Chapter 4 - Carbon and Its Compounds Class 10 Science (Ncert Solutions)
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NCERT Solutions for Chapter 4 Carbon and Its Compounds Class 10 Science (Question/Answers, Activity & Projects)
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QUESTIONS
Q 1. Make a list of ten things you have used or consumed since the morning
Ans 1: List of Ten Items:
- Smartphone
- Coffee
- Paper
- Plastic Water Bottle
- Laptop
- Cotton Shirt
- Glass of Juice
- Pen
- Bread
- Wooden Desk
Q 2. Compile this list with the lists made by your classmates and then sort the items into the adjacent Table
| Things Made of Metal | Things Made of Glass/Clay | Others |
Ans 2: A List of items into two categories: “Things made of metal” and “Things made of glass/clay,” with an additional category for “Others.” Here’s an example of how the sorting might look in the adjacent table format:
| Things made of metal | Things made of glass/clay | Others |
| Spoon | Glass cup | Wooden chair |
| Fork | Clay pot | Plastic bottle |
| Knife | Glass vase | Paper book |
| Coin | Clay plate | Fabric cloth |
| Car | Glass bottle | Rubber ball |
Q 3. If there are items which are made up of more than one material, put them into both the relevant columns of the table:
Ans 3: Here’s a table categorizing items made of metal, glass/clay, or both:
| Things Made of Metal | Things Made of Glass/Clay | Things Made of Both Metal and Glass/Clay |
| Spoon | Mug | Glass lamp with a metal base |
| Fork | Vase | Metal frame with glass panels |
| Knife | Plate | Stained glass window with metal framing |
| Key | Pot | Ceramic teapot with metal lid |
| Coins | Jar | Clay stove with metal parts |
| Nail | Bottle | Glass door with metal handle |
| Lock | Glass cup | Metal sculpture with glass components |
QUESTIONS
Q 1. What would be the electron dot structure of carbon dioxide which has the formula ?
Ans 1: the structure of CO₂
Q 2. What would be the electron dot structure of a molecule of Sulphur which is made up of eight atoms of Sulphur? (Hint –The eight atoms of Sulphur are joined together in the form of a ring.)
Ans 2: The electron dot structure of a molecule of sulfur (S₈) involves eight sulfur atoms arranged in a ring. Each sulfur atom has six valence electrons, and they form bonds by sharing electrons. In the structure, each sulfur atom shares two electrons with its adjacent sulfur atoms, creating a single bond between them. The remaining four electrons on each sulfur atom form lone pairs, which are not involved in bonding. The overall structure is a ring with each sulfur atom bonded to two others, with lone pairs on each atom.
Eight atoms (– S₈) of Sulphur molecule
Activity 4.2
Q 1. Calculate the difference in the formulae and molecular masses for
- CH₃OH and C₂H₅OH
Ans 1:
(a): CH₃OH and C₂H₅OH:
- Formula Difference: C₂H₅OH – CH₃OH → C₂H₅OH has 1 more carbon, 2 more hydrogens.
- Molecular Mass Difference:
- CH₃OH = 32 g/mol
- C₂H₅OH = 46 g/mol
- Difference = 46 – 32 = 14 g/mol
B): C₂H₅OH and C₃H₇OH:
Ans (b) C₂H₅OH and C₃H₇OH:
- Formula Difference: C₃H₇OH – C₂H₅OH → C₃H₇OH has 1 more carbon, 2 more hydrogens.
- Molecular Mass Difference:
- C₂H₅OH = 46 g/mol
- C₃H₇OH = 60 g/mol
- Difference = 60 – 46 = 14 g/mol
(c) C₃H₇OH and C₄H₉OH:
Ans (c) C₃H₇OH and C₄H₉OH:
- Formula Difference: C₄H₉OH – C₃H₇OH → C₄H₉OH has 1 more carbon, 2 more hydrogens.
- Molecular Mass Difference:
- C₃H₇OH = 60 g/mol
- C₄H₉OH = 74 g/mol
- Difference = 74 – 60 = 14 g/mol
Q 2. Is there any similarity in these three?
Ans 2: In all three comparisons, the molecular mass increases by 14 g/mol as the number of carbon and hydrogen atoms increases by 1 carbon and 2 hydrogens.
Q 3. Arrange these alcohols in the order of increasing carbon atoms to get a family. Can we call this family a homologous series?
Ans 3: Family of Alcohols:
- Order of increasing carbon atoms: CH₃OH, C₂H₅OH, C₃H₇OH, C₄H₉OH.
- These alcohols form a homologous series, as they have the same functional group (-OH) and differ by a CH₂ unit, showing a consistent pattern of molecular mass increase.
Q 4. Generate the homologous series for compounds containing up to four carbons for the other functional groups given in Table 4.3.
