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Solved 2025 Specimen Paper ICSE Class 10 Chemistry

Solutions for Chemistry, Class 10, ICSE

Section A 40 Marks

20 questions

Question 1(i)

An aqueous solution of copper sulphate turns colourless on electrolysis.

Which of the following could be the electrodes?

P. anode: copper; cathode: copper

Q. anode: platinum; cathode: copper

R. anode: copper; cathode: platinum

only P
only Q
only R
both Q and R

Section A 40 Marks

Answer:

only Q

Reason — If platinum anode is used the blue colour of CuSO₄ solution fades since the blue Cu²⁺ ions which are discharged at the cathode are not replaced at the anode.

Question 1(ii)

A compound P is heated in a test tube with sodium hydroxide solution. A red litmus paper held at the mouth of the test tube turns blue. Which of the following could compound P be?

zinc sulphate
copper sulphate
ferrous sulphate
ammonium sulphate

Section A 40 Marks

Answer:

ammonium sulphate

Reason — Ammonium sulphate reacts with sodium hydroxide to produce sodium sulphate, ammonia gas, and water. The red litmus turns blue due to the basic ammonia gas. The reaction is:

(NH₄)₂SO₄ + 2NaOH ⟶ 2NH₃(g) + Na₂SO₄ +2H₂O

Question 1(iii)

Which of the following would weigh the least?
(Atomic masses C=12, O=16, Na=23)

2 gram atoms of oxygen
one mole of sodium
22.4 litres of carbon dioxide at STP
6.023 x 10²² atoms of carbon

Section A 40 Marks

Answer:

6.023 x 10²² atoms of carbon

Reason — To determine which of the options weighs the least, let's calculate the mass of each:

2 gram atoms of oxygen:
- A gram atom of oxygen refers to 1 mole of oxygen atoms, which has a mass of its atomic mass in grams.
- Oxygen has an atomic mass of 16 g/mol.
- Therefore, 2 gram atoms of oxygen would weigh 2 x 16 g = 32 grams.
One mole of sodium:
- The atomic mass of sodium is 23 g/mol.
- Thus, one mole of sodium weighs 23 grams.
22.4 litres of carbon dioxide at STP:
- At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 liters.
- The molecular formula for carbon dioxide is CO₂.
- The molar mass of CO₂ is 12 (for C) + 2 x 16 (for O) = 44 grams/mol.
- Therefore, 22.4 litres of carbon dioxide weighs 44 grams.
6.023 x 10²² atoms of carbon:
- Avogadro's number is 6.023 x 10²³, which is the number of atoms in 1 mole.
- Therefore, 6.023 x 10²² atoms of carbon is $\dfrac{1}{10}$ of a mole of carbon.
- The atomic mass of carbon is 12 g/mol.
- Therefore, $\dfrac{1}{10}$ mole of carbon has a mass of $\dfrac{12}{10}$ = 1.2 grams.

Comparing the masses:

2 gram atoms of oxygen = 32 grams
One mole of sodium = 23 grams
22.4 litres of carbon dioxide at STP = 44 grams
6.023 x 10²² atoms of carbon = 1.2 grams

Hence, out of the given options, 6.023 x 10²² atoms of carbon will weigh the least.

Question 1(iv)

The equation below shows the reaction between element 'X' and dilute sulphuric acid.

X(s) + H₂SO₄ (aq.) ⟶ XSO₄ (aq.) + H₂ (g)

Which particles are responsible for conducting electricity in dilute sulphuric acid and compound XSO₄?

Electrons
Only positive ions
Only negative ions
Both positive and negative ions

Section A 40 Marks

Answer:

Both positive and negative ions

Reason — Sulphuric acid will dissociate into H⁺ and SO₄^2-. Also, in aqueous solution, the compound XSO₄ would dissociate into X⁺ and SO₄^2-. In both cases, both positive and negative ions are responsible for conducting electricity.

Question 1(v)

Assertion (A): Dry hydrogen chloride gas is collected by the upward displacement of air.

Reason (R): Hydrogen chloride gas is lighter than air.

Both A and R are true and R is the correct explanation of A
Both A and R are true but R is not the correct explanation of A
A is true but R is false
A is false but R is true

Section A 40 Marks

Answer:

A is true but R is false

Reason — Dry hydrogen chloride gas is collected by the upward displacement of air as it is heavier than air.

