Set A

1. Water exists in all the three states. Discuss.

   Water exists in three states: solid (ice), liquid (water), and gas (steam or vapor). The state of water changes with temperature. Below 0°C, water is in its solid form as ice. Between 0°C and 100°C, water is in its liquid form. Above 100°C, water turns into steam, which is its gaseous form.

2. Why is water considered a compound?

   Water is considered a compound because it is made up of two different elements, hydrogen and oxygen, chemically bonded together in a fixed ratio of 2:1, forming the molecular formula H₂O.

3. (a) Why does temperature in Mumbai and Chennai not fall as low as it does in Delhi?

   Mumbai and Chennai have coastal climates, which are moderated by the presence of the sea. The sea absorbs heat during the day and releases it during the night, maintaining a relatively stable temperature. Delhi, being inland, does not benefit from this moderating effect, resulting in more extreme temperatures.

   (b) Give the properties of water responsible for controlling the temperature of our body.

   Water has a high specific heat capacity, which means it can absorb a lot of heat before its temperature rises. This property helps in regulating body temperature by absorbing excess heat produced by metabolic activities and preventing rapid temperature changes.

4. ‘Water is a universal solvent’. Comment.

   Water is called a universal solvent because it can dissolve a wide variety of substances due to its polar nature. The positive and negative ends of water molecules attract different ions and molecules, allowing them to dissolve.

5. What causes the violence associated with torrential rain?

   The violence associated with torrential rain is caused by intense and rapid precipitation, which can lead to flooding, landslides, and strong, gusty winds. The large volume of water falling in a short period overwhelms drainage systems and saturates the ground, causing these violent effects.

6. (a) Which property of water enables it to modify the climate?

   Water’s high specific heat capacity enables it to modify the climate. It can absorb and store large amounts of heat, which helps to moderate temperature variations in the environment.

   (b) Density of water varies with temperature. What are its consequences?

   As water cools, its density increases until it reaches 4°C. Below this temperature, water becomes less dense and expands upon freezing. This causes ice to float on water, insulating the liquid water below and providing a stable environment for aquatic life during winter.

   (c) State the melting point, boiling point, specific heat capacity, specific latent heat of fusion, and specific latent heat of vaporization of water.

   • Melting point: 0°C
   • Boiling point: 100°C
   • Specific heat capacity: 4.18 J/g°C
   • Specific latent heat of fusion: 334 J/g
   • Specific latent heat of vaporization: 2260 J/g

7. How do fishes and aquatic animals survive when the pond gets covered with thick ice?

   When a pond is covered with ice, the water below the ice remains at a temperature of around 4°C, which is dense enough to stay liquid. This insulated environment allows fishes and aquatic animals to survive even when the surface is frozen.

8. The properties of water are different from the properties of the elements of which it is formed. Discuss.

   Water has properties that are significantly different from hydrogen and oxygen. Hydrogen is a flammable gas, and oxygen supports combustion, but water is a liquid that extinguishes fires. The combination of these elements in a compound results in unique properties not found in the individual elements.

9. How is aquatic life benefitted by the fact that water has maximum density at 4°C?

   Water’s maximum density at 4°C ensures that during winter, colder water (which is less dense) stays on top and freezes, while the denser 4°C water remains at the bottom, providing a stable environment for aquatic life to survive.

10. Describe an experiment to show:

    (a) Tap water contains dissolved salts.

    Take a sample of tap water and evaporate it using a heat source. Observe the residue left behind, indicating the presence of dissolved salts.

    (b) Tap water contains dissolved gases.

    Take a sample of tap water and heat it slowly without boiling. Observe the formation of small bubbles on the sides of the container, indicating the release of dissolved gases.

11. State the importance of the solubility of CO₂ and O₂ in water.

    The solubility of CO₂ in water is important for aquatic plants, as they use it for photosynthesis. The solubility of O₂ is crucial for the respiration of aquatic animals.

12. How is air dissolved in water different from ordinary air?

    Air dissolved in water has a higher concentration of oxygen compared to ordinary air because oxygen is more soluble in water than nitrogen, the main component of air.

13. Why do rivers and lakes not freeze easily?

    Rivers and lakes do not freeze easily due to the large volume of water, which retains heat. Additionally, the movement of water in rivers prevents ice formation.

14. What is the importance of dissolved salts in water?

    Dissolved salts in water are essential for various biological processes, including nutrient transport and electrical conductivity, which are crucial for the survival of aquatic life.

15. Explain why:

    (a) Boiled or distilled water tastes flat.

    Boiled or distilled water tastes flat because it lacks dissolved gases and minerals that contribute to the taste of water.

    (b) Ice at zero degree centigrade has a greater cooling effect than water at 0°C.

    Ice at 0°C absorbs more heat as it melts, providing a greater cooling effect compared to liquid water at the same temperature.

    (c) Burns caused by steam are more severe than burns caused by boiling water.

    Steam contains more heat energy due to the latent heat of vaporization, causing more severe burns upon contact.

