.86 \times \frac{25}{M}$
First compute $0.40 \times 1.86 = 0.744$ .

Then $0.744 \times 25 = 18.6$ .

So: …
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Q.How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.

Concept understanding — Colligative Properties Association

Colligative Properties: The Intuition First

The Precise Statement

The Crucial Distinction: Association vs. Dissociation

The van't Hoff Factor $i$

A Concrete Example

Why this formula?

Colligative Properties & Association: Why the Formula Holds

The Core Idea: What Are Colligative Properties?

The van't Hoff Factor: The Bridge

Example: Dimerization of Benzoic Acid in Benzene

Deriving the Modified Formula

Step 1: Start with the Normal Colligative Formula

Step 2: Replace $m$ with Effective Molality

Step 3: Express $i$ in Terms of Degree of Association

For a dimerization ( $n=2$ ):

General formula for association of $n$ molecules:

Why This Makes Physical Sense …

Concept: Freezing Point Depression (Colligative Property)

Method: Application of Raoult’s Law to Solutions

Mistake 1: Confusing the Phenomenon with "Boiling Point Elevation"

Mistake 2: Forgetting the "Why" — The Colligative Nature

Mistake 3: Ignoring the "Hilly Areas" Context

Mistake 4: Using the Wrong Salt or Wrong Mechanism

1. Take a 100 g basis of the solution

2. Moles of each solute

3. van't Hoff factors (partial ionisation)

4. Total effective (particle) moles

5. Effective molality

6. Depression of freezing point …

Q.How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.

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