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ICSE • Class 8 • Science • Ch 16
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Water

In ICSE Class 8 Science (Chemistry), "Water" provides an authoritative, experimentally rigorous master study guide investigating the chemistry of water, anomalous expansion, solutions, solubility curves, and hard vs soft water treatment. This comprehensive chapter explores Chemical Identity of Water (Universal solvent, molecular formula $\text{H}_2\text{O}$, bent polar geometry, hydrogen-to-oxygen volume ratio of $2 : 1$ and mass ratio of $1 : 8$; Verification of composition: Henry Cavendish synthesis and Electrolytic decomposition in Hoffmann's Voltameter: $2\text{H}_2\text{O} \to 2\text{H}_2 \uparrow \text{ [Cathode]} + \text{O}_2 \uparrow \text{ [Anode]}$), Physical Properties of Pure Water (Freezing point: $0^{\circ}\text{C}$, Boiling point: $100^{\circ}\text{C}$ at $76\text{ cm Hg}$; Latent heat of fusion: $336\text{ J/g} = 80\text{ cal/g}$; Latent heat of vaporisation: $2268\text{ J/g} = 540\text{ cal/g}$; Specific heat capacity: $4.186\text{ J/g}\cdot^{\circ}\text{C}$; Anomalous expansion between $0^{\circ}\text{C}$ and $4^{\circ}\text{C}$), Water as a Universal Solvent (Solutions: solute, solvent, saturated solution, unsaturated solution, supersaturated solution; Solubility: maximum grams of solute dissolved in $100\text{ g}$ of solvent at a given temperature; Solubility Curves; Effect of temperature on solubility: solids usually increase, gases decrease [thermal pollution]), Water of Crystallisation & Hydrated Salts (Fixed number of water molecules chemically bonded in crystal lattice: blue vitriol $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$, green vitriol $\text{FeSO}_4 \cdot 7\text{H}_2\text{O}$, washing soda $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$, Epsom salt $\text{MgSO}_4 \cdot 7\text{H}_2\text{O}$, gypsum $\text{CaSO}_4 \cdot 2\text{H}_2\text{O}$), Efflorescence, Deliquescence & Hygroscopy, and Hardness of Water (Temporary hardness [calcium and magnesium bicarbonates: $\text{Ca(HCO}_3)_2, \text{Mg(HCO}_3)_2$] removed by boiling or Clark's method; Permanent hardness [calcium and magnesium sulphates and chlorides: $\text{CaCl}_2, \text{CaSO}_4, \text{MgCl}_2, \text{MgSO}_4$] removed by washing soda process or Permutit / Ion-Exchange zeolite resins) aligned with the 2026–27 CISCE ICSE curriculum.

Why Does a Deep Lake in the Freezing Arctic Tundra Freeze Completely Solid on Top While Fish Swim Happily in Liquid Water Below?

In the dead of winter in Siberia, air temperatures plummet to a lethal $-50^{\circ}\text{C}$. Soil freezes rock-solid to a depth of meters. Every lake develops a thick sheet of white ice across its surface. Yet, if you drill a hole through that two-foot ice barrier and peer into the water, you will see fish, frogs, and aquatic weeds swimming peacefully in liquid water! Why didn't the lake freeze solid from the bottom up like an ice cube tray in a freezer? The answer is the miraculous ANOMALOUS EXPANSION OF WATER! Almost every liquid in the universe contracts and gets denser as it cools. But water is an eccentric rebel: as water cools down to $4^{\circ}\text{C}$, it contracts normally, reaching its maximum density of $1000\text{ kg/m}^3$ and sinking to the lake bottom. But as it cools below $4^{\circ}\text{C}$ toward $0^{\circ}\text{C}$, it abruptly expands! The colder, freezing water becomes lighter and floats to the surface, freezing into an insulating floating ice crust that traps heat below! What is the difference between Efflorescence and Deliquescence? How do ion-exchange beads turn hard, scale-forming water into pure soft water? Let's master water.

Why This Chapter Matters

Water chemistry governs planetary life support, industrial steam boiler operations, municipal potable water sanitation, hydrometallurgical processing, and agricultural crop irrigation. Mastering solubility curves, water of crystallisation, and hardness treatment is a core ICSE chemistry milestone.

