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Concise Chemistry Part II - Selina Solution for Class 10 Chemistry Chapter 11 - Sulphuric Acid

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Selina Textbook Solutions Chapter 11 - Sulphuric Acid

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Chemistry exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 11 - Sulphuric Acid.

All Selina textbook questions of Chapter 11 - Sulphuric Acid solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Chemistry will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Chemistry exam.

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Concise Chemistry Part II - Selina Solution for Class 10 Chemistry Chapter 11 - Sulphuric Acid Page/Excercise Intext 1

Solution 1

(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.

Solution 2

(a) Two balanced equations to obtain SO2 is:

(i) 4FeS2 + 11O2 2Fe2O3 +8SO2

(ii) S +O2 SO2


(b) The conditions for the oxidation of SO2 are:

(i) The temperature should be as low as possible. The yield has been found to be maximum at about 4100C-450oC

(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 SO3 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 SO3 is concentrated sulphuric acid. The product formed is oleum.

(e) Hydrogen sulphide.

Solution 3

Impurity of ARSENIC poisons the catalyst [i.e. deactivates the catalyst]. So, it must be removed before passing the mixture of SO2 air through the catalytic chamber.

Solution 4

(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) SO3 + H2SO4H2S2O7

(ii) H2S2O7 + H2O 2H2SO4

(c)The substance that will liberate sulphur dioxide in step E is dilute H2SO4.

(d) The equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F is:




Concise Chemistry Part II - Selina Solution for Class 10 Chemistry Chapter 11 - Sulphuric Acid Page/Excercise 1

Solution 2015

(a) In the contact process for the manufacture of sulphuric acid, the equations for the conversion of sulphur trioxide to sulphuric acid are

 SO3 + H2SO4 H2S2O7

  (oleum or pyrosulphuric acid)


 H2S2O7 + H2O 2H2SO4


(i) Action of sulphuric acid on potassium hydrogen carbonate

 2KHCO3+ H2SO4 K2SO4+ 2H2O + 2CO2 

(ii) Action of sulphuric acid on sulphur

 S + 2H2SO4 3SO2 + 2H2O


(i) Concentrated sulphuric acid

(ii) Concentrated sulphuric acid

Concise Chemistry Part II - Selina Solution for Class 10 Chemistry Chapter 11 - Sulphuric Acid Page/Excercise 11

Solution 1

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.

Solution 2

Balanced reactions are:

(a) Acidic nature:

(i) Dilute H2SO4 reacts with basic oxides to form sulphate and water.

2 NaOH+H2SO4 Na2SO4+2H2O

(ii) CuO+H2SO4 CuSO4+H2O

(iii) It reacts with carbonate to produce CO2.


(b) Oxidising agent:

H2SO4 H2O +SO2 +[O]

Nascent oxygen oxidizes non-metals, metals and inorganic compounds.

For example,

Carbon to carbon dioxide

C+H2SO4 CO2 +H2O +2SO2

Sulphur to sulphur dioxide

S +H2SO4 3SO2 +2H2O

(c) Hygroscopic nature:

It has great affinity for water. It readily absorbs moisture from atmospheric air.


C6H12O6 6C +6H2O

(d) Non-volatile nature:

It has a high boiling point (356oC) 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 + H2SO4NaHSO4 +HCl

KCl + H2SO4 KHSO4 +HCl

Solution 3

(a)Bring a glass rod dipped in Ammonia solution near the mouth of each test tubes containing dil. Hel and dil. H2SO4each.

Dil HCl

Dil. H2SO4

White fumes of ammonium chloride

No such fumes



1. Dilute sulphuric acid treated with zinc gives Hydrogen gas which bums with pop sound.

Concentrated H2SO4 gives SO2 gas with zinc and the gas turns Acidified potassium dichromate paper green.

2.Barium chloride solution gives white ppt. with dilute H2SO4, This white ppt. is insoluble in all acids.

Concentrated H2SO4 and NaCl mixture when heated gives dense white fumes if glass rod dipped in Ammonia solution is brought near it.

Solution 4

(a) When sulphuric acid reacts with sulphur the product formed is Sulphur dioxide is formed.

S +2H2SO4 3SO2 + 2H2O

(b) When sulphuric acid reacts with sodium hydroxide it neutralizes base to form sodium sulphate.

2NaOH + H2SO4 Na2SO4 + 2H2O

(c) When sulphuric acid reacts with sugar it forms carbon

C12 H22O1112C + 11H2O

(d) When sulphuric acid reacts with carbon it forms carbon dioxide and sulphur dioxide gas.

C +2H2SO4 CO2 + 2H2O + 2SO2

(e) When sulphuric acid reacts with copper it forms copper sulphate and sulphur dioxide.

Cu + H2SO4 CuSO4 + 2H2O + SO2

Solution 5

(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 H2S because it reacts with H2S to form sulphur.

H2SO4+H2S 2H2O+SO2+S

(c) Concentrated sulphuric acid has high boiling point (356oC). 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.



Solution 6

(a) Due to its reducing property. i.e, it is a non-volatile acid.

NaCl+ H2SO4NaHSO4 + HCl


(b) It is a dehydrating agent.


(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 + H2SO4 MgSO4+H2

(d) Due to its oxidizing character

Cu +H2SO4 CuSO4 +2H2O +SO2

(e) Due to its oxidizing property Hydrogen sulphide gas is passed through concentrated sulphuric acid to liberate sulphur dioxide and sulphur is formed.

H2S + H2SO4 S + 2H2O + SO2

Solution 7

The name of the salt of

(a) Hydrogen sulphites and Sulphites.