    (d) Rainwater does not leave behind concentric rings when boiled.

    Rainwater is relatively pure and lacks dissolved salts and minerals that would form residue rings when boiled.

    (e) Air dissolved in water contains a higher proportion of oxygen.

    Water dissolves oxygen more readily than nitrogen, resulting in a higher proportion of oxygen in dissolved air.

    (f) If distilled water is kept in a sealed bottle for a long time, it leaves etchings on the surface of the glass.

    Distilled water can dissolve small amounts of silica from the glass, causing etchings over time.

16. Explain what you understand from the following diagram:

    The diagram shows the phases of matter (solid, liquid, gas) and the processes involved in phase changes:

    • Latent heat of fusion: The heat required to change a solid into a liquid without changing its temperature.
    • Latent heat of solidification: The heat released when a liquid turns into a solid.
    • Latent heat of vaporization: The heat required to change a liquid into a gas without changing its temperature.
    • Latent heat of condensation: The heat released when a gas turns into a liquid.

Set B

Concentration of a Solution

The concentration of a solution is the amount of solute dissolved in a given quantity of that solution. It can be expressed in many ways.

(i) Mass Percent

Mass percent is defined as the mass of solid solute in grams present in 100 grams of the solution. It is mostly used when the solute is solid and the solvent is liquid.

Formula:

\[ \text{Mass percent} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution (solute + solvent)}} \right) \times 100 \]

Example:

If 10 grams of sodium chloride is added to 90 grams of water, the concentration is calculated as follows:

\[ \text{Concentration} = \left( \frac{10}{10 + 90} \right) \times 100 = \left( \frac{10}{100} \right) \times 100 = 10\% \]

(ii) Volume Percent

Volume percent is defined as the volume of solute in milliliters present in 100 mL of a solution. It is mostly used when both solute and solvent are liquids.

Formula:

\[ \text{Volume percent} = \left( \frac{\text{Volume of solute}}{\text{Volume of solute + Volume of solvent}} \right) \times 100 \]

Example:

If 30 mL of alcohol is mixed with 70 mL of water to get 100 mL of solution, volume percent is calculated as follows:

\[ \text{Volume percent} = \left( \frac{30}{30 + 70} \right) \times 100 = \left( \frac{30}{100} \right) \times 100 = 30\% \]

Solved Examples

Example 1:

2.5 liters of alcohol is present in 10 liters of aqueous solution of alcohol. Calculate the volume percent.

\[
\text{Volume of solute} = 2.5 \, \text{liters} \\
\text{Volume of solution} = 10.0 \, \text{liters} \\
\text{Volume percent} = \left( \frac{2.5}{10} \right) \times 100 = 25\%
\]

Example 2:

50 grams of sugar is dissolved in 2.45 kg of water. Calculate the concentration of the solution.

\[
\text{Mass of solute} = 50 \, \text{grams} \\
\text{Mass of solvent} = 2450 \, \text{grams} \\
\text{Mass of solution} = \text{mass of solute + mass of solvent} = 2500 \, \text{grams} \\
\text{Mass percent} = \left( \frac{50}{2500} \right) \times 100 = 2\%
\]

Additional Numericals for Practice

Numerical 1:

Calculate the mass percent of a solution prepared by dissolving 25 grams of salt in 225 grams of water.

\[
\text{Mass of solute} = 25 \, \text{grams} \\
\text{Mass of solvent} = 225 \, \text{grams} \\
\text{Mass of solution} = 25 + 225 = 250 \, \text{grams} \\
\text{Mass percent} = \left( \frac{25}{250} \right) \times 100 = 10\%
\]

Numerical 2:

If 15 mL of ethanol is mixed with 85 mL of water to prepare 100 mL of solution, calculate the volume percent.

\[
\text{Volume of solute (ethanol)} = 15 \, \text{mL} \\
\text{Volume of solution} = 100 \, \text{mL} \\
\text{Volume percent} = \left( \frac{15}{100} \right) \times 100 = 15\%
\]

Set C

Solubility and Crystallization

Formulas and Concepts

Solubility

1. Solubility Definition: The maximum amount of solute that can dissolve in a specific amount of solvent at a given temperature.

2. Formula for Solubility:

\[ \text{Solubility} = \left( \frac{\text{Mass of solute}}{\text{Mass of solvent}} \right) \times 100 \]

Crystallization

1. Crystallization Concept: When a saturated solution is cooled, solute comes out of the solution in the form of crystals.

2. Calculation for Crystals Obtained:

\[ \text{Amount of crystals} = \text{Solubility at higher temperature} – \text{Solubility at lower temperature} \]

3. Adjustment for Different Masses:

\[ \text{Amount of solute required} = \left( \frac{\text{Amount of solute for given crystals}}{\text{Amount of crystals obtained}} \right) \times \text{Desired amount of crystals} \]

Solved Examples

Example 1:

12 g of a saturated solution of potassium chloride at 20°C, when evaporated to dryness, leaves a solid residue of 3 g. Calculate the solubility of potassium chloride.

