Selina Concise Solutions for ICSE Class 10 Chemistry Chapter 11 Sulphuric Acid

Question 1. Comment, sulphuric acid is referred to as:
(a) King of chemicals
(b) Oil of vitriol
Answer:
(a) Sulphuric acid is called King of Chemicals because there is no other manufactured compound which is used by such a large number of key industries.
(b) Sulphuric acid is referred to as Oil of vitriol as it was obtained as an oily viscous liquid by heating crystals of green vitriol.

📝 Teacher's Note: Use historical context to help students remember these names — show them how ancient chemists discovered these properties. Connect "vitriol" to the glass-like appearance of crystals.

🎯 Exam Tip: Remember both names and their reasons. Questions often ask for explanations, so don't just memorize the names.

Question 2. Sulphuric acid is manufactured by contact process
(a) Give two balanced equations to obtain \( SO_2 \) in this process.
(b) Give the conditions for the oxidation of \( SO_2 \)
Answer:
(a) Two balanced equations to obtain \( SO_2 \) is:
(i) \( 4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2 \)
(ii) \( S + O_2 \rightarrow SO_2 \)
(b) The conditions for the oxidation of \( SO_2 \) are:
(i) The temperature should be as low as possible. The yield has been found to be maximum at about 410°C - 450°C
(ii) High pressure (2 atm) is favoured because the product formed has less volume than reactant.
(iii) Excess of oxygen increases the production of sulphur trioxide.
(iv) Vanadium pentoxide or platinised asbestos is used as catalyst.
(c) Sulphuric acid is not obtained directly by reacting \( SO_3 \) with water because the reaction is highly exothermic which produce the fine misty droplets of sulphuric acid that is not directly absorbed by water.
(d) The chemical used to dissolve \( SO_3 \) is concentrated sulphuric acid. The product formed is oleum.
(e) Main reactions of this process are:
\( S + O_2 \rightarrow SO_2 \)
\( 2SO_2 + O_2 \xrightarrow[450°C]{V_2O_5} 2SO_3 \)
\( SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)
\( H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4 \)

📝 Teacher's Note: Emphasize that contact process has multiple steps and optimal conditions are crucial for maximum yield. Use Le Chatelier's principle to explain why certain conditions are chosen.

🎯 Exam Tip: Learn all reaction equations and conditions. Questions often test the complete process, not just individual steps.

Question 3. Why is water not added to concentrated \( H_2SO_4 \) in order to dilute it?
Answer: Water is not added to concentrated acid since it is an exothermic reaction. If water is added to the acid, there is a sudden increase in temperature and the acid being in bulk tends to spurt out with serious consequences.
In simple words: Adding water to concentrated acid creates so much heat so quickly that the acid can splash and burn you badly.

📝 Teacher's Note: Always demonstrate the correct dilution method — add acid to water slowly while stirring. Use the memory aid "Do as you oughta, add acid to water."

🎯 Exam Tip: Mention both the exothermic nature and the safety hazard of spurting acid for full marks.

Question 4. Why the impurity of arsenic oxide must be removed before passing the mixture of \( SO_4 \) and air through the catalytic chamber?
Answer: Impurity of ARSENIC poisons the catalyst [i.e. deactivates the catalyst]. So, it must be removed before passing the mixture of \( SO_2 \) air through the catalytic chamber.
In simple words: Arsenic acts like poison for the catalyst, making it stop working properly, so it must be cleaned out first.

📝 Teacher's Note: Explain catalyst poisoning with everyday analogies — like how certain foods can make us sick. Emphasize that even small amounts of poison can ruin the entire process.

🎯 Exam Tip: Use the term "catalyst poisoning" and explain that arsenic deactivates the catalyst for full marks.

Question 5. Give two balanced reactions of each type to show the following properties of sulphuric acid:
(a) Acidic nature.
(b) Oxidising agent
(c) Hygroscopic nature
(d) Non-volatile nature
Answer:
Balanced reactions are:
(a) Acidic nature:
(i) Dilute \( H_2SO_4 \) reacts with basic oxides to form sulphate and water.
\( 2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \)
(ii) \( CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O \)
(iii) It reacts with carbonate to produce \( CO_2 \).
\( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)
(b) Oxidising agent:
\( H_2SO_4 \rightarrow H_2O + SO_2 + [O] \)
Nascent oxygen oxidizes non-metals, metals and inorganic compounds.
For example,
Carbon to carbon dioxide
\( C + H_2SO_4 \rightarrow CO_2 + H_2O + 2SO_2 \)
Sulphur to sulphur dioxide
\( S + H_2SO_4 \rightarrow 3SO_2 + 2H_2O \)
(c) Hygroscopic nature:
It has great affinity for water. It readily absorbs moisture from atmospheric air.
\( HCOOH \xrightarrow{conc. H_2SO_4} CO + H_2O \)
\( C_6H_{12}O_6 \xrightarrow{conc. H_2SO_4} 6C + 6H_2O \)
(d) Non-volatile nature:
It has a high boiling point (356°C) so it is considered to be non-volatile. Therefore, it is used for preparing volatile acids like hydrochloric acid, nitric acid from their salts by double decomposition reaction.
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \)
\( KCl + H_2SO_4 \rightarrow KHSO_4 + HCl \)

📝 Teacher's Note: Group these properties logically and show how the same acid can behave differently under different conditions. Use demonstrations for hygroscopic and dehydrating properties.

🎯 Exam Tip: Learn at least two reactions for each property. Questions often ask for specific examples, and variety in answers shows thorough understanding.

Question 6. Give a chemical test to distinguish between:
(a) dilute sulphuric acid and dilute hydrochloric acid
(b) dilute sulphuric acid and conc. Sulphuric acid
Answer:
(a) Bring a glass rod dipped in Ammonia solution near the mouth of each test tubes containing dil. HCl and dil. \( H_2SO_4 \) each.

