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.

COMEDK 2026Set 2026-A1 markMCQ
$Q. The system that forms maximum boiling azeotrope is: (A) Benzene-toluene (B) Carbon-di-sulphide-acetone (C) Ethyl alcohol-water (D) Acetone-chloroform$
›Reveal solutionSolution
A maximum‑boiling azeotrope forms when the solution’s vapour pressure is lower than Raoult’s law predicts (negative deviation), so the boiling point is higher than either pure component. Among the given pairs, only acetone‑chloroform shows strong negative deviation due to hydrogen bonding, making (D) the correct answer.

Concept and Intuition

An azeotrope is a mixture that boils at a constant composition because the vapour has the same composition as the liquid. For a maximum‑boiling azeotrope, the boiling point of the mixture is higher than that of either pure component. This happens when the intermolecular attractions between the two different molecules are stronger than those in the pure liquids — a situation called negative deviation from Raoult’s law. The stronger cross‑attractions make it harder for molecules to escape into the vapour, lowering the vapour pressure and raising the boiling point.

The key is to identify which pair forms such strong cross‑interactions. Let’s examine each option.

Step‑by‑Step Reasoning

Option (A): Benzene‑toluene

Benzene and toluene are both non‑polar hydrocarbons with similar structures. Their intermolecular forces (dispersion forces) are nearly identical, so the mixture behaves almost ideally — no significant deviation from Raoult’s law. No azeotrope forms; the boiling point lies between those of the pure components.

→ Not a maximum‑boiling azeotrope.

Option (B): Carbon disulphide‑acetone

Carbon disulphide ( $\mathrm{CS_2}$ ) is non‑polar, while acetone ( $\mathrm{CH_3COCH_3}$ ) is polar. Mixing them disrupts the dipole‑dipole interactions in acetone, making it easier for molecules to escape. This gives a positive deviation from Raoult’s law (higher vapour pressure, lower boiling point). Such systems form minimum‑boiling azeotropes, not maximum‑boiling ones.

→ Incorrect.

Option (C): Ethyl alcohol‑water

Ethanol and water both form strong hydrogen bonds. However, when mixed, the hydrogen‑bond network in each pure liquid is partially broken, and the cross‑hydrogen bonds (ethanol‑water) are weaker than the sum of the pure‑liquid interactions. This leads to a positive deviation (higher vapour pressure, lower boiling point). Ethanol‑water indeed forms a minimum‑boiling azeotrope (boiling point ~78.2 °C, lower than both pure ethanol and water).

→ Incorrect.

Option (D): Acetone‑chloroform

Acetone ( $\mathrm{CH_3COCH_3}$ ) has a carbonyl oxygen with lone pairs; chloroform ( $\mathrm{CHCl_3}$ ) has a hydrogen attached to three electronegative chlorines. The hydrogen in chloroform is sufficiently acidic to form a hydrogen bond with the oxygen of acetone:

$\mathrm{CHCl_3 \cdots O=C(CH_3)_2}$
This cross‑hydrogen bond is stronger than the dipole‑dipole interactions in pure acetone or pure chloroform. As a result, the vapour pressure of the mixture is lower than predicted by Raoult’s law (negative deviation), and the boiling point is higher than either pure component. The system forms a maximum‑boiling azeotrope at about 64.5 °C (pure acetone boils at 56 °C, pure chloroform at 61 °C).

→ Correct.

Watch out

A common mistake is to assume that any mixture of polar compounds with hydrogen‑bonding ability will give a maximum‑boiling azeotrope. In fact, ethanol‑water (both hydrogen‑bond donors and acceptors) gives a minimum‑boiling azeotrope because the cross‑bonds are weaker. The key is whether the cross‑interaction is stronger than the average of the pure‑component interactions — as in acetone‑chloroform, where one molecule is a strong donor and the other a strong acceptor.

Tip

A quick mental check: maximum‑boiling azeotropes often involve a proton donor (like chloroform, $\mathrm{CHCl_3}$ , or water) and a proton acceptor (like acetone, a ketone) that form a specific, strong hydrogen bond. If the mixture shows negative deviation from Raoult’s law, it’s a candidate.