Solution:

\[
\text{Weight of water in solution} = 12 \, \text{g} – 3 \, \text{g} = 9 \, \text{g} \\
9 \, \text{g of H}_2\text{O} \text{dissolves} 3 \, \text{g of solid} \\
\text{Solubility} = \left( \frac{3}{9} \right) \times 100 = 33.3 \, \text{g/100 g water}
\]

Answer: Solubility of KCl in H₂O at 20°C = 33.3 g

Example 2:

Find the weight of sodium nitrate required to prepare 60 g pure crystals from its saturated solution at 70°C. Solubility of sodium nitrate is 140 g at 70°C and 100 g at 25°C.

Solution:

\[
\text{Solubility at 70°C} = 140 \, \text{g} \\
\text{Solubility at 25°C} = 100 \, \text{g} \\
\text{Amount of crystals obtained when the solution is cooled from 70°C to 25°C} = 140 \, \text{g} – 100 \, \text{g} = 40 \, \text{g} \\
\text{To obtain 40 g of crystals, sodium nitrate taken is 140 g} \\
\text{To obtain 60 g crystals, sodium nitrate required will be:} \\
\left( \frac{140}{40} \right) \times 60 = 210 \, \text{g}
\]

Answer: 210 g

Additional Numericals for Practice

Question 1:

20 g of a saturated solution of copper sulfate at 30°C, when evaporated to dryness, leaves a solid residue of 5 g. Calculate the solubility of copper sulfate.

Solution:

\[
\text{Weight of water in solution} = 20 \, \text{g} – 5 \, \text{g} = 15 \, \text{g} \\
15 \, \text{g of water dissolves} 5 \, \text{g of solid} \\
\text{Solubility} = \left( \frac{5}{15} \right) \times 100 = 33.3 \, \text{g/100 g water}
\]

Question 2:

Determine the weight of potassium nitrate required to prepare 80 g pure crystals from its saturated solution at 60°C. Solubility of potassium nitrate is 150 g at 60°C and 80 g at 20°C.

Solution:

\[
\text{Solubility at 60°C} = 150 \, \text{g} \\
\text{Solubility at 20°C} = 80 \, \text{g} \\
\text{Amount of crystals obtained when the solution is cooled from 60°C to 20°C} = 150 \, \text{g} – 80 \, \text{g} = 70 \, \text{g} \\
\text{To obtain 70 g of crystals, potassium nitrate taken is 150 g} \\
\text{To obtain 80 g crystals, potassium nitrate required will be:} \\
\left( \frac{150}{70} \right) \times 80 = 171.4 \, \text{g}
\]

Question 3:

25 g of a saturated solution of sodium chloride at 25°C, when evaporated to dryness, leaves a solid residue of 7 g. Calculate the solubility of sodium chloride.

Solution:

\[
\text{Weight of water in solution} = 25 \, \text{g} – 7 \, \text{g} = 18 \, \text{g} \\
18 \, \text{g of water dissolves} 7 \, \text{g of solid} \\
\text{Solubility} = \left( \frac{7}{18} \right) \times 100 = 38.9 \, \text{g/100 g water}
\]

Question 4:

Find the weight of ammonium chloride required to prepare 50 g pure crystals from its saturated solution at 50°C. Solubility of ammonium chloride is 120 g at 50°C and 40 g at 10°C.

Solution:

\[
\text{Solubility at 50°C} = 120 \, \text{g} \\
\text{Solubility at 10°C} = 40 \, \text{g} \\
\text{Amount of crystals obtained when the solution is cooled from 50°C to 10°C} = 120 \, \text{g} – 40 \, \text{g} = 80 \, \text{g} \\
\text{To obtain 80 g of crystals, ammonium chloride taken is 120 g} \\
\text{To obtain 50 g crystals, ammonium chloride required will be:} \\
\left( \frac{120}{80} \right) \times 50 = 75 \, \text{g}
\]

Set D

Determination of Water of Crystallization

Water of crystallization refers to the water molecules that are part of the crystal structure of a compound. These water molecules are essential for maintaining the structure and properties of the crystals. The determination of water of crystallization involves heating the crystals to remove the water and measuring the weight loss.

Steps to Determine Water of Crystallization:

1. Take a known weight of crystals in a china dish.
2. Heat the crystals above 100°C until the weight of the residue becomes constant.
3. Measure the weight of the crystals before and after heating.

Formulas:

Let the weight of crystals at room temperature be \(a\) grams.
Let the weight of crystals after heating be \(b\) grams.
Weight of water lost = \(a – b\) grams.
Percentage of water of crystallization:
\[ \% \text{ of water of crystallization} = \left( \frac{a – b}{a} \right) \times 100 \]

Solved Example

Example:

Given:
Weight of crystals at room temperature (\(a\)) = 20 g
Weight of crystals after heating (\(b\)) = 15 g
Solution:
Weight of water lost = \(a – b = 20 – 15 = 5\) g
Percentage of water of crystallization:
\[ \% \text{ of water of crystallization} = \left( \frac{5}{20} \right) \times 100 = 25\% \]

Practice Questions

Question 1:

A sample of hydrated copper sulfate (\( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)) weighs 25 g. After heating, the weight of anhydrous copper sulfate (\( \text{CuSO}_4 \)) is found to be 16 g. Calculate the percentage of water of crystallization.