Question 1(vi)

The structures of four hydrocarbons are shown below:

How many isomers of butene are there?

Section A 40 Marks

Answer:

Reason — 2-methyl propene and But-1-ene are two isomers of butene present in the four hydrocarbons shown above.

Question 1(vii)

Element 'P' has electronic configuration 2,8,8,1. The number of chlorine atoms present in the chloride of 'P' is:

Section A 40 Marks

Answer:

Reason — P has 1 electron in its outermost shell. Since each chlorine atom can accept one electron, the chloride of P will have one chlorine atom for one electron lost by P. Therefore, the number of chlorine atoms in the chloride of 'P' will be 1.

Question 1(viii)

₁H² is an isotope of hydrogen. In the modern Periodic Table it will:

be placed before hydrogen
be placed after hydrogen
be placed at the same position as hydrogen
not have any position in the Periodic Table

Section A 40 Marks

Answer:

be placed at the same position as hydrogen

Reason — The position of an element in the Periodic Table is determined by its atomic number. Therefore, isotopes of an element are not listed separately; instead, they are all represented under the same element symbol.

Question 1(ix)

A nitrate which forms a precipitate with ammonium hydroxide and is also soluble in excess of it:

ferrous nitrate
ferric nitrate
lead nitrate
copper nitrate

Section A 40 Marks

Answer:

copper nitrate

Reason — When copper nitrate reacts with ammonium hydroxide NH₄OH, it initially forms a light blue precipitate of copper(II) hydroxide (Cu(OH)₂).
This precipitate is soluble in excess ammonium hydroxide due to the formation of a deep blue complex, tetraamminecopper(II) ([Cu(NH₃)₄]²⁺), making it soluble.

Question 1(x)

Which of the following electronic configuration represents the most electropositive element?

2, 1
2, 8, 1
2, 2
2, 8, 2

Section A 40 Marks

Answer:

2, 8, 1

Reason — Electropositivity increases while traversing down a group and decreases from left to right in period. 2, 1 and 2, 2 belong to period 2 as they have 2 shells whereas 2, 8, 1 and 2, 8, 2 belong to period 3. Between 2, 8, 1 and 2, 8, 2, element with electronic configuration 2, 8, 1 will be more electropositive as it lies to the left of 2, 8, 2 in period 3.

Question 1(xi)

Assertion (A): Alkali metals do not form dipositive ions.

Reason (R): After loss of one electron alkali metals achieve stable electronic configuration of noble gases.

Both A and R are true and R is the correct explanation of A
Both A and are true but R is not the correct explanation of A
A is true but R is false
A is false but R is true

Section A 40 Marks

Answer:

Both A and R are true and R is the correct explanation of A

Reason — Alkali metals (such as lithium, sodium, and potassium) do not form dipositive ions. Instead, they form monovalent cations by losing one electron. This is because they have a single electron in their outermost shell, and losing this one electron achieves the stable noble gas configuration.

Question 1(xii)

The ratio between the volumes occupied by 4.4 grams of carbon dioxide and 2 grams of hydrogen gas is:

2.2 : 1
1 : 2.2
1 : 10
10 : 1

Section A 40 Marks

Answer:

1 : 10

Reason —

Molar mass of CO₂ = 12 + 2 x 16 = 44 g/mole

∴ Moles of 4.4g CO₂ = 0.1

Molar mass of H₂ = 2 x 1 = 2 g/mole

∴ Moles of 2g H₂ = 1

Under the same conditions of temperature and pressure, the volume of a gas is directly proportional to the number of moles. Therefore, the volume ratio will be the same as the mole ratio.

Volume ratio = $\dfrac{\text{Vol. of CO}_2}{\text{Vol. of H}_2}$ = $\dfrac{0.1}{1}$ = 1 : 10

Question 1(xiii)

Aqueous lead (II) nitrate can be distinguished from aqueous zinc nitrate by adding any of the following solution in excess, except:

aqueous potassium chloride
aqueous sodium sulphate
dilute sulphuric acid
sodium hydroxide solution

Section A 40 Marks

Answer:

sodium hydroxide solution

Reason — Let's analyze each option:

Aqueous potassium chloride: This will precipitate lead(II) chloride (PbCl₂) from lead (II) nitrate, whereas zinc chloride is soluble. Thus, this can distinguish between them, as lead will form a precipitate while zinc will not.
Aqueous sodium sulphate: This will precipitate lead (II) sulphate (PbSO₄) from lead (II) nitrate since PbSO₄ is poorly soluble, while zinc sulphate is soluble. Thus, this can also distinguish between them.
Dilute sulphuric acid: Similar to aqueous sodium sulphate, adding dilute H₂SO₄ will precipitate lead (II) sulphate from lead (II) nitrate but not zinc sulphate from zinc nitrate. So, this can distinguish between them.
Sodium hydroxide solution: When added to both solutions in excess, this will not distinguish them because both lead (II) hydroxide and zinc hydroxide will initially precipitate but then both dissolve upon further addition of NaOH. Lead hydroxide forms a plumbate ion, and zinc hydroxide forms a zincate ion, both of which are soluble.

Question 1(xiv)

Which of the following about oxides is correct?

A basic oxide is an oxide of a non-metal
Acidic oxides contain ionic bonds
Amphoteric oxides contain a metal
Basic oxides are always gases

Section A 40 Marks

Answer:

Amphoteric oxides contain a metal

Reason — Let's analyze each option:

A basic oxide is an oxide of a non-metal: This is incorrect. Basic oxides are typically oxides of metals. For example, calcium oxide (CaO) and magnesium oxide (MgO) are basic oxides, and both contain metals.
Acidic oxides contain ionic bonds: This is incorrect. Acidic oxides are usually covalent compounds. They are generally formed by non-metals, such as sulphur dioxide (SO₂) and carbon dioxide (CO₂), and tend to form acids when dissolved in water.
Amphoteric oxides contain a metal: This is correct. Amphoteric oxides, such as aluminum oxide (Al₂O₃) and zinc oxide (ZnO), contain metals and can react both as acids and bases.
Basic oxides are always gases: This is incorrect. Basic oxides are typically solid at room temperature. For example, calcium oxide (CaO) and magnesium oxide (MgO) are solid basic oxides.

Question 1(xv)

A student takes Cu, Al, Fe and Zn strips, separately in four test tubes labeled as I, II, III and IV respectively. He adds 10 ml of freshly prepared ferrous sulphate solution to each test tube and observes the colour of the metal residue in each case.

He would observe a black residue in the test tubes:

(I) and (II)
(I) and (III)
(II) and (III)
(II) and (IV)

Section A 40 Marks

Answer:

(II) and (IV)

Reason — Zinc and aluminium are more reactive than iron. Therefore they will displace iron which will be seen as black residue.

Question 2(i)

Electroplating steel objects with silver involves a three-step process.

Step 1 — A coating of copper is applied to the object.

Step 2 — A coating of nickel is applied to the object.

Step 3 — The coating of silver is applied to the object.

(a) A diagram of the apparatus used for step 1 is shown.

Electroplating steel objects with silver involves a three-step process. ICSE 2025 Chemistry Solved Question Paper.

The chemical process taking place on the surface of the object is
Cu²⁺(aq) + 2e^- ⟶ Cu (s)
What is the observation seen on the surface of the object?
Explain why the concentration of copper ions in the electrolyte remains constant throughout step 1.

(b) Give two changes which would be needed in order to coat nickel onto the object in step 2.

Section A 40 Marks

Answer:

(a)

The object will develop a coating of copper. Visually, this results in the object taking on a reddish-brown appearance due to the deposition of metallic copper. Also there is an increase in mass of the steel object.
As copper ions are reduced and deposited onto the object, an equal amount of copper metal is oxidised from the anode to maintain the concentration of Cu²⁺ in the electrolyte solution.

(b) Two changes which would be needed in order to coat nickel onto the object in step 2 are:

Replace the copper sulphate solution with a nickel-containing electrolyte solution, such as nickel sulphate.
Replace the copper anode with a nickel anode.

Question 2(ii)

Match the following Column A with Column B.

Column A	Column B
Aluminium	Covalent compound
Sulphuric acid	Carbonate ore
Calcination	Hall Heroult’s process
Calcium Chloride	Contact Process
Carbon tetrachloride	Electrovalent compound

Section A 40 Marks

Answer:

The matched columns are shown below:

Column A	Column B
Aluminium	Hall Heroult’s process
Sulphuric acid	Contact Process
Calcination	Carbonate ore
Calcium Chloride	Electrovalent compound
Carbon tetrachloride	Covalent compound

Question 2(iii)

Complete the following by choosing the correct answers from the bracket:

(a) If an element has one electron in the outermost shell, then it is likely to have the ............... [smallest/largest] atomic size amongst all the elements in the same period.