(b)
1. Dilute sulphuric acid treated with zinc gives Hydrogen gas which burns with pop sound. Concentrated \( H_2SO_4 \) gives \( SO_2 \) gas with zinc and the gas turns Acidified potassium dichromate paper green.
2. Barium chloride solution gives white ppt. with dilute \( H_2SO_4 \). This white ppt. is insoluble in all acids. Concentrated \( H_2SO_4 \) and NaCl mixture when heated gives dense white fumes if glass rod dipped in Ammonia solution is brought near it.

📝 Teacher's Note: Demonstrate these tests in the lab. The ammonia test is particularly striking and helps students remember the difference between these acids.

🎯 Exam Tip: Describe both the test procedure and expected results clearly. Mention specific observations like "white fumes" or "pop sound."

Question 7. Name the products formed when hot and concentrated sulphuric acid reacts with the following:
(a) Sulphur
(b) NaOH
(c) Sugar
(d) Carbon
(e) Copper
Answer:
(a) When sulphuric acid reacts with sulphur the product formed is Sulphur dioxide is formed.
\( S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O \)
(b) When sulphuric acid reacts with sodium hydroxide it neutralizes base to form sodium sulphate.
\( 2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \)
(c) When sulphuric acid reacts with sugar it forms carbon
\( C_{12}H_{22}O_{11} \xrightarrow{conc. H_2SO_4} 12C + 11H_2O \)
(d) When sulphuric acid reacts with carbon it forms carbon dioxide and sulphur dioxide gas.
\( C + 2H_2SO_4 \rightarrow CO_2 + 2H_2O + 2SO_2 \)
(e) When sulphuric acid reacts with copper it forms copper sulphate and sulphur dioxide.
\( Cu + H_2SO_4 \rightarrow CuSO_4 + 2H_2O + SO_2 \)

📝 Teacher's Note: The sugar-acid reaction is dramatic and memorable for students. Emphasize how the same acid shows different properties (acidic, oxidizing, dehydrating) with different substances.

🎯 Exam Tip: Write balanced equations along with product names. Questions often test both the products and the chemical equations.

Question 8. Why is:
(a) Concentrated sulphuric acid kept in air tight bottles?
(b) \( H_2SO_4 \) not a drying agent for \( H_2S \)?
(c) Sulphuric acid used in the preparation of HCl and \( HNO_3 \)? Give equations in both cases.
Answer:
(a) Concentrated sulphuric acid is hygroscopic substance that absorbs moisture when exposed to air. Hence, it is stored in air tight bottles.
(b) Sulphuric acid is not a drying agent for \( H_2S \) because it reacts with \( H_2S \) to form sulphur.
\( H_2SO_4 + H_2S \rightarrow 2H_2O + SO_2 + S \)
(c) Concentrated sulphuric acid has high boiling point (356°C). So, it is considered to be non-volatile. Hence, it is used for preparing volatile acids like Hydrochloric acid and Nitric acids from their salts by double decomposition.
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \)
\( NaNO_3 + H_2SO_4 \rightarrow NaHSO_4 + HNO_3 \)

📝 Teacher's Note: Connect the hygroscopic property to everyday observations — why car batteries need sealed caps, why concentrated acid bottles feel heavy with moisture.

🎯 Exam Tip: For part (c), always mention the "non-volatile nature" and provide both equations. This shows complete understanding of the principle.

Question 9. What property of conc. \( H_2SO_4 \) is made use of in each of in each of the following cases? Give an equation for the reaction on each case:
(a) in the production of HCl gas when it reacts with a chloride
(b) in the preparation of CO and HCOOH
(c) as a source of hydrogen by diluting it and adding a strip of magnesium
(d) in the preparation of sulphur dioxide by warming a mixture of conc. Sulphuric acid and copper – turnings
(e) Hydrogen sulphide gas is passed through concentrated sulphuric acid
Answer:
(a) Due to its reducing property. i.e, it is a non-volatile acid.
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \) (Conc.)
(b) It is a dehydrating agent.
\( HCOOH \xrightarrow{conc. H_2SO_4} CO + H_2O \)
(c) Magnesium is present above hydrogen in the reactivity series so sulphuric acid is able to liberate hydrogen gas by reacting with magnesium strip.
\( Mg + H_2SO_4 \rightarrow MgSO_4 + H_2 \)
(d) Due to its oxidizing character
\( Cu + H_2SO_4 \rightarrow CuSO_4 + 2H_2O + SO_2 \)
(e) Due to its oxidizing property Hydrogen sulphide gas is passed through concentrated sulphuric acid to liberate sulphur dioxide and sulphur is formed.
\( H_2S + H_2SO_4 \rightarrow S + 2H_2O + SO_2 \)

📝 Teacher's Note: This question beautifully shows how one compound can have multiple properties. Create a chart showing property vs. application to help students organize this information.

🎯 Exam Tip: Each part asks for both the property AND the equation. Don't forget either component — both are needed for full marks.

Question 10. What is the name given to the salts of:
(a) sulphurous acid
(b) sulphuric acid?
Answer:
The name of the salt of
(a) Hydrogen sulphites and Sulphites.
(b) Sulphate and bisulphate.
In simple words: Different acids form different types of salts — sulphurous acid makes sulphites, while sulphuric acid makes sulphates.

📝 Teacher's Note: Help students distinguish between sulphites (from \( H_2SO_3 \)) and sulphates (from \( H_2SO_4 \)). Use examples of common salts they might know.

🎯 Exam Tip: Remember both normal salts and acid salts (bisulphates/hydrogen sulphites) for each acid.