Solution:
Weight of crystals at room temperature (\(a\)) = 25 g
Weight of crystals after heating (\(b\)) = 16 g
Weight of water lost = \(25 – 16 = 9\) g
Percentage of water of crystallization:
\[ \% \text{ of water of crystallization} = \left( \frac{9}{25} \right) \times 100 = 36\% \]

Question 2:

A 30 g sample of hydrated magnesium sulfate (\( \text{MgSO}_4 \cdot 7\text{H}_2\text{O} \)) is heated until all the water is removed, leaving 18 g of anhydrous magnesium sulfate (\( \text{MgSO}_4 \)). Calculate the percentage of water of crystallization.

Solution:
Weight of crystals at room temperature (\(a\)) = 30 g
Weight of crystals after heating (\(b\)) = 18 g
Weight of water lost = \(30 – 18 = 12\) g
Percentage of water of crystallization:
\[ \% \text{ of water of crystallization} = \left( \frac{12}{30} \right) \times 100 = 40\% \]

Question 3:

A 40 g sample of hydrated sodium carbonate (\( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \)) is heated to remove all water, resulting in 22 g of anhydrous sodium carbonate (\( \text{Na}_2\text{CO}_3 \)). Calculate the percentage of water of crystallization.

Solution:
Weight of crystals at room temperature (\(a\)) = 40 g
Weight of crystals after heating (\(b\)) = 22 g
Weight of water lost = \(40 – 22 = 18\) g
Percentage of water of crystallization:
\[ \% \text{ of water of crystallization} = \left( \frac{18}{40} \right) \times 100 = 45\% \]

Set E

Dehydrating Agent vs. Drying Agent

Dehydrating Agent:

• Definition: Removes chemically combined elements of water (hydrogen and oxygen) from a compound.
• Uses:
  • Prepares substances like carbon monoxide and sugar charcoal.
  • Represents a chemical change.
• Example: Concentrated sulfuric acid (\(H_2SO_4\)).

Drying Agent:

• Definition: Removes moisture (water) from other substances.
• Uses:
  • Used to dry gases like chlorine, sulfur dioxide, hydrogen chloride.
  • Keeps substances dry in desiccators.
  • Represents a physical change.
• Examples: Phosphorus pentoxide (\(P_2O_5\)), fused calcium chloride (\(CaCl_2\)), calcium oxide (\(CaO\)), concentrated sulfuric acid (\(H_2SO_4\)).

Tabulated Differences:

Solutions to the Exercise Questions

1. Explain the terms:

   (a) Solution: A homogeneous mixture of two or more substances.
   (b) Solute: The substance that is dissolved in a solution.
   (c) Solvent: The substance in which the solute is dissolved.

2. Explain why a hot saturated solution of potassium nitrate forms crystals as it cools.

   As the temperature decreases, the solubility of potassium nitrate decreases, leading to the formation of crystals.

3. Give three factors that affect the solubility of a solid solute in a solvent.

   • Temperature
   • Nature of the solute and solvent
   • Pressure (for gases)

4. How would you prepare a saturated solution of copper sulfate at room temperature?

   Dissolve copper sulfate in water while stirring until no more dissolves. To confirm saturation, add a small amount more copper sulfate; it should not dissolve.

5. (a) Define Henry’s law and crystallization.

   (a) Henry’s Law: The solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.
   (b) Crystallization: The process of forming solid crystals from a homogeneous solution.

   (b) State the different methods of crystallization.

   • Evaporation
   • Cooling of a hot saturated solution
   • Precipitation from a solution

6. Observation when heating crystals of copper (II) sulfate and iron (II) sulfate:

   – Copper (II) sulfate turns white as it loses water.
   – Iron (II) sulfate turns brown and releases water vapor.

7. Why table salt becomes sticky on exposure to humid air:

   Table salt absorbs moisture from the air due to its hygroscopic nature.

8. Effect of temperature on the solubility of \(KNO_3\) and \(CaSO_4\):

   – Solubility of \(KNO_3\) increases with temperature.
   – Solubility of \(CaSO_4\) decreases with temperature.

9. Solubility of NaCl at 40°C is 36.5 g. What does this mean?

   36.5 grams of NaCl can dissolve in 100 grams of water at 40°C to form a saturated solution.

10. Test to find if a solution is saturated, unsaturated, or supersaturated:

    Add a small amount of solute. If it dissolves, the solution is unsaturated. If it does not dissolve, the solution is saturated. If crystals form, the solution is supersaturated.

11. Effect of pressure on the solubility of gases:

    Increased pressure increases the solubility of gases (e.g., carbonated drinks).

12. What are solubility curves?

    Graphs showing the solubility of a substance at different temperatures.