✓Final answer
$The correct option is (D) . ANSWER: D$
COMEDK 2026Set 2026-A1 markMCQ
$Q. When 1 mole of benzene is mixed with 1 mole of toluene, the vapours will contain (A) Higher percentage of benzene (B) Equal mass of benzene and toluene since it is an ideal solution (C) Higher percentage of toluene (D) Equal number of moles of benzene and toluene since it is an ideal solution$
›Reveal solutionSolution
In an ideal solution of benzene and toluene, the more volatile component (benzene) enriches the vapour phase, so the vapour contains a higher percentage of benzene. The correct option is (A).

Concept & Intuition

Benzene and toluene form an ideal solution (Raoult’s law holds). At a given temperature, benzene has a higher vapour pressure than toluene — it is more volatile. When a liquid mixture evaporates, the vapour is richer in the more volatile component. This is the key idea behind distillation: the vapour composition differs from the liquid composition. Here, with equal moles in the liquid, the vapour will not be 50:50; it will be skewed toward benzene.

Step-by-step reasoning

Identify the vapour pressures

At 25 °C, pure benzene has vapour pressure $P_B^0 \approx 95.1$  mm Hg, and pure toluene has $P_T^0 \approx 28.4$  mm Hg. (Exact values depend on temperature, but the key is $P_B^0 > P_T^0$ .)

Apply Raoult’s law for the liquid mixture

For 1 mole each, mole fractions in the liquid are $x_B = 0.5$ and $x_T = 0.5$ .

Partial pressures above the liquid:

$P_B = x_B P_B^0 = 0.5 \times 95.1 = 47.55\ \text{mm Hg}$ $P_T = x_T P_T^0 = 0.5 \times 28.4 = 14.2\ \text{mm Hg}$

Find the vapour composition The total pressure $P_{\text{total}} = P_B + P_T = 61.75$  mm Hg. Mole fraction of benzene in the vapour ( $y_B$ ) is given by Dalton’s law:

$y_B = \frac{P_B}{P_{\text{total}}} = \frac{47.55}{61.75} \approx 0.77$
So the vapour is about 77 % benzene and 23 % toluene.

Interpret the result Even though the liquid had equal moles, the vapour contains a higher percentage of benzene. This is a general property: the vapour is always enriched in the more volatile component.

Watch out

A common mistake is to assume that an ideal solution gives identical liquid and vapour compositions. That is false — ideality means Raoult’s law holds, not that compositions are equal. The vapour composition depends on the relative volatilities.

Tip

For a binary ideal solution, the vapour mole fraction of the more volatile component is always greater than its liquid mole fraction. This is the basis of fractional distillation.

✓Final answer
$The correct option is (A) . ANSWER: A$
COMEDK 2025Set 2025-E1 markMCQ
$P_X^0=1400 \quad P_Y^0=600$
›Reveal solutionSolution
Using Raoult’s law for an ideal solution, we set up two equations from the two given conditions (initial and after adding more X) and solve for the pure vapour pressures. The result is $P_X^0 = 1200$ mm and $P_Y^0 = 600$ mm, which corresponds to option (A).

Concept & Intuition

Raoult’s law states that for an ideal solution, the partial vapour pressure of each component equals its mole fraction times its pure vapour pressure. The total vapour pressure is the sum of these partial pressures. Here we have two different compositions (before and after adding extra X), each giving a total pressure. That gives two linear equations in the two unknowns $P_X^0$ and $P_Y^0$ . Solving them is straightforward — no tricks, just careful algebra.

Step-by-step solution

Initial condition Moles of X = 1, moles of Y = 1 → total moles = 2. Mole fractions:

$x_X = \frac{1}{2}, \quad x_Y = \frac{1}{2}$
Total vapour pressure $P_{\text{total}} = 1000$ mm.