Before You Begin (Prerequisites)

  • Hydrogen from Chapter 15.
  • Anomalous expansion of water from Chapter 2 and 6.
  • Solutions, mixtures, and physical changes from Chapters 10 and 11.

What You Will Learn (Core Objectives)

  • Explain the composition of water by volume ($2 : 1$) and by mass ($1 : 8$).
  • Describe the electrolysis of acidified water using Hoffmann's Voltameter.
  • Define solubility and interpret solubility temperature curves.
  • Define water of crystallisation, hydrated salts, and anhydrous substances.
  • Distinguish between efflorescent, deliquescent, and hygroscopic substances.
  • Differentiate temporary and permanent water hardness and explain removal methods (boiling, washing soda, permutit ion-exchange).

Chapter Roadmap & Progression

1 1. Composition of Water: Electrolys...
2 2. Solubility & Solubility Curves
3 3. Water of Crystallisation, Efflor...
4 4. Hard vs Soft Water & Softening M...

Complete Concept Guide (100% Curriculum Coverage)

1. Composition of Water: Electrolysis in Hoffmann's Voltameter

Understand
A. Electrolytic Splitting of Water:

Pure water is an extremely weak electrolyte (almost non-conducting). To conduct electricity, a few drops of dilute sulphuric acid ($\text{H}_2\text{SO}_4$) are added. When electric current is passed through Hoffmann's voltameter with platinum electrodes:

$$\mathbf{2\text{H}_2\text{O (l)} \xrightarrow{\text{Electric Current}} 2\text{H}_2\text{ (g) [Cathode]} + \text{O}_2\text{ (g) [Anode]}}$$
  • At Cathode (Negative Electrode): Hydrogen gas collects.
  • At Anode (Positive Electrode): Oxygen gas collects.
  • Volume Ratio: $\mathbf{V_{\text{Hydrogen}} : V_{\text{Oxygen}} = 2 : 1}$.
  • Mass Ratio: $2(2) : 32 = 4 : 32 = \mathbf{1 : 8}$ by mass!

2. Solubility & Solubility Curves

Solutions & Solubility
A. Definition of Solubility ($S$):

The maximum mass of solute in grams that can dissolve in $100\text{ grams}$ of solvent at a specified temperature to form a saturated solution:

$$\mathbf{\text{Solubility } S = \frac{\text{Mass of Solute}}{\text{Mass of Solvent}} \times 100}$$
B. Temperature Dependence:
  • Solids: For most salts ($\text{KNO}_3, \text{CuSO}_4, \text{NaCl}$), solubility increases with temperature.
  • Gases: Solubility of gases in water decreases sharply as temperature rises (e.g., boiled water tastes flat because dissolved $\text{O}_2$ and $\text{CO}_2$ escape; thermal pollution harms aquatic fish by de-oxygenating rivers).

3. Water of Crystallisation, Efflorescence & Deliquescence

Hydrates & Moisture
A. Water of Crystallisation:

The fixed number of water molecules chemically bonded in loose combination to the formula unit of a salt in its crystalline lattice:

  • Blue Vitriol: Copper(II) sulphate pentahydrate: $\mathbf{\text{CuSO}_4 \cdot 5\text{H}_2\text{O}}$ (loses water on heating to form white anhydrous $\text{CuSO}_4$).
  • Washing Soda: Sodium carbonate decahydrate: $\mathbf{\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}}$.
  • Green Vitriol: Ferrous sulphate heptahydrate: $\mathbf{\text{FeSO}_4 \cdot 7\text{H}_2\text{O}}$.
  • Gypsum: Calcium sulphate dihydrate: $\mathbf{\text{CaSO}_4 \cdot 2\text{H}_2\text{O}}$.
B. Moisture Phenomena:
  1. Efflorescence: When a hydrated crystal loses its water of crystallisation to dry air, crumbling into a powdery form (e.g., $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \to \text{Na}_2\text{CO}_3 \cdot \text{H}_2\text{O} + 9\text{H}_2\text{O}$).
  2. Deliquescence: When a solid absorbs moisture from moist air, dissolves in it, and turns into a liquid solution (e.g., $\text{CaCl}_2, \text{NaOH, KOH, FeCl}_3$).
  3. Hygroscopy: Substances that absorb moisture from air without dissolving into solutions (e.g., concentrated $\text{H}_2\text{SO}_4$, quicklime $\text{CaO}$, silica gel).

