Solved 2025 Specimen Paper ICSE Class 10 Chemistry

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.

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

6.023 x 10²² atoms of carbon

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

1. 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.
2. One mole of sodium:
   • The atomic mass of sodium is 23 g/mol.
   • Thus, one mole of sodium weighs 23 grams.
3. 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.
4. 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:

1. 2 gram atoms of oxygen = 32 grams
2. One mole of sodium = 23 grams
3. 22.4 litres of carbon dioxide at STP = 44 grams
4. 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.

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.

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.

2

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

1

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.

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.

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.

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.

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.

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

sodium hydroxide solution

Reason — Let's analyze each option:

1. 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.

2. 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.

3. 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.

4. 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.

Amphoteric oxides contain a metal

Reason — Let's analyze each option:

1. 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.

2. 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.

3. 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.

4. 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.

(II) and (IV)

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

(a)

1. 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.
2. 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:

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

(c) Ag (s) ⟶ Ag⁺(aq) + e^-

The matched columns are shown below:

(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.

(a) Coordinate bond

(b) Normal salt

(c) Substitution

(d) Ionization potential

(e) Alloy

(a) Structural Diagram:

1. Propanoic acid :

2. Pentan-2-ol

3. 2,2 dibromo butane

(b) IUPAC names are:

1. 1-propanol

2. pent-1-ene

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

OR

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

(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.

(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.