By Raoult’s law:
$P_{\text{total}} = x_X P_X^0 + x_Y P_Y^0 = \frac{1}{2}P_X^0 + \frac{1}{2}P_Y^0$
Multiply by 2:
$P_X^0 + P_Y^0 = 2000 \quad \text{(Equation 1)}$

After adding 2.0 moles of X New moles: X = 1 + 2 = 3, Y = 1 (unchanged) → total moles = 4. New mole fractions:

$x_X' = \frac{3}{4}, \quad x_Y' = \frac{1}{4}$
Vapour pressure increases by 200 mm, so new total pressure = $1000 + 200 = 1200$ mm.

Raoult’s law gives:
$1200 = \frac{3}{4}P_X^0 + \frac{1}{4}P_Y^0$
Multiply by 4:
$4800 = 3P_X^0 + P_Y^0 \quad \text{(Equation 2)}$

Solve the system Subtract Equation 1 from Equation 2:

$(3P_X^0 + P_Y^0) - (P_X^0 + P_Y^0) = 4800 - 2000$ $2P_X^0 = 2800 \quad \Rightarrow \quad P_X^0 = 1400 \text{ mm}$
Substitute into Equation 1:
$1400 + P_Y^0 = 2000 \quad \Rightarrow \quad P_Y^0 = 600 \text{ mm}$
Watch out

A common mistake is to forget that the total pressure increases by 200 mm, not becomes 200 mm. Always read “increases by” as addition to the original value.

Tip

Notice that the sum $P_X^0 + P_Y^0$ is fixed by the first condition (2000 mm). The second condition then gives a weighted sum that lets you isolate each value cleanly.

✓Final answer
$The correct option is (A) . ANSWER: A$
COMEDK 2025Set 2025-M1 markMCQ
$a / 20$
›Reveal solutionSolution
Using Raoult’s law for an ideal solution and Dalton’s law for the vapour phase, the mole fraction of Y in the vapour is found to be 0.75, which equals $a/20$ ; solving gives $a = 15$ .

Concept & Intuition

For an ideal solution of two volatile liquids, the vapour above the solution is not simply the same composition as the liquid. Instead, the more volatile component (higher pure vapour pressure) enriches the vapour. Raoult’s law gives the partial pressure of each component in the vapour:
$P_X = x_X P_X^\circ, \quad P_Y = x_Y P_Y^\circ$
where $x$ is the mole fraction in the liquid and $P^\circ$ is the pure vapour pressure. Dalton’s law then tells us the total pressure is the sum of partial pressures, and the mole fraction of a component in the vapour is its partial pressure divided by the total pressure. So we can compute the vapour composition directly from the liquid composition and pure vapour pressures.

Step-by-step solution

Identify given data

Pure vapour pressures: $P_X^\circ = 100\ \text{mm Hg}$ , $P_Y^\circ = 200\ \text{mm Hg}$

Mole fraction of X in the liquid: $x_X = 0.4$

Since it’s a binary solution, $x_Y = 1 - x_X = 0.6$

Mole fraction of Y in the vapour is given as $\frac{a}{20}$ .

Apply Raoult’s law to find partial pressures

$P_X = x_X P_X^\circ = 0.4 \times 100 = 40\ \text{mm Hg}$ $P_Y = x_Y P_Y^\circ = 0.6 \times 200 = 120\ \text{mm Hg}$

Find total vapour pressure

$P_{\text{total}} = P_X + P_Y = 40 + 120 = 160\ \text{mm Hg}$

Compute mole fraction of Y in the vapour By Dalton’s law, the mole fraction of Y in the vapour phase is:

$y_Y = \frac{P_Y}{P_{\text{total}}} = \frac{120}{160} = 0.75$

Relate to the given expression The problem states $y_Y = \frac{a}{20}$ . Therefore:

$\frac{a}{20} = 0.75 \quad \Rightarrow \quad a = 20 \times 0.75 = 15$
Tip

A quick check: Since Y is more volatile (higher $P^\circ$ ), its vapour mole fraction (0.75) is indeed larger than its liquid mole fraction (0.6) — always true for the more volatile component in an ideal solution.

Watch out

A common mistake is to confuse liquid mole fractions with vapour mole fractions. Here, $x_Y = 0.6$ is not the vapour composition; you must compute $y_Y$ via partial pressures.