4. Hard vs Soft Water & Softening Methods

Water Hardness
A. Definitions:
  • Soft Water: Lathers readily and profusely with soap (e.g., distilled water, rainwater).
  • Hard Water: Does not lather easily with soap, producing an insoluble greyish scum (curd) of calcium/magnesium stearate.
B. Types of Hardness & Removal:
Type of HardnessCause (Chemical Salts)Removal Technique & Equation
Temporary HardnessDissolved Bicarbonates of Ca and Mg: $\mathbf{\text{Ca(HCO}_3)_2, \text{Mg(HCO}_3)_2}$Boiling: Decomposes soluble bicarbonates into insoluble carbonates:
$$\text{Ca(HCO}_3)_2 \xrightarrow{\Delta} \mathbf{\text{CaCO}_3 \downarrow} + \text{H}_2\text{O} + \text{CO}_2 \uparrow$$
Permanent HardnessDissolved Sulphates & Chlorides of Ca and Mg: $\mathbf{\text{CaCl}_2, \text{CaSO}_4, \text{MgCl}_2, \text{MgSO}_4}$Washing Soda Process: Precipitates ions as carbonates:
$$\text{CaSO}_4 + \text{Na}_2\text{CO}_3 \to \mathbf{\text{CaCO}_3 \downarrow} + \text{Na}_2\text{SO}_4$$
Permutit (Zeolite) Resin: Ion exchange.

Key Formulas, Reactions & Definitions

Solubility Formula
$$S = \frac{\text{Mass of Solute (g)}}{\text{Mass of Solvent (g)}} \times 100$$
Grams of solute per 100 grams of solvent at temperature T.
Temporary Hardness Removal by Boiling
$$\text{Ca(HCO}_3)_2 \xrightarrow{\Delta} \text{CaCO}_3 \downarrow + \text{H}_2\text{O} + \text{CO}_2 \uparrow$$
Soluble calcium bicarbonate decomposes into insoluble calcium carbonate.

Chemistry: Electrolysis of Water & Hydrated Crystals

Water Chemistry: Hoffmann's Voltameter & Hardness Softening ELECTROLYSIS (HOFFMANN'S VOLTAMETER) 2 Vol H2 Cathode (-) 1 Vol O2 Anode (+) 2H2O → 2H2 (Cathode) + O2 (Anode) Volume Ratio: H2 : O2 = 2 : 1 Mass Ratio: Hydrogen : Oxygen = 1 : 8 WATER HARDNESS & REMOVAL TEMPORARY HARDNESS: Bicarbonates Ca(HCO3)2 → CaCO3↓ + H2O + CO2↑ (REMOVED BY BOILING) PERMANENT HARDNESS: Sulphates / Chlorides CaSO4 + Na2CO3 → CaCO3↓ + Na2SO4 (WASHING SODA) • Efflorescence: Crystal loses water to dry air • Deliquescence: Solid absorbs moisture & dissolves • Blue Vitriol: CuSO4•5H2O • Washing Soda: Na2CO3•10H2O H2:O2 VOL = 2:1 • MASS = 1:8 • BOILING REMOVES TEMPORARY • WASHING SODA REMOVES PERMANENT

