On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following ca

NCERT Exemplar · Q2

Q. On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?

(i) Sugar crystals in cold water.

(ii) Sugar crystals in hot water.

(iii) Powdered sugar in cold water.

(iv) Powdered sugar in hot water.

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Concept understanding — Types Of Solutions

Types of Solutions: From Everyday Life to Chemistry

You already know what a solution is — sugar dissolved in water, salt in water, even the air you breathe. But not all solutions behave the same way. Some dissolve easily, some refuse to dissolve beyond a point, and some can hold more solute than they normally should. That difference is what we classify as types of solutions based on how much solute is dissolved.

The Intuition: A Cup of Tea

Imagine making a cup of tea. You add one spoon of sugar — it dissolves completely. You add a second spoon — still dissolves. A third spoon — maybe it dissolves, maybe it doesn't. At some point, no matter how much you stir, the sugar just sits at the bottom.

That moment — when no more sugar dissolves — is the saturation point. Before that, you have an unsaturated solution. At that exact point, you have a saturated solution. And if you carefully heat the tea, dissolve more sugar, then cool it down without disturbing it — you might get a supersaturated solution, where more sugar stays dissolved than should be possible at that temperature.

That's the entire idea. Three types, defined by how much solute is dissolved relative to the maximum possible.

The Precise Statement

A solution is a homogeneous mixture of a solute (the substance being dissolved) and a solvent (the substance doing the dissolving). Based on the amount of solute dissolved relative to its solubility at a given temperature, solutions are classified into three types:

Types of Solutions (by saturation) Unsaturated solution — contains less solute than the maximum that can be dissolved at that temperature. Saturated solution — contains exactly the maximum amount of solute that can be dissolved at that temperature. Supersaturated solution — contains more solute than the maximum normally possible at that temperature (a metastable state).

Breaking Down Each Type

Unsaturated solution — the most common type. You can still add more solute and it will dissolve. The concentration is below the solubility limit. If you have a glass of water at room temperature and add a pinch of salt, you get an unsaturated solution. Add more salt — still unsaturated, until you hit the limit.

Saturated solution — the solute and undissolved solute are in dynamic equilibrium. At the molecular level, the rate at which solute particles dissolve equals the rate at which they crystallize out. No net change. If you keep adding salt to water and it stops dissolving, the liquid above the undissolved salt is a saturated solution. The concentration is fixed at the solubility value for that temperature.

Watch out

A common mistake: thinking a saturated solution is always "thick" or "concentrated." Not true. Saturation depends on the solute's solubility. Lead(II) chloride saturates at about 0.45 g per 100 mL water — that's a very dilute saturated solution. Saturation ≠ high concentration.

Supersaturated solution — this is a tricky one. You create it by heating the solvent, dissolving more solute than normally possible, then carefully cooling it. The excess solute stays dissolved because there's no nucleation site (no scratch, no dust particle) to trigger crystallization. It's unstable — the slightest disturbance (a dust speck, a scratch on the glass, even a sudden jolt) causes the excess solute to crystallize out instantly.

Tip

Supersaturated solutions are the reason "hot ice" (sodium acetate) hand warmers work. You click a metal disc inside, which creates a nucleation site, and the entire solution crystallizes in seconds, releasing heat.

A Quick Comparison

Type	Solute amount vs. solubility	Can more solute dissolve?	Stability
Unsaturated	Less than maximum	Yes	Stable
Saturated	Equal to maximum	No (at equilibrium)	Stable
Supersaturated	More than maximum	No (excess will crystallize)	Metastable

Why This Matters

In exams, you'll often be asked to identify the type of solution from a given scenario — like "50 g of salt dissolved in 100 g water at 30°C, given solubility is 36 g per 100 g water." That's a supersaturated solution (50 > 36). Or you might be asked what happens when you add a seed crystal to a supersaturated solution — it triggers crystallization.

The key is always: compare the actual amount dissolved to the solubility at that temperature. That single comparison gives you the type.

Important

Solubility is temperature-dependent. A solution that is saturated at 20°C becomes unsaturated if heated to 50°C (because solubility usually increases with temperature). Always check the temperature condition given in the problem.