(b) ............... [sulphuric acid/hydrochloric acid] does not form an acid salt.

(c) A ............... [reddish brown/dirty green] coloured precipitate is formed when ammonium hydroxide is added to a solution of ferrous chloride.

(d) Alkynes undergo ............... [addition/substitution] reactions.

(e) An ............... [alkaline/acidic] solution will turn methyl orange solution pink or red.

Section A 40 Marks

Answer:

(a) If an element has one electron in the outermost shell, then it is likely to have the largest atomic size amongst all the elements in the same period.

(b) Hydrochloric acid does not form an acid salt.

(c) A dirty green coloured precipitate is formed when ammonium hydroxide is added to a solution of ferrous chloride.

(d) Alkynes undergo addition reactions.

(e) An acidic solution will turn methyl orange solution pink or red.

Question 2(iv)

Identify the following:

(a) A bond formed between two atoms by sharing of a pair of electrons, with both electrons being provided by the same atom.

(b) A salt formed by the complete neutralization of an acid by a base.

(d) The energy required to remove an electron from a neutral gaseous atom.

(e) A homogenous mixture of two or more metals or a metal and a non-metal in a definite proportion in their molten state.

Section A 40 Marks

Answer:

(a) Coordinate bond

(b) Normal salt

(d) Ionization potential

(e) Alloy

Question 2(v)

(a) Draw the structural diagram for the following compounds:

propanoic acid
pentan-2-ol
2,2 dibromo butane

(b) Give the IUPAC name of the following organic compounds:

Draw the structural diagram for the following compounds: ICSE 2025 Chemistry Solved Question Paper.

Section A 40 Marks

Answer:

(a) Structural Diagram:

1. Propanoic acid :

Draw the structural formula for Propanoic acid. ICSE 2025 Specimen Chemistry Solved Question Paper.

2. Pentan-2-ol

3. 2,2 dibromo butane

(b) IUPAC names are:

1-propanol
pent-1-ene

Section B 40 Marks

23 questions

Question 3(i)

Identify the reactant and write the balanced equation for the following:

Nitric acid reacts with compound Q to give a salt Ca(NO₃)₂, water and carbon dioxide.

Section B 40 Marks

Answer:

The compound 'Q' in this reaction is calcium carbonate or calcium bicarbonate.

When nitric acid reacts with calcium carbonate, it produces calcium nitrate, water and carbon dioxide.

CaCO₃ + 2HNO₃ ⟶ Ca(NO₃)₂ + H₂O + CO₂

Ca(HCO₃)₂ + 2HNO₃ ⟶ Ca(NO₃)₂ + 2H₂O + CO₂

Question 3(ii)

What property of Sulphuric acid is exhibited in each of the following cases:

(a) In the preparation of HCl gas when it reacts with Sodium chloride.

(b) When concentrated Sulphuric acid reacts with Copper to produce Sulphur dioxide gas.

Section B 40 Marks

Answer:

(a) Non-volatile nature
Reason — Concentrated sulphuric acid has a high boiling point (338°C) and so it is considered to be a non-volatile acid. It is, therefore, used for preparing volatile acids like HCl.

(b) Oxidising property
Reason — The oxidising property of concentrated sulphuric acid is due to the fact that on thermal decomposition, it yields nascent oxygen which oxidises metals.

Question 3(iii)

The electron affinity of an element X is greater than that of element Y.

(a) How is the oxidising power of X likely to compare with that of Y?

(b) How is the electronegativity of X likely to compare with that of Y?

Section B 40 Marks

Answer:

(a) X has more oxidising power compared to Y.
Reason — Oxidising Power refers to an element's ability to gain electrons and thus oxidize other substances. Given that element X has a greater electron affinity than element Y, X is more eager to accept electrons. This suggests that X has a stronger tendency to act as an oxidizing agent compared to Y.

(b) X will be more electronegative than Y.
Reason — A higher electron affinity correlates with higher electronegativity because both properties reflect an element's tendency to attract electrons. Since X has a greater electron affinity than Y, X is likely to have a higher electronegativity compared to Y.