✓Final answer
$The correct option is (B) . ANSWER: B$
COMEDK 2025Set 2025-M1 markMCQ
$60 \%$
›Reveal solutionSolution
The key idea is that the relative lowering of vapour pressure equals the mole fraction of the solute. Using the given 60% reduction, we set up the mole fraction equation and solve for the molar mass of A, obtaining 60 g/mol.

Concept & Intuition

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution is lower than that of the pure solvent. Raoult’s law tells us that the relative lowering of vapour pressure is exactly equal to the mole fraction of the solute. Here, the vapour pressure is reduced to 60% of its original value — meaning the lowering is 40% of the original. That fraction directly gives the mole fraction of solute, and from there we can find its molar mass.

Step-by-step solution

Interpret the vapour pressure reduction Let $P^0$ be the vapour pressure of pure xylene. The problem says the vapour pressure is reduced to $60\%$ of $P^0$ . So the new vapour pressure $P = 0.60\,P^0$ . The lowering of vapour pressure is:

$\Delta P = P^0 - P = P^0 - 0.60\,P^0 = 0.40\,P^0$
Hence the relative lowering is:
$\frac{\Delta P}{P^0} = 0.40$

Apply Raoult’s law for a non-volatile solute For a dilute solution of a non-volatile solute, Raoult’s law states:

$\frac{\Delta P}{P^0} = x_{\text{solute}}$
where $x_{\text{solute}}$ is the mole fraction of the solute. Therefore:
$x_{\text{solute}} = 0.40$

Express mole fraction in terms of masses and molar masses

Let $M_A$ be the molar mass of solute A (in g/mol).

Mass of solute = 60 g → moles of solute = $\frac{60}{M_A}$ .

Mass of solvent (xylene) = 212 g, molar mass of xylene = 106 g/mol → moles of solvent = $\frac{212}{106} = 2.00$ mol.

The mole fraction of solute is:

$x_{\text{solute}} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} = \frac{60/M_A}{\frac{60}{M_A} + 2.00}$

Set up the equation and solve We know $x_{\text{solute}} = 0.40$ , so:

$\frac{60/M_A}{\frac{60}{M_A} + 2} = 0.40$
Multiply both sides by the denominator:
$\frac{60}{M_A} = 0.40 \left( \frac{60}{M_A} + 2 \right)$ $\frac{60}{M_A} = \frac{24}{M_A} + 0.80$
Subtract $\frac{24}{M_A}$ from both sides:
$\frac{36}{M_A} = 0.80$
Hence:
$M_A = \frac{36}{0.80} = 45 \ \text{g/mol}$
Watch out

A common mistake is to think “reduced to 60%” means the lowering is 60%. But “reduced to 60%” means the final pressure is 60% of the original, so the lowering is 40%. Using 60% as the mole fraction would give a different (wrong) answer.

Tip

Notice that the moles of solvent came out as a nice round number (2.00 mol). This often happens in well-designed problems — it’s a clue that you’re on the right track.

✓Final answer
$The correct option is (D) . ANSWER: D$
COMEDK 2024Set 2024-A1 markMCQ
$Q. The system that forms minimum boiling azeotrope is : (A) Phenol-Aniline (B) Ethyl alcohol - Water (C) Acetone-Chloroform (D) Nitric Acid - Water$
›Reveal solutionSolution
A minimum-boiling azeotrope arises from a positive deviation from Raoult's law. Ethanol–water is the textbook example; phenol–aniline, acetone–chloroform and nitric acid–water all show negative deviations and form maximum-boiling azeotropes.

Minimum-boiling azeotropes form in solutions showing positive deviation (weaker A–B interactions than A–A/B–B, higher vapour pressure, lower boiling point).

(A) Phenol–aniline: strong O–H···N hydrogen bonding between components → negative deviation → maximum-boiling.

(B) Ethyl alcohol–water: positive deviation → minimum-boiling azeotrope (≈95.6% ethanol by mass).

(C) Acetone–chloroform: new H-bond (C–H···O) between them → negative deviation → maximum-boiling.