Question 11. Give reasons for the following:
(a) Sulphuric acid forms two types of salts with NaOH
(b) Red brown vapours are produced when concentrated sulphuric acid is added to hydrogen bromide
(c) A piece of wood becomes black when concentrated sulphuric acid is poured on it
(d) Brisk effervescence is seen when oil of vitriol is added to sodium carbonate
Answer:
(a) Sulphuric acid is dibasic acid (can release two \( H^+ \) ions), so it can form two types of salts: normal salt (\( Na_2SO_4 \)) when both hydrogen atoms are replaced, and acid salt (\( NaHSO_4 \)) when only one hydrogen is replaced.
(b) Concentrated sulphuric acid acts as an oxidizing agent and oxidizes hydrogen bromide to bromine, which appears as red-brown vapours.
(c) Wood contains cellulose and other organic compounds. Concentrated sulphuric acid acts as a dehydrating agent, removing water from these organic molecules and leaving behind black carbon.
(d) Oil of vitriol is another name for sulphuric acid. When it reacts with sodium carbonate, carbon dioxide gas is produced, which causes brisk effervescence: \( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)

📝 Teacher's Note: These observations are perfect for demonstrations. The wood-blackening experiment particularly impresses students and shows the dehydrating power vividly.

🎯 Exam Tip: For each observation, identify which specific property of sulphuric acid is responsible. This shows analytical thinking beyond mere memorization.

Question. (a) Two types of salts are formed when sulphuric acid reacts with NaOH because sulphuric acid is dibasic. (b) When hydrogen bromide reacts with sulphuric acid the bromine gas is obtained which produce red brown vapours. (c) A piece of wood becomes black when concentrated sulphuric acid is poured on it because it gives a mass of carbon. (d) When sulphuric acid is added to sodium carbonate it liberates carbon dioxide which produces brisk effervescence.
Answer: (a) Two types of salts are formed when sulphuric acid reacts with NaOH because sulphuric acid is dibasic.
NaOH + \( H_2SO_4 \rightarrow NaHSO_4 + H_2O \)
2NaOH + \( H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O \)
(b) When hydrogen bromide reacts with sulphuric acid the bromine gas is obtained which produce red brown vapours.
2KBr + \( 3H_2SO_4 \rightarrow 2KHSO_4 + SO_2 + Br_2 + 2H_2O \)
(c) A piece of wood becomes black when concentrated sulphuric acid is poured on it because it gives a mass of carbon.
(d) When sulphuric acid is added to sodium carbonate it liberates carbon dioxide which produces brisk effervescence.
\( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)
In simple words: Sulphuric acid can react in different ways - it can form different types of salts, remove water from substances like wood turning them to carbon, and release gases when mixed with carbonates.

📝 Teacher's Note: Demonstrate each property with simple experiments - show students how concentrated sulphuric acid turns sugar black, and how it fizzes with baking soda. This helps them understand the different behaviors of the same acid.

🎯 Exam Tip: Always write balanced chemical equations for each reaction described, and mention specific observations like "red brown vapours" or "brisk effervescence" to score full marks.

Question. Copy and complete the following table:

Answer:

In simple words: Different substances react with sulphuric acid to produce different gases - metals like zinc give hydrogen, carbonates give carbon dioxide, and bleaching powder gives chlorine gas.

📝 Teacher's Note: Use the activity series to explain why zinc produces hydrogen gas, and connect this to students' knowledge of acid-metal reactions from earlier chapters.

🎯 Exam Tip: Remember the pattern - metals produce hydrogen, carbonates produce carbon dioxide, and chlorides of active metals produce chlorine when treated with sulphuric acid.

Question. Name a gas that can be oxidized to sulphur.
Answer: Hydrogen sulphide (\( H_2S \)) can be oxidized to sulphur.
In simple words: When hydrogen sulphide gas is oxidized, it loses hydrogen and forms solid sulphur as one of the products.

📝 Teacher's Note: Show students the rotten egg smell of hydrogen sulphide and explain how oxidation removes the hydrogen, leaving behind yellow sulphur powder.

🎯 Exam Tip: Write the chemical formula \( H_2S \) along with the name "hydrogen sulphide" to show complete understanding.

Question. Give the odour of the gas evolved and name the gas produced when sodium sulphide is added to solution of HCI in water.
Answer: When sodium sulphide is added to solution of HCl, Hydrogen sulphide gas is produced. It has rotten egg like smell.
In simple words: When these two chemicals mix, they produce a gas that smells exactly like rotten eggs - this gas is hydrogen sulphide.

📝 Teacher's Note: This is a safe way to demonstrate hydrogen sulphide production in the lab. Always emphasize the characteristic smell as an identification method.

🎯 Exam Tip: Always mention both the name of the gas AND its characteristic odour when asked - examiners look for complete identification.

Question. (a) Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C. (b) In the contact process for the manufacture of sulphuric acid, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two-steps procedure is used. Write the equations for the two steps involved in D. (c) What type of substance will liberate sulphur dioxide from sodium sulphite in step E? (d) Write the equation for the reaction by which sulphure dioxide is converted to sodium sulphite in step F.
Answer: (a) The catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C is Vanadium pentoxide.
(b) The two steps for the conversion of sulphur trioxide to sulphuric acid is:
(i) \( SO_3 + H_2SO_4 \rightarrow H_2S_2O_7 \)
(ii) \( H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4 \)
(c) The substance that will liberate sulphur dioxide in step E is dilute \( H_2SO_4 \).
(d) The equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F is:
\( SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O \) Or
\( Na_2O + SO_2 \rightarrow Na_2SO_3 \)
In simple words: The contact process uses vanadium pentoxide as a catalyst and a special two-step method to make sulphuric acid safely, avoiding the dangerous direct reaction with water.

📝 Teacher's Note: Draw the contact process flowchart on the board and explain why direct reaction of \( SO_3 \) with water is avoided - it's too violent and produces acid mist.

🎯 Exam Tip: Remember "V2O5" as the catalyst and the two-step oleum method - these are frequently asked details in the contact process.

