Topic – 1.1
Which row describes the arrangement and motion of the particles in a liquid?
| arrangement | motion | |
|---|---|---|
| A | random and particles are touching | moving slowly |
| B | random with space between all particles | moving slowly |
| C | an ordered lattice with all particles touching | moving slowly |
| D | an ordered lattice with space between all particles | moving quickly |
▶️ Answer/Explanation
Ans: A
In a liquid, particles are close together (touching) but arranged randomly, not in a lattice. They move slowly compared to gas particles but faster than solid particles. Option A correctly describes this arrangement and motion.
Option B is incorrect because there isn’t space between all particles in a liquid. Option C describes a solid (ordered lattice). Option D describes a gas (space between particles and fast motion).
Topic – 1.2
Which gas has the lowest rate of diffusion at room temperature and pressure?
A) the gas produced when ammonium chloride is heated with aqueous sodium hydroxide
B) the gas which makes up approximately 78% of clean, dry air
C) the gas produced when sodium carbonate is added to dilute hydrochloric acid
D) the gas produced when zinc is added to dilute sulfuric acid
▶️ Answer/Explanation
Ans: C
The rate of diffusion is inversely proportional to the square root of the molar mass (Graham’s Law). Therefore, the gas with the highest molar mass will diffuse the slowest.
Let’s identify each gas:
A) Ammonia (NH₃, M = 17 g/mol) – from heating NH₄Cl with NaOH
B) Nitrogen (N₂, M = 28 g/mol) – 78% of air
C) Carbon dioxide (CO₂, M = 44 g/mol) – from Na₂CO₃ + HCl
D) Hydrogen (H₂, M = 2 g/mol) – from Zn + H₂SO₄
CO₂ has the highest molar mass (44 g/mol), so it will have the lowest rate of diffusion.
Topic – 2.2
Which diagram represents one helium atom?
A) 
B) 
C) 
D) 
▶️ Answer/Explanation
Ans: B
A helium atom has an atomic number of 2, meaning it has 2 protons in its nucleus. The most common isotope (helium-4) has 2 neutrons as well, giving it a mass number of 4. It has 2 electrons in its first and only electron shell.
The correct diagram (B) would show:
– A nucleus containing 2 protons and 2 neutrons
– 2 electrons in the first electron shell
Other options likely show incorrect configurations (wrong number of particles or incorrect electron arrangement).
Topic – 2.4
The diagram shows part of an ionic lattice structure.
Which compound does the diagram represent?
A) potassium bromide
B) sodium oxide
C) magnesium chloride
D) carbon monoxide
▶️ Answer/Explanation
Ans: A
The question implies the diagram shows a simple 1:1 ionic lattice (one positive ion for each negative ion). Let’s analyze the options:
A) KBr – potassium forms 1+ ions, bromine forms 1- ions → 1:1 ratio
B) Na₂O – sodium forms 1+ ions, oxygen forms 2- ions → 2:1 ratio
C) MgCl₂ – magnesium forms 2+ ions, chlorine forms 1- ions → 1:2 ratio
D) CO – covalent molecular compound, not ionic
Only potassium bromide (KBr) forms a simple 1:1 ionic lattice among the options.
Topic – 2.5
Which statement about nitrogen molecules and ethene molecules is correct?
A) A nitrogen molecule has 2 more shared electrons than an ethene molecule.
B) An ethene molecule has 3 more shared electrons than a nitrogen molecule.
C) A nitrogen molecule has 4 more shared electrons than an ethene molecule.
D) An ethene molecule has 6 more shared electrons than a nitrogen molecule.
▶️ Answer/Explanation
Ans: D
First, let’s determine the number of shared electrons in each molecule:
1. Nitrogen molecule (N₂): Triple bond between two nitrogen atoms (N≡N), which involves 6 shared electrons (3 pairs).
2. Ethene molecule (C₂H₄): Double bond between carbon atoms (4 shared electrons) plus 4 single C-H bonds (4 × 2 = 8 shared electrons). Total = 12 shared electrons.
Difference: Ethene (12) – Nitrogen (6) = 6 more shared electrons in ethene.