Ans 4: Here’s the homologous series for compounds containing up to four carbon atoms for each of the functional groups mentioned:
1. Haloalkanes (Chloro/Bromo)
General Formula: CₙH₂ₙ₊₁X (where X is Cl or Br)
- Methane (CH₄) → Methyl chloride (CH₃Cl) / Methyl bromide (CH₃Br)
- Ethane (C₂H₆) → Ethyl chloride (C₂H₅Cl) / Ethyl bromide (C₂H₅Br)
- Propane (C₃H₈) → Propyl chloride (C₃H₇Cl) / Propyl bromide (C₃H₇Br)
- Butane (C₄H₁₀) → Butyl chloride (C₄H₉Cl) / Butyl bromide (C₄H₉Br)
2. Alcohols (–OH)
General Formula: CₙH₂ₙ₊₁OH
- Methanol (CH₃OH)
- Ethanol (C₂H₅OH)
- Propanol (C₃H₇OH)
- Butanol (C₄H₉OH)
3. Aldehydes (–CHO)
General Formula: CₙH₂ₙO
- Methanal (CH₂O) (Formaldehyde)
- Ethanal (C₂H₄O) (Acetaldehyde)
- Propanal (C₃H₆O)
- Butanal (C₄H₈O)
4. Ketones (–CO)
General Formula: CₙH₂ₙO
- Methanone (CH₂O) (Formaldehyde) — Usually, methanone does not exist in isolation, as methanol would be oxidized into an aldehyde.
- Ethanone (C₂H₄O) (Acetone)
- Propanone (C₃H₆O)
- Butanone (C₄H₈O)
5. Carboxylic Acids (–COOH)
General Formula: CₙH₂ₙ₋₁COOH
- Methanoic acid (HCOOH) (Formic acid)
- Ethanoic acid (C₂H₄O₂) (Acetic acid)
- Propanoic acid (C₃H₆O₂)
- Butanoic acid (C₄H₈O₂)
Each of these groups follows a typical homologous series where the functional group remains constant, and the rest of the molecule changes in a predictable pattern as the number of carbon atoms increases.
QUESTIONS
Q 1. How many structural isomers can you draw for pentane?
Ans 1: we can draw three structural isomers of pentane
A): n-Pentane (straight-chain pentane): This is the unbranched form where all five carbon atoms are connected in a continuous chain.
B): Isopentane (2-methylbutane): This is a branched isomer where one methyl group (-CH₃) is attached to the second carbon of a four-carbon chain.
C): Neopentane (2,2-dimethylpropane): This is a more highly branched isomer where two methyl groups are attached to the second carbon of a three-carbon chain.
Q 2. What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Ans 2: The two key properties of carbon that lead to the vast number of carbon compounds are:
- Covalent Bonding: Carbon has four electrons in its outer shell, allowing it to form four strong covalent bonds with other atoms. This bonding flexibility enables carbon to create a variety of complex structures, including chains, rings, and branched molecules.
- Catenation: Carbon has the unique ability to bond with other carbon atoms, forming long chains or rings. This property, called catenation, allows carbon to create vast networks of molecules, leading to an immense variety of organic compounds.
Together, these properties make carbon central to the diversity of life and the complexity of organic chemistry.
Q 3. What will be the formula and electron dot structure of cyclopentane?
Ans 3: Cyclopentane is a cyclic hydrocarbon with the molecular formula C₅H₁₀. It consists of five carbon atoms forming a pentagon-shaped ring, with each carbon bonded to two hydrogen atoms.
Q 4. Draw the structures for the following compounds.
(i). Ethanoic acid
(ii). Bromo pentane
(iii). Butanone
(iv). Hexanal
Are structural isomers possible for bromo pentane?
Ans 4: (i) Ethanoic acid (CH₃COOH):
(ii). Bromo pentane (C₅H₁₁Br):
(iii) Butanone (C₄H₈O):
(iv) Hexanal (C₆H₁₂O):
Yes, there are three structural isomers possible for bromo pentane.
(i)
(ii)
(iii)
Q 5. How would you name the following compounds?
Ans 5: (i). bromo-ethane
(ii). Methanal
(iii). Hex 1-yne
Activity 4.3
CAUTION: This Activity needs the teacher’s assistance.
- Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatula and burn them.
- Observe the nature of the flame and note whether smoke is produced.
- Place a metal plate above the flame. Is there a deposition on the plate in case of any of the compounds?
Solution: Activity 4.3 [students do yourself with the help of teachers]
[Hint solution]: This activity demonstrates the combustion of different carbon compounds and helps observe their characteristics.
Here’s a step-by-step guide for the experiment:
- Materials Needed:
- Naphthalene
- Camphor
- Alcohol (like ethanol or isopropyl alcohol)
- Spatula
- Metal plate (e.g., a metal sheet or aluminum foil)
- Flame source (e.g., a Bunsen burner or a lighter)
- Procedure:
- Step 1: Take a small amount of each carbon compound (naphthalene, camphor, alcohol) on a spatula.
- Step 2: Burn each compound one by one using the flame. Observe and note the characteristics of the flame, such as its color, size, and whether there’s any smoke produced.
- Naphthalene: Typically burns with a yellow, sooty flame and produces smoke due to incomplete combustion.
- Camphor: Burns with a bright, clean flame, producing little to no smoke.
- Alcohol: Burns with a clean, blue flame, producing minimal or no smoke.
- Step 3: Hold a metal plate above the flame during the burning process.
- Observation: After burning the compounds, check if there’s any deposition of soot or other residues on the metal plate.
