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આ MCQ મોડ્યુલ આના પર આધારિત છે: Types Concentration Solubility
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Types Concentration Solubility
🎓 Class 12
Chemistry
CBSE
Theory
Ch 1 – Solutions
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આ મૂલ્યાંકન આના પર આધારિત હશે: Types Concentration Solubility
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Types Concentration Solubility
Introduction: Why Solutions Matter
Step into a pharmacy, a chemistry lab, or even your own bloodstream, and you will find yourself surrounded by solutions. Almost every reaction inside a living organism — from digestion to nerve signalling — happens in aqueous solution. The salinity of sea water, the taste of a soft drink, the strength of a disinfectant, the therapeutic dose of an IV fluid, all rest on ideas we will develop in this chapter.
A striking example: in drinking water, a fluoride concentration of about 1 part per million (ppm) is desirable — it strengthens tooth enamel. Push that figure up to about 1.5 ppm and the same ion causes dental fluorosis, the brown mottling of teeth seen in many regions of India. A difference of a few milligrams per tonne separates a healthy smile from a dental disaster, and chemistry gives us the vocabulary to describe it.
Definition: A solution is a homogeneous mixture of two or more chemically non-reacting components. The component present in larger amount is the solvent; the component(s) present in smaller amount are the solute(s).
1.1 Types of Solutions
A solution containing two components is called a binary solution; three components give a ternary solution, and so on. Depending on the physical state of the solvent (and hence the resulting solution), binary solutions fall into three broad families — gaseous, liquid and solid solutions — each with three sub-cases, giving nine types in all.
| Type | Solute | Solvent | Example |
|---|---|---|---|
| Gaseous | Gas | Gas | Air (mainly N₂ + O₂) |
| Gaseous | Liquid | Gas | Humid air (water vapour in air) |
| Gaseous | Solid | Gas | Sublimed iodine mixed with air |
| Liquid | Gas | Liquid | Dissolved oxygen in rivers; CO₂ in cold drinks |
| Liquid | Liquid | Liquid | Ethanol dissolved in water |
| Liquid | Solid | Liquid | Common salt or sugar in water |
| Solid | Gas | Solid | Hydrogen absorbed in palladium metal |
| Solid | Liquid | Solid | Sodium amalgam (Na dissolved in Hg solidifies) |
| Solid | Solid | Solid | Brass (Cu + Zn), bronze (Cu + Sn), 22-carat gold (Au + Cu) |
Table 1.1: The nine types of binary solutions.
1.2 Expressing the Concentration of Solutions
"How much solute is dissolved in how much solvent?" — this simple question has several answers depending on how we choose to measure the amounts. Each method has its own strengths and limitations.
1.2.1 Mass Percent — (w/w)
\(\text{Mass \%} \;=\; \dfrac{\text{mass of component}}{\text{total mass of solution}} \times 100\)
A 10 % w/w glucose solution means 10 g of glucose is present in 100 g of solution (i.e., 90 g water).
1.2.2 Volume Percent — (v/v)
\(\text{Volume \%} \;=\; \dfrac{\text{volume of component}}{\text{total volume of solution}} \times 100\)
A 10 % v/v ethanol–water solution contains 10 mL of ethanol in every 100 mL of solution. This is the unit typically used for antifreeze and rubbing alcohols.
1.2.3 Mass-by-Volume — (w/v)
Common in medicine: mass of solute (in grams) dissolved per 100 mL of solution. A 0.9 % w/v saline bag used as IV fluid has 0.9 g NaCl per 100 mL.
1.2.4 Parts per Million (ppm)
\(\text{ppm} \;=\; \dfrac{\text{number of parts of component}}{\text{total parts of all components}} \times 10^6\)
Used for trace amounts — e.g., fluoride in drinking water, pollutants in air, dissolved oxygen (DO) in river water. Concentrations in ppm can be expressed by mass/mass, volume/volume or mass/volume.
1.2.5 Mole Fraction (x)
\(x_{A} \;=\; \dfrac{n_{A}}{n_{A}+n_{B}}, \qquad x_{B} \;=\; \dfrac{n_{B}}{n_{A}+n_{B}}, \qquad x_{A}+x_{B}=1\)
Useful in gas-phase chemistry and when describing vapour pressures (Section 1.4). Being a ratio of moles, mole fraction is dimensionless and temperature-independent.