(b) Sulphate and bisulphate.

Solution 8

(a) Two types of salts are formed when sulphuric acid reacts with NaOH because sulphuric acid is dibasic.

NaOH + H2SO4 NaHSO4 + H2O

2NaOH + H2SO4Na2SO4 + 2H2O

(b) When hydrogen bromide reacts with sulphuric acid the bromine gas is obtained which produce red brown vapours.

2KBr+3H2SO4 2KHSO4+SO2+Br2 +2H2O

(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.

Na2CO3+H2SO4Na2SO4 +H2O+CO2

Solution 9

Column 1

Substance reacted with acid

Column 2

Dilute or concentrated acid

Column 3


Substance reacted with acid

Dilute or concentrated sulphuric acid



Dilute sulphuric acid


Calcium carbonate

Concentrated sulphuric acid

Carbon dioxide

Bleaching power CaOCl2

Dilute sulphuric acid

only chlorine


Solution 10

When sodium sulphide is added to solution of HCl, Hydrogen sulphide gas is produced. It has rotten egg like smell.

Solution 11(a)

Sulphuric acid is powerful dehydrating agent on account of its strong affinity towards water.

Solution 11(b)

Concentrated sulphuric acid as

 i. Oxidising agent:

 The oxidising property of conc. sulphuric acid its due to the fact that on thermal decomposition, it yeilds nacent oxygen [O].

 H2SO4 H2O + SO2 + [O] 

 ii. Non-volatile acid:

 conc. sulphuric acid has high boiling point (338°C) that why it is said to be a non volitile compound, therefore it is used for preparing volatile acids like hydrochloric acids, nitric acids from there salts by double decomposition

  H2SO4 + NaCl NaHSO4 + HCl

Solution 12

(i) B

(ii) D

(iii) C

(iv) A

(v) A

Solution 13

(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.

Solution 2008

a. (C) Lead nitrate

b. Liquid E is Ethanol.


Name of process



Equation for catalyzed reaction output




Contact process

Sulphur dioxide + oxygen

Platinum or V2O5

2SO2 + O2  2SO3

Sulphuric acid


 i. Zn + dil. H2SO4 ZnSO4 + H2 

 ii. Na2CO3 + dil. H2SO4 Na2SO4 + H2O + CO2

 iii. Pb(NO3) + dil. H2SO4 PbSO4  + 2HNO3

 iv. Zn + dil. H2SO4 ZnSO4 + H2

ZnSO4 + Na2CO3 ZnCO3 + Na2SO4


 i. The property of concentrated sulphuric acid which allows it to be used in is used in the action when sugar turns black in its presence is its dehydrating property. 

 ii. The property of concentrated sulphuric acid which allows it to be used in the preparation of hydrogen chloride and nitric acid is its non-volatility.

H2SO4 + BaCl2   BaSO4 + 2HCl

Solution 2009

Hydrogen Chloride

Solution 2010


 i. S + H2SO4 3SO2 +2H2O.

 ii. C12H22O11  + Conc. H2SO4 6C + 6H2O

b. ZnO + H2SO4 ZnSO4 + H2O. 

c. C) Dilute sulphuric acid. 

Solution 2011

a. Charring of sugar takes place. Sulphuric acid dehydrates sugar leaving behind carbon which is black in colour.

b. i. Hydrogen sulphide

c. i.



 i. Non-volatile nature

 ii. as an oxidising agent

Solution 2012

a. Hydrogen Sulphide (H2S).


 i. (B) Dehydrating agent

 ii. (D) Oxidising agent

 iii. (C) Non-volatile acid

 iv. (A) Dilute acid

 v. (D) Oxidising agent

c. ZnS + dil.H2SO4 ZnSO4 + H2S

Solution 2013

a. when Conc. H2SO4 is added to a crystal of hydrated copper sulphate,it removes water of crystalisation from salt.

b. ii. Oxidising agent 

c. C12H22O11 + Conc. H2SO4 6C + 6H2O

d. Sulphuric acid (H2SO4)

Solution 2014

a. C + H2SO4 CO2 + 2H2O + 2SO2.

b. Sulphuric acid precipitates the insoluble sulphate of barium from the solution of barium chloride.

BaCl2 + H2SO4 BaSO4 + 2HCl

Dilute HCl does not react with barium chloride solution, and thus, no precipitate is produced in the reaction.

c. Two conditions for the conversion of sulphur dioxide to sulphur trioxide is as follows: 

Temperature: 450-500° C Catalyst: V2O5



 i. Dehydrating property of sulphuric acid:

H2SO4 has a great affinity for water, and therefore, it acts as a dehydrating agent. 


 ii. Acidic nature of sulphuric acid:

 It acts as a strong dibasic acid.

 H2SO4 2Hi+ + SO42-

 It reacts with metals, metal oxides, metal hydroxides, carbonates etc. to form metallic sulphates and hydrogen at ordinary temperature.

 Mg + H2SO4 MgSO4 + H2 

 CuO + H2SO4 CuSO4 + H2O

 2NaOH + H2SO4 Na2SO4 + 2H2O

 ZnCO3 + H2SO4 ZnSO4 + H2O + CO2 

 iii. As a non-volatile acid:  

It has a high boiling point, so it is used to prepare volatile acids such as HCl, HNO3 and acetic acid from their salts.

NaCl + H2SO4 NaHSO4 + HCl

NaNO3 + H2SO4 NaHSO4 + HNO3


TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 11 - Sulphuric Acid  for ICSE Class 10 Chemistry free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Chemistry textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well. 

Text Book Solutions

ICSE X - Chemistry

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