- Naphthalene: Likely to leave a soot deposition on the metal plate because of incomplete combustion.
- Camphor: May leave minimal or no deposition, as it generally burns more completely.
- Alcohol: Generally, burns cleanly, leaving little to no deposition.
- Conclusion:
- The nature of the flame and the amount of smoke or soot produced depend on how completely the substance combusts. Naphthalene tends to produce more smoke and soot due to incomplete combustion, while camphor and alcohol usually burn more cleanly with minimal residue.
Please ensure this activity is done with proper supervision due to the use of an open flame.
Activity 4.4
- Light a Bunsen burner and adjust the air hole at the base to get different types of flames/presence of smoke.
- When do you get a yellow, sooty flame?
- When do you get a blue flame?
Solution: Here’s how to complete Activity 4.4 with a Bunsen burner to observe different types of flames and the presence of smoke:
- Light the Bunsen burner: Ensure you follow safety protocols—wear goggles, keep flammable items away, and light the Bunsen burner with a spark lighter or matchstick while the air hole is fully closed.
- Adjust the air hole at the base of the burner to experiment with different flame types:
- Yellow, sooty flame (Luminous flame):
- You will observe a yellow, sooty flame when the air hole is closed or only slightly open. This flame is less hot and produces smoke because it has incomplete combustion. The limited oxygen supply causes the flame to burn inefficiently, leading to soot formation (carbon particles) and a lower temperature.
- Blue flame (non-luminous flame):
- You will see a blue flame when the air hole is fully open. This flame is much hotter and does not produce smoke because it allows for complete combustion with adequate oxygen supply. The blue flame is more efficient and is commonly used for heating substances because it produces consistent, high heat without soot.
- Yellow, sooty flame (Luminous flame):
This activity demonstrates how the adjustment of oxygen intake affects the combustion process and flame characteristics.
Activity 4.5
- Take about 3 mL of ethanol in a test tube and warm it gently in a water bath.
- Add a 5% solution of alkaline potassium permanganate drop by drop to this solution.
- Does the colour of potassium permanganate persist when it is added initially?
- Why does the colour of potassium permanganate not disappear when excess is added?
Q 1. Does the colour of potassium permanganate persist when it is added initially?
Ans 1: The color of potassium permanganate will continue to fade as long as there is ethanol available to be oxidized. Once all the ethanol is oxidized, any additional potassium permanganate added will retain its color, as there are no more ethanol molecules left to react with it.
Q 2. Why does the colour of potassium permanganate not disappear when excess is added?
Ans 2: In this activity, ethanol is being oxidized by potassium permanganate, which acts as an oxidizing agent. When potassium permanganate is added drop by drop, it reacts with ethanol, causing the purple color to disappear as it is reduced to manganese dioxide (MnO₂), which is brown. This process continues until all the ethanol has been oxidized.
However, if you add an excess of potassium permanganate beyond the amount needed to react with the ethanol, there will be no more ethanol left for further reactions. As a result, the purple color of the potassium permanganate will persist in the solution because there is no remaining ethanol to react with it.
Questions
Q 1. Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Ans 1: The conversion of ethanol (CH₃CH₂OH) to ethanoic acid (CH₃COOH) is an oxidation reaction because:
- The number of hydrogen atoms decreases in the molecule.
- The number of oxygen atoms increases due to the addition of oxygen during the reaction.
Q 2. A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?
Ans 2: A mixture of oxygen and ethyne is used for welding because it produces a high-temperature flame (around 3000°C), which is essential for melting metals. In contrast, ethyne and air do not burn at such high temperatures because air contains only about 21% oxygen, which is insufficient for complete combustion. This results in a cooler flame, making it unsuitable for welding.
Activity 4.6
Teacher’s demonstration –
- Drop a small piece of sodium, about the size of a couple of grains of rice, into ethanol (absolute alcohol).
- What do you observe?
- How will you test the gas evolved?
Q 1. What do you observe?
Ans 1: Observation:
- Physical Reaction:
- When a small piece of sodium is added to ethanol, it moves rapidly across the surface of the liquid.
- Fizzing or bubbling is observed as hydrogen gas is produced.
- Thermal and Visual Effects:
- Heat is released during the reaction, which may cause the hydrogen gas to ignite, producing a small flame.
- Gas Evolution:
- Hydrogen gas () is evolved during the reaction.
These observations confirm the reaction between sodium and ethanol, forming sodium ethoxide and hydrogen gas.
Q 2. How will you test the gas evolved?
Ans 2: To test the gas evolved in the reaction between sodium and ethanol, you should:
- Bring a lit splint near the mouth of the container where the gas is being produced.
- Observe the reaction: If the gas is hydrogen (H₂), it will ignite with a distinct ‘pop’ sound when it comes into contact with the flame.
This ‘pop’ sound is a clear indication that hydrogen gas is present.
Activity 4.7
- Compare the pH of dilute acetic acid and dilute hydrochloric acid using both litmus paper and universal indicator.
- Are both acids indicated by the litmus test?
- Does the universal indicator show them as equally strong acids?