1.2.6 Molarity (M)
\(M \;=\; \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}\)
Unit: mol L⁻¹. Extremely popular in laboratory practice because you simply weigh the solute and top up to a mark in a volumetric flask. The catch — volume changes with temperature, so molarity is T-dependent.
1.2.7 Molality (m)
\(m \;=\; \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}\)
Unit: mol kg⁻¹. Mass of solvent does not depend on T, so molality is temperature-independent and is preferred whenever we will later change the temperature (e.g., in colligative-property calculations).
1.2.8 Normality (N) — brief mention
Normality = number of gram-equivalents of solute per litre of solution. It is related to molarity through the n-factor (number of H⁺, OH⁻ or e⁻ exchanged per formula unit). N = n × M. Though no longer preferred by IUPAC, you will still meet it in acid–base titrations.
Compare and remember: Percentages and molarity involve volume or total mass, so they change with temperature. Mole fraction and molality involve only masses / moles, so they are independent of temperature.
Worked Examples — Concentration Terms
Example 1.1 — Mass percent of benzene in CCl₄
A solution contains 22 g of benzene (C₆H₆) dissolved in 122 g of carbon tetrachloride (CCl₄). Find the mass percent of benzene.
\(\text{Total mass} = 22\,\text{g} + 122\,\text{g} = 144\,\text{g}\)
\(\text{Mass \%}_{C_6H_6} = \dfrac{22}{144}\times 100 = 15.28\,\%\)
Answer: The solution is 15.28 % benzene by mass.
Example 1.2 — Molarity calculation
Calculate the molarity of a solution prepared by dissolving 5.00 g of NaOH in enough water to make 450 mL of solution.
Step 1 — moles of NaOH (MNaOH = 23 + 16 + 1 = 40 g mol⁻¹):
\(n_{NaOH} = \dfrac{5.00}{40} = 0.125\,\text{mol}\)
Step 2 — volume of solution in L: 450 mL = 0.450 L.
\(M = \dfrac{0.125}{0.450} = 0.278\,\text{mol L}^{-1}\)
Answer: 0.278 M NaOH.
Example 1.3 — Mole fraction from masses
Calculate the mole fraction of ethylene glycol (C₂H₆O₂, M = 62 g mol⁻¹) in a solution containing 20 % (by mass) of glycol in water.
Take 100 g of solution → 20 g glycol + 80 g water (MH₂O = 18).
\(n_{glycol}=\dfrac{20}{62}=0.3226\,\text{mol};\quad n_{water}=\dfrac{80}{18}=4.444\,\text{mol}\)
\(x_{glycol}=\dfrac{0.3226}{0.3226+4.444}=\dfrac{0.3226}{4.7666}=0.0677\)
Answer: xglycol ≈ 0.068, xwater ≈ 0.932.
Example 1.4 — Molality
Calculate the molality of a solution in which 2.5 g of ethanoic acid (CH₃COOH, M = 60) is dissolved in 75 g of benzene.
\(n_{CH_3COOH}=\dfrac{2.5}{60}=0.04167\,\text{mol}\)
\(m = \dfrac{0.04167\,\text{mol}}{0.075\,\text{kg}} = 0.556\,\text{mol kg}^{-1}\)
Answer: m ≈ 0.556 mol kg⁻¹.
Example 1.5 — Convert molality to mole fraction
A 1.00 m aqueous solution of urea (NH₂CONH₂) contains 1 mol urea per 1 kg water. Find xurea.
\(n_{water}=\dfrac{1000}{18}=55.56\,\text{mol}\)
\(x_{urea}=\dfrac{1}{1+55.56}=\dfrac{1}{56.56}=0.01768\)
Answer: xurea ≈ 0.0177.
1.3 Solubility
The solubility of a substance is the maximum amount of it that can dissolve in a specified quantity of solvent at a given temperature. Every solid–liquid or gas–liquid pair has its own solubility, and that number tells us whether a solution is unsaturated, saturated, or supersaturated.
1.3.1 Solubility of a Solid in a Liquid
When a solid is added to a liquid, two opposing processes begin: dissolution (solute → solution) and crystallisation (solution → solute). At a fixed temperature an equilibrium is reached at which the two rates are equal — the solution is said to be saturated. The concentration at this point is the solubility.
"Like dissolves like"
A polar solute (sugar, salt, urea) dissolves easily in a polar solvent (water); a non-polar solute (naphthalene, fats) dissolves in a non-polar solvent (benzene, hexane). Mismatched pairs do not mix — oil and water famously separate because water's hydrogen-bonded network has no favourable interaction with an oil molecule's non-polar chain.