(c) X will be placed to the right of Y.
Reason — Elements with higher electron affinities are generally found in the upper right part of the periodic table. Thus, if X has a greater electron affinity than Y, X is likely to be positioned further to the right compared to Y in the periodic table.

Question 3(iv)

You are provided with the list of chemicals mentioned below in the box:

Sodium hydroxide solution, copper carbonate, zinc, hydrochloric acid, copper, dilute sulphuric acid, chlorine, iron

Using suitable chemicals from the list given, write balanced chemical equation for the preparation of the salts mentioned below:

(a) copper sulphate

(b) sodium zincate

Section B 40 Marks

Answer:

(a) CuCO₃ + H₂SO₄ ⟶ CuSO₄ + H₂O + CO₂

(b) Zn + 2NaOH ⟶ Na₂ZnO₂ + H₂

Question 4(i)

The following questions relate to the extraction of Aluminium by electrolysis.

(a) Name the other aluminium containing compound added to alumina.

(b) Give a balanced equation for the reaction that takes place at the cathode.

Section B 40 Marks

Answer:

(a) Cryolite.

(b) Al³⁺ + 3e^- ⟶ Al

Question 4(ii)

Pratik heated 11.2 grams of element ‘M’ (atomic weight 56) with 4.8 grams of element ‘N’ (atomic weight 16) to form a compound. Find the empirical formula of the compound obtained by Pratik.

Section B 40 Marks

Answer:

Gram atom = $\dfrac{\text{Mass of element}}{\text{Atomic mass}}$

∴ g atom of M = $\dfrac{11.2}{56}$ = 0.2

and, g atom of N = $\dfrac{4.8}{16}$ = 0.3

simple ratio of gram-atoms of M to gram-atoms of N

= $\dfrac{0.2}{0.3}$ = $\dfrac{2}{3}$ = M : N

So, the formula is M₂N₃

Question 4(iii)

Give balanced equations for each of the following:

(a) Action of warm water on Aluminium nitride.

(b) Oxidation of carbon with conc. Nitric acid.

Section B 40 Marks

Answer:

(a) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃

(b) C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

Question 4(iv)

The diagram given below is a representation of the Industrial preparation of Nitric acid by Ostwald’s process. With respect to the process answer the following questions:

(a) Write the temperature and the catalyst required during the catalytic oxidation of ammonia.

(b) Give balanced chemical equation for the reaction occurring during the conversion of nitrogen dioxide to nitric acid.

Section B 40 Marks

Answer:

(a) 800°C, Platinum

(b) 4NO₂ + 2H₂O + O₂ ⟶ 4HNO₂

Question 5(i)

(a) Ranjana wants to prove that ammonia is a reducing agent. To demonstrate this, she passes ammonia gas over heated copper oxide. What will she observe?

(b) Write a balanced chemical equation for the above reaction.

Section B 40 Marks

Answer:

(a) The black copper oxide (CuO) will change to reddish-brown copper metal (Cu).
Reason — On passing ammonia gas over heated copper oxide, Ammonia acts as a reducing agent and reduces copper oxide (CuO) to copper metal (Cu). During this process, ammonia is oxidized to nitrogen gas and water.

(b) 3CuO + 2NH₃ ⟶ 3Cu + 3H₂O +N₂↑

Question 5(ii)

Name the alloy which is made up of:

(a) Copper, Zinc and Tin

(b) Lead and Tin

Section B 40 Marks

Answer:

(a) Bronze

(b) Solder

Question 5(iii)

Abhishek was given a salt 'X' which was white in colour for analysis. On strong heating it produced a yellow residue, a colourless gas and also a reddish-brown gas. The solution of the salt ‘X’ when tested with excess of ammonium hydroxide produced a chalky white insoluble precipitate.

(a) Name the coloured gas evolved when Abhishek heated the salt strongly.

(b) Which cation was present in the sample given to Abhishek?

Section B 40 Marks

Answer:

(a) Nitrogen dioxide
Reason — The reddish-brown colour indicates that the gas is Nitrogen dioxide.

(b) Pb²⁺ (Lead)
Reason — The chalky white insoluble precipitate with excess of ammonium hydroxide indicates that the cation present is Pb²⁺.

