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.

Telangana TsbieShort· 2mImportance★★★★★

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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.

TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$Q. The relative lowering of vapour pressure of an aqueous solution containing a non-volatile solute is 0.0125. The molality of the solution is (A) 0.65 m (B) 0.35 m (C) 0.70 m (D) 0.30 m$
›Reveal solutionSolution
The relative lowering of vapour pressure equals the mole fraction of the solute; converting mole fraction to molality for a dilute aqueous solution gives the answer. The molality is approximately 0.70 m, so the correct option is (C).

Concept & Intuition

Raoult’s law for a non‑volatile solute tells us that the relative lowering of vapour pressure is exactly equal to the mole fraction of the solute in the solution:
$\frac{P^0 - P}{P^0} = x_{\text{solute}}$
Here the given value is 0.0125, so $x_{\text{solute}} = 0.0125$ .

We need molality (moles of solute per kg of solvent). For a dilute aqueous solution, the mole fraction of solute is small, so the mole fraction of water is nearly 1. This lets us convert directly:
$x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}} \approx \frac{n_{\text{solute}}}{n_{\text{water}}}$
since $n_{\text{solute}} \ll n_{\text{water}}$ . Then molality $m = \frac{n_{\text{solute}}}{\text{kg of water}}$ . For water, 1 kg contains $1000/18 \approx 55.56$ moles. So $n_{\text{water}} \approx 55.56$ per kg. Thus:
$m \approx x_{\text{solute}} \times 55.56$
Step‑by‑step calculation

Write the exact relation

$x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}} = 0.0125$

Express in terms of molality Let $m$ be molality (mol solute per kg water). For 1 kg water, $n_{\text{water}} = \frac{1000}{18} = 55.555\ldots$ mol. Then $n_{\text{solute}} = m$ . So:

$0.0125 = \frac{m}{m + 55.555}$

Solve for $m$ Multiply both sides:

$0.0125 (m + 55.555) = m$ $0.0125 m + 0.69444 = m$
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TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$_6$
(g) and H $_2$ O (l) are $-409$ , $-393$ and $-286$ kJ mol $^{-1}$ respectively. Enthalpy of combustion of benzoic acid (in kJ mol $^{-1}$ ) is (A) $-1600$ (B) $+1600$ (C) $-3200$ (D) $+4800$
›Reveal solutionSolution
The enthalpy of combustion is found by applying Hess’s law: the combustion reaction is written, then the enthalpy change is calculated as the sum of the enthalpies of formation of products minus that of the reactant. The result is –3200 kJ mol⁻¹, option (C).

The key idea here is that enthalpy of combustion is just a special case of enthalpy of reaction. For any reaction,
$\Delta H_{\text{reaction}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$
where $\Delta H_f^\circ$ is the standard enthalpy of formation. Combustion means burning in excess oxygen to give CO₂(g) and H₂O(l) as the only products. So we first write the balanced combustion equation for benzoic acid, then plug in the given formation enthalpies.

Write the balanced combustion reaction Benzoic acid is C₆H₅COOH. Its molecular formula is C₇H₆O₂. Complete combustion with O₂ gives CO₂ and H₂O:

$\text{C}_7\text{H}_6\text{O}_2(s) + \frac{15}{2}\,\text{O}_2(g) \rightarrow 7\,\text{CO}_2(g) + 3\,\text{H}_2\text{O}(l)$
Check: left has 7 C, 6 H, 2 + 15 = 17 O; right has 7×2 + 3×1 = 17 O. Balanced.

Apply Hess’s law The enthalpy of combustion is $\Delta H_{\text{comb}} = \Delta H_{\text{reaction}}$ for this equation. Using formation enthalpies:

$\Delta H_{\text{comb}} = \big[7 \cdot \Delta H_f^\circ(\text{CO}_2) + 3 \cdot \Delta H_f^\circ(\text{H}_2\text{O})\big] - \big[1 \cdot \Delta H_f^\circ(\text{C}_7\text{H}_6\text{O}_2) + \frac{15}{2} \cdot \Delta H_f^\circ(\text{O}_2)\big]$
Note: $\Delta H_f^\circ$ for any element in its standard state (like O₂ gas) is zero by definition. So the O₂ term vanishes.

Substitute the given values $\Delta H_f^\circ(\text{CO}_2) = -393$ kJ/mol, $\Delta H_f^\circ(\text{H}_2\text{O}) = -286$ kJ/mol, $\Delta H_f^\circ(\text{C}_7\text{H}_6\text{O}_2) = -409$ kJ/mol.

