Q.A beaker contains a solution of substance 'A'. Precipitation of substance 'A' takes place when small amount of 'A' is added to the solution. The solution is _________.
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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.
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:
Psolution=xsolvent⋅Psolvent0
Why does this hold?
Imagine a pure solvent surface. The vapour pressure P0 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 (xsolvent).
- Therefore, the rate of escape (vapour pressure) is proportional to that fraction:
Psolution∝xsolvent
- At the limit xsolvent=1, Psolution=P0, so the constant is P0.
Key insight: Raoult's law is a surface-area argument, not a volume argument.
3. Relative Lowering of Vapour Pressure
Formula:
P0P0−P=xsolute
Derivation in one line:
From Raoult's law:
P=xsolvent⋅P0
Since xsolvent+xsolute=1,
P=(1−xsolute)P0
⇒P0−P=xsolute⋅P0
⇒P0P0−P=xsolute
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:
ΔTb=Kb⋅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 ΔTb is proportional to the molality m (moles of solute per kg of solvent), because:
- More solute → greater vapour pressure lowering → more temperature needed.
- Kb (ebullioscopic constant) is a property of the solvent only.
5. Depression of Freezing Point
Formula:
ΔTf=Kf⋅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, ΔTf∝m, and Kf depends only on the solvent.
Common exam trap: Both ΔTb and ΔTf are colligative — they depend on number of solute particles, not their nature.
6. Osmotic Pressure
Formula:
Π=i⋅C⋅R⋅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). …
The key idea is supersaturation: a solution that holds more solute than its equilibrium solubility at that temperature. When even a tiny amount of additional solute is added, the excess solute immediately precipitates out until the solution becomes saturated.
Reasoning:
- If the solution were unsaturated, adding more solute would simply dissolve it — no precipitation.
- If it were saturated, adding more solute would leave it undissolved, but precipitation of the existing solute would not occur spontaneously. …
The key idea is that precipitation upon adding a small amount of solute indicates the solution is holding more dissolved solute than its equilibrium solubility - it is supersaturated. The correct option is (ii).
Why this works: The concept of saturation
- Unsaturated: More solute can still dissolve. Adding a little more simply dissolves it - no precipitation.
- Saturated: The solution holds exactly the maximum solute. Adding extra remains undissolved, but the existing dissolved amount stays unchanged.
- Supersaturated: The solution holds more solute than the equilibrium solubility - a metastable state. The slightest disturbance (adding a seed crystal or a tiny amount of solute) triggers rapid precipitation until the concentration drops to the saturation level.
The question says precipitation occurs when a small amount of 'A' is added - the classic behaviour of a supersaturated solution.
Step-by-step reasoning
- Rule out unsaturated (iii). Added solute would simply dissolve - no precipitation. …
Concept: Saturation and Supersaturation
When a small amount of solute is added to a solution, the behaviour tells us about the solution's state:
- Unsaturated: added solute dissolves completely.
- Saturated: added solute does not dissolve — it just settles.
- Supersaturated: added solute triggers immediate precipitation of excess solute already present.
Method: Observation of Solute Addition
Steps:
- Observe the effect — here, adding a small amount of 'A' causes precipitation of 'A'.
- Interpret the result: …
Here’s a breakdown of the common mistakes students make on this question, along with how to avoid each.
Mistake 1: Confusing "adding solute" with "adding a seed crystal"
- The error: Many students think that if you add more solute and it dissolves, the solution must be unsaturated. Here, the question says precipitation takes place when a small amount of 'A' is added. Students often miss the key trigger: the added solid acts as a seed or nucleation site.
- Why it’s wrong: In a supersaturated solution, the solute is already present in an amount greater than its solubility at that temperature. It is metastable. Adding even a tiny crystal of the solute provides a surface for the excess solute to crystallize out immediately.
- How to avoid: Always read the action carefully. If adding a small amount of the solid causes precipitation, the solution was holding more solute than it normally can — that’s the definition of supersaturated.
Mistake 2: Choosing "saturated" because precipitation occurs
- The error: Students recall that in a saturated solution, adding more solute causes it to settle at the bottom. They think this matches the description.
- Why it’s wrong: In a saturated solution, adding more solute does not cause precipitation of the original solute — the added solid simply does not dissolve and remains as a separate solid. The question says "precipitation of substance 'A' takes place," meaning the existing dissolved solute comes out of solution. That only happens in a supersaturated solution.