Therefore, option D is correct – an ethene molecule has 6 more shared electrons than a nitrogen molecule.
Topic – 2.5
Sulfur is a simple molecule with the formula \( S_8 \).
Which row describes and explains the melting point of sulfur?
| melting point | explanation | |
|---|---|---|
| A | high | the covalent bonds between sulfur atoms are strong |
| B | high | the covalent bonds between sulfur molecules are strong |
| C | low | the forces of attraction between sulfur atoms are weak |
| D | low | the forces of attraction between sulfur molecules are weak |
▶️ Answer/Explanation
Ans: D
Sulfur (\( S_8 \)) exists as simple molecular structures. The covalent bonds within each \( S_8 \) molecule are strong, but the intermolecular forces (van der Waals forces) between different \( S_8 \) molecules are weak. This explains why sulfur has a relatively low melting point – it doesn’t take much energy to overcome these weak intermolecular forces.
Option A is incorrect because while the covalent bonds within molecules are strong, they don’t break during melting. Option B is wrong because there are no covalent bonds between molecules. Option C is incorrect because it refers to forces between atoms within molecules rather than between molecules.
Topic – 2.6
Which row identifies a property and an explanation of the property for both diamond and silicon(IV) oxide?
| property | explanation of property | |
|---|---|---|
| A | very hard | diamond has a giant covalent structure and silicon(IV) oxide has a giant ionic structure |
| B | high melting point | both have giant covalent structures with many strong bonds between the atoms |
| C | good lubricant | both have layers of atoms, which can slide over each other |
| D | poor conductor | both contain only non-metal elements and are simple molecules |
▶️ Answer/Explanation
Ans: B
Both diamond and silicon(IV) oxide (silica) have giant covalent structures with strong covalent bonds between their atoms. This explains their high melting points as a lot of energy is needed to break these strong bonds.
Option A is incorrect because silicon(IV) oxide has a giant covalent structure, not ionic. Option C is wrong because neither has layered structures (graphite does, but not diamond or silica). Option D is incorrect because while they are poor conductors, it’s not because they’re simple molecules – they’re actually giant structures.
Topic – 9.1
Which statement about the structure of metals explains why metals are malleable?
A) The electrons can move freely throughout the lattice.
B) The layers of metal ions can slide over each other.
C) The metal ions are positively charged.
D) There is a strong force of attraction between the metal ions and the electrons.
▶️ Answer/Explanation
Ans: B
Metals are malleable because their structure consists of layers of positive ions that can slide over one another without breaking the metallic bonds. The delocalized electrons (sea of electrons) can move to accommodate the new positions of the ions, maintaining the metallic bonding throughout the structure.
While options A, C, and D describe true aspects of metallic bonding, only option B directly explains malleability. The free electrons (A) contribute to conductivity, the positive ions (C) are part of the structure, and the strong attraction (D) explains the strength but not specifically the malleability.
Topic – 3.1
What is the formula of iron(III) oxide?
A) FeO
B) Fe3O4
C) FeO2
D) Fe2O3
▶️ Answer/Explanation
Ans: D
Iron(III) oxide means iron has an oxidation state of +3 (Fe³⁺) and oxygen has -2 (O²⁻). To balance the charges, we need two Fe³⁺ ions (total +6 charge) and three O²⁻ ions (total -6 charge), giving the formula Fe2O3.
Option A (FeO) is iron(II) oxide. Option B (Fe3O4) is a mixed oxide containing both Fe²⁺ and Fe³⁺. Option C (FeO2) is not a common iron oxide.
Topic – 3.3
Calcium carbonate is heated. Calcium oxide and carbon dioxide gas are formed.
The equation for the reaction is shown.
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
225 kg of calcium carbonate is heated until there is no further change in mass.
The yield of calcium oxide is 85 kg.
What is the percentage yield?