Effect of temperature
Apply Le Chatelier's principle to the dissolution equilibrium:
- If dissolution is endothermic (ΔsolH > 0), heating the saturated solution increases solubility. Most solids behave this way — KNO₃, sugar, NH₄Cl.
- If dissolution is exothermic (ΔsolH < 0), heating decreases solubility. Examples: Ce₂(SO₄)₃, Li₂CO₃.
Effect of pressure
Pressure has almost no effect on the solubility of a solid in a liquid — both the solid and the liquid are essentially incompressible.
1.3.2 Solubility of a Gas in a Liquid
Unlike solids, gases are very compressible, so pressure has a large effect on how much gas dissolves.
Henry's Law
Henry's law: At a constant temperature, the partial pressure of a gas (p) over a solution is proportional to the mole fraction (x) of the dissolved gas in the liquid.
\(p \;=\; K_{H}\, x\)
where KH is Henry's-law constant, which depends on the gas, the solvent and the temperature.A larger KH means less gas dissolves at the same partial pressure. For oxygen in water at 293 K, KH = 34.86 kbar; for carbon dioxide, only 1.67 kbar — which is why CO₂ dissolves much more easily than O₂.
Effect of temperature
Dissolution of a gas in a liquid is always exothermic (gas molecules lose kinetic energy when they join the solution). By Le Chatelier's principle, raising T decreases gas solubility. This is why boiled water tastes "flat" and why warm lake surfaces hold less dissolved oxygen for fish.
Real-life applications
- Soft drinks: bottlers seal the drink under high CO₂ pressure (several bar) so that p × x = high value on the left side of Henry's law. Open the bottle → p drops → x must drop → bubbles rush out.
- Scuba diving ("bends"): under the high pressure of deep water, more N₂ dissolves in the diver's blood. A too-rapid ascent lowers p quickly; dissolved N₂ is released as bubbles in the tissues, causing agonising pain. Divers therefore breathe helium-enriched air (less soluble).
- Mountain climbers: at altitude the partial pressure of O₂ is lower, so less O₂ dissolves in the blood — this is the root cause of anoxia ("mountain sickness").
Worked Examples — Solubility & Henry's Law
Example 1.6 — Henry's law for CO₂ in soda water
The partial pressure of CO₂ above a soft drink at 298 K is 2.5 atm. If KH for CO₂ in water at that temperature is 1.67 × 10³ atm, find the mole fraction of CO₂ in the drink.
\(x_{CO_2}=\dfrac{p_{CO_2}}{K_{H}}=\dfrac{2.5}{1670}=1.497\times10^{-3}\)
Answer: xCO₂ ≈ 1.50 × 10⁻³. When the bottle is opened, p drops to ~0.0003 atm (atmospheric partial pressure of CO₂) and almost all of this CO₂ escapes as fizz.
Example 1.7 — Oxygen solubility in a lake
At 293 K, the Henry's-law constant for O₂ in water is 34.86 kbar. Atmospheric O₂ has a partial pressure of 0.21 atm (= 2.127 × 10⁻⁴ kbar). Estimate the mole fraction of O₂ dissolved.
\(x_{O_2}=\dfrac{2.127\times10^{-4}}{34.86}=6.10\times10^{-6}\)
This tiny mole fraction still corresponds to ~8 mg of O₂ per litre of water — just enough to support aquatic life, and one reason why warm, stagnant water kills fish (lower x as T rises).
Activity 1.1 — Prepare a 0.1 M NaCl Solution in the Kitchen
L3 Apply
Predict: If you dissolve 5.85 g (one level teaspoon) of common salt in 1 litre of water, will the solution be exactly 0.1 M? What could cause a small deviation?
- Weigh 5.85 g NaCl on a kitchen balance (MNaCl = 58.5 g mol⁻¹, so 5.85 g = 0.100 mol).
- Transfer to a 1-L measuring jug and add distilled water up to the 1-L mark.
- Stir to dissolve.
- Using the formula \(M=n/V\), compute the molarity.
- Now warm the jug on a stove (do not boil) for a few minutes. Is the molarity still exactly 0.1 M? Why?
Expected: At room temperature M ≈ 0.100 mol L⁻¹. On warming, the volume of the solution expands slightly, so the molarity falls by ~0.2 %. The molality (mol per kg solvent) however does not change — the kitchen-scale demonstration of why chemists prefer molality when the temperature may change (e.g., in boiling-point elevation).