13. Explain why:

    (a) Water is an excellent liquid to use in cooling systems.
    – High specific heat capacity.
    (b) A solution is always clear and transparent.
    – Homogeneous mixture with particles that are too small to scatter light.
    (c) Lakes and rivers do not suddenly freeze in the winter.
    – Water has a high specific heat capacity and density anomaly at 4°C.
    (d) The solute cannot be separated from a solution by filtration.
    – The solute particles are too small to be trapped by a filter.

14. Solubility changes with temperature:

    (a) Substance with solubility increasing rapidly: \(KNO_3\)
    (b) Substance with solubility increasing gradually: \(NaCl\)
    (c) Substance with solubility increasing slightly: \(CaSO_4\)
    (d) Substance with solubility initially increasing then decreasing: \(NH_4Cl\)

15. What are drying or desiccating agents? Examples.

    Substances that remove moisture from other materials. Examples: \(P_2O_5\), \(CaCl_2\), \(CaO\).

16. Complete the table:

    Common Name	Chemical Name	Formula	Acid, Base, or Salt	Efflorescent, Hygroscopic, or Deliquescent
    Solid caustic potash	Potassium hydroxide	\(KOH\)	Base	Deliquescent
    Quick lime	Calcium oxide	\(CaO\)	Base	Hygroscopic
    Oil of vitriol	Sulfuric acid	\(H_2SO_4\)	Acid	Hygroscopic
    Washing soda	Sodium carbonate	\(Na_2CO_3 \cdot 10H_2O\)	Salt	Efflorescent
    Solid caustic soda	Sodium hydroxide	\(NaOH\)	Base	Deliquescent
    Blue vitriol	Copper sulfate	\(CuSO_4 \cdot 5H_2O\)	Salt	Efflorescent

17. In which of the following substances will there be:

    (a) Increase in mass: Iron (due to rusting)
    (b) Decrease in mass: Sodium chloride (if it absorbs moisture and dissolves)
    (c) No change in mass when exposed to air: Conc. sulfuric acid
    Substances:
    1. Sodium chloride
    2. Iron
    3. Conc. sulfuric acid
    4. Table salt
    5. Sodium carbonate crystals

18. To make a saturated solution, 136 g of a salt is dissolved in 500 g of water at 293 K. Find its solubility at this temperature.

    Solubility = (Mass of solute / Mass of solvent) × 100
    Solubility = (136 g / 500 g) × 100 = 27.2 g/100 g water

19. (a) A solution contains 15 g of sodium chloride in 285 g of water. Calculate the concentration of the solution.

    Concentration = (Mass of solute / Mass of solution) × 100
    Mass of solution = 15 g (solute) + 285 g (solvent) = 300 g
    Concentration = (15 g / 300 g) × 100 = 5%

    (b) 4 liters of an organic compound, acetone, is present in 90 liters of an aqueous solution. Calculate its volume percent.

    Volume percent = (Volume of solute / Volume of solution) × 100
    Volume percent = (4 L / 90 L) × 100 = 4.44%

20. Using the table provided for solubility at different temperatures:

    (a) What mass of \( KNO_3 \) would be needed to produce a saturated solution of \( KNO_3 \) in 50 grams of water at 313 K?

    Solubility of \( KNO_3 \) at 313 K = 62 g/100 g water
    Mass of \( KNO_3 \) for 50 g water = (62 g / 100 g) × 50 g = 31 g

    (b) If a saturated solution of \( KCl \) is made at 353 K and then cooled to room temperature, what would you observe? Explain.

    At 353 K, the solubility of \( KCl \) is high. When cooled to room temperature, \( KCl \) will crystallize out as its solubility decreases.

    (c) Find the solubility of each salt at 293 K.

    \( KNO_3 \): 32 g/100 g water
    \( NaCl \): 36 g/100 g water
    \( KCl \): 35 g/100 g water
    \( NH_4Cl \): 37 g/100 g water

    (d) Which salt has the lowest solubility at 283 K?

    \( KNO_3 \): 21 g/100 g water

    (e) What is the effect of change of temperature on the solubility of a salt?

    Solubility of salts generally increases with temperature.

21. (a) Find the solubility of \( KNO_3 \) at 20°C when the mass of the empty dish is 50 g, the mass of the dish and solution is 65 g, while the mass of dish and residue is 54.3 g.

    Mass of solution = 65 g – 50 g = 15 g
    Mass of residue (solute) = 54.3 g – 50 g = 4.3 g
    Mass of solvent = 15 g – 4.3 g = 10.7 g
    Solubility = (Mass of solute / Mass of solvent) × 100
    Solubility = (4.3 g / 10.7 g) × 100 = 40.18 g/100 g water

    (b) What weight of sodium nitrate will separate when a saturated solution containing 50 g of water is cooled from 50°C to 30°C? The solubility of \( NaNO_3 \) at 50°C and 30°C is 114 g and 86 g respectively.