Searches such as "types of solutions saturated unsaturated supersaturated" and "solutions class 12 chemistry notes" align directly with the Solutions chapter of the NCERT/CBSE Class 12 Chemistry curriculum. Identifying which type a given scenario describes is a common short-answer question in board exams.

Why this formula?

Types of Solutions: Why the Key Formulae Hold

Understanding why the formulae work is essential for Indian exams (JEE, NEET, CBSE). Let's break down the reasoning behind the most important relationships.

1. The Basic Classification: What Makes a Solution?

A solution is a homogeneous mixture of two or more substances. The key idea is intermolecular forces between solute and solvent particles.

Ideal Solution: Solute-solvent interactions are identical to solute-solute and solvent-solvent interactions. Why? No net energy change on mixing — the molecules "fit" perfectly.
Non-Ideal Solution: Interactions differ, leading to deviation from Raoult's law.

2. Raoult's Law: The Foundation

Formula:

P_{\text{solution}} = x_{\text{solvent}} \cdot P_{\text{solvent}}^0

Why does this hold?

Imagine a pure solvent surface. The vapour pressure $P^0$ comes from molecules escaping the liquid. When you add a non-volatile solute, solute molecules occupy some surface area, blocking solvent molecules from escaping.

The fraction of surface available to solvent = mole fraction of solvent ( $x_{\text{solvent}}$ ).
Therefore, the rate of escape (vapour pressure) is proportional to that fraction:

P_{\text{solution}} \propto x_{\text{solvent}}

At the limit $x_{\text{solvent}} = 1$ , $P_{\text{solution}} = P^0$ , so the constant is $P^0$ .

Key insight: Raoult's law is a surface-area argument, not a volume argument.

3. Relative Lowering of Vapour Pressure

Formula:

\frac{P^0 - P}{P^0} = x_{\text{solute}}

Derivation in one line:

From Raoult's law:

P = x_{\text{solvent}} \cdot P^0

Since $x_{\text{solvent}} + x_{\text{solute}} = 1$ ,

P = (1 - x_{\text{solute}}) P^0

\Rightarrow P^0 - P = x_{\text{solute}} \cdot P^0

\Rightarrow \frac{P^0 - P}{P^0} = x_{\text{solute}}

Why is this useful?

It depends only on the mole fraction of solute, not on its identity — making it a colligative property.

4. Elevation of Boiling Point

Formula:

\Delta T_b = K_b \cdot m

Why does boiling point rise?

Boiling occurs when vapour pressure = atmospheric pressure.
Adding a non-volatile solute lowers vapour pressure (Raoult's law).
To reach atmospheric pressure again, you must raise the temperature.
The shift $\Delta T_b$ $Δ T_{b}$ is proportional to the molality $m$ $m$ (moles of solute per kg of solvent), because:
- More solute → greater vapour pressure lowering → more temperature needed.
- $K_b$ (ebullioscopic constant) is a property of the solvent only.

5. Depression of Freezing Point

Formula:

\Delta T_f = K_f \cdot m

Why does freezing point drop?

At the freezing point, solid and liquid solvent are in equilibrium.
Adding solute disrupts this equilibrium — solute molecules interfere with the orderly crystal formation of the solvent.
To re-establish equilibrium, you must lower the temperature.
Again, $\Delta T_f \propto m$ , and $K_f$ depends only on the solvent.

Common exam trap: Both $\Delta T_b$ and $\Delta T_f$ are colligative — they depend on number of solute particles, not their nature.

6. Osmotic Pressure

Formula:

\Pi = i \cdot C \cdot R \cdot T

Why does this hold?

Osmosis is the net movement of solvent from low solute concentration to high solute concentration across a semipermeable membrane.
The solvent moves to dilute the higher concentration — this is a entropy-driven process (mixing increases disorder).
Osmotic pressure $\Pi$ is the external pressure needed to stop this flow.
It behaves like an ideal gas law for solute particles:

\Pi V = nRT \quad \Rightarrow \quad \Pi = \frac{n}{V}RT = CRT

The van't Hoff factor $i$ accounts for dissociation/association of solute (e.g., NaCl gives $i \approx 2$ ).