Ans: Activity 4.7: Comparing pH of Dilute Acetic Acid and Dilute Hydrochloric Acid
1. Litmus Paper Test:
- Acetic Acid (CH₃COOH): When a piece of litmus paper (either red or blue) is dipped into dilute acetic acid, the litmus paper will turn red since acetic acid is an acid.
- Hydrochloric Acid (HCl): When a piece of litmus paper is dipped into dilute hydrochloric acid, it will also turn red because hydrochloric acid is a strong acid.
Conclusion: Both acids are indicated as acids by the litmus test, as both turn blue litmus paper red.
2. Universal Indicator Test:
- Acetic Acid (CH₃COOH): The universal indicator will show a slightly acidic color (around pH 4-6), which is more toward yellow or orange.
- Hydrochloric Acid (HCl): The universal indicator will show a strongly acidic color (around pH 1-3), typically red.
Conclusion: No, the universal indicator does not show them as equally strong acids. Hydrochloric acid is a strong acid (lower pH) and will show a red color, while acetic acid is a weak acid (higher pH) and will show an orange or yellowish color.
Summary of Results:
- Litmus test: Both acetic acid and hydrochloric acid are indicated as acids (turn blue litmus paper red).
- Universal indicator: Hydrochloric acid is a stronger acid than acetic acid, with a lower pH and a red color, while acetic acid shows a weaker acidic color.
Activity 4.8
- Take 1 mL ethanol (absolute alcohol) and 1 mL glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
- Warm in a water-bath for at least five minutes as shown in Fig. 4.11.
- Pour into a beaker containing 20-50 mL of water and smell the resulting mixture.
Solution: Activity 4.8: The activity you’re describing seems to be part of an experiment for preparing ethyl acetate (an ester) through an esterification reaction. Here’s a breakdown of the procedure and the reactions involved:
Fig 4.11 Formation of ester
Esterification Reaction:
The reaction between ethanol (ethyl alcohol) and acetic acid in the presence of concentrated sulfuric acid produces ethyl acetate and water. This is a typical esterification reaction, which can be written as:
Ethanol (C₂H₅OH) + Acetic Acid (CH₃COOH) → Ethyl Acetate (CH₃COOC₂H₅) + Water (H₂O)
Step-by-Step Procedure:
- Combine the reactants:
- Add 1 mL of absolute ethanol (C₂H₅OH) and 1 mL of glacial acetic acid (CH₃COOH) into a clean test tube.
- Add a few drops of concentrated sulfuric acid (H₂SO₄). The sulfuric acid acts as a catalyst and helps in removing water from the reaction, which shifts the equilibrium towards the production of more ester.
- Heating:
- Warm the mixture in a water bath for about 5 minutes. The heating accelerates the reaction, allowing the esterification process to occur. The reaction typically produces heat.
- Add water:
- After the reaction has been allowed to proceed for 5 minutes, pour the mixture into a beaker containing 20-50 mL of water. This helps to dilute the mixture and can facilitate the separation of the ester from the reaction mixture.
- Smell the mixture:
- Carefully smell the resulting mixture. You should detect the characteristic fruity odor of ethyl acetate. This is because the ester has a distinct smell often described as sweet or fruity.
Important Notes:
- Safety: Make sure to conduct this experiment in a well-ventilated area or under a fume hood due to the volatile nature of ethanol, acetic acid, and sulfuric acid. Wear gloves and safety goggles to protect from any potential splashes or fumes.
- Observations: The product (ethyl acetate) will have a fruity smell, which is typical of many esters. The reaction also produces water, but the sulfuric acid will help to remove it, ensuring the ester is formed efficiently.
This experiment demonstrates the process of esterification and the formation of an ester, which is a key reaction in organic chemistry.
Activity 4.9
- Set up the apparatus as shown in Chapter 2, Activity 2.5.
- Take a spatula full of sodium carbonate in a test tube and add 2 mL of dilute ethanoic acid.
Q 1. What do you observe?
Ans 1: Observations:
- Reaction with Sodium Carbonate: When sodium carbonate reacts with dilute ethanoic acid, you should observe the evolution of gas, which is carbon dioxide (CO₂). This can cause the formation of bubbles in the solution. The reaction is as follows:
Na₂CO₃(aq) + 2CH₃COOH(aq) → 2CH₃COONa(aq) + H₂O(l) + CO₂(g)
Q 2. Pass the gas produced through freshly prepared lime-water. What do you observe?
Ans 2: Pass Gas Through Lime-water: When the gas (carbon dioxide) produced is passed through freshly prepared lime-water (a solution of calcium hydroxide), you will observe that the lime-water turns milky due to the formation of calcium carbonate (CaCO₃). This indicates the presence of carbon dioxide gas:
CO₂(g) + Ca(OH)₂(aq) → CaCO₃(s) + H₂O(l)
Q 3. Can the gas produced by the reaction between ethanoic acid and sodium carbonate be identified by this test?
Ans 3: Yes, the gas produced by the reaction between ethanoic acid and sodium carbonate can be identified by this test. When carbon dioxide (CO₂) is passed through lime water (calcium hydroxide solution), it reacts with the lime water to form a milky white precipitate of calcium carbonate (CaCO₃). This confirms the presence of carbon dioxide.
