Match the items given in Column I with the type of solutions given in Column II. Column I: (i) Soda water; (ii

NCERT Exemplar · Q48

Q. Match the items given in Column I with the type of solutions given in Column II. Column I:

(i) Soda water;

(ii) Sugar solution;

(iii) German silver;

(iv) Air;

(v) Hydrogen gas in palladium.
Column II:

(a) A solution of gas in solid.

(b) A solution of gas in gas.

(d) A solution of solid in solid.

(e) A solution of gas in liquid.

(f) A solution of liquid in solid.

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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. …

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). …

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The key idea is to classify each mixture by the physical state of its solute and solvent. The correct matches are: (i)→(e), (ii)→(c), (iii)→(d), (iv)→(b), (v)→(a).

Concept and Intuition

A solution is a homogeneous mixture of two or more substances. The substance present in the larger amount is called the solvent, and the substance present in the smaller amount is called the solute. The type of solution is described by the phrase "a solution of [solute state] in [solvent state]".

To classify correctly, you must identify which component is the solute (the one being dissolved) and which is the solvent (the one doing the dissolving). The physical state of each at room temperature (or the given conditions) determines the classification.

Watch out

A common mistake is to reverse the solute and solvent. For example, in "soda water", water is the solvent (liquid) and carbon dioxide is the solute (gas). The correct classification is "a solution of gas in liquid", not "a solution of liquid in gas".

Step-by-Step Matching

1. (i) Soda water

Soda water is carbon dioxide gas dissolved in water. The solvent is water (liquid), and the solute is carbon dioxide (gas).

→ This is a solution of gas in liquid.

Match: (i) → (e)

2. (ii) Sugar solution

Sugar solution is sugar (solid) dissolved in water (liquid). The solvent is water (liquid), and the solute is sugar (solid).

→ This is a solution of solid in liquid.

Match: (ii) → (c)

3. (iii) German silver

German silver is an alloy of copper, zinc, and nickel. Alloys are solid solutions where one metal is dissolved in another. Here, all components are solids.

→ This is a solution of solid in solid.

Match: (iii) → (d)

4. (iv) Air …

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Here’s a breakdown of the common mistakes students make when matching types of solutions, along with how to avoid each.

Mistake 1: Confusing the solute and solvent in alloys (German silver)

What students do wrong:

Students often think German silver is a solution of solid in liquid (because it’s a metal alloy) or misidentify it as a solution of liquid in solid.

Why it happens:

They forget that an alloy is a solid-solid solution — both components are solids at room temperature.

How to avoid:

Remember: Alloys = solid in solid.
German silver is an alloy of copper, zinc, and nickel — all solids.
So the correct match is (iii) → (d).

Mistake 2: Thinking “soda water” is a solution of liquid in liquid

What students do wrong:

They see “water” and assume it’s a liquid-liquid solution, ignoring the dissolved gas.

Why it happens:

They focus on the solvent (water) and forget the solute (carbon dioxide gas).

How to avoid:

Identify the solute first: In soda water, CO₂ gas is dissolved in water.
So it’s gas in liquid → match (i) → (e).
Tip: If a drink fizzes, it contains dissolved gas.

Mistake 3: Misclassifying “air” as a solution of gas in liquid or solid

What students do wrong:

They sometimes match air with “gas in solid” or “gas in liquid” because they think of dust or moisture.

Why it happens:

They overcomplicate — air is primarily a mixture of gases (N₂, O₂, etc.) with no liquid or solid phase dominating.

How to avoid:

Air is a homogeneous mixture of gases → gas in gas.
Match (iv) → (b).
Ignore trace impurities unless the question specifies them.

Mistake 4: Confusing “hydrogen gas in palladium” with a gas-gas solution

What students do wrong:

They see “hydrogen gas” and “palladium” and think both are gases, or they match it with “gas in liquid”.

Why it happens:

They don’t know that palladium is a solid metal that can absorb hydrogen gas.

How to avoid:

Learn this classic example: Hydrogen in palladium is a gas in solid solution.
Match (v) → (a).
Remember: Palladium is a solid, so the solute (H₂ gas) is trapped in the solid lattice.

Mistake 5: Matching “sugar solution” with gas in liquid or solid in solid

What students do wrong: …

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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. …
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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. …
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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. …
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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. …
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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. …
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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: $\Delta_{mix}V\approx0$ . …
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Q.Match the items given in Column I with the type of solutions given in Column II. Column I:

Concept and Intuition

Step-by-Step Matching

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Q.Match the items given in Column I with the type of solutions given in Column II.
Column I:

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