    Solubility at 50°C = 114 g/100 g water
    Solubility at 30°C = 86 g/100 g water
    Amount of \( NaNO_3 \) in solution at 50°C = (114 g / 100 g) × 50 g = 57 g
    Amount of \( NaNO_3 \) in solution at 30°C = (86 g / 100 g) × 50 g = 43 g
    Amount of \( NaNO_3 \) that will separate = 57 g – 43 g = 14 g

22. To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.

    Concentration = (Mass of solute / Mass of solution) × 100
    Mass of solution = 36 g (solute) + 100 g (solvent) = 136 g
    Concentration = (36 g / 136 g) × 100 = 26.47%

Set F

Stalagmites and Stalactites Explained

Detailed Explanation

In limestone caves, you can see cone-shaped pillars called stalagmites and stalactites.

Stalactites:

• Hang from the roof of the cave.
• Formed by water dripping from cracks in the rocks.
• Contain dissolved calcium hydrogen carbonate.
• When water drips, it loses pressure and converts calcium hydrogen carbonate into calcium carbonate (solid).
• Over time, calcium carbonate deposits form stalactites.

Stalagmites:

• Rise from the floor of the cave.
• Formed similarly to stalactites, but the deposits build up on the ground where the drips land.

Chemical Reaction:

\[ \text{Ca(HCO}_3\text{)}_2 \rightarrow \text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]

This reaction shows calcium hydrogen carbonate turning into calcium carbonate, carbon dioxide, and water.

Additional Notes:

• Growth Rate:
  • Stalagmites and stalactites grow very slowly.
  • Some grow less than a centimeter per year, while others can grow as tall as 100 cm over many years.
• Formation Process:
  • Calcium Carbonate Formation: Water with dissolved calcium carbonate drips and forms solid calcium carbonate as it loses carbon dioxide and water.
  • Magnesium Carbonate Formation: Similar process occurs with magnesium carbonate.
• Chemical Reaction for Magnesium Carbonate:

\[ \text{MgCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{Mg(HCO}_3\text{)}_2 \]

• Effect of Gypsum:
  • If water flows over gypsum beds (\( \text{CaSO}_4 \cdot 2\text{H}_2\text{O} \)), gypsum dissolves in water making it hard.

Differences Between Stalagmites and Stalactites

Set G

Understanding Hard Water, Soap, and Detergents

Hard Water and Furring of Tea Kettles

Furring of Tea Kettles:

• Cause: The furring or scaling of a tea kettle is due to the sediment formed on its walls from boiling hard water. This sediment is primarily composed of calcium and magnesium carbonates.
• Process: When hard water is boiled, the soluble bicarbonates of calcium and magnesium are converted into insoluble carbonates, which precipitate out and form a scale on the kettle walls.

Hard Water:

• Definition: Hard water contains high levels of dissolved calcium and magnesium ions.
• Effect on Washing:
  • Hard water is not suitable for washing purposes because it reacts with soap to form an insoluble substance known as soap-curd or scum.
  • Soap molecules react with calcium and magnesium ions in hard water, preventing the soap from lathering and effectively cleaning.

Soap and Detergents

Soap:

• Composition: Soap is chemically a sodium salt of stearic acid, an organic acid with the formula \( \text{C}_{17}\text{H}_{35}\text{COOH} \). The formula for soap is \( \text{C}_{17}\text{H}_{35}\text{COONa} \).
• Reaction with Hard Water:
  • When soap is used in hard water, it forms insoluble calcium and magnesium salts (soap-curd), making it difficult to form lather.
  • Chemical Reaction:
    \[
    2\text{NaSt} + \text{Ca(HCO}_3\text{)}_2 \rightarrow \text{CaSt}_2 \downarrow + 2\text{NaHCO}_3
    \]
    Here, \( \text{NaSt} \) represents sodium stearate (soap), and \( \text{CaSt}_2 \) represents calcium stearate (soap-curd).

Detergents:

• Composition: Detergents are synthetic cleaning agents made by sulfonating higher alkenes (organic compounds) with sulfuric acid and converting them into their sodium salts.
• Advantages:
  • Detergents are more effective in hard water because they do not form insoluble scum with calcium and magnesium ions.
  • Detergents are more soluble in water and have better cleaning properties.

Cleansing Action of Soaps and Detergents

Micelles: When soap or detergent is dissolved in water, the molecules gather into clusters called micelles. The hydrophobic (water-repelling) tails of the molecules stick inward, while the hydrophilic (water-attracting) heads stick outward.

Mechanism:

• The hydrocarbon tail of the soap/detergent attaches to oily dirt.
• When water is stirred, the oily dirt lifts off the surface and breaks into smaller fragments, allowing other tails to attach to it.
• The micelles surround the oil, forming small globules suspended in water, which are washed away, leaving the surface clean.