7. The van't Hoff Factor $i$

Formula:

i = \frac{\text{observed colligative property}}{\text{expected colligative property}}

Why is $i$ needed?

Colligative properties depend on number of particles.
If a solute dissociates (e.g., $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$ ), the effective particle count doubles.
If it associates (e.g., benzoic acid in benzene forms dimers), the count halves.
$i$ corrects for this:

\Delta T_f = i \cdot K_f \cdot m

Quick Summary Table

Property	Formula	Why it works
Raoult's law	$P = x_{\text{solvent}} P^0$	Surface area blocking by solute
Relative lowering	$\frac{\Delta P}{P^0} = x_{\text{solute}}$	Direct algebraic consequence
Boiling point elevation	$\Delta T_b = K_b m$	Need higher temp to overcome vapour pressure drop
Freezing point depression	$\Delta T_f = K_f m$	Solute disrupts crystal formation
Osmotic pressure	$\Pi = iCRT$	Analogy to ideal gas law for solute particles

Final takeaway: Every formula in "Types of Solutions" flows from Raoult's law (for vapour pressure) and the particle-counting principle (for colligative properties). Understand these two roots, and you can reconstruct the rest.

The rate of dissolution is increased by higher temperature and greater surface area. Therefore, powdered sugar in hot water will dissolve most rapidly.

When sugar dissolves in water, the sugar molecules separate from the solid crystal lattice and disperse into the water. The observation that the solution feels cool to touch indicates that the dissolution of sugar in water is an endothermic process - the system absorbs heat from its surroundings (your hand) as the sugar dissolves.

Two factors matter here:

Temperature: Increasing the temperature increases the rate of dissolution. Higher temperatures give solvent molecules greater kinetic energy, so they collide more frequently and forcefully with the solute, dislodging it faster.
Surface Area: Finely divided (powdered) solute presents a much larger total surface area than crystals, letting more solvent molecules interact simultaneously with more solute molecules.

Evaluating each option:

(i) Sugar crystals in cold water - small surface area, low temperature: slowest.
(ii) Sugar crystals in hot water - small surface area, but high temperature: faster than (i).
(iii) Powdered sugar in cold water - large surface area, but low temperature: faster than (i), comparable to (ii).
(iv) Powdered sugar in hot water - large surface area AND high temperature: both factors favourable.

Comparing all options, the combination of high temperature and large surface area gives the most rapid dissolution.

✓Final answer

The dissolution of sugar will be most rapid with (iv) Powdered sugar in hot water .

Here’s a breakdown of the common mistakes students make on this question and how to avoid each.

Mistake 1: Ignoring the “cool to touch” clue and picking (ii) without thinking

Why it happens:

Students see “sugar dissolves faster in hot water” as a memorised fact and immediately choose Sugar crystals in hot water (ii). They forget the question is about most rapid dissolution, not just “faster than cold”.

How to avoid:

Always read the full question. The “cool to touch” hint tells you that dissolving sugar is an endothermic process (it absorbs heat). Hot water provides more heat energy, which speeds up dissolution. But that’s only one factor — you must also consider surface area.

Correct reasoning:

Hot water → faster dissolution than cold water.
Powdered sugar → much larger surface area than crystals → even faster dissolution.
So the fastest is powdered sugar in hot water (iv).

Mistake 2: Choosing (iii) — Powdered sugar in cold water — because “powder dissolves faster”

Why it happens:

Students over-focus on surface area and forget that temperature also matters. They think “powdered sugar always dissolves fastest” regardless of temperature.

How to avoid:

Remember: Both factors matter.

Surface area increases rate.
Temperature increases rate.
The combination (hot + powder) is faster than either alone.

Quick check:

If you had to dissolve sugar in 10 seconds, would you use cold water + powder or hot water + crystals? Hot water + crystals is faster than cold + powder because temperature has a stronger effect than surface area in many cases. But hot + powder beats both.