$\Delta H_{\text{comb}} = \big[7(-393) + 3(-286)\big] - \big[(-409)\big]$
Compute stepwise: …
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$4$
›Reveal solutionSolution
The key idea is to treat the two-dimensional box as a grid where each atom occupies a square of side equal to the nearest neighbour distance in the bcc lattice. For bcc, the nearest neighbour distance is $\frac{\sqrt{3}}{2}a$ , and with $a = 4$ Å that gives $2\sqrt{3} \approx 3.46$ Å. The number of atoms that fit along each side is $\left\lfloor \frac{35}{3.46} \right\rfloor = 10$ , so the total is $10 \times 10 = 100$ . The correct option is (C).

The problem asks how many metal atoms can be placed in a two-dimensional open box of length 35 Å and width 35 Å, given that the metal crystallizes in a bcc lattice with edge length 4 Å. The phrase "two-dimensional open box" means we are arranging atoms in a flat, square region — essentially a grid — and we need to know the spacing between atoms in the bcc structure when projected onto a plane.

In a bcc lattice, atoms touch along the body diagonal, not along the edge. The nearest neighbour distance is half the body diagonal: $\frac{\sqrt{3}}{2}a$ . For $a = 4$ Å, this distance is $\frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3} \approx 3.464$ Å. This is the centre-to-centre distance between adjacent atoms in the lattice. When placing atoms in a box, each atom effectively occupies a square of side equal to this distance (since we cannot overlap atoms and they must be placed at lattice positions).

Now, we simply see how many such spacings fit along each dimension of the box.

Find the spacing between atoms.

In bcc, the nearest neighbour distance $d = \frac{\sqrt{3}}{2}a = 2\sqrt{3}$ Å. This is the minimum centre-to-centre separation.

Number of atoms along one side.

Along a length of 35 Å, the number of atoms that can be placed is the integer part of $35 / d$ , because you need one full spacing per atom (the first atom can be at one edge, then each subsequent atom requires a full $d$ distance). So
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$0.40$
›Reveal solutionSolution
Using Raoult’s law for an ideal solution, the vapour mole fraction of toluene is found to be 0.20, so the correct option is (D).

Concept & Intuition

This problem is about vapour composition above an ideal liquid mixture. Raoult’s law tells us that the partial pressure of each component in the vapour equals its mole fraction in the liquid times its pure vapour pressure. The total pressure is the sum of these partial pressures. The vapour composition is then the ratio of a component’s partial pressure to the total pressure. The key insight: even if the liquid has equal amounts of two substances, the vapour will be richer in the more volatile component (the one with higher pure vapour pressure). Here, benzene is more volatile than toluene, so the vapour should contain more benzene than the liquid does.

Step-by-step solution

Identify the given data

Temperature: $T = 300\ \text{K}$

Pure vapour pressure of toluene: $P_\text{toluene}^\circ = 3.63\ \text{kPa}$

Pure vapour pressure of benzene: $P_\text{benzene}^\circ = 9.7\ \text{kPa}$

Mole fraction of toluene in the liquid: $x_\text{toluene} = 0.4$

Since it’s a binary mixture, mole fraction of benzene in the liquid:

$x_\text{benzene} = 1 - x_\text{toluene} = 0.6$

Apply Raoult’s law to find partial pressures Raoult’s law: $P_i = x_i P_i^\circ$

Partial pressure of toluene:

$P_\text{toluene} = 0.4 \times 3.63 = 1.452\ \text{kPa}$

Partial pressure of benzene:

$P_\text{benzene} = 0.6 \times 9.7 = 5.82\ \text{kPa}$

Calculate total vapour pressure

$P_\text{total} = P_\text{toluene} + P_\text{benzene} = 1.452 + 5.82 = 7.272\ \text{kPa}$

Find vapour composition …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$^\circ$
›Reveal solutionSolution
Using Raoult’s law for a non‑volatile solute, the mole fraction of benzene in the solution equals the ratio of its vapour pressure to that of pure benzene. From that, the required moles of solute per mole of benzene is found to be 0.605, corresponding to option (C).

Concept & Intuition

When a non‑volatile solute is dissolved in a volatile solvent, the vapour pressure of the solvent above the solution is lowered. Raoult’s law states that the vapour pressure of the solvent in the solution ( $p$ ) equals the vapour pressure of the pure solvent ( $p^\circ$ ) multiplied by its mole fraction in the liquid phase:
$p = x_{\text{solvent}} \, p^\circ$
Here, the solute does not contribute to the vapour, so the entire vapour pressure comes from benzene. The problem asks for the number of moles of solute per mole of benzene — that is, the ratio $n_{\text{solute}} / n_{\text{benzene}}$ . If we let $n_B = 1$ mole of benzene, then the mole fraction of benzene is
$x_B = \frac{1}{1 + n_{\text{solute}}}$
and we can solve for $n_{\text{solute}}$ .