Question. (a) Name the process used for the large-scale manufacture of sulphuric acid. (b) Which property of sulphuric acid accounts for its use as a dehydrating agent? (c) Concentrated sulphuric acid is both an oxidizing agent and a non-volatile acid. Write one equation each to illustrate the above mentioned properties of sulphuric acid.
Answer: (a) The process used for the large-scale manufacture of sulphuric acid is Contact process.
(b) Sulphuric acid has great affinity for water. It readily removes element of water from other compound. Thus it acts as a dehydrating agent.
(c) Concentrated acid is non-volatile thus it is used for the preparation of volatile acids:
\( NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl \)
Concentrated acid act as an oxidizing agent:
\( C + 2H_2SO_4 \rightarrow CO_2 + 2H_2O + 2SO_2 \)
In simple words: Sulphuric acid is made by the contact process, it loves water so much that it removes water from other substances, and it can also give oxygen to other substances in reactions.

📝 Teacher's Note: Demonstrate dehydration with sugar turning black, and explain how non-volatile acids can displace volatile acids from their salts.

🎯 Exam Tip: For property-based questions, always write the specific equation that demonstrates that exact property mentioned in the question.

Question. Some properties of sulphuric acid are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v) some properties may be repeated. A. Acid B. Dehydrating agent C. Non-volatile acid D. Oxidizing agent (i) \( C_{12}H_{22}O_{11} + nH_2SO_4 \rightarrow 12C + 11H_2O + nH_2SO_4 \) (ii) \( S + 2H_2SO_4 \rightarrow 3SO_2 + 2H_2O \) (iii) \( NaCI + H_2SO_4 \rightarrow NaHSO_4 + HCI \) (iv) \( CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O \) (v) \( Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 \)
Answer: (i) B
(ii) D
(iii) C
(iv) A
(v) A
In simple words: Each reaction shows a different property - removing water from sugar (dehydrating), giving oxygen to sulphur (oxidizing), displacing volatile acid (non-volatile), and neutralizing base and carbonate (acidic).

📝 Teacher's Note: Create a chart showing each property with 2-3 example reactions. This helps students quickly identify which property is being demonstrated.

🎯 Exam Tip: Look for key clues - water removal = dehydrating, element oxidation = oxidizing agent, volatile acid formation = non-volatile acid property, neutralization = acidic property.

Question. (a) Name the acid formed when sulphur dioxide dissolves in water (b) Name the gas released when sodium carbonate is added to a solution of sulphur dioxide.
Answer: (a) The acid formed when sulphur dioxide dissolves in water is sulphurous acid.
(b) Carbondioxide gas is released when sodium carbonate is added to solution of sulphur dioxide.
In simple words: When sulphur dioxide gas dissolves in water, it forms a weak acid called sulphurous acid. When this acidic solution meets sodium carbonate, it releases carbon dioxide gas.

📝 Teacher's Note: Show students that \( SO_2 \) forms a weaker acid (\( H_2SO_3 \)) compared to \( SO_3 \) which forms stronger \( H_2SO_4 \). This pattern helps in understanding acid strength.

🎯 Exam Tip: Don't confuse sulphurous acid (\( H_2SO_3 \)) with sulphuric acid (\( H_2SO_4 \)) - the first has one less oxygen atom and is much weaker.

Question. (a) What is the property of concentrated sulphuric acid which allows it to be used in the preparation of hydrogen chloride and nitric acid? (b) What property of concentrated sulphuric acid is in action when sugar turns black in its presence?
Answer: (a) Concentrated sulphuric acid is non-volatile; hence it is used for the preparation of higher volatile acids.
(b) Due to its dehydrating nature sugar turns black in the presence of concentrated sulphuric acid.
In simple words: Sulphuric acid doesn't evaporate easily, so it can push out other acids that do evaporate easily. It also removes water from sugar, leaving behind black carbon.

📝 Teacher's Note: Demonstrate with a drop of concentrated sulphuric acid on sugar. The dramatic blackening shows dehydration, while displacement reactions show non-volatile nature.

🎯 Exam Tip: "Non-volatile" is the key word for acid preparation, "dehydrating" is the key word for organic substance reactions like sugar turning black.

Question 1. Comment, sulphuric acid is referred to as:
(a) King of chemicals
(b) Oil of vitriol
Answer:
(a) Sulphuric acid is named the "King of Chemicals" since it serves as a crucial raw material across an incredibly wide range of major chemical industries, with no other manufactured substance matching its extensive industrial utility.
(b) It is designated as "Oil of Vitriol" because historical alchemists first prepared it as a thick, greasy fluid by distilling or heating green vitriol crystals (hydrated iron(II) sulphate).
In simple words: Sulphuric acid is called the "King of Chemicals" because almost every factory uses it to make other things. It is called "Oil of vitriol" because scientists long ago made it by heating green crystals, and it looked like a thick, slippery oil.

Exam Tip: State the connection to green vitriol crystals (hydrated iron sulphate) clearly when explaining the origin of its historical name.

Question 2. Sulphuric acid is manufactured by contact process:
(a) Give two balanced equations to obtain \( \text{SO}_2 \) in this process.
(b) Give the conditions for the oxidation of \( \text{SO}_2 \)
Answer:
(a) Two balanced chemical equations for producing \( \text{SO}_2 \) are:
(i) \( 4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2 \)
(ii) \( \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \)
(b) The essential parameters required for oxidising \( \text{SO}_2 \) to \( \text{SO}_3 \) are:
(i) An optimum temperature between \( 410^\circ\text{C} - 450^\circ\text{C} \) is maintained because lower temperatures favor higher yield, but a minimum warmth is needed for the reaction to proceed.
(ii) A pressure of about \( 2\text{ atm} \) is preferred as the forward reaction leads to a decrease in gas volume.
(iii) An abundance of oxygen (excess air) drives the equilibrium to increase the yield of sulphur trioxide.
(iv) \( \text{V}_2\text{O}_5 \) (Vanadium pentoxide) or platinised asbestos acts as the catalyst.
(c) Directly mixing \( \text{SO}_3 \) with water is avoided because this hydration is extremely exothermic. It generates a dense acid mist that does not readily dissolve or condense into liquid water.
(d) Concentrated sulphuric acid is employed to absorb \( \text{SO}_3 \), yielding a heavy chemical liquid known as oleum (\( \text{H}_2\text{S}_2\text{O}_7 \)).
(e) The key chemical equations involved in this manufacturing route are:
\( \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \)

\( 2\text{SO}_2 + \text{O}_2 \xrightarrow[450^\circ\text{C}]{\text{V}_2\text{O}_5} 2\text{SO}_3 \)

\( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \)

\( \text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4 \)
In simple words: To make sulphuric acid, we first burn sulphur to get sulphur dioxide. Then, we use a helper chemical called vanadium pentoxide, some extra oxygen, and a warm temperature of around 450 degrees to turn it into sulphur trioxide. We do not mix this gas directly with water because it creates a dangerous, hot acid fog; instead, we dissolve it in concentrated acid first to make oleum, which we then safely dilute with water.