Mistake 3: Confusing “dissolution rate” with “solubility”

Why it happens:

Students think “hot water dissolves more sugar” means it dissolves faster. Actually, solubility (maximum amount) increases with temperature, but rate (how quickly it dissolves) also increases — but they are different concepts.

How to avoid:

Solubility = how much can dissolve at a given temperature.
Rate of dissolution = how fast it dissolves.
Both increase with temperature, but the question asks about rate (most rapid), not amount.

Example:

Even if cold water could eventually dissolve the same amount, hot water does it much faster.

Mistake 4: Not knowing that powdered sugar has more surface area

Why it happens:

Some students don’t connect “powdered” with “larger surface area”. They think “crystals are bigger so they dissolve faster” — which is wrong.

How to avoid:

Memorise: Smaller particles → larger total surface area → more contact with water → faster dissolution.

Sugar cube vs powdered sugar: powdered dissolves in seconds, cube takes minutes.
Same logic applies here.

Final Answer (for reference)

Correct option: (iv) Powdered sugar in hot water.

Why:

Hot water provides more kinetic energy and heat (endothermic process).
Powdered sugar has maximum surface area.
Both factors together give the most rapid dissolution.

Key takeaway:

Always check both temperature and surface area when comparing dissolution rates. Don’t rely on a single memorised fact.

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

AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$\Delta H_{mix} = 0$
›Reveal solutionSolution
Ideal solutions are defined by obeying Raoult's law across the full concentration range with zero enthalpy and volume change on mixing; "does not obey Raoult's law" describes a non-ideal solution instead, so it's the one that does not belong.

Concept and Intuition

In an ideal solution, the intermolecular forces between unlike molecules (A–B) are essentially identical in strength to those between like molecules (A–A and B–B). Because mixing doesn't change the net interaction energy or the packing, there's no enthalpy change ( $\Delta H_{mix}=0$ ) and no volume change ( $\Delta V_{mix}=0$ ) on mixing, and every component's vapour pressure follows Raoult's law ( $p_i = x_i p_i^0$ ) at every composition, not just at the dilute limit. Any solution that deviates from Raoult's law (positive or negative deviation) is, by definition, non-ideal — it will typically show $\Delta H_{mix}\neq0$ and $\Delta V_{mix}\neq0$ as well.

Step-by-Step Solution

Recall the three defining conditions for an ideal solution: obeys Raoult's law over the whole composition range, $\Delta H_{mix}=0$ , $\Delta V_{mix}=0$ .

Check (A) $\Delta H_{mix}=0$ — a genuine ideal-solution property. Correctly belongs.

Check (B) $\Delta V_{mix}=0$ — also a genuine ideal-solution property. Correctly belongs.

Check (C) obeying Raoult's law over the entire range — this is literally the definition. Correctly belongs.

Check (D) "does not obey Raoult's law" — this is the opposite of the defining condition; it describes non-ideal solutions. This is the one that does not belong to an ideal solution.

Common Mistakes

Mixing up "obeys Raoult's law over the entire range" (ideal) with "obeys Raoult's law only in the dilute limit" (which is actually true of any solution, ideal or not, as the solvent approaches purity) — the entire-range obedience is what's special to ideal solutions.

Misreading the negative phrasing of option (D) and picking a positive-sounding property instead.

✓Final answer
$The correct option is (D) — Does not obey Raoult's law.$

ANSWER: D
AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$Q. Which of the following form an ideal solution? (I) Chloroethane and bromoethane (II) Benzene and toluene (III) n - Hexane and n - heptane (IV) Phenol and aniline (A) I & II only (B) I, II & III only (C) II, III & IV only (D) I & IV only$
›Reveal solutionSolution
An ideal solution requires that solute-solvent interactions closely resemble solute-solute and solvent-solvent interactions; similar nonpolar/weakly-polar homologues satisfy this, but phenol-aniline's strong specific H-bonding interaction breaks ideality.

Concept and Intuition

Raoult's law (ideal solution behaviour) holds best when the two components are structurally and electronically similar, so that molecules of A and B interact with each other about as strongly as A-A and B-B do — no new, unusually strong or weak interaction is introduced by mixing.