- How to avoid: Distinguish between:
- Saturated: Added solid stays undissolved (no change in dissolved amount).
- Supersaturated: Added solid triggers crystallization of dissolved solute.
Mistake 3: Thinking "concentrated" means the same as "supersaturated"
- The error: Students pick "concentrated" because they think a lot of solute is present.
- Why it’s wrong: "Concentrated" is a relative term (compared to a dilute solution) and does not imply that the solution is unstable or that precipitation will occur upon seeding. A concentrated solution can be unsaturated, saturated, or supersaturated. The key property here is metastability, not just concentration. …
- TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQQ.Which of the following does not show Tyndall effect? (A) Clouds (B) Milk (C) Sugar solution (D) Suspension
›Reveal solutionSolution
The Tyndall effect is the scattering of light by colloidal particles. Clouds, milk, and suspensions are colloids that scatter light, while a true solution like sugar solution does not. Therefore, the answer is (C).
The Tyndall effect is the visible scattering of light when a beam passes through a medium containing particles large enough to scatter it — typically in the colloidal size range (1–1000 nm). This is why a projector beam cuts through fog or why a laser becomes visible in a glass of milk. The key idea is that the particles must be of a size comparable to the wavelength of visible light; if they are too small (as in a true solution), the light passes through without scattering, and the beam remains invisible from the side.
Let’s examine each option:
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Clouds — Clouds are made of tiny water droplets or ice crystals suspended in air. These droplets are in the colloidal range, so they scatter sunlight strongly. That’s why clouds appear white (all colours scattered equally) and why you see a distinct beam of light through a gap in the clouds. Clouds definitely show the Tyndall effect.
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Milk — Milk is a classic example of a colloid: it contains fat globules and protein micelles dispersed in water. These particles are large enough to scatter light, which is why milk looks opaque and white. A beam of light passing through milk is clearly visible from the side. Milk shows the Tyndall effect. …
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- TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQQ.A liquid mixture is an ideal solution, ifa) It obeys ideal gas equationb) It obeys Raoult’s law at all concentrationsc) Solute – solute, solute – solvent and solvent – solvent interactions are similar (A) a only (B) a, b only (C) b, c only (D) c only
›Reveal solutionSolution
An ideal solution is defined by obedience to Raoult’s law at all concentrations, which in turn requires that all intermolecular interactions (solute-solute, solute-solvent, solvent-solvent) are similar. The correct option is (C).
The concept here is what makes a liquid mixture "ideal" in the thermodynamic sense. Unlike an ideal gas, which is about gas-phase behavior, an ideal solution is about how the components mix at the molecular level. The key condition is that the mixture obeys Raoult’s law — that is, the partial vapor pressure of each component is proportional to its mole fraction in the liquid phase. This law holds exactly only when the intermolecular forces between all pairs of molecules are identical, so that there is no net energy change or volume change upon mixing. Let’s examine each statement.
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Statement (a): "It obeys ideal gas equation"
This is irrelevant. The ideal gas equation (PV=nRT) describes the behavior of gases, not liquid mixtures. A liquid mixture is not a gas, so this condition has nothing to do with an ideal solution. Statement (a) is false.
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Statement (b): "It obeys Raoult’s law at all concentrations"
This is the defining criterion. Raoult’s law states that for a component i, Pi=xiPi∗, where xi is its mole fraction in the liquid and Pi∗ is its vapor pressure when pure. An ideal solution obeys this law exactly over the entire range of composition. Statement (b) is true.
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Statement (c): "Solute – solute, solute – solvent and solvent – solvent interactions are similar" …
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- TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQQ.Which of the following are correct for an ideal solution?a) ΔVmix=0b) Vsolvent+Vsolute=Vsolutionc) ΔHmix=0d) H2O+CO2→H2CO3 is an example of ideal solution. (A) a, b only (B) b, c only (C) a, b, c only (D) a, b, c, d
›Reveal solutionSolution
An ideal solution is defined by zero volume change and zero enthalpy change upon mixing, so statements a and c are correct; statement b is a restatement of a, and statement d describes a chemical reaction, not a solution. The correct option is (C).
The key concept is that an ideal solution is one where the intermolecular forces between all molecules (solute–solute, solvent–solvent, and solute–solvent) are identical. This means that when you mix the components, there is no net energy change and no net volume change — the mixture behaves as if the molecules simply “replace” each other without any interaction effects.