Q 4. Repeat this Activity with sodium hydrogen carbonate instead of sodium carbonate.
Ans 4: To repeat the activity with sodium hydrogen carbonate (NaHCO₃) instead of sodium carbonate (Na₂CO₃):
- Take a spatula full of sodium hydrogen carbonate (NaHCO₃) and place it in a test tube.
- Add 2 mL of dilute ethanoic acid (CH₃COOH) to the test tube.
- Observe effervescence (bubbling) due to the release of gas.
- Pass the gas through freshly prepared lime water.
- The lime water will turn milky, confirming the presence of carbon dioxide (CO₂).
The reaction with sodium hydrogen carbonate is:
NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
This confirms the production of carbon dioxide, which turns lime water milky.
Chapter 4 – Carbon and Its Compounds Class 10 Science Solutions Pdf Download
Updated Solution 2024-2025
Questions
Q 1. How would you distinguish experimentally between an alcohol and a carboxylic acid?
Ans 1: To distinguish experimentally between an alcohol and a carboxylic acid, you can perform the following tests:
- Reaction with Sodium Bicarbonate (NaHCO₃):
- Carboxylic acids react with sodium bicarbonate, producing carbon dioxide gas (effervescence).
- Alcohols do not react with sodium bicarbonate.
- Reaction with Oxidizing Agent (e.g., Potassium Dichromate):
- Alcohols can be oxidized to aldehydes or carboxylic acids (depending on the type of alcohol).
- Carboxylic acids do not undergo oxidation under normal conditions.
These tests help identify the functional groups present in the compounds.
Q 2. What are oxidizing agents?
Ans 2: Oxidizing agents are substances that can accept electrons from another substance during a chemical reaction, causing the other substance to undergo oxidation. In the process, the oxidizing agent itself gets reduced. Examples of oxidizing agents include oxygen (O₂), chlorine (Cl₂), hydrogen peroxide (H₂O₂), and potassium permanganate (KMnO₄).
Activity 4.10
- Take about 10 mL of water each in two test tubes.
- Add a drop of oil (cooking oil) to both the test tubes and label them as A and B.
- To test tube B, add a few drops of soap solution.
- Now shake both the test tubes vigorously for the same period of time.
- Can you see the oil and water layers separately in both the test tubes immediately after you stop shaking them?
- Leave the test tubes undisturbed for some time and observe. Does the oil layer separate out? In which test tube does this happen first?
Ans 4.10: In this experiment, you’re exploring the behavior of oil, water, and soap solution under different conditions. Here’s what you can expect to observe:
- Immediately after shaking:
- In Test Tube A (with only water and oil), the oil and water will likely form two separate layers after shaking stops. The oil, being less dense, will float on top of the water.
- In Test Tube B (with water, oil, and soap solution), the soap acts as an emulsifier, breaking up the oil into small droplets and allowing the oil to mix with water temporarily. After shaking, you may notice the oil doesn’t separate into a distinct layer immediately.
- After leaving the test tubes undisturbed:
- In Test Tube A, you should observe the oil layer separating out from the water as the oil floats on top again, since no emulsifier is present.
- In Test Tube B, the oil may stay mixed with the water for a longer period due to the emulsifying effect of the soap. However, eventually, over time, the oil may still separate slowly, but it will take longer compared to Test Tube A.
Conclusion: The oil separates faster in Test Tube A, where no soap is present. The soap in Test Tube B slows down the separation process by preventing the oil from forming large droplets, keeping it mixed with water for a while.
Activity 4.11
- Take about 10 mL of distilled water (or rain water) and 10 mL of hard water (from a tubewell or hand-pump) in separate test tubes.
- Add a couple of drops of soap solution to both.
- Shake the test tubes vigorously for an equal period of time and observe the amount of foam formed.
- In which test tube do you get more foam?
- In which test tube do you observe a white curdy precipitate?
Note for the teacher: If hard water is not available in your locality, prepare some hard water by dissolving hydrogen carbonates/sulphate/chlorides of calcium or magnesium in water.
Ans 4.11: In this experiment, you will observe the difference in soap behavior in distilled (or rain) water versus hard water.
Procedure and Observations:
- Add Soap Solution:
- In separate test tubes, take 10 mL of distilled (or rain) water and 10 mL of hard water.
- Add a couple of drops of soap solution to each test tube.
- Shake the Test Tubes:
- Shake both test tubes vigorously for the same period of time.
- Observations:
- Foam Formation:
- The distilled or rainwater (which is soft water) will produce more foam as the soap molecules form lather easily.
- The hard water will produce less foam due to the presence of calcium and magnesium ions that react with soap, reducing its foaming ability.
- White Curdy Precipitate:
- In the hard water, you will observe the formation of a white curdy precipitate. This happens because the calcium or magnesium ions in hard water react with soap to form an insoluble calcium or magnesium soap, which appears as a curd-like precipitate.
- Foam Formation:
Conclusion:
- More foam will form in the distilled water, and a white curdy precipitate will form in the hard water due to the formation of insoluble calcium or magnesium soaps.
Activity 4.12
- Take two test tubes with about 10 mL of hard water in each.