Exam Tip: Memorize the multi-step Contact Process equations, particularly the catalyst (\( \text{V}_2\text{O}_5 \)) and the temperature range (\( 410^\circ\text{C} - 450^\circ\text{C} \)), as these are high-yield exam questions.

Question 3. Why is water not added to concentrated \( \text{H}_2\text{SO}_4 \) in order to dilute it?
Answer: Diluting concentrated sulphuric acid is a highly exothermic process that liberates an immense amount of heat. If water is poured directly into the concentrated acid, the sudden generation of intense heat instantly vaporizes the water, causing the acid to violently splash and spurt out of the container, which can lead to severe chemical burns.
In simple words: When water meets concentrated acid, it releases a lot of heat very fast. If you pour water into acid, it will boil instantly and splash hot acid onto your skin or eyes.

Exam Tip: Remember the rule: always add acid to water slowly with constant stirring, never water to acid ("A to W" is safe).

Question 4. Why the impurity of arsenic oxide must be removed before passing the mixture of \( \text{SO}_4 \) and air through the catalytic chamber?
Answer: Traces of arsenic oxide act as a catalytic poison, which permanently deactivates the catalyst (such as platinised asbestos or vanadium pentoxide). To prevent this loss of catalytic activity, the gas stream must be thoroughly purified of arsenic before entering the reaction chamber.
In simple words: Arsenic acts like a poison to the helper chemical (catalyst) and stops it from working. We must clean it out first so the reaction can keep running.

Exam Tip: The term "catalytic poisoning" or "deactivating the catalyst" is a key technical phrase that must be included in your answer to secure maximum marks.

Question 5. Give two balanced reactions of each type to show the following properties of sulphuric acid:
(a) Acidic nature.
(b) Oxidising agent,
(c) Hygroscopic nature,
(d) Non-volatile nature
Answer:
(a) Acidic nature:
(i) Dilute \( \text{H}_2\text{SO}_4 \) neutralizes soluble bases to yield a salt and water:
\( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
(ii) Dilute \( \text{H}_2\text{SO}_4 \) reacts with insoluble metal oxides to form a salt and water:
\( \text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} \)
(iii) It decomposes metallic carbonates to release carbon dioxide gas:
\( \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \uparrow \)

(b) Oxidising agent:
Concentrated \( \text{H}_2\text{SO}_4 \) decomposes to release nascent oxygen, which oxidizes non-metals:
\( \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{O} + \text{SO}_2 + [\text{O}] \)
(i) Oxidation of carbon to carbon dioxide:
\( \text{C} + 2\text{H}_2\text{SO}_4 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 2\text{SO}_2 \)
(ii) Oxidation of sulphur to sulphur dioxide:
\( \text{S} + 2\text{H}_2\text{SO}_4 \rightarrow 3\text{SO}_2 + 2\text{H}_2\text{O} \)

(c) Hygroscopic / Dehydrating nature:
Concentrated sulphuric acid has an extreme affinity for water and chemically removes hydrogen and oxygen as water molecules from compounds:
(i) Dehydration of methanoic acid (formic acid):
\( \text{HCOOH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CO} + \text{H}_2\text{O} \)
(ii) Dehydration of glucose (charring of sugar):
\( \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{conc. H}_2\text{SO}_4} 6\text{C} + 6\text{H}_2\text{O} \)

(d) Non-volatile nature:
With its high boiling point of \( 356^\circ\text{C} \), concentrated \( \text{H}_2\text{SO}_4 \) is highly non-volatile and can displace more volatile acids from their respective metal salts:
(i) Preparation of hydrogen chloride gas:
\( \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \)
(ii) Preparation of nitric acid (represented here with potassium chloride):
\( \text{KCl} + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{HCl} \)
In simple words: Sulphuric acid shows four key habits: it behaves like a normal acid by neutralising bases, it acts as an oxidising agent by giving away oxygen, it works as a dehydrator by pulling out water from sugars, and because it does not evaporate easily (non-volatile), it can be used to make other lighter acids.

Exam Tip: For the dehydrating property, clearly distinguish between "hygroscopic" (absorbing physical moisture) and "dehydrating" (removing chemically combined water, such as the elements of water from glucose or formic acid).

Question 6. Give a chemical test to distinguish between:
(a) dilute sulphuric acid and dilute hydrochloric acid
(b) dilute sulphuric acid and conc. Sulphuric acid
Answer:
(a) Introduce a glass rod moistened with an aqueous ammonia solution to the opening of separate test tubes containing dilute \( \text{HCl} \) and dilute \( \text{H}_2\text{SO}_4 \).