(D) Nitric acid–water: negative deviation → maximum-boiling azeotrope (≈68% HNO $_3$ ).

✓Final answer
$The correct option is (B) — Ethyl alcohol - Water$
COMEDK 2023Set 2023-M1 markMCQ
$Q. Which of the following shows negative deviation from Raoult's law? (A) Benzene-acetone (B) Benzene-chloroform (C) Benzene-ethanol (D) Benzene-carbon tetrachloride$
›Reveal solutionSolution
Negative deviation occurs when the new A–B interactions are stronger than the average of A–A and B–B interactions. Among the pairs listed, benzene–chloroform develops extra attraction (weak C–H··· $\pi$ interaction), lowering vapour pressure — a negative deviation.

Negative deviation from Raoult's law $\Rightarrow$ observed vapour pressure lower than ideal, arising when solute–solvent attractions exceed solvent–solvent and solute–solute attractions.

Evaluating the mixtures:

Benzene–acetone: breaks acetone's dipole interactions; weaker new forces $\Rightarrow$ positive deviation.

Benzene–chloroform: chloroform's acidic C–H forms a weak attractive interaction with the benzene $\pi$ -cloud, giving stronger A–B forces $\Rightarrow$ negative deviation.

Benzene–ethanol: disrupts ethanol H-bonding $\Rightarrow$ positive deviation.

Benzene– $\mathrm{CCl_4}$ : similar non-polar molecules $\Rightarrow$ nearly ideal / slight positive deviation.

Hence benzene–chloroform is the negative-deviation system.

✓Final answer
$The correct option is (B) — Benzene-chloroform.$
COMEDK 2022Set 20221 markMCQ
$Q. Which among the following forms minimum boiling azeotropes?$
(i) Heptane + Octane
(ii) Water + Nitric acid
(iii) Ethanol + Water
(iv) Acetone + Carbon dioxide (A) (i), (ii),
(iv) (B) (i),
(ii) only (C) (i), (iii),
(iv) (D)
(iv) only
›Reveal solutionSolution
So the genuine minimum-boiling pairs are (iii) and (iv). Option (ii) is definitely excluded (it is maximum boiling), which rules out options (A) and (B). Option (D) omits ethanol + water, the canonical minimum-boiling azeotrope. The only option containing both (iii) and (iv) is (C).

Concept: Azeotropes. A minimum-boiling azeotrope is formed by solutions showing a LARGE POSITIVE deviation from Raoult's law; a maximum-boiling azeotrope by solutions showing a large negative deviation.

Heptane + octane: chemically very similar, near-ideal solution - strictly no azeotrope at all.

Water + nitric acid: strong H-bonding/interaction -> negative deviation -> MAXIMUM boiling azeotrope (not minimum).

Ethanol + water: the textbook example of a positive deviation -> MINIMUM boiling azeotrope (95% ethanol, 351.1 K). YES.

Acetone + carbon disulphide (printed here as 'carbon dioxide' - the standard NCERT pair is acetone + CS2): positive deviation -> MINIMUM boiling azeotrope. YES. So the genuine minimum-boiling pairs are (iii) and (iv). Option (ii) is definitely excluded (it is maximum boiling), which rules out options (A) and (B). Option (D) omits ethanol + water, the canonical minimum-boiling azeotrope. The only option containing both (iii) and (iv) is (C).
✓Final answer
$The correct option is (C) — (i), (iii), (iv)$
ANSWER: C
COMEDK 2021Set 20211 markMCQ
$_2$
›Reveal solutionSolution
Chlorobenzene + bromobenzene: nearly ideal (very similar structures) - no azeotrope - C2H5Br + C2H5Cl: nearly ideal - no azeotrope - n-hexane + n-heptane: classic ideal solution - no azeotrope - CCl4 + CHCl3: shows a measurable (positive) deviation from Raoult's law because CHCl3-CHCl3 / CCl4-CCl4 interactions differ from the mixed ones; this pair is the non-ideal one and forms an azeotropic mixture.