- Add five drops of soap solution to one and five drops of detergent solution to the other.
- Shake both test tubes for the same period.
- Do both test tubes have the same amount of foam?
- In which test tube is a curdy solid formed?
Ans 4.12 Activity: In this activity, you’re comparing the behavior of soap and detergent in hard water.
- Foam formation: The test tube with the detergent solution will likely produce more foam compared to the one with the soap solution. This is because detergents are designed to work well in hard water, where soap tends to lose its effectiveness due to the presence of calcium and magnesium ions.
- Curdy solid formation: The test tube with the soap solution will likely form a curdy solid. This is because soap reacts with the calcium and magnesium ions in hard water to form insoluble compounds, leading to the formation of a precipitate or curdy solid. Detergents, on the other hand, are formulated to avoid this reaction, so no curdy solid will form in the detergent solution.
In summary:
- Foam: More foam in the detergent solution.
- Curdy solid: Formed in the soap solution.
Questions
Q 1. Would you be able to check if water is hard by using a detergent?
Ans 1: Yes, you can get an idea of water hardness by using detergent. Hard water contains high levels of calcium and magnesium ions, which can affect the lathering ability of soaps and detergents. If you use a detergent or soap in hard water, it will tend to form less lather compared to soft water. Here’s a basic way to check for hardness using detergent:
- Fill a Bottle: Take a clean, clear plastic bottle and fill it halfway with the water you want to test.
- Add Detergent: Add a few drops of liquid detergent or soap to the bottle.
- Shake the Bottle: Cap the bottle tightly and shake it vigorously for a few seconds.
- Observe the Lather: If the water is soft, you should see a lot of foam or lather. If the water is hard, you’ll notice less foam or it may not lather well at all.
While this method can give you a rough idea of water hardness, it’s not as precise as using a water hardness test kit or sending a sample to a lab for detailed analysis.
Q 2. People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Ans 2: Agitation is necessary for cleaning clothes because it helps in the physical process of removing dirt, stains, and oils from fabrics. Here’s why it’s important:
- Loosens Dirt and Debris: Agitation helps break the bond between the fabric and dirt particles by creating friction. When clothes are beaten or stirred, dirt that is stuck in the fibers is loosened and released, making it easier for the detergent to wash it away.
- Increases Detergent Effectiveness: When clothes are agitated, the detergent is evenly distributed and works more efficiently to break down oils, grease, and stains on the fabric. It also helps the detergent penetrate deep into the fibers of the fabric.
- Boosts Cleaning Power: The motion helps create a better flow of water and soap through the clothes, allowing the detergent to lift the dirt and oil away from the fabric. This action increases the overall cleaning power, especially for tougher stains.
- Improves Rinsing: Agitation ensures that the detergent is thoroughly rinsed out, preventing soap buildup on the fabric and helping to remove all the residues.
In short, agitation is a key part of the mechanical process that assists in thoroughly cleaning clothes, removing dirt, stains, and detergent residues from the fabric.
Exercise
Q 1. Ethane, with the molecular formula C₂H₆, has:
(a) 6 covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds
Ans 1: (b) 7 covalent bonds.
Q 2. Butanone is a four-carbon compound with the functional group
(a) carboxylic acid
(b) aldehyde
(c) ketone
(d) alcohol
Ans 2: (c) ketone
Q 3. While cooking, if the bottom of the vessel is getting blackened on the outside, it means that
(a) the food is not cooked completely
(b) the fuel is not burning completely
(c) the fuel is wet
(d) the fuel is burning completely
Ans 3: (b) the fuel is not burning completely
Q 4. Explain the nature of the covalent bond using the bond formation in CH₃Cl.
Ans 4: The nature of a covalent bond in CH₃Cl (methyl chloride) can be explained by the sharing of electrons:
- C-H Bonds: Carbon (C) shares one electron each with three hydrogen (H) atoms, forming three single covalent bonds (C-H). This occurs because hydrogen has one valence electron and needs one more to achieve a stable configuration.
- C-Cl Bond: Carbon shares one electron with chlorine (Cl), forming a single covalent bond (C-Cl). Chlorine has seven valence electrons and shares one to complete its octet.
The resulting molecule has a tetrahedral structure with a covalent bond between each pair of shared electrons, ensuring stability for all atoms involved.
Q 5. Draw the electron dot structures for
(a) Ethanoic acid
(b) H₂S
(c) Propanone
(d) F₂
Ans 5: (a) Ethanoic acid
(b) H₂S
(c) Propanone
(d) F₂
Q 6. What is a homologous series? Explain with an example.
Ans 6: A homologous series is a group of organic compounds that have a similar structure, the same functional group, and follow a regular pattern in their molecular formula, differing by a constant CH₂ unit. These compounds have similar chemical properties but show gradual changes in physical properties (e.g., boiling point).
Example:
The alkane series (CnH₂n+₂): Methane (CH₄), Ethane (C₂H₆), Propane (C₃H₈). Here, each compound differs by a CH₂ unit.
Q 7. How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?