(b) These two forms can be distinguished by the following tests:
1. Reaction with active metals (e.g., Zinc): Dilute sulphuric acid reacts with zinc granules to produce hydrogen gas (\( \text{H}_2 \)), which burns with a characteristic pop sound. In contrast, concentrated sulphuric acid acts as an oxidizing agent, reacting with zinc to yield sulphur dioxide gas (\( \text{SO}_2 \)), which turns orange acidified potassium chromate paper green.
2. Barium Chloride Test and Salt Decomposition: Dilute sulphuric acid reacts with barium chloride (\( \text{BaCl}_2 \)) solution to form a dense white precipitate of barium sulphate (\( \text{BaSO}_4 \)) that is completely insoluble in mineral acids. Concentrated sulphuric acid, when heated with solid sodium chloride (\( \text{NaCl} \)), releases hydrogen chloride gas, which forms thick white fumes when exposed to a glass rod dipped in ammonia solution.
In simple words: You can tell dilute hydrochloric acid from dilute sulphuric acid because only the hydrochloric acid will produce white smoke when exposed to ammonia. You can tell dilute sulphuric acid from concentrated sulphuric acid because dilute acid makes popping hydrogen gas with zinc, while concentrated acid makes sulphur dioxide gas instead.

Exam Tip: In distinction tests, always state the reagents, the specific observations for both substances being tested, and the balanced equations where applicable to secure full marks.

Question 7. Name the products formed when hot and concentrated sulphuric acid reacts with the following:
(a) Sulphur
(b) NaOH,
(c) Sugar
(d) Carbon
(e) Copper.
Answer:
(a) Sulphur: It is oxidized to produce sulphur dioxide gas and water vapor.
\( \text{S} + 2\text{H}_2\text{SO}_4 \rightarrow 3\text{SO}_2 + 2\text{H}_2\text{O} \)
(b) Sodium hydroxide (NaOH): It neutralizes the alkali to yield sodium sulphate (salt) and water.
\( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
(c) Sugar (Sucrose): The acid dehydrates sugar, leaving behind a black mass of carbon (sugar charcoal) and water vapor.
\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \xrightarrow{\text{conc. H}_2\text{SO}_4} 12\text{C} + 11\text{H}_2\text{O} \)
(d) Carbon: It is oxidized to yield carbon dioxide, sulphur dioxide, and water.
\( \text{C} + 2\text{H}_2\text{SO}_4 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 2\text{SO}_2 \uparrow \)
(e) Copper: It is oxidized to produce copper(II) sulphate, water, and sulphur dioxide gas.
\( \text{Cu} + 2\text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + 2\text{H}_2\text{O} + \text{SO}_2 \uparrow \br />In simple words: When mixed with hot, strong sulphuric acid, sulphur turns into sulphur dioxide gas; sodium hydroxide is neutralised into salt water; sugar turns into black carbon; carbon turns into carbon dioxide and sulphur dioxide; and copper turns into blue copper sulphate salt and sulphur dioxide gas.

Exam Tip: Be sure to write the physical states or include upward arrows (\( \uparrow \)) for gaseous products like \( \text{SO}_2 \) and \( \text{CO}_2 \) to make your equations look professional and complete.

Question 8. Why is:
(a) Concentrated sulphuric acid kept in air tight bottles?
(b) \( \text{H}_2\text{SO}_4 \) not a drying agent for \( \text{H}_2\text{S} \)?
(c) Sulphuric acid used in the preparation of \( \text{HCl} \) and \( \text{HNO}_3 \)? Give equations in both cases.
Answer:
(a) Concentrated sulphuric acid behaves as a highly hygroscopic liquid, meaning it readily pulls in and absorbs water vapor from the surrounding atmosphere. If left open, it dilutes itself over time, which is why it must be preserved in sealed, airtight containers.
(b) Concentrated \( \text{H}_2\text{SO}_4 \) is a strong oxidizing agent, whereas hydrogen sulphide (\( \text{H}_2\text{S} \)) is a powerful reducing agent. Instead of merely drying the gas, the acid oxidizes \( \text{H}_2\text{S} \) to form elemental sulphur:
\( \text{H}_2\text{SO}_4 + \text{H}_2\text{S} \rightarrow 2\text{H}_2\text{O} + \text{SO}_2 + \text{S} \downarrow \)
(c) Concentrated sulphuric acid has an exceptionally high boiling point of \( 356^\circ\text{C} \), making it highly non-volatile. As a result, it can readily displace volatile acids like hydrochloric acid and nitric acid from their respective metal salts via double decomposition:
\( \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \)
\( \text{NaNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HNO}_3 \)
In simple words: Strong sulphuric acid must be kept sealed because it drinks moisture out of the air. It cannot dry hydrogen sulphide gas because it reacts with it to make yellow sulphur. Finally, it helps make other acids because it does not evaporate easily.

Exam Tip: Remember that a drying agent must be chemically inert towards the gas it dries. Since \( \text{H}_2\text{SO}_4 \) reacts with \( \text{H}_2\text{S} \), it cannot be used for this purpose.

Question 9. What property of conc. \( \text{H}_2\text{SO}_4 \) is made use of in each of the following cases? Give an equation for the reaction on each case:
(a) in the production of \( \text{HCl} \) gas when it reacts with a chloride,
(b) in the preparation of \( \text{CO} \) and \( \text{HCOOH} \),
(c) as a source of hydrogen by diluting it and adding a strip of magnesium.
(d) in the preparation of sulphur dioxide by warming a mixture of conc. Sulphuric acid and copper - turnings,
(e) Hydrogen sulphide gas is passed through concentrated sulphuric acid.
Answer:
(a) Non-volatile nature: It allows the displacement of the more volatile hydrochloric acid from chloride salts.
\( \text{NaCl} + \text{H}_2\text{SO}_4\text{ (conc.)} \rightarrow \text{NaHSO}_4 + \text{HCl} \uparrow \)

(b) Dehydrating property: It extracts water elements from organic compounds like methanoic acid.
\( \text{HCOOH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CO} \uparrow + \text{H}_2\text{O} \)

(c) Acidic character of dilute acid: Since magnesium lies above hydrogen in the reactivity series, dilute acid easily displaces hydrogen gas upon reacting with the metal.
\( \text{Mg} + \text{H}_2\text{SO}_4\text{ (dil.)} \rightarrow \text{MgSO}_4 + \text{H}_2 \uparrow \)