Removal of Hardness

Temporary Hardness:

• Method 1: By Boiling
  • Process: Boiling hard water drives off carbon dioxide and converts soluble hydrogen carbonates into insoluble carbonates, which can be removed by filtration or decantation.
  • Chemical Reaction:
    \[
    \text{Ca(HCO}_3\text{)}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O} + \text{CO}_2 \uparrow
    \]
  • Effect: Calcium carbonate and magnesium carbonate precipitate out, leaving the water soft.
• Method 2: By Addition of Lime (Clark’s Process)
  • Process: Adding a calculated quantity of slaked lime to hard water. The lime reacts with soluble hydrogen carbonates to form insoluble carbonates, which can be removed by filtration.
  • Chemical Reactions:
    \[
    \text{Ca(HCO}_3\text{)}_2 + \text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3 \downarrow + 2\text{H}_2\text{O}
    \]
    \[
    \text{Mg(HCO}_3\text{)}_2 + \text{Ca(OH)}_2 \rightarrow \text{MgCO}_3 \downarrow + \text{CaCO}_3 \downarrow + 2\text{H}_2\text{O}
    \]
  • Effect: The lime causes the calcium and magnesium bicarbonates to precipitate out as carbonates, leaving the water soft.
  • Procedure: Lime is mixed with water in a tank, and the mixture is fed into another tank containing hard water. The resulting precipitate is removed by filtration.

Summary Points

• Furring of Tea Kettle: Caused by calcium and magnesium carbonates from hard water.
• Hard Water: Contains calcium and magnesium ions, not suitable for washing due to soap scum formation.
• Soap: Forms insoluble scum with hard water, making it difficult to lather.
• Detergents: More effective than soap in hard water; not affected by hard water.
• Cleansing Action: Involves micelle formation and removal of oily dirt.
• Removal of Hardness:
  • Boiling: Converts soluble hydrogen carbonates to insoluble carbonates.
  • Clark’s Process: Uses slaked lime to precipitate out carbonates and remove hardness.

Set H

Removal of Permanent Hardness in Water

Simple Explanation

Permanent Hardness:

• Caused by the presence of dissolved calcium and magnesium salts (like sulfates and chlorides) in water.
• Cannot be removed by boiling.

Methods to Remove Permanent Hardness:

Using Soda Ash (Sodium Carbonate):

• Process: Adding soda ash (\( \text{Na}_2\text{CO}_3 \)) to hard water.
• Chemical Reactions:
  \[
  \text{CaSO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 \downarrow + \text{Na}_2\text{SO}_4
  \]
  \[
  \text{CaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 \downarrow + 2\text{NaCl}
  \]
• Result: Calcium carbonate (\( \text{CaCO}_3 \)) precipitates out, and the water becomes soft.

Using Lime (Calcium Hydroxide):

• Process: If magnesium sulfate is present, lime (\( \text{Ca(OH)}_2 \)) is added along with soda ash.
• Chemical Reactions:
  \[
  \text{MgSO}_4 + \text{Na}_2\text{CO}_3 + \text{Ca(OH)}_2 \rightarrow \text{Mg(OH)}_2 \downarrow + \text{CaCO}_3 \downarrow + \text{Na}_2\text{SO}_4
  \]
• Result: Magnesium hydroxide (\( \text{Mg(OH)}_2 \)) and calcium carbonate precipitate out, leaving the water soft.

Permutit Process:

• Material: Permutit is an artificial zeolite, chemically a hydrated sodium aluminium orthosilicate.
• Formula: \( \text{Na}_2\text{Al}_2\text{Si}_2\text{O}_8 \cdot \text{XH}_2\text{O} \)
• Process: Hard water passes through lumps of permutit in a tall cylinder.
  • Ion Exchange: Sodium ions from permutit replace calcium and magnesium ions in water, softening it.
  • Regeneration: When permutit is no longer active, it is regenerated by washing with a concentrated brine solution (\( \text{NaCl} \)).
    \[
    \text{CaP} + 2\text{NaCl} \rightarrow \text{Na}_2\text{P} + \text{CaCl}_2
    \]

Ion Exchange Resins:

• Usage: For more scientific purposes where complete removal of all ions is required.
• Process: Water passes through synthetic resins that exchange unwanted ions for harmless ones.
  • Cation Exchange: Sodium ions replaced by hydrogen ions (\( \text{H}^+ \)).
  • Anion Exchange: Bicarbonate (\( \text{HCO}_3^- \)) and sulfate (\( \text{SO}_4^{2-} \)) ions replaced by hydroxide ions (\( \text{OH}^- \)).
• Result: Water becomes deionized, removing all ion content.

Summary Points

• Soda Ash Method: Removes calcium and magnesium ions by forming insoluble carbonates.
• Lime Addition: Enhances the precipitation of magnesium when combined with soda ash.
• Permutit Process: Uses artificial zeolite to replace calcium and magnesium with sodium ions; can be regenerated with brine.
• Ion Exchange Resins: Completely deionizes water by exchanging all ions for hydrogen and hydroxide ions.