Chapter Summary & 10 Key Takeaways

Takeaway 1
Water has formula H2O with volume ratio H:O = 2:1 and mass ratio 1:8.
Takeaway 2
Electrolysis in Hoffmann's voltameter produces 2 volumes of H2 at the cathode and 1 volume of O2 at the anode.
Takeaway 3
Water exhibits anomalous expansion between 0 and 4 degrees C, reaching maximum density at 4 degrees C.
Takeaway 4
Solubility is the maximum grams of solute dissolved in 100 grams of solvent at a specific temperature.
Takeaway 5
Water of crystallisation is chemically bound water in a salt crystal lattice (e.g., CuSO4 * 5H2O).
Takeaway 6
Efflorescent crystals lose water of crystallisation to dry air, crumbling into powder.
Takeaway 7
Deliquescent solids absorb moisture from air and dissolve into a liquid solution (e.g., CaCl2, NaOH).
Takeaway 8
Hygroscopic substances absorb moisture from air without dissolving (e.g., concentrated H2SO4, silica gel).
Takeaway 9
Temporary hardness is caused by bicarbonates of calcium and magnesium and is removed by boiling.
Takeaway 10
Permanent hardness is caused by chlorides and sulphates of calcium and magnesium and is removed with washing soda.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Describe Hoffmann's Voltameter experiment for the electrolysis of water. State the gases evolved at the cathode and anode and their volume ratio.
Reveal Answer & Explanation
Answer:
  1. Hoffmann's Voltameter consists of three interconnected vertical glass tubes with platinum electrodes at the base of the two outer limbs.
    2. The apparatus is filled with acidified water (water containing a few drops of dilute $\text{H}_2\text{SO}_4$ to provide conducting ions).
    3. When DC electric current is passed through the electrodes:
    • At the Cathode (Negative Electrode): Hydrogen gas ($\text{H}_2$) evolves.
    • At the Anode (Positive Electrode): Oxygen gas ($\text{O}_2$) evolves.
    4. The volume of gas collected in the cathode limb is observed to be EXACTLY TWICE the volume of gas collected in the anode limb:

$$\mathbf{V_{\text{Hydrogen}} : V_{\text{Oxygen}} = 2 : 1}$$


Equation: $2\text{H}_2\text{O} \xrightarrow{\text{electrolysis}} 2\text{H}_2 \uparrow + \text{O}_2 \uparrow$.


Hydrogen evolves at the cathode, oxygen at the anode. The volume ratio is $2 : 1$.
2
Differentiate between Temporary Hardness and Permanent Hardness of water. Explain how boiling removes temporary hardness with a chemical equation.
Reveal Answer & Explanation
Answer:

• Temporary Hardness: Caused by dissolved bicarbonates of calcium and magnesium: $\text{Ca(HCO}_3)_2$ and $\text{Mg(HCO}_3)_2$. It can be easily removed by simple boiling.
• Permanent Hardness: Caused by dissolved chlorides and sulphates of calcium and magnesium: $\text{CaCl}_2, \text{CaSO}_4, \text{MgCl}_2, \text{MgSO}_4$. It cannot be removed by boiling.
• Removal of Temporary Hardness by Boiling:
Heating decomposes soluble calcium bicarbonate into insoluble calcium carbonate ($\text{CaCO}_3$), which precipitates out and is filtered off:

$$\mathbf{\text{Ca(HCO}_3)_2\text{ (aq)} \xrightarrow{\Delta} \text{CaCO}_3 \downarrow \text{ (white ppt)} + \text{H}_2\text{O (l)} + \text{CO}_2\text{ (g)} \uparrow}$$

.


Temporary: bicarbonates, removed by boiling. Permanent: chlorides/sulphates, requires chemical softening.
3
Differentiate between Efflorescence and Deliquescence with two chemical examples of each.
Reveal Answer & Explanation
Answer:

• Efflorescence:
The phenomenon whereby a hydrated crystalline salt loses its water of crystallisation when exposed to dry air, crumbling into an amorphous powder.
Examples:
1. Washing soda: $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \to \text{Na}_2\text{CO}_3 \cdot \text{H}_2\text{O} + 9\text{H}_2\text{O}$
2. Glauber's salt: $\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} \to \text{Na}_2\text{SO}_4 + 10\text{H}_2\text{O}$
• Deliquescence:
The phenomenon whereby certain water-soluble substances absorb moisture from the atmosphere, dissolve in that absorbed water, and turn into a liquid solution.
Examples:
1. Solid Sodium Hydroxide ($\text{NaOH}$ pellets)
2. Anhydrous Calcium Chloride ($\text{CaCl}_2$)
3. Ferric chloride ($\text{FeCl}_3$).