Concept: Azeotropes form only from NON-IDEAL solutions (those showing appreciable positive or negative deviation from Raoult's law). Ideal solutions never form azeotropes.

Chlorobenzene + bromobenzene: nearly ideal (very similar structures) - no azeotrope

C2H5Br + C2H5Cl: nearly ideal - no azeotrope

n-hexane + n-heptane: classic ideal solution - no azeotrope

CCl4 + CHCl3: shows a measurable (positive) deviation from Raoult's law because CHCl3-CHCl3 / CCl4-CCl4 interactions differ from the mixed ones; this pair is the non-ideal one and forms an azeotropic mixture.

✓Final answer
$_4$

ANSWER: D

KCET 2019Set A-11 markMCQ

Q. Solution 'A' contains acetone dissolved in chloroform and solution 'B' contains acetone dissolved in carbon disulphide. The type of deviations from Raoult's law shown by solutions A and B, respectively are (A) positive and positive (B) negative and negative (C) positive and negative (D) negative and positive

›Reveal solution

Compare the A–B interaction with the A–A and B–B interactions: stronger A–B ⇒ negative deviation (acetone/chloroform), weaker A–B ⇒ positive deviation (acetone/ $CS_2$ ).

Step 1 — The rule behind Raoult's-law deviations.

For a solution of A and B, let $A\!-\!B$ be the new solute–solvent interaction and $A\!-\!A$ , $B\!-\!B$ the interactions in the pure liquids.

Comparison	Escaping tendency	Vapour pressure	Deviation	$\Delta_{mix}H$ , $\Delta_{mix}V$
$A\!-\!B$ stronger	decreases	$p < p_{\text{Raoult}}$	Negative	$<0$ (exothermic), $<0$
$A\!-\!B$ weaker	increases	$p > p_{\text{Raoult}}$	Positive	$>0$ (endothermic), $>0$

Step 2 — Solution A: acetone + chloroform.

Chloroform's C–H bond is strongly polarised by its three electronegative chlorine atoms, so its hydrogen is acidic enough to form a genuine hydrogen bond with the lone pair on the carbonyl oxygen of acetone:

\ce{Cl3C-H \cdots O=C(CH3)2}

This $A\!-\!B$ attraction is stronger than the dipole–dipole forces present in either pure liquid. The molecules are therefore held back from the vapour phase, the total vapour pressure falls below the Raoult prediction ⇒ negative deviation (this pair also forms a maximum-boiling azeotrope).

Step 3 — Solution B: acetone + carbon disulphide.

$CS_2$ is a linear, non-polar molecule (its two C=S bond dipoles cancel), so it can offer only weak dispersion forces to acetone. Mixing it with acetone breaks the relatively strong dipole–dipole attractions between acetone molecules and replaces them with much weaker acetone– $CS_2$ interactions. Molecules escape more easily, the vapour pressure exceeds the Raoult prediction ⇒ positive deviation (this pair forms a minimum-boiling azeotrope).

Step 4 — Read off the order asked.

Solution A ⇒ negative; Solution B ⇒ positive.

✓Final answer

The correct option is (D) negative and positive .

ANSWER: D

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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?

Concept understanding — Types Of Solutions

Types of Solutions: From Everyday Life to Chemistry

The Intuition: A Cup of Tea

The Precise Statement

Breaking Down Each Type

A Quick Comparison

Why This Matters

Why this formula?

Types of Solutions: Why the Key Formulae Hold

1. The Basic Classification: What Makes a Solution?

2. Raoult's Law: The Foundation

Why does this hold?

3. Relative Lowering of Vapour Pressure

Derivation in one line:

4. Elevation of Boiling Point

Why does boiling point rise?

5. Depression of Freezing Point

Why does freezing point drop?

6. Osmotic Pressure

Why does this hold?

7. The van't Hoff Factor $i$

Why is $i$ needed?

Quick Summary Table

Concept: Factors Affecting the Rate of Dissolution

Method: Comparative Analysis of Dissolution Rate Factors

Steps:

Final Answer:

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

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

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

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

Final Answer (for reference)

Concept and Intuition

Step‑by‑Step Reasoning

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?

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