A) 37.8%
B) 47.2%
C) 67.5%
D) 85.0%
▶️ Answer/Explanation
Ans: C
First calculate the theoretical yield:
1. Molar mass of CaCO3 = 40 + 12 + (16×3) = 100 g/mol
2. Moles of CaCO3 = 225,000 g / 100 g/mol = 2,250 mol
3. From the equation, 1 mol CaCO3 produces 1 mol CaO
4. Molar mass of CaO = 40 + 16 = 56 g/mol
5. Theoretical yield = 2,250 mol × 56 g/mol = 126,000 g = 126 kg
Now calculate percentage yield:
Percentage yield = (actual yield / theoretical yield) × 100
= (85 kg / 126 kg) × 100 ≈ 67.5%
Topic – 4.1
The apparatus used for electrolysis is shown.
Which statement is correct?
A) Copper forms at the anode in some electrolysis reactions.
B) Hydrogen forms at the cathode in some electrolysis reactions.
C) Oxygen forms at the cathode in some electrolysis reactions.
D) Sodium forms at the anode in some electrolysis reactions.
▶️ Answer/Explanation
Ans: B
During electrolysis, reduction occurs at the cathode where positively charged ions gain electrons. Hydrogen ions (H⁺) are commonly reduced to form hydrogen gas (H₂) at the cathode in many electrolysis reactions, especially in aqueous solutions. This makes option B correct.
Option A is incorrect because copper forms at the cathode, not the anode, in copper electrolysis. Option C is wrong because oxygen forms at the anode, not cathode, in water electrolysis. Option D is incorrect because sodium ions are reduced at the cathode, not oxidized at the anode.
Topic – 4.1
Which statement about the electrolysis of aqueous copper(II) sulfate is correct?
A) When copper electrodes are used, the solution turns from blue to colorless.
B) When graphite electrodes are used, bubbles of gas are formed at the cathode.
C) When copper electrodes are used, the anode gets smaller.
D) When graphite electrodes are used, the color of the solution does not change.
▶️ Answer/Explanation
Ans: C
In the electrolysis of copper(II) sulfate using copper electrodes, copper from the anode dissolves into the solution as Cu²⁺ ions, causing the anode to get smaller. This makes option C correct.
Option A is incorrect because the blue color (from Cu²⁺ ions) remains as copper is deposited at the cathode and dissolved from the anode in equal amounts. Option B is wrong because with graphite electrodes, hydrogen gas forms at the cathode (not oxygen). Option D is incorrect because with graphite electrodes, the blue color fades as copper is deposited and not replaced.
Topic – 4.2
Which statement describes an advantage of using a hydrogen-oxygen fuel cell in a car compared to a gasoline engine?
A) The hydrogen is difficult to store.
B) The hydrogen is highly flammable.
C) The hydrogen used is made from hydrocarbons.
D) The only chemical product is water.
▶️ Answer/Explanation
Ans: D
The main advantage of hydrogen-oxygen fuel cells is their clean operation, producing only water as a byproduct (H₂ + ½O₂ → H₂O). This makes them environmentally friendly compared to gasoline engines that produce CO₂ and other pollutants.
Options A and B describe disadvantages, not advantages. Option C is incorrect because while hydrogen can be made from hydrocarbons, it can also be produced through electrolysis of water using renewable energy.
Topic – 5.1
Two reaction pathway diagrams are shown.
Which arrow represents the activation energy for a reaction which releases thermal energy?
▶️ Answer/Explanation
Ans: A
For an exothermic reaction (which releases thermal energy), the activation energy is represented by the energy difference between the reactants and the transition state (peak of the curve). This is shown by the arrow from reactants to the peak in the first diagram where products are at lower energy than reactants.
The other options represent either the enthalpy change (B) or refer to an endothermic reaction (C and D).
Topic – 6.3
Which statements about the Haber process are correct?
- A high temperature is used because the reaction is slow at room temperature.
- A high pressure is used because there are more moles of gaseous reactants than moles of gaseous product.
- A nickel catalyst is used to increase the rate of reaction.
- An iron catalyst is used to increase the equilibrium yield of ammonia.
A) 1 and 2
B) 1 and 4
C) 2 and 3
D) 4 only
▶️ Answer/Explanation
Ans: A
In the Haber process (N₂ + 3H₂ ⇌ 2NH₃):
1. Correct – High temperature (450°C) increases the rate as the reaction is too slow at room temperature.
2. Correct – High pressure (200 atm) favors the side with fewer gas moles (4 → 2).
3. Incorrect – An iron catalyst is used, not nickel.
4. Incorrect – Catalysts don’t affect equilibrium yield, only the rate.
Therefore, only statements 1 and 2 are correct.