Step-by-Step Solution

Chloroethane & bromoethane: nearly identical structure and polarity (differ only by halogen), classic ideal-solution pair. Ideal.

Benzene & toluene: both aromatic, very similar size/polarity/intermolecular forces (dispersion-dominated), textbook ideal-solution example. Ideal.

n-Hexane & n-heptane: both nonpolar straight-chain alkanes differing by one $CH_2$ , essentially identical intermolecular forces. Ideal.

Phenol & aniline: phenol's $-OH$ and aniline's $-NH_2$ engage in strong, specific hydrogen bonding/acid-base type interaction between unlike molecules, stronger than the like-like interactions — this is a large negative deviation from Raoult's law, not ideal.

So only I, II, and III qualify as ideal solutions.

Common Mistakes

Assuming any two liquids that mix completely (miscible) automatically form an ideal solution — miscibility and ideality are different concepts.

Overlooking the strong intermolecular H-bond/acid-base interaction unique to the phenol-aniline pair.

✓Final answer
$The correct option is (B) — I, II & III only.$

ANSWER: B
AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$\Delta_{mix}H = 0$
›Reveal solutionSolution
Ideal solutions ( $\Delta_{mix}H=0$ , $\Delta_{mix}V=0$ ) form only from liquids with very similar molecular structure/polarity; ethyl chloride and ethyl bromide fit this best among the given pairs.

Concept and Intuition

An ideal solution obeys Raoult's law over the whole composition range, which requires that solute-solvent (A-B) intermolecular forces be essentially the same as solute-solute (A-A) and solvent-solvent (B-B) forces. When this holds, mixing causes no net enthalpy change and no volume change, since the molecules "don't notice" whether they're surrounded by like or unlike neighbours.

Step-by-Step Solution

Check each pair for structural/polarity similarity.

Ethanol + acetone: very different functional groups (H-bonding alcohol vs. non-H-bonding ketone) — strong negative deviation, not ideal.

Phenol + aniline: phenol H-bonds strongly with itself; mixing with aniline changes H-bonding pattern significantly — not ideal.

Chloroform + acetone: chloroform's H can H-bond with acetone's carbonyl oxygen, causing strong negative deviation (a classic non-ideal pair) — not ideal.

Ethyl chloride + ethyl bromide: nearly identical size, shape and polarity (both are simple haloethanes), so A-A, B-B, A-B forces are essentially equal — ideal solution.

Common Mistakes

Assuming any two liquids that mix completely form an ideal solution — miscibility alone doesn't guarantee ideality.

Confusing chloroform-acetone (classic negative-deviation example) with an ideal pair.

✓Final answer
$The correct option is (A) — ethyl chloride, ethyl bromide.$

ANSWER: A
AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
$C_2H_5OH\ \&\ H_2O$
›Reveal solutionSolution
An ideal solution requires nearly identical A–A, B–B, and A–B intermolecular interactions; benzene + toluene fit this best among the given pairs.

Concept and Intuition

An ideal solution obeys Raoult's law across the whole concentration range, which physically requires that the solute-solute, solvent-solvent, and solute-solvent interactions all be of very similar strength (so mixing causes no significant enthalpy or volume change). This happens when the two components are chemically and structurally very similar — same functional groups, similar size and polarity.

Step-by-Step Solution

$C_2H_5OH \& H_2O$ (option A): ethanol and water show strong, dissimilar H-bonding patterns and significant negative/positive deviations — not ideal.

$HNO_3 \& H_2O$ (option B): strong acid-base/ionization interactions dominate — large negative deviation, not ideal.

$CHCl_3 \& CH_3COCH_3$ (option C): chloroform and acetone form a strong H-bond ( $\ce{C-H\cdots O=C}$ ) leading to significant negative deviation — a classic non-ideal pair, not ideal.

$C_6H_6 \& C_6H_5CH_3$ (option D): benzene and toluene are both non-polar aromatic hydrocarbons of similar size/shape with only van der Waals interactions of comparable strength — this pair is the standard example of a near-ideal solution.