Ans 7: Here’s a table comparing ethanol and ethanoic acid based on their physical properties:
| Property | Ethanol (C₂H₅OH) | Ethanoic Acid (CH₃COOH) |
| State at room temperature | Liquid | Liquid |
| Color | Colorless | Colorless |
| Boiling Point | 78.37°C | 118.1°C |
| Melting Point | -114.1°C | 16.6°C |
| Solubility in Water | Highly soluble | Highly soluble |
| pH (in solution) | Neutral or slightly acidic | Acidic (pH ~2.4) |
Here’s a table differentiating ethanol and ethanoic acid based on their chemical properties:
| Chemical Property | Ethanol (C₂H₅OH) | Ethanoic Acid (CH₃COOH) |
| Reaction with Na | Produces hydrogen gas; forms sodium ethoxide. | Produces hydrogen gas; forms sodium acetate. |
| Reaction with Na₂CO₃ | No reaction. | Produces CO₂ gas (effervescence). |
| Reaction with Base | No reaction (neutral). | Reacts to form salt and water (acidic). |
| Reaction with FeCl₃ | No reaction. | No specific reaction (unlike phenols). |
| Smell | Pleasant alcoholic smell. | Sharp vinegar-like smell. |
Q 8. Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Ans 8: Micelle formation happens when soap is added to water because soap molecules have two parts: a hydrophobic (water-repelling) tail and a hydrophilic (water-attracting) head. In water, the soap molecules arrange themselves in a spherical shape, with the hydrophobic tails facing inward, away from the water, and the hydrophilic heads facing outward. This forms a structure called a micelle, which helps trap and remove dirt or grease.
In other solvents like ethanol, micelles might not form because ethanol is a polar solvent, but it doesn’t have the same properties as water. The soap’s hydrophobic tails may not be able to arrange themselves in the same way, making micelle formation less likely.
Q 9. Why are carbon and its compounds used as fuels for most applications?
Ans 9: Carbon and its compounds, like coal, oil, and natural gas, are used as fuels because they contain a lot of energy. When burned, they react with oxygen in the air to produce heat and light, which makes them perfect for power generation, transportation, and heating. These fuels are easy to store and transport, and they release a lot of energy per unit, making them efficient and cost-effective for most applications.
Q 10. Explain the formation of scum when hard water is treated with soap.
Ans 10: When hard water is treated with soap, scum is formed due to the presence of dissolved minerals, mainly calcium (Ca²⁺) and magnesium (Mg²⁺) ions. Soap molecules have two parts: a hydrophilic (water-attracting) head and a hydrophobic (water-repelling) tail.
- In hard water, soap reacts with calcium and magnesium ions:
Ca²⁺ + 2RCOONa → Ca(RCOO)₂ (scum) + 2Na⁺
where RCOONa is the soap (sodium salt of fatty acid), and is the insoluble scum formed.
- Similarly, magnesium ions react in a similar way:
Mg²⁺ + 2RCOONa → Mg(RCOO)₂ (scum) + 2Na⁺
The scum is a solid, insoluble substance that forms when these metal ions bind with the soap molecules. This scum reduces the effectiveness of soap, as it doesn’t lather properly, and can also leave deposits on surfaces.
Q 11. What change will you observe if you test soap with litmus paper (red and blue)?
Ans 11: When you test soap with litmus paper, you’ll notice a change because soap is usually alkaline (basic).
- Red litmus paper: It will turn blue, indicating the soap’s alkalinity.
- Blue litmus paper: It will stay blue, showing it’s not acidic.
This happens because the soap contains hydroxide ions, which make it basic.
Q 12. What is hydrogenation? What is its industrial application?
Ans 12: Hydrogenation is a chemical process in which hydrogen is added to a substance, usually in the presence of a catalyst, to make it more saturated. This process is commonly used to convert liquid oils into solid fats, such as turning vegetable oil into margarine.
Industrial application: Hydrogenation is widely used in the food industry to create products like margarine and shortening. It also plays a role in making fuels and in the production of various chemicals used in manufacturing.
Q 13. Which of the following hydrocarbons undergo addition reactions: C₂H₆, C₃H₈, C₃H₆, C₂H₂, CH₄
Ans 13: Addition reactions occur in unsaturated hydrocarbons (those with double or triple bonds). Here’s the analysis:
- C₂H₆ (ethane): Saturated (single bonds) → No addition reaction.
- C₃H₈ (propane): Saturated (single bonds) → No addition reaction.
- C₃H₆ (propene): Unsaturated (one double bond) → Undergoes addition reaction.
- C₂H₂ (ethyne): Unsaturated (one triple bond) → Undergoes addition reaction.
- CH₄ (methane): Saturated (single bonds) → No addition reaction.
Final Answer: The hydrocarbons that undergo addition reactions are C₃H₆ and C₂H₂.
Q 14. Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
Ans 14: To differentiate between saturated and unsaturated hydrocarbons, use the Bromine Water Test:
- Procedure: Add bromine water (reddish-brown solution) to the hydrocarbon.
- Observation:
- Unsaturated hydrocarbons (alkenes/alkynes) decolorize bromine water due to an addition reaction: CₙH₂ₙ + Br₂ → CₙH₂ₙBr₂
- Saturated hydrocarbons (alkanes) do not react; the solution remains brown.