Topic – 6.2
Which substance is a raw material used to manufacture sulfuric acid?
A) vanadium(V) oxide
B) sulfur
C) sulfur dioxide
D) sulfur trioxide
▶️ Answer/Explanation
Ans: B
Sulfur is the primary raw material used in the Contact Process for manufacturing sulfuric acid. The process begins by burning sulfur in air to produce sulfur dioxide (SO₂). While vanadium(V) oxide is used as a catalyst in the process, it’s not a raw material. Sulfur dioxide and sulfur trioxide are intermediate products in the process, not the starting materials.
Topic – 7.1
Which colours are seen when litmus and methyl orange are added to separate samples of aqueous sodium hydroxide?
| litmus | methyl orange | |
|---|---|---|
| A | blue | orange |
| B | blue | yellow |
| C | purple | orange |
| D | purple | yellow |
▶️ Answer/Explanation
Ans: B
Sodium hydroxide is a strong alkali. Litmus turns blue in alkaline solutions (pH > 8). Methyl orange turns yellow in alkaline solutions (pH > 4.4). Purple is the neutral color for litmus, which would be incorrect for NaOH. Orange for methyl orange would indicate an acidic solution, which is also incorrect for NaOH.
Topic – 7.2
Information about the solubility in water of four oxides is shown.
Which oxide, when added to water, gives a solution with a pH less than pH 7?
| name of oxide | solubility in water | |
|---|---|---|
| A | nitrogen dioxide | soluble |
| B | copper(II) oxide | insoluble |
| C | silicon(IV) oxide | insoluble |
| D | barium oxide | soluble |
▶️ Answer/Explanation
Ans: A
Nitrogen dioxide (NO₂) dissolves in water to form a mixture of nitrous acid (HNO₂) and nitric acid (HNO₃), both of which are acidic (pH < 7). Copper(II) oxide and silicon(IV) oxide are insoluble so wouldn’t affect pH. Barium oxide is soluble but forms barium hydroxide, an alkaline solution (pH > 7).
Topic – 7.3
Copper(II) sulfate is made when copper(II) carbonate reacts with dilute sulfuric acid.
\[ \text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2 \]
Pure copper(II) sulfate crystals are obtained.
Which reagent is in excess and how are the crystals obtained?
| reagent in excess | how the crystals are obtained | |
|---|---|---|
| A | copper(II) carbonate | filter and evaporate the solution to dryness |
| B | copper(II) carbonate | filter, evaporate the solution to crystallising point and then cool |
| C | dilute sulfuric acid | evaporate the solution to dryness |
| D | dilute sulfuric acid | evaporate the solution to crystallising point and then cool |
▶️ Answer/Explanation
Ans: B
Copper(II) carbonate is insoluble, so it’s typically added in excess to ensure all the acid reacts. The excess solid can then be removed by filtration. To obtain pure crystals, the solution should be evaporated to the crystallising point (not to dryness, which would produce an impure powder) and then cooled to allow crystals to form. This method gives larger, purer crystals than evaporating to dryness.
Topic – 8.2
Which statement about elements in Group I or Group VII of the Periodic Table is correct?
A) Bromine reacts with potassium chloride to produce chlorine.
B) Iodine is a monatomic non-metal.
C) Lithium has a higher melting point than potassium.
D) Sodium is more reactive with water than potassium.
▶️ Answer/Explanation
Ans: C
In Group I (alkali metals), melting points decrease down the group, so lithium (higher in the group) has a higher melting point than potassium. Option A is incorrect because bromine is less reactive than chlorine and can’t displace it. Option B is wrong as iodine exists as diatomic I₂ molecules. Option D is incorrect because reactivity increases down Group I, making potassium more reactive than sodium.
Topic – 8.3
Some information about an element from Group VII of the Periodic Table is shown.
| melting point/°C | –7 |
| boiling point/°C | 59 |
What is the element?