Step‑by‑step solution

Write Raoult’s law for the solution Given: $p^\circ = 268$ torr (pure benzene vapour pressure) $p = 167$ torr (solution vapour pressure)

$p = x_B \, p^\circ \quad \Rightarrow \quad 167 = x_B \times 268$

Solve for the mole fraction of benzene

$x_B = \frac{167}{268}$
Simplify the fraction:
$x_B = \frac{167}{268} \approx 0.62313$

Express $x_B$ in terms of moles Let $n_B = 1$ mole of benzene and let $n_s$ = moles of solute. Then

$x_B = \frac{n_B}{n_B + n_s} = \frac{1}{1 + n_s}$
Equate to the value from step 2:
$\frac{1}{1 + n_s} = \frac{167}{268}$

Solve for $n_s$ Invert both sides:

$1 + n_s = \frac{268}{167}$ $n_s = \frac{268}{167} - 1 = \frac{268 - 167}{167} = \frac{101}{167}$
Compute the decimal:
$n_s \approx 0.60479 \approx 0.605$

Interpret the result …
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TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
$0~^\circ \mathrm{C}$
›Reveal solutionSolution
Heavy water ( $D_2O$ ) has a higher freezing point than normal water ( $H_2O$ ) because the heavier deuterium atoms lead to stronger hydrogen bonds, requiring more energy to break the crystal lattice. Its freezing point is approximately $3.8~^\circ \mathrm{C}$ .

Concept and Intuition

Water, as we commonly know it, is $H_2O$ , where $H$ represents protium, the most common isotope of hydrogen with one proton and no neutrons. Heavy water, on the other hand, is $D_2O$ , where $D$ represents deuterium, an isotope of hydrogen with one proton and one neutron. This makes deuterium approximately twice as heavy as protium.

This difference in mass, though seemingly small, significantly affects the physical properties of water. The heavier deuterium atoms lead to stronger intermolecular forces, specifically hydrogen bonds, in $D_2O$ compared to $H_2O$ . This is primarily because the vibrational energy of the O-D bond is lower than that of the O-H bond (due to the higher reduced mass), which translates to a lower zero-point energy. A lower zero-point energy means the bond is effectively stronger and requires more energy to break. Consequently, more thermal energy is needed to disrupt the ordered structure of ice and transition it into liquid water. This results in higher melting (freezing) and boiling points for heavy water.

Step-by-Step Solution

Recall the freezing point of normal water:

Normal water ( $H_2O$ ) freezes at $0~^\circ \mathrm{C}$ at $1$ atm pressure. This is a fundamental reference point for understanding the properties of water.

Understand the nature of heavy water ( $D_2O$ ):

Heavy water contains deuterium ( $D$ ) instead of protium ( $H$ ). Deuterium is an isotope of hydrogen with an atomic mass of approximately $2$ amu, while protium has an atomic mass of approximately $1$ amu. This difference in mass affects the physical properties of the molecule.

Identify the freezing point of heavy water: …
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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
$^\circ$
›Reveal solutionSolution
Molarity depends on volume, which expands with temperature. Since the solution is prepared at 25 °C but measured at 90 °C, the volume increases, so molarity decreases. The final molarity is less than 0.25 M, making option (A) correct.

The key concept here is that molarity is defined as moles of solute per litre of solution. Moles of NaOH are fixed once dissolved, but the volume of the solution changes with temperature because liquids expand when heated. At 90 °C, the solution occupies a larger volume than at 25 °C, so the molarity drops.

Let’s work through it step by step.

Find the moles of NaOH.

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.

Given mass = 2 g.

Moles of NaOH = $\frac{2}{40} = 0.05$ mol.

Calculate molarity at 25 °C.

Volume at 25 °C = 200 mL = 0.200 L.

Molarity at 25 °C = $\frac{0.05}{0.200} = 0.25$ M.

What happens at 90 °C?

The number of moles of NaOH stays the same — no solute is added or removed. But the solution expands. Water (and dilute aqueous solutions) has a coefficient of volume expansion of roughly $2.1 \times 10^{-4}$  °C⁻¹. Over a 65 °C rise (from 25 °C to 90 °C), the volume increases by about

$\Delta V = V_0 \cdot \gamma \cdot \Delta T = 200 \times 2.1\times10^{-4} \times 65 \approx 2.73$ mL.

So the new volume ≈ 202.73 mL = 0.20273 L.