Let’s examine each statement:
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Statement a: ΔVmix=0
In an ideal solution, the volume of the mixture is exactly the sum of the volumes of the pure components before mixing. There is no contraction or expansion because the molecular packing is unchanged. This is a defining property. So a is correct.
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Statement b: Vsolvent+Vsolute=Vsolution
This is simply another way of saying ΔVmix=0 — the total volume after mixing equals the sum of the volumes before mixing. So b is also correct (it’s equivalent to a).
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Statement c: ΔHmix=0
Because the intermolecular forces are all the same, no heat is absorbed or released when mixing. The enthalpy change is zero. This is the other defining property of an ideal solution. So c is correct.
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Statement d: H2O+CO2→H2CO3 is an example of an ideal solution. …
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- TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQQ.A and B on mixing form an ideal solution at room temperature. Which of the following options is correct for this process? (A) ΔG System − \quad ΔS System + \quad ΔS Surroundings + \quad ΔH + (B) ΔG System + \quad ΔS System + \quad ΔS Surroundings 0 \quad ΔH + (C) ΔG System − \quad ΔS System + \quad ΔS Surroundings 0 \quad ΔH 0 (D) ΔG System − \quad ΔS System − \quad ΔS Surroundings + \quad ΔH +
›Reveal solutionSolution
For mixing of ideal solutions at room temperature, the process is spontaneous (ΔG<0) and driven by an increase in entropy of the system (ΔSsys>0), while enthalpy change is zero (ΔH=0) and surroundings experience no entropy change (ΔSsurr=0). The correct option is (C).
The key idea here is that an ideal solution is defined by having no change in enthalpy or volume upon mixing — the intermolecular forces between unlike molecules are identical to those between like molecules. So mixing is purely an entropy-driven process.
When two pure substances A and B are mixed, the molecules have more possible arrangements in the mixture than they did in the separate pure states. This increase in microstates means the entropy of the system increases: ΔSsys>0.
Since ΔH=0 for ideal mixing, and the process occurs at constant temperature and pressure, the entropy change of the surroundings is given by ΔSsurr=−TΔH=0. The surroundings neither gain nor lose heat.
Spontaneity at constant T and P is governed by the Gibbs free energy change: ΔG=ΔH−TΔSsys. With ΔH=0 and ΔSsys>0, we get ΔG=−TΔSsys<0. The process is spontaneous.
Let’s match these signs to the options:
- ΔGsys: Negative (−) — spontaneous mixing.
- ΔSsys: Positive (+) — increased disorder.
- ΔSsurr: Zero (0) — no heat exchange with surroundings. …
- TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQQ.Which of the following mixture form an ideal solution? (A) CCl4+C7H8 (B) CHCl3+C6H6 (C) H2O+CH3OH (D) n−C6H14+n−C7H16
›Reveal solutionSolution
An ideal solution forms when the intermolecular forces between unlike molecules are nearly identical to those between like molecules. This happens for structurally similar, non-polar hydrocarbons. The correct pair is n-hexane and n-heptane, option (D).
An ideal solution obeys Raoult's law at all concentrations and temperatures. The key condition is that the solute-solvent interactions (A–B) must be equal in strength to the pure component interactions (A–A and B–B). When this holds, there is no volume change or enthalpy change on mixing — the solution is "ideal."
Let’s examine each pair.
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Option (A): CCl4+C7H8 (carbon tetrachloride + toluene)
CCl4 is a non-polar, symmetrical molecule. Toluene (C7H8) is also non-polar but has a slightly polarizable aromatic ring. While both are non-polar, their molecular shapes and sizes differ enough that the intermolecular forces are not perfectly matched. In fact, CCl4 and toluene show slight positive deviation from Raoult’s law — not ideal.
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Option (B): CHCl3+C6H6 (chloroform + benzene)
Chloroform has a polar C–H bond and can form weak hydrogen bonds with benzene’s π-electron cloud. This creates stronger A–B interactions than the pure A–A or B–B interactions. The result is a negative deviation from Raoult’s law — definitely not ideal.
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Option (C): H2O+CH3OH (water + methanol)
Both are strongly hydrogen-bonded. However, water’s hydrogen-bond network is more structured than methanol’s. When mixed, the interactions are not identical — there is a significant enthalpy change and volume contraction. This mixture shows positive deviation and is far from ideal.
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Option (D): n−C6H14+n−C7H16 (n-hexane + n-heptane) …
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