Common Mistakes

Assuming any two liquids that mix completely (miscible) form an ideal solution — miscibility doesn't imply ideality; the interaction strengths must also match closely.

✓Final answer
$C_6H_6\ \&\ C_6H_5CH_3$

ANSWER: D
AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$Q. Which of the following is not an ideal solution? (A) Benzene and Toluene (B) Chloro-benzene and 1,2-dichloro benzene (C) Methyl iodide and Isopropanol (D) Ethyl bromide and Methyl bromide$
›Reveal solutionSolution
Ideal solutions need near-identical A–A, B–B, and A–B intermolecular interactions; mixing polar, non-H-bonding methyl iodide with H-bonded isopropanol breaks the alcohol's hydrogen bonds, so this pair is non-ideal.

Concept and Intuition

A solution behaves ideally (obeys Raoult's law over the whole composition range, $\Delta H_{mix}=0$ , $\Delta V_{mix}=0$ ) when the two components are so structurally alike that molecules of A and B interact with each other exactly as they interact with themselves. Classic ideal pairs are structural analogues: benzene/toluene (same ring, one extra methyl), chlorobenzene/1,2-dichlorobenzene (same ring, one extra Cl), ethyl bromide/methyl bromide (same halide, homologous alkyl chain). Isopropanol, however, is strongly hydrogen-bonded to itself; introducing methyl iodide (which cannot hydrogen-bond) breaks some of these O–H···O interactions, weakening net attractive forces and causing the mixture to show positive deviation from Raoult's law — a hallmark of non-ideal behaviour.

Step-by-Step Solution

Compare each pair for structural/chemical similarity.

Benzene-toluene, chlorobenzene/1,2-dichlorobenzene, and ethyl bromide/methyl bromide are each homologous or near-identical structurally → ideal.

Methyl iodide and isopropanol differ fundamentally: one is a non-associated polar halide, the other a hydrogen-bonded alcohol → mixing disrupts H-bonding → non-ideal (positive deviation).

Common Mistakes

Assuming any two liquids that mix completely must form an ideal solution — miscibility alone doesn't guarantee ideality.

✓Final answer
$The correct option is (C) — Methyl iodide and Isopropanol.$

ANSWER: C
AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$\Delta_{mix}H = 0\ ;\ \Delta_{mix}V < 0$
›Reveal solutionSolution
n-Hexane and n-heptane are chemically very similar nonpolar hydrocarbons, so their solution is essentially ideal: zero heat of mixing and zero volume change on mixing.

Concept and Intuition

An ideal solution is defined by having solute-solvent (here, hexane-heptane) interactions that are essentially identical in strength to the solute-solute and solvent-solvent interactions. n-Hexane and n-heptane are both straight-chain nonpolar alkanes differing by only one $CH_2$ unit — their van der Waals interactions with each other are nearly indistinguishable from their interactions with themselves, so mixing causes no net energy change and no net volume change (molecules pack together just as efficiently as in the pure liquids).

Step-by-Step Solution

Both liquids are nonpolar hydrocarbons with very similar molecular size and intermolecular (London dispersion) forces.

Since A–B interactions (hexane–heptane) closely match A–A and B–B interactions in strength, the enthalpy of mixing is essentially zero: $\Delta_{mix}H\approx0$ .

With no significant differences in molecular packing or interaction strength, the volume of the mixture equals the sum of the pure component volumes: $\Delta_{mix}V\approx0$ .

This combination (zero $\Delta H$ , zero $\Delta V$ ) is the defining signature of an ideal solution, obeying Raoult's law across the full composition range.

Common Mistakes

Assuming any hydrocarbon mixture must show some volume contraction or expansion — that's only true for mixtures of dissimilar-sized or dissimilar-polarity molecules.

Confusing this ideal-solution case with a real/non-ideal solution (e.g. ethanol-water) where hydrogen bonding differences cause noticeable $\Delta_{mix}H$ and $\Delta_{mix}V$ .

✓Final answer
$\Delta_{mix}H = 0\ ;\ \Delta_{mix}V = 0$

ANSWER: B

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