A) fluorine
B) chlorine
C) bromine
D) iodine
▶️ Answer/Explanation
Ans: C
To identify the element from Group VII (halogens) with these properties:
1. Fluorine (F) has mp = -220°C and bp = -188°C – too low
2. Chlorine (Cl) has mp = -101°C and bp = -34°C – too low
3. Bromine (Br) has mp = -7°C and bp = 59°C – matches exactly
4. Iodine (I) has mp = 114°C and bp = 184°C – too high
The melting and boiling points increase down Group VII. Bromine is the only halogen that is liquid near room temperature, with these specific values.
Topic – 8.4
Manganese(IV) oxide, MnO2, is a black solid.
The equation for the reaction between manganese(IV) oxide and dilute hydrochloric acid is shown.
\[ \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2 \]
The reaction produces a pale pink solution.
Which properties of transition elements does this reaction show?
- They can act as catalysts.
- They form coloured compounds.
- They have high melting points.
- They have variable oxidation numbers.
A) 1 and 3
B) 1 and 4
C) 2 and 3
D) 2 and 4
▶️ Answer/Explanation
Ans: D
Let’s analyze each property:
1. Catalytic ability – Not demonstrated in this reaction (MnO2 is a reactant, not a catalyst)
2. Colored compounds – The pale pink solution (Mn2+ ions) shows this property
3. High melting points – While true, this isn’t shown in the reaction
4. Variable oxidation numbers – Mn changes from +4 (MnO2) to +2 (MnCl2)
Therefore, only properties 2 and 4 are demonstrated in this reaction.
Topic – 9.5
Part of a steel ship is protected from rusting using a sacrificial metal.
What is a suitable sacrificial metal?
A) copper
B) zinc
C) silver
D) potassium
▶️ Answer/Explanation
Ans: B
Sacrificial protection requires a metal that is more reactive than iron (in steel):
1. Copper (less reactive than iron) would actually accelerate rusting
2. Zinc is more reactive than iron and commonly used for this purpose (galvanization)
3. Silver is less reactive than iron
4. While potassium is more reactive, it’s too reactive (reacts violently with water) for practical use
Zinc is the standard choice as it’s sufficiently reactive to protect iron but stable enough for practical applications.
Topic – 9.2
Which row gives a use for the named metal and two properties which both explain this use?
| metal | use | property 1 | property 2 | |
|---|---|---|---|---|
| A | aluminium | aircraft construction | high density | resistant to corrosion |
| B | copper | electrical wiring | good electrical conductivity | ductile |
| C | aluminium | food containers | resistant to corrosion | not malleable |
| D | copper | aircraft construction | malleable | low density |
▶️ Answer/Explanation
Ans: B
Let’s evaluate each option:
A: Incorrect – Aluminium is used in aircraft but because of its low density, not high density
B: Correct – Copper is ideal for wiring because:
- Excellent conductivity (essential for wiring)
- Ductility (can be drawn into thin wires)
C: Incorrect – Aluminium is malleable (can be shaped into containers)
D: Incorrect – Copper isn’t used in aircraft (too dense) and doesn’t have low density
Only option B correctly matches the metal, its use, and two relevant properties.
Topic – 9.6
The apparatus used for the extraction of aluminium by electrolysis is shown.
Which equation represents the reaction at the anode?
A) \( O + 2e^- \rightarrow O^{2-} \)
B) \( 2O^{2-} \rightarrow O_2 + 4e^- \)
C) \( Al^{3-} \rightarrow Al + 3e^- \)
D) \( Al^{3+} + 3e^- \rightarrow Al \)
▶️ Answer/Explanation
Ans: B
In aluminium extraction:
At the cathode: Al3+ + 3e– → Al (reduction)
At the anode: Oxygen ions are oxidized:
2O2- → O2 + 4e– (option B)
Other options are incorrect because:
A: Shows reduction (gaining electrons) at anode – wrong
C: Incorrect ion (Al3- doesn’t exist) and shows oxidation at cathode – wrong
D: Shows cathode reaction, not anode
The carbon anode reacts with the oxygen produced, forming CO2, but the initial oxidation is of oxide ions.