(d) Oxidizing property: Warm concentrated acid behaves as a strong oxidizing agent, converting copper metal to copper sulphate while being reduced to sulphur dioxide.
\( \text{Cu} + 2\text{H}_2\text{SO}_4\text{ (conc.)} \rightarrow \text{CuSO}_4 + 2\text{H}_2\text{O} + \text{SO}_2 \uparrow \)

(e) Oxidizing property: It oxidizes hydrogen sulphide gas to deposit solid sulphur, while the acid gets reduced to sulphur dioxide.
\( \text{H}_2\text{S} + \text{H}_2\text{SO}_4\text{ (conc.)} \rightarrow \text{S} \downarrow + 2\text{H}_2\text{O} + \text{SO}_2 \uparrow \)
In simple words: Concentrated sulphuric acid uses its non-volatile nature to make hydrochloric gas, its dehydrating strength to pull water from formic acid, its acidic strength to release hydrogen with magnesium, and its strong oxidizing power to react with copper and hydrogen sulphide.

Exam Tip: Be careful with the state of the acid — dilute sulphuric acid acts as a typical acid to release hydrogen gas, whereas concentrated sulphuric acid acts as a powerful oxidizing agent.

Question 10. What is the name given to the salts of:
(a) sulphurous acid
(b) sulphuric acid?
Answer:
(a) Salts derived from sulphurous acid (\( \text{H}_2\text{SO}_3 \)) are classified as sulphites (normal salts) and hydrogen sulphites or bisulphites (acid salts).
(b) Salts derived from sulphuric acid (\( \text{H}_2\text{SO}_4 \)) are classified as sulphates (normal salts) and hydrogen sulphates or bisulphates (acid salts).
In simple words: Sulphurous acid creates salts called sulphites, while sulphuric acid makes salts called sulphates. Both can also make acid salts with "bisulphite" or "bisulphate" in their names.

Exam Tip: Remember that dibasic acids like \( \text{H}_2\text{SO}_3 \) and \( \text{H}_2\text{SO}_4 \) always form two series of salts: normal salts and acid salts.

Question 11. Give reasons for the following:
(a) Sulphuric acid forms two types of salts with NaOH,
(b) Red brown vapours are produced when concentrated sulphuric acid is added to hydrogen bromide.
(c) A piece of wood becomes black when concentrated sulphuric acid is poured on it,
(d) Brisk effervescence is seen when oil of vitriol is added to sodium carbonate.
Answer:
(a) Sulphuric acid is a dibasic acid containing two replaceable hydrogen atoms per molecule, enabling it to undergo step-wise neutralization to yield both an acid salt (sodium bisulphate) and a normal salt (sodium sulphate):
\( \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O} \)
\( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)

(b) Concentrated sulphuric acid acts as a strong oxidizing agent that oxidizes hydrogen bromide (generated from bromide salts like \( \text{KBr} \)) to produce reddish-brown bromine gas (\( \text{Br}_2 \)):
\( 2\text{KBr} + 3\text{H}_2\text{SO}_4 \rightarrow 2\text{KHSO}_4 + \text{SO}_2 \uparrow + \text{Br}_2 \uparrow + 2\text{H}_2\text{O} \)

(c) Wood consists of cellulose (a carbohydrate). Concentrated sulphuric acid behaves as a powerful dehydrating agent, stripping away water molecules (hydrogen and oxygen elements) and leaving behind a black mass of carbon (charring):
Wood \( \xrightarrow{\text{conc. H}_2\text{SO}_4} \) Carbon (Black mass) + \( \text{H}_2\text{O} \)

(d) When sulphuric acid reacts with sodium carbonate, it decomposes the carbonate to release carbon dioxide gas, which escapes rapidly as bubbles, causing brisk effervescence:
\( \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \uparrow \)
In simple words: Sulphuric acid makes two salts because it has two acid parts to give away. It turns hydrogen bromide into red-brown bromine gas because it is an oxidiser. It turns wood black because it sucks out all the water, leaving pure black charcoal. It causes fizzing with sodium carbonate because it makes carbon dioxide gas escape quickly.

Exam Tip: Use the term "dibasic" to explain the formation of two salts, and "charring" or "dehydration" to explain why organic substances like wood or paper turn black.

Question 12. Copy and complete the following table:

Answer: The complete and balanced interactions are tabulated above.
In simple words: This table shows which type of sulphuric acid is mixed with different chemicals to produce hydrogen, carbon dioxide, or chlorine gas.

Exam Tip: Be specific about whether the acid used is dilute or concentrated, as the reaction products and the type of gas evolved depend entirely on the concentration of the acid.

Question 1. Name a gas that can be oxidized to sulphur.
Answer: Hydrogen sulphide (\( \text{H}_2\text{S} \)) is oxidized to yield elemental sulphur.
In simple words: Hydrogen sulphide is the gas that can be changed into solid sulphur.

Exam Tip: Always write both the name "Hydrogen sulphide" and its formula \( \text{H}_2\text{S} \) to show complete understanding.

Question 2(2004). Give the odour of the gas evolved and name the gas produced when sodium sulphide is added to solution of \( \text{HCl} \) in water.
Answer: Adding sodium sulphide to dilute hydrochloric acid releases hydrogen sulphide (\( \text{H}_2\text{S} \)) gas, which carries a highly distinct, unpleasant odor resembling rotten eggs.
In simple words: The gas produced is hydrogen sulphide, and it smells like bad, rotten eggs.

Exam Tip: The classic "rotten egg smell" is the standard chemical identification test for hydrogen sulphide gas and sulfide salts.

Question 3. (a) Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C.
(b) In the contact process for the manufacture of sulphuric acid, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two-steps procedure is used. Write the equations for the two steps involved in D.
(c) What type of substance will liberate sulphur dioxide from sodium sulphite in step E?
(d) Write the equation for the reaction by which sulphure dioxide is converted to sodium sulphite in step F.