Efflorescence: loses crystal water to dry air (washing soda). Deliquescence: absorbs air moisture and liquefies ($ ext{NaOH}, ext{CaCl}_2$).
4
What is Water of Crystallisation? What happens when blue crystals of Copper(II) Sulphate pentahydrate are heated strongly in a dry test tube?
Reveal Answer & Explanation
Answer:

• Water of Crystallisation: The fixed number of water molecules chemically combined in a definite stoichiometric ratio to a salt formula unit, forming its regular crystalline shape and color.
• Heating Blue Vitriol ($\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$):
1. When heated strongly, the salt loses all its water of crystallisation:

$$\mathbf{\text{CuSO}_4 \cdot 5\text{H}_2\text{O [blue]} \xrightarrow{\Delta} \text{CuSO}_4\text{ [white anhydrous]} + 5\text{H}_2\text{O (vapor)}}$$


2. The blue crystalline solid crumbles into an amorphous white powder of anhydrous copper sulphate.
3. Steam condenses into water droplets on the cool upper walls of the test tube.
4. Adding a drop of water back turns the powder bright blue again!


Chemically bound water molecules. Heating blue $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ turns it into white anhydrous $\text{CuSO}_4$.
5
Define Solubility. $24\text{ grams}$ of a salt dissolves in $80\text{ grams}$ of water at $25^{\circ}\text{C}$ to form a saturated solution. Calculate its solubility at this temperature.
Reveal Answer & Explanation
Answer:

• Solubility: The maximum mass of solute in grams that can dissolve in $100\text{ grams}$ of solvent at a specific temperature to produce a saturated solution.
• Calculation:

$$\text{Solubility } S = \frac{\text{Mass of Solute}}{\text{Mass of Solvent}} \times 100$$


$$S = \frac{24\text{ g}}{80\text{ g}} \times 100 = \frac{3}{10} \times 100 = \mathbf{30\text{ grams}}$$

.
The solubility of the salt at $25^{\circ}\text{C}$ is $30\text{ grams per 100 g of water}$.


$\text{Solubility} = (24 / 80) \times 100 = 30\text{ g}$.
6
How does the Permutit (Zeolite) process remove both temporary and permanent hardness from water?
Reveal Answer & Explanation
Answer:

• Permutit (Zeolite): An artificial hydrated sodium aluminum silicate (formula $\text{Na}_2\text{Z}$ or $\text{Na}_2\text{Al}_2\text{Si}_2\text{O}_8 \cdot x\text{H}_2\text{O}$).
• Hard water containing $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ ions is passed through a cylinder packed with a thick bed of permutit.
• An Ion-Exchange Reaction takes place: the zeolite replaces the hardness-causing calcium and magnesium ions with soluble sodium ions:

$$\mathbf{\text{Na}_2\text{Z} + \text{Ca}^{2+} \to \text{CaZ} + 2\text{Na}^+}$$


$$\mathbf{\text{Na}_2\text{Z} + \text{Mg}^{2+} \to \text{MgZ} + 2\text{Na}^+}$$


• The water emerging from the base is completely soft.
• The exhausted column is easily regenerated by backwashing with concentrated brine solution ($10\%\text{ NaCl}$).


Zeolite trades its sodium ions ($ ext{Na}^+$) for the hard calcium and magnesium ions ($ ext{Ca}^{2+}, ext{Mg}^{2+}$).
7
Why does boiled water taste flat and insipid compared to fresh spring water?
Reveal Answer & Explanation
Answer:

• Fresh natural water contains small quantities of dissolved gases (Oxygen and Carbon Dioxide) as well as beneficial dissolved mineral salts, which impart a pleasant, brisk, refreshing taste to water.
• The solubility of gases in water decreases dramatically with rising temperature.
• When water is boiled, all the dissolved oxygen and carbon dioxide gases escape into the atmosphere.
• The absence of these dissolved gases makes boiled water taste flat and unappetizing.


Boiling expels dissolved oxygen and carbon dioxide gases that provide natural taste.
8
What is a Saturated Solution? What happens when a hot saturated solution of potassium nitrate is allowed to cool down to room temperature?
Reveal Answer & Explanation
Answer:

• Saturated Solution: A solution that cannot dissolve any more solute at a given specific temperature in the presence of excess undissolved solute.
• Cooling a Hot Saturated Solution:
The solubility of potassium nitrate ($\text{KNO}_3$) decreases sharply as temperature decreases.
As the hot solution cools, the excess dissolved solute that can no longer remain in solution precipitates out of the liquid in the form of beautiful, pure needle-like crystals of potassium nitrate.


Cannot dissolve more solute at that temperature. Cooling decreases solubility, precipitating pure crystals.
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