Topic – 8.5
Which gas is both an element and present in clean, dry air?
A) argon
B) carbon dioxide
C) chlorine
D) water vapour
▶️ Answer/Explanation
Ans: A
Argon is a noble gas (element) and makes up about 0.93% of clean, dry air. Carbon dioxide (B) and water vapor (D) are compounds, not elements. Chlorine (C) is an element but not normally present in clean, dry air.
Topic – 10.3
Oxides of nitrogen formed in a car’s engine are removed using a catalytic converter.
What happens to the oxides of nitrogen in the catalytic converter?
A) They are hydrated.
B) They are neutralised.
C) They are oxidised.
D) They are reduced.
▶️ Answer/Explanation
Ans: D
In a catalytic converter, harmful nitrogen oxides (NOx) are reduced to nitrogen (N2) and oxygen (O2). This is a reduction reaction where the nitrogen gains electrons (its oxidation state decreases). The other options are incorrect because hydration (A) involves adding water, neutralization (B) involves acids and bases, and oxidation (C) would make the nitrogen oxides even more harmful.
Topic – 10.1
What is the equation for photosynthesis?
A) \(6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2\)
B) \(2CO_2 + 2H_2O \rightarrow 2C_2H_5OH + 3O_2\)
C) \(C_6H_{12}O_6 \rightarrow 2CO_2 + 2C_2H_5OH\)
D) \(C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\)
▶️ Answer/Explanation
Ans: A
Photosynthesis is the process where plants convert carbon dioxide and water into glucose and oxygen using sunlight energy. Option A correctly shows this balanced equation. Option B shows fermentation, option C shows anaerobic respiration, and option D shows combustion of ethanol.
Topic – 11.1
Four statements about members of the same homologous series are listed.
- They have the same volatility.
- They have the same molecular formula.
- They have the same functional group.
- They have the same general formula.
Which statements are correct?
A) 1 and 2
B) 1 and 4
C) 2 and 3
D) 3 and 4
▶️ Answer/Explanation
Ans: D
Members of a homologous series have the same functional group (3) and the same general formula (4). They do not have the same molecular formula (2 is incorrect) as each successive member differs by CH2. Volatility (1) decreases as chain length increases, so this is incorrect. Therefore, the correct statements are 3 and 4.
Topic – 11.5
Ethene reacts with steam to produce ethanol.
Which row describes each compound?
| ethene | ethanol | |
|---|---|---|
| A | saturated | saturated |
| B | saturated | unsaturated |
| C | unsaturated | saturated |
| D | unsaturated | unsaturated |
▶️ Answer/Explanation
Ans: C
Ethene (C2H4) is unsaturated because it contains a C=C double bond. When it reacts with steam (H2O), the product ethanol (C2H5OH) is saturated because all carbon-carbon bonds are single bonds. Therefore, the correct description is ethene = unsaturated, ethanol = saturated, which is option C.
Topic – 11.4
Which process is used to make an alkene from a long-chain alkane?
A) combustion
B) condensation
C) cracking
D) polymerisation
▶️ Answer/Explanation
Ans: C
Cracking is the process where long-chain alkanes are broken down into smaller molecules, including alkenes. This is typically done by heating the alkane in the presence of a catalyst. The other options don’t fit:
– Combustion (A) burns hydrocarbons to produce CO₂ and H₂O
– Condensation (B) joins molecules together with the loss of a small molecule like water
– Polymerisation (D) joins small molecules (like alkenes) to form large polymer chains
Only cracking produces alkenes from alkanes by breaking carbon-carbon bonds in the long chain.
Topic – 11.3
Which fraction obtained from petroleum has the lowest boiling point?
A) diesel oil
B) fuel oil
C) kerosene
D) naphtha
▶️ Answer/Explanation
Ans: D
In fractional distillation of petroleum, fractions are separated based on their boiling points, with the lowest boiling point fractions collected at the top of the fractionating column. The order from lowest to highest boiling point is:
1. Refinery gases (not listed)
2. Gasoline/petrol (not listed)
3. Naphtha (D) – lowest boiling point among the options
4. Kerosene (C)
5. Diesel oil (A)
6. Fuel oil (B) – highest boiling point among the options
Therefore, naphtha has the lowest boiling point among the given choices.