(c) Lead nitrate [Pb(NO₃)₂].
Reason — The given salt is a nitrate salt as nitrogen dioxide (NO₂), is a common product obtained from heating nitrates. As the cation is lead, hence the salt is Lead nitrate.

Question 5(iv)

Given below in column A is a schematic diagram of the electrolytic reduction of alumina. Identify the parts labelled as A, B and C with the correct options from the Column B.

Section B 40 Marks

Answer:

A → Cathode
B → Anode
C → Electrolytic mixture

Question 6(i)

Element 'X' forms an oxide with the formula X₂O₃ which is a solid with high melting point. 'X' would most likely be placed in the group of the Periodic Table as:

(a) Na
(b) Mg
(c) Al
(d) Si
Justify your answer in the above question (1).

Section B 40 Marks

Answer:

(c) Al
The formula X₂O₃ indicates that the element 'X' forms a trivalent oxide. Thus, 'X' has a valency of 3 like that of Al (Valency of Na, Mg and Si are 1, 2 and 4, respectively). In Group 13, the elements typically form oxides with the general formula X₂O₃, where 'X' represents the element. The oxides of Group 13 elements, such as Al₂O₃ (aluminum oxide), have high melting points due to the strong ionic bonds between the metal cations and the oxide anions.

Question 6(ii)

A student was asked to perform two experiments in the laboratory based on the instructions given:

Observe the picture given below and state one observation for each of the Experiments 1 and 2 that you would notice on mixing the given solutions.

(a) Experiment 1

A student was asked to perform two experiments in the laboratory based on the instructions given: ICSE 2025 Chemistry Solved Question Paper.

(b) Experiment 2

Section B 40 Marks

Answer:

(a) White precipitate of barium sulphate is formed.
Reason — When a solution of zinc sulphate is added to a solution of barium chloride, a double displacement reaction occurs. This reaction involves the exchange of ions between the two compounds and a white precipitate of barium sulphate is formed.
ZnSO₄ (aq) + BaCl₂ (aq) ⟶ BaSO₄ (s) + ZnCl₂ (aq).

(b) Blue precipitate dissolves to form an inky blue/deep blue solution.
Reason — The light blue precipitate of copper(II) hydroxide (Cu(OH)₂) dissolves in excess of ammonium hydroxide due to the formation of a deep blue complex, tetraamminecopper(II) ([Cu(NH₃)₄]²⁺), making it soluble.

Question 6(iii)

Copper sulphate solution is electrolysed using copper electrodes.

(a) Which electrode [cathode or anode] is the oxidizing electrode? Why?

(b) Write the equation for the reaction occurring at the above electrode.

Section B 40 Marks

Answer:

(a) In the electrolysis of copper sulphate solution using copper electrodes the anode is the oxidizing electrode. Copper atoms lose electrons at the anode and go into the solution as copper ions. Thus, the anode causes oxidation, making it the oxidizing electrode.

(b) Cu ⟶ Cu⁺² + 2e^-

Question 6(iv)

X [2, 8, 7] and Y [2, 8, 2] are two elements. Using this information complete the following:

(a) ............... is the metallic element.

(b) Metal atoms tend to have a maximum of ............... electrons in the outermost shell.

Section B 40 Marks

Answer:

(a) Y is the metallic element.

(b) Metal atoms tend to have a maximum of 3 electrons in the outermost shell.

Reason

(a) As Y has two valence electrons, hence it is a metallic element.

(b) Metal atoms tend to have a maximum of 3 electrons in the outermost shell.

(c) The reducing agent is Y because it can easily lose its two valence electrons to form a stable ion, thus reducing other substances by donating electrons.

Question 7(i)

One variety of household fuel is a mixture of propane (60%) and butane (40%). If 20 litres of this mixture is burnt, find the total volume of carbon dioxide added to the atmosphere. The combination reactions can be represented as:

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

Section B 40 Marks

Answer:

Given, 20 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 12 litres and butane is 8 litres

$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} \\ 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} \\ \end{matrix}$

1 Vol. C₃H₈ produces carbon dioxide = 3 Vol

So, 12 litres C₃H₈ will produce carbon dioxide = 3 x 12 = 36 litres

2 Vol. C₄H₁₀ produces carbon dioxide = 8 Vol

So, 8 litres C₄H₁₀ will produce carbon dioxide = $\dfrac{8}{2}$ x 8 = 32 litres

Hence, 68 (i.e., 36 + 32) litres of CO₂ is produced.