This test works because unsaturated hydrocarbons have double/triple bonds that react with bromine.
Q 15. Explain the mechanism of the cleaning action of soaps.
Ans 15: Mechanism of Cleaning Action of Soap:
- Soap Molecule Structure:
- Soap has two parts:
- Hydrophilic head (water-attracting).
- Hydrophobic tail (oil/grease-attracting).
- Soap has two parts:
- Cleaning Process:
- Step 1: Hydrophobic tails bind to oil/grease, while hydrophilic heads remain in water.
- Step 2: Soap molecules surround the grease particle, forming a spherical structure called a micelle.
- Step 3: Micelles lift grease particles off the surface and disperse them in water.
- Step 4: Grease is washed away with water.
Chemical Equation for Soap Formation:
Fat / Oil + Alkali (NaOH/KOH) → Soap + Glycerol
Group activity
Q 1. Use molecular model kits to make models of the compounds you have learnt in this Chapter.
Ans 1: {Students do yourself with the help of teachers}
[hint: to make molecular model follow these steps]
Using molecular model kits is an excellent way to visualize and understand the structure of chemical compounds. These kits typically include different colored balls (representing atoms) and connectors (representing bonds) that help illustrate molecular geometry. Here’s how you can proceed to create models for the compounds you’ve learned:
Steps to Build Molecular Models:
- Identify the Compound
Choose a compound you want to model, such as methane (CH₄), water (H₂O), carbon dioxide (CO₂), or any other compound covered in the chapter. - Understand the Structure
Refer to the molecular formula and the Lewis structure to know the number of atoms and bonds involved. - Gather the Correct Components
Use the model kit to select appropriate balls:- Carbon (C): Typically, black
- Hydrogen (H): White
- Oxygen (O): Red
- Nitrogen (N): Blue
Use the connectors to represent single, double, or triple bonds.
- Assemble the Model
Connect the atoms according to the compound’s molecular structure:- For single bonds, use a single connector.
- For double bonds, use two connectors.
- For triple bonds, use three connectors.
- Check Molecular Geometry
Arrange the atoms to reflect the 3D geometry based on VSEPR theory (e.g., tetrahedral, linear, trigonal planar).
Examples of Compounds to Model:
1. Methane (CH₄)
- Central atom: Carbon (C)
- 4 hydrogen atoms attached in a tetrahedral shape.
2. Water (H₂O)
- Central atom: Oxygen (O)
- 2 hydrogen atoms attached with a bent structure.
3. Carbon Dioxide (CO₂)
- Central atom: Carbon (C)
- Two oxygen atoms double-bonded in a linear structure.
4. Ammonia (NH₃)
- Central atom: Nitrogen (N)
- 3 hydrogen atoms attached in a trigonal pyramidal structure.
5. Ethanol (C₂H₅OH)
- A more complex molecule with two carbons, one oxygen, and six hydrogens.
Benefits of Using Model Kits:
- Visualize the spatial arrangement of atoms.
- Understand bond angles and molecular geometry.
- Recognize isomerism (structural and geometric).
- Simplify learning and make it interactive.
Q 2. (a). Take about 20 mL of castor oil/cotton seed oil/linseed oil/soyabean oil in a beaker. Add 30 mL of 20 % sodium hydroxide solution. Heat the mixture with continuous stirring for a few minutes till the mixture thickens. Add 5-10 g of common salt to this. Stir the mixture well and allow it to cool.
Ans 2 (a). This is an example of the process of soap preparation or saponification, which is a chemical reaction between a fat or oil and a base (such as sodium hydroxide) to produce soap and glycerol. Here’s a detailed explanation for both parts:
(a) Explanation of the process
- Ingredients:
- Oil: Castor oil, cotton seed oil, linseed oil, or soybean oil serves as the fat component.
- Sodium hydroxide (NaOH): Acts as the alkali needed for saponification.
- Common salt (NaCl): Helps in the precipitation of soap.
- Steps:
- Mixing: Combine 20 mL of the oil with 30 mL of 20% NaOH solution in a beaker.
- Heating and Stirring: Heat the mixture gently while continuously stirring. This promotes the reaction where triglycerides (fats) in the oil react with NaOH, breaking into soap (sodium salt of fatty acid) and glycerol.
- Thickening: The mixture thickens as the reaction proceeds, forming soap.
- Adding Salt: Stirring in 5–10 g of common salt precipitates the soap from the solution. This step is called “salting out.”
- Cooling: Allow the mixture to cool, during which the soap solidifies.
(b). You can cut out the soap in fancy shapes. You can also add perfume to the soap before it sets.
Ans (b). (b) Customization of Soap
- Shaping:
- Once cooled and partially set, the soap can be molded into fancy shapes using molds or by cutting it with a knife.
- Adding Perfume:
- Before the soap completely solidifies, add a few drops of perfume or essential oils to impart fragrance.
Chemical Reaction:
The general reaction can be written as:
Fat/Oil (Triglycerides) + NaOH → Soap (Sodium salt of fatty acids) + Glycerol
This experiment illustrates a fundamental process of soap-making, which can be modified for both industrial and homemade soaps.