Molarity at 90 °C = $\frac{0.05}{0.20273} \approx 0.2466$ M, which is less than 0.25 M. …
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TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
$Q. The intramolecular hydrogen bonding is present in (A) phenol (B) Benzoic acid (C) para-Nitrophenol (D) 2-Hydroxybenzoic Acid$
›Reveal solutionSolution
Intramolecular hydrogen bonding requires a hydrogen donor and an acceptor in close proximity within the same molecule, forming a stable five- or six-membered ring. Among the given options, only 2-Hydroxybenzoic acid (salicylic acid) can form such a bond, making (D) the correct answer.

The key concept here is the geometry of hydrogen bonding. For a hydrogen bond to form within a single molecule, the donor group (like –OH) and the acceptor atom (like a carbonyl oxygen) must be positioned so that a stable, usually five- or six-membered ring can close. This is only possible when the groups are on adjacent carbon atoms in a rigid structure like a benzene ring.

Let’s examine each option:

Phenol (A) – Phenol has an –OH group attached to a benzene ring. The only nearby acceptor atoms are the ring carbons, which are not electronegative enough to form a strong hydrogen bond. The oxygen of the –OH can act as an acceptor for another molecule, but within the same molecule, there is no suitable acceptor in the right position. So, no intramolecular H-bond.

Benzoic acid (B) – The –COOH group has both a donor (–OH) and an acceptor (C=O). However, these are part of the same functional group and are already bonded to each other. The carbonyl oxygen is too far from the –OH of another benzoic acid molecule to form an intramolecular bond; instead, benzoic acid forms intermolecular dimers (two molecules linked by two H-bonds). No intramolecular bond here.

para-Nitrophenol (C) – The –OH group is at the para position relative to the –NO₂ group. The distance between them is too large (four carbon atoms apart) to allow a stable ring. The nitro group’s oxygen is a good acceptor, but the geometry forces the H-bond to be intermolecular. So, no intramolecular bond. …
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TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
$K_c$
›Reveal solutionSolution
For a weak acid, the equilibrium constant $K_c$ is related to the degree of dissociation $\alpha$ and initial concentration $C$ by $K_c = C\alpha^2$ . With $C = 0.1$ M and $\alpha = 0.05$ , we get $K_c = 2.5 \times 10^{-4}$ , which rounds to option (C) $2.6 \times 10^{-4}$ .

The key idea here is that a weak acid does not dissociate completely. The degree of dissociation, $\alpha$ , tells you what fraction of the original acid molecules have broken into ions at equilibrium. For a monoprotic acid like HA, the equilibrium constant $K_c$ (also called the acid dissociation constant $K_a$ ) is directly linked to $\alpha$ and the initial concentration $C$ through a simple relation — but only if we make the standard approximation that $\alpha$ is small.

Let’s walk through it.

Set up the equilibrium. For the reaction

$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$
start with initial concentration $C = 0.1$ M of HA, and zero of the ions. At equilibrium, a fraction $\alpha$ of HA has dissociated. So:

$[\text{HA}] = C(1 - \alpha)$

$[\text{H}^+] = C\alpha$

$[\text{A}^-] = C\alpha$

Write the expression for $K_c$ .

$K_c = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$

Plug in the numbers. Here $C = 0.1$ and $\alpha = 5\% = 0.05$ .

$K_c = \frac{0.1 \times (0.05)^2}{1 - 0.05} = \frac{0.1 \times 0.0025}{0.95} = \frac{0.00025}{0.95}$

Simplify. $\frac{0.00025}{0.95} = \frac{2.5 \times 10^{-4}}{0.95} \approx 2.6316 \times 10^{-4}$ …
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$_2$
›Reveal solutionSolution
Colligative properties depend on the number of particles in solution. For the same concentration, BaCl₂ and MgCl₂ each give 3 ions, while NaCl gives 2 — so the trend is MgCl₂ = BaCl₂ > NaCl.

Colligative properties — like boiling point elevation, freezing point depression, and osmotic pressure — depend only on the number of solute particles in a given amount of solvent, not on the chemical identity of those particles. This is the core idea: more particles means a larger effect.

For ionic compounds dissolved in water, the number of particles per formula unit equals the number of ions produced upon dissociation. So the key is to count ions.

NaCl dissociates as:

$\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$

That gives 2 ions per formula unit.

MgCl₂ dissociates as:

$\text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^-$

That gives 3 ions per formula unit.

BaCl₂ dissociates as:

$\text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^-$

That also gives 3 ions per formula unit. …
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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

Concept and Intuition

Step-by-Step Solution

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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