Interactive: Concentration Converter L3 Apply
Enter one concentration for a water-based solution and read off the others. Assumes solute molar mass is user-supplied and density of dilute solution ≈ 1 g mL⁻¹.
Solute molar mass (g/mol):
Input type:
Enter values and press Convert.
Competency-Based Questions
A chemistry student prepares three aqueous solutions at 298 K: (i) 36 g glucose (C₆H₁₂O₆, M = 180 g mol⁻¹) in water to make 500 mL of solution, (ii) 5.85 g NaCl in 1 kg water, and (iii) a bottle of cola which has a CO₂ partial pressure of 2.5 atm in the gas space above the liquid (KH = 1.67 × 10³ atm).
Q1. L1 Remember The quantity "moles of solute per kilogram of solvent" is called:
Answer: B. Molality. Molarity uses volume of solution; normality uses equivalents; mole fraction is dimensionless.
Q2. L3 Apply Compute the molarity of the glucose solution (i). (2 marks)
n = 36/180 = 0.20 mol. V = 0.500 L. M = 0.20/0.500 = 0.40 mol L⁻¹.
Q3. L3 Apply Find the molality and mole fraction of NaCl in solution (ii). (3 marks)
nNaCl = 5.85/58.5 = 0.10 mol. Mass of water = 1 kg → m = 0.10 mol kg⁻¹. nwater = 1000/18 = 55.56 mol. xNaCl = 0.10/(0.10+55.56) = 1.80 × 10⁻³.
Q4. L3 Apply Using Henry's law, find xCO₂ in the cola. When opened, p falls to 0.0004 atm — compute the new xCO₂ and comment on the fizz. (3 marks)
Before opening: x = 2.5/1670 = 1.50 × 10⁻³. After opening: x = 4 × 10⁻⁴/1670 = 2.4 × 10⁻⁷. The solution is hugely supersaturated relative to the new pressure, so CO₂ rushes out as visible bubbles (the familiar fizz). Equilibrium is re-established only when x drops to the new Henry's-law value.
Q5. L4 Analyse The student left the glucose solution on a sunny windowsill; its temperature rose from 298 K to 313 K. Explain why its molarity decreased but its molality remained the same. (3 marks)
On heating, the solution's volume expands (water expands above 277 K). Molarity = n/V, so with V increasing and n constant, M decreases. Molality = n/mass of solvent; mass is T-independent, so m is unchanged. This is why molality is the preferred measure in boiling-point elevation and freezing-point depression experiments, where T is deliberately varied.
Assertion-Reason Questions
Assertion (A): A cold bottle of soda "fizzes" more violently when opened than a warm one.
Reason (R): Gases are less soluble in warmer liquids, so a warm bottle already holds less CO₂ in solution.
Answer: A. The cold bottle holds more dissolved CO₂ (small KH favours high x); the warm one holds less. When opened, the cold one has more gas to release — hence the louder fizz.
Assertion (A): The molality of a solution is independent of temperature.
Reason (R): Molality is defined per kilogram of solvent, and mass does not change with temperature.
Answer: A. Mass is invariant with T, while volume is not; therefore molality is preferred for T-dependent studies.
Assertion (A): Oil does not dissolve in water.
Reason (R): Non-polar oil molecules cannot form favourable interactions with polar water molecules, so the "like dissolves like" rule fails.
Answer: A. Oil is non-polar; water is strongly hydrogen-bonded. The two cannot interchange partners favourably — separate phases form.
Frequently Asked Questions - Types Concentration Solubility
What is the main concept covered in Types Concentration Solubility?
In NCERT Class 12 Chemistry Chapter 1 (Solutions), "Types Concentration Solubility" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Types Concentration Solubility useful in real-life or applied chemistry?
Real-life applications of "Types Concentration Solubility" from NCERT Class 12 Chemistry Chapter 1 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Types Concentration Solubility?
Key reactions in "Types Concentration Solubility" (NCERT Class 12 Chemistry Chapter 1 Solutions) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 1?
NCERT Class 12 Chemistry Chapter 1 (Solutions) is structured so each part builds chemical understanding sequentially. "Types Concentration Solubility" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Types Concentration Solubility?
CBSE board questions from "Types Concentration Solubility" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Types Concentration Solubility" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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