Answer:
(a) The catalyst utilized to convert sulphur dioxide into sulphur trioxide during reaction C is Vanadium pentoxide (\( \text{V}_2\text{O}_5 \)) or platinised asbestos.
(b) The two consecutive equations indicating how sulphur trioxide is converted into sulphuric acid are:
(i) \( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \) (Oleum production)
(ii) \( \text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4 \) (Dilution of oleum)
(c) To evolve sulphur dioxide from sodium sulphite in step E, dilute sulphuric acid (\( \text{dil. H}_2\text{SO}_4 \)) or dilute hydrochloric acid is used.
(d) The chemical reaction showing how sulphur dioxide gas is converted into sodium sulphite in step F is:
\( \text{SO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_3 + \text{H}_2\text{O} \)
Or:
\( \text{Na}_2\text{O} + \text{SO}_2 \rightarrow \text{Na}_2\text{SO}_3 \)
In simple words: Vanadium pentoxide is the catalyst used in step C. We turn sulphur trioxide into acid by first dissolving it in concentrated acid to make oleum, then mixing it with water. Dilute acid is used in step E to get sulphur dioxide back. Finally, sulphur dioxide is reacted with sodium hydroxide to make sodium sulphite.

Exam Tip: Flowchart questions are extremely common in ICSE Class X chemistry exams. Practice identifying each reagent and reaction condition represented by the letters A to F.

Question 1. (a) Name the process used for the large-scale manufacture of sulphuric acid.
(b) Which property of sulphuric acid accounts for its use as a dehydrating agent?
(c) Concentrated sulphuric acid is both an oxidizing agent and a non-volatile acid. Write one equation each to illustrate the above mentioned properties of sulphuric acid.
Answer:
(a) The commercial method employed for the bulk production of sulphuric acid is the Contact Process.
(b) Concentrated sulphuric acid possesses an extraordinarily high affinity for water, which allows it to chemically abstract the elements of water (hydrogen and oxygen in a 2:1 ratio) from various compounds.
(c) 1. Non-volatile property: Since it has a high boiling point, it can decompose metal salts of volatile acids:
\( \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \)
2. Oxidizing property: It readily oxidizes non-metals such as carbon to carbon dioxide:
\( \text{C} + 2\text{H}_2\text{SO}_4 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 2\text{SO}_2 \)
In simple words: Sulphuric acid is made on a large scale by the Contact Process. It dries things by chemically pulling out water molecules. Because it is non-volatile, it can make hydrochloric gas, and because it is a strong oxidiser, it can turn carbon into carbon dioxide.

Exam Tip: When writing the displacement reaction of concentrated sulphuric acid with sodium chloride, ensure the temperature condition is maintained below \( 200^\circ\text{C} \) for safety and fuel efficiency.

Question 1. Some properties of sulphuric acid are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v) some properties may be repeated.
A. Acid
B. Dehydrating agent
C. Non-volatile acid
D. Oxidizing agent
(i) \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \xrightarrow{\text{conc. H}_2\text{SO}_4} 12\text{C} + 11\text{H}_2\text{O} \)
(ii) \( \text{S} + 2\text{H}_2\text{SO}_4 \rightarrow 3\text{SO}_2 + 2\text{H}_2\text{O} \)
(iii) \( \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \)
(iv) \( \text{CuO} + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} \)
(v) \( \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \)
Answer:
(i) B (Dehydrating agent) - The acid extracts water elements from sucrose, leaving behind carbon.
(ii) D (Oxidizing agent) - Concentrated sulphuric acid oxidizes sulphur to sulphur dioxide.
(iii) C (Non-volatile acid) - It decomposes sodium chloride to yield hydrogen chloride gas.
(iv) A (Acid) - It neutralizes the basic metallic oxide (copper oxide) to form salt and water.
(v) A (Acid) - It decomposes the carbonate salt to release carbon dioxide gas.
In simple words: (i) is dehydrating because it dries up sugar, (ii) is oxidizing because it gives oxygen to sulphur, (iii) is non-volatile because it creates hydrochloric acid, and (iv) and (v) show normal acidic reactions where salts are formed.

Exam Tip: Be sure you can associate every distinct chemical equation of \( \text{H}_2\text{SO}_4 \) with its specific chemical property, as this matching format is extremely common in board exams.

Question 2. (a) Name the acid formed when sulphur dioxide dissolves in water
(b) Name the gas released when sodium carbonate is added to a solution of sulphur dioxide.
Answer:
(a) The chemical produced upon dissolving sulphur dioxide gas in water is sulphurous acid (\( \text{H}_2\text{SO}_3 \)).
(b) When sodium carbonate is added to this acidic solution, carbon dioxide (\( \text{CO}_2 \)) gas is released with rapid bubbling (effervescence).
In simple words: (a) Dissolving sulphur dioxide in water makes sulphurous acid. (b) Adding sodium carbonate to this acid releases carbon dioxide gas.

Exam Tip: Distinguish carefully between "sulphurous acid" (\( \text{H}_2\text{SO}_3 \)) and "sulphuric acid" (\( \text{H}_2\text{SO}_4 \)) as they have different chemical formulas and properties.

Question 1. (a) What is the property of concentrated sulphuric acid which allows it to be used in the preparation of hydrogen chloride and nitric acid?
(b) What property of concentrated sulphuric acid is in action when sugar turns black in its presence?
Answer:
(a) Concentrated sulphuric acid is highly non-volatile, meaning it possesses a high boiling point and does not evaporate easily, allowing it to displace more volatile acids from their respective salts.
(b) The dehydrating property (or charring action) is responsible, as the acid pulls out hydrogen and oxygen from sugar, leaving behind a black mass of carbon.
In simple words: (a) The acid does not turn into vapor easily (non-volatile), so it can help make lighter acids. (b) It removes all water from sugar, leaving behind black carbon.

Exam Tip: In questions about the "turning black" of sugar, always mention the key term "dehydration" or "charring" to secure full marks.