Set I

Exercise 3(C) Solutions

1. What do you understand by:

   (a) Soft water: Water that does not contain significant amounts of dissolved minerals, such as calcium and magnesium ions. It easily forms lather with soap.
   (b) Hard water: Water that contains high concentrations of dissolved minerals, primarily calcium and magnesium ions. It does not easily form lather with soap.
   (c) Temporary hard water: Hard water that contains dissolved bicarbonates of calcium and magnesium. It can be softened by boiling.
   (d) Permanent hard water: Hard water that contains dissolved sulfates or chlorides of calcium and magnesium. It cannot be softened by boiling.

2. What are the causes for:

   (a) Temporary hardness: Caused by the presence of dissolved calcium bicarbonate (\( \text{Ca(HCO}_3\text{)}_2 \)) and magnesium bicarbonate (\( \text{Mg(HCO}_3\text{)}_2 \)) in water.
   (b) Permanent hardness: Caused by the presence of dissolved calcium sulfate (\( \text{CaSO}_4 \)) and magnesium sulfate (\( \text{MgSO}_4 \)), as well as calcium chloride (\( \text{CaCl}_2 \)) and magnesium chloride (\( \text{MgCl}_2 \)).

3. What are the advantages of:

   (i) Soft water:

   • Easily forms lather with soap, making it more effective for cleaning and washing.
   • Does not form scale in boilers and heating systems, preventing damage and inefficiency.

   (ii) Hard water:

   • Contains essential minerals like calcium and magnesium, which are beneficial for health.
   • Often tastes better due to the presence of these minerals.

4. What are stalagmites and stalactites? How are they formed?

   Stalagmites: Conical formations rising from the floor of a cave, formed by the deposition of calcium carbonate from dripping water.
   Stalactites: Icicle-like formations hanging from the roof of a cave, formed by the deposition of calcium carbonate from dripping water.
   Formation Process: Water containing dissolved calcium hydrogen carbonate drips from the roof of a cave. When the water loses carbon dioxide, calcium carbonate is deposited, forming stalactites. When the drips hit the floor, they form stalagmites.

5. Name the substances which make water:

   (i) Temporarily hard: Calcium bicarbonate (\( \text{Ca(HCO}_3\text{)}_2 \)), magnesium bicarbonate (\( \text{Mg(HCO}_3\text{)}_2 \)).
   (ii) Permanently hard: Calcium sulfate (\( \text{CaSO}_4 \)), magnesium sulfate (\( \text{MgSO}_4 \)), calcium chloride (\( \text{CaCl}_2 \)), magnesium chloride (\( \text{MgCl}_2 \)).

6. Give equations to show what happens when temporary hard water is:

   (a) Boiled:
   \[
   \text{Ca(HCO}_3\text{)}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O} + \text{CO}_2 \uparrow
   \]
   (b) Treated with slaked lime:
   \[
   \text{Ca(HCO}_3\text{)}_2 + \text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3 \downarrow + 2\text{H}_2\text{O}
   \]

7. State the disadvantages of using hard water.

   • Difficult to form lather with soap, leading to higher soap consumption.
   • Forms scale in boilers, pipes, and kettles, reducing efficiency and causing potential damage.
   • Can cause damage to appliances and increase maintenance costs.
   • Leaves stains on clothes and fixtures.

8. What is a soap, what is it used for?

   Soap: A sodium or potassium salt of fatty acids. It is used for cleaning and washing purposes.
   Use: Soap molecules emulsify fats and oils, allowing them to be washed away with water.

9. What is the advantage of a detergent over soap?

   • Detergents are more effective in hard water because they do not form insoluble scum with calcium and magnesium ions.
   • Detergents are more soluble in water and have better cleaning properties.

10. Why does the hardness of water render it unfit for use in:

    (i) Boiler:

    • Forms scale and deposits inside the boiler, reducing efficiency and causing potential damage.

    (ii) Washing purposes:

    • Prevents soap from lathering properly, leading to higher soap consumption and less effective cleaning.

11. Explain with equation, what is noticed when permanent hard water is treated with:

    (a) Slaked lime:
    \[
    \text{MgSO}_4 + \text{Na}_2\text{CO}_3 + \text{Ca(OH)}_2 \rightarrow \text{Mg(OH)}_2 \downarrow + \text{CaCO}_3 \downarrow + \text{Na}_2\text{SO}_4
    \]
    (b) Washing soda:
    \[
    \text{CaSO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 \downarrow + \text{Na}_2\text{SO}_4
    \]

12. Explain the permutit method for softening hard water.

    Permutit: An artificial zeolite, a hydrated sodium aluminium orthosilicate.
    Process: Hard water passes through lumps of permutit in a cylinder.

    • Sodium ions from permutit replace calcium and magnesium ions in the water.
    • Water becomes soft as calcium and magnesium are removed.
    • When permutit is no longer effective, it is regenerated by washing with a concentrated brine solution (\( \text{NaCl} \)), which replaces calcium and magnesium ions with sodium ions again.
    • Chemical Reaction:
      \[
      \text{CaP} + 2\text{NaCl} \rightarrow \text{Na}_2\text{P} + \text{CaCl}_2
      \]