Question 7(ii)

Rohit has solution X, Y and Z that has pH 2, 7 and 13 respectively.
Which solution:

(a) will liberate sulphur dioxide gas when heated with sodium sulphite

(b) will liberate ammonia gas when reacted with ammonium chloride

Section B 40 Marks

Answer:

(a) X
Reason — Sodium sulphite can react with acids to liberate sulphur dioxide gas. So, the solution X with pH 2 (acidic) will react with sodium sulphite to liberate sulphur dioxide gas.
Na₂SO₃ + 2H^- ⟶ SO₂ + H₂O + 2Na⁺

(b) Z
Reason — Ammonium chloride is a salt that, in the presence of a base, will release ammonia gas. Therefore, a basic solution (with a high pH) will react with ammonium chloride to liberate ammonia gas. Thus, the solution Z with pH 13 (basic) will liberate ammonia gas when reacted with ammonium chloride.
NH₄Cl + OH^- ⟶ NH₃ + H₂O + Cl^-

Question 7(iii)

8.2 grams of calcium nitrate is decomposed by heating according to the equation

2Ca (NO₃)₂ ⟶ 2CaO +4NO₂ + O₂

Calculate the following:

(a) Volume of nitrogen dioxide obtained at STP

(b) Mass of CaO formed

(Atomic weights : Ca- 40, N-14, O-16)

Section B 40 Marks

Answer:

$\begin{matrix} \text{2Ca}(\text{NO}_3)_2 & \longrightarrow & 2\text{CaO}& +&4\text{NO}_2 & +& \text{O}_2 \\ 2[(40) + 2(14) +6(16)] & & 2(40 + 16) & & 4(14)+8(16) & & 1 \text{mole} \\ = 328 \text{g} & & = 112\text{g} & & =184 & & =32 \text{g} \\ \end{matrix}$

(a) 328 g of calcium nitrate liberated 4 x 22.4 lit of NO₂

∴ 8.2 g of calcium nitrate liberated 4 x 22.4 x $\dfrac{8.2}{328}$ = 0.224 lit of NO₂

(b) 328 g of calcium nitrate produces 2 x 56 g of CaO

∴ 8.2 g of calcium nitrate will produce 2 x 56 x $\dfrac{8.2}{328}$ = 2.8 g of CaO.

Question 8(i)

State giving reasons if:

(a) zinc and aluminium can be distinguished by heating the metal powder with concentrated sodium hydroxide solution.

(b) calcium nitrate and lead nitrate can be distinguished by adding ammonium hydroxide solution to the salt solution.

Section B 40 Marks

Answer:

(a) No, zinc and aluminium cannot be distinguished by heating the metal powder with concentrated sodium hydroxide solution because both will form a white precipitate which is soluble in excess of sodium hydroxide.

(b) Yes, calcium nitrate and lead nitrate can be distinguished by adding ammonium hydroxide solution to the salt solution because white precipitate will be formed with lead nitrate but no precipitate is formed with calcium nitrate.

Question 8(ii)

Draw the electron dot diagram of ammonium ion.

Section B 40 Marks

Answer:

Electron dot diagram of ammonium ion is shown below:

Draw the electron dot structure of ammonium ion. ICSE 2025 Specimen Chemistry Solved Question Paper

Question 8(iii)

Give balanced equations for the following:

(a) Laboratory preparation of ethyne from calcium carbide.

(b) Conversion of acetic acid to ethyl acetate.

Section B 40 Marks

Answer:

(a) CaC₂ + 2H₂O ⟶ Ca(OH)₂ + C₂H₂

(b) CH₃COOH + C₂H₅OH ⟶ CH₃COOC₂H₅ + H₂O

Question 8(iv)

The structures of six organic compounds are shown:

(a) Identify two of the compounds that are members of the same homologous series but are not isomers.

(b) Which two compounds are isomers of each other?

Section B 40 Marks

Answer:

(a) A and F

(b) C and E

Reason

(a) Butene (C₄H₈) and Ethene (C₂H₄) are members of homologous series of alkenes and are not isomers.

(b) Both C and E have the same molecular formula (C₄H₁₀) but different structure. C is n-butane and E is iso-butane.