Topic – 11.4
Alkanes undergo substitution reactions with chlorine in the presence of ultraviolet light.
Which equation shows a reaction of this type?
A) \( C_3H_6 + Cl_2 \rightarrow C_3H_6Cl_2 \)
B) \( C_3H_8 + Cl_2 \rightarrow C_3H_6Cl_2 + H_2 \)
C) \( C_3H_8 + Cl_2 \rightarrow C_3H_7Cl + HCl \)
D) \( C_3H_6 + Cl_2 \rightarrow C_3H_5Cl + HCl \)
▶️ Answer/Explanation
Ans: C
The question asks about substitution reactions of alkanes with chlorine. Key points:
1. Alkanes have the general formula \( C_nH_{2n+2} \), so \( C_3H_8 \) is an alkane while \( C_3H_6 \) is not (it’s an alkene).
2. In substitution reactions, a hydrogen atom is replaced by a chlorine atom, forming HCl as a byproduct.
Option C shows this correctly: \( C_3H_8 \) (propane) reacts with \( Cl_2 \) to form \( C_3H_7Cl \) (chloropropane) and HCl.
The other options are incorrect because:
– A shows addition to an alkene (not substitution)
– B shows formation of \( H_2 \) (not typical for chlorine substitution)
– D involves an alkene (\( C_3H_6 \)) rather than an alkane
Topic – 11.5
Information about two reactions of ethene is listed.
- Reaction 1 requires a nickel catalyst.
- Reaction 2 requires an acid catalyst.
Which substance reacts with ethene in each reaction?
| reaction 1 | reaction 2 | |
|---|---|---|
| A) | bromine | steam |
| B) | bromine | hydrogen |
| C) | hydrogen | bromine |
| D) | hydrogen | steam |
▶️ Answer/Explanation
Ans: D
Let’s analyze the reactions:
Reaction 1 (nickel catalyst): This is hydrogenation, where hydrogen adds across the double bond of ethene to form ethane. Nickel is a common hydrogenation catalyst.
Reaction 2 (acid catalyst): This is hydration, where steam (water) adds to ethene to form ethanol. Phosphoric acid is typically used as the catalyst.
Therefore, the correct combination is:
– Reaction 1: hydrogen (with nickel catalyst)
– Reaction 2: steam (with acid catalyst)
This matches option D. The other options are incorrect because:
– Bromine reactions with ethene don’t require these catalysts (they occur readily at room temperature)
– Hydrogen doesn’t react with ethene without a nickel catalyst
Topic – 11.7
Which process converts \( CH_3CH_2OH \) to \( CH_3COOH \)?
A) bacterial oxidation
B) fermentation
C) catalytic addition of steam
D) catalytic addition of hydrogen
▶️ Answer/Explanation
Ans: A
The question involves converting ethanol (\( CH_3CH_2OH \)) to ethanoic acid (\( CH_3COOH \)), which is an oxidation process.
Let’s evaluate each option:
A) Bacterial oxidation: Correct. When wine is left exposed to air, bacteria oxidize the ethanol to ethanoic acid (vinegar).
B) Fermentation: Incorrect. Fermentation converts sugars to ethanol, not ethanol to ethanoic acid.
C) Catalytic addition of steam: Incorrect. This would hydrate an alkene to form an alcohol, not oxidize an alcohol to a carboxylic acid.
D) Catalytic addition of hydrogen: Incorrect. This would reduce a compound, not oxidize it.
Therefore, only bacterial oxidation (A) correctly describes the conversion of ethanol to ethanoic acid.
Topic – 11.7
The structure of an ester is shown.
Which row identifies the name of the ester and the two compounds from which it is made?
| name | compound 1 | compound 2 | |
|---|---|---|---|
| A | ethyl propanoate | ethanol | propanoic acid |
| B | ethyl propanoate | propanol | ethanoic acid |
| C | propyl ethanoate | ethanol | propanoic acid |
| D | propyl ethanoate | propanol | ethanoic acid |
▶️ Answer/Explanation
Ans: A
Esters are named based on the alcohol and carboxylic acid from which they are formed. The first part of the ester name comes from the alcohol (alkyl group), and the second part comes from the carboxylic acid (with -oate ending).
The structure shown has 2 carbons from the alcohol portion (ethanol) and 3 carbons from the acid portion (propanoic acid). Therefore, the correct name is ethyl propanoate, formed from ethanol and propanoic acid.
Option B incorrectly pairs propanol with ethanoic acid. Options C and D have the ester name reversed (propyl ethanoate would be from propanol and ethanoic acid).
Topic – 11.8
Which statements about monomers or polymers are correct?
- Monomers are always joined together by addition reactions.
- A polymer can be formed from a single type of monomer.
- A polymer can be formed by joining two different types of monomer.
- Water is always produced when monomer molecules join together.
A) 1 and 2
B) 1 and 4
C) 2 and 3
D) 3 and 4
▶️ Answer/Explanation
Ans: C
Let’s analyze each statement:
1. False – Monomers can join by addition (like in polyethene) or condensation reactions (like in nylon formation).
2. True – Many polymers like polyethene are made from a single monomer type (ethene).
3. True – Polymers like nylon are formed from two different monomers.
4. False – Water is only produced in condensation polymerization, not in addition polymerization.
Therefore, the correct statements are 2 and 3, making option C correct.
Topic – 12.3
The diagram shows the structure of a naturally occurring polymer, Q.
What is Q?
A) an amino acid
B) nylon
C) a protein
D) PET
▶️ Answer/Explanation
Ans: C
The structure shown represents the peptide bond (-CO-NH-) that links amino acids together in proteins.
Option A is incorrect because an amino acid is a monomer, not a polymer.
Option B (nylon) is incorrect because while nylon does contain similar amide linkages, it is synthetic, not naturally occurring.
Option D (PET) is incorrect as it contains ester linkages, not amide linkages.
Proteins are natural polymers formed from amino acids through condensation reactions, creating peptide bonds as shown in the diagram.
Topic – 12.4
Which row shows how the boiling point and the melting point of water change when a soluble impurity is added to the water?
| boiling point | melting point | |
|---|---|---|
| A | increases | increases |
| B | decreases | decreases |
| C | increases | decreases |
| D | decreases | increases |
▶️ Answer/Explanation
Ans: C
When a soluble impurity is added to water:
1. Boiling point increases – This is called boiling point elevation. The impurity particles interfere with the water molecules’ ability to escape into the gas phase, requiring more energy (higher temperature) for boiling to occur.
2. Melting point decreases – This is called freezing point depression. The impurity disrupts the formation of the regular ice crystal lattice, making it harder for water to freeze, thus requiring a lower temperature for freezing to occur.
This phenomenon is why salt is spread on icy roads (lowers melting point) and why adding salt to cooking water makes it boil at a slightly higher temperature.
Topic – 12.5
X is a white powder. The following tests are done on X.
- When a few drops of aqueous sodium hydroxide are added to a solution of X, no precipitate is seen.
- When X is heated with aqueous sodium hydroxide, no gas is formed.
- X gives a lilac colour when put into a flame.
- When acidified aqueous silver nitrate is added to a solution of X, a yellow precipitate is seen.
What is X?
A) ammonium bromide
B) ammonium iodide
C) potassium bromide
D) potassium iodide
▶️ Answer/Explanation
Ans: D
Let’s analyze each test result:
1. No precipitate with NaOH – Rules out most metal cations except Group I (like potassium).
2. No gas when heated with NaOH – Rules out ammonium compounds (which would release ammonia gas).
3. Lilac flame test – Characteristic of potassium ions (K⁺).
4. Yellow precipitate with acidified AgNO₃ – Characteristic of iodide ions (I⁻), forming silver iodide (AgI). Bromide would give cream precipitate.
Therefore, X must be potassium iodide (KI).
Option A and B are ammonium salts (ruled out by test 2). Option C is potassium bromide but would give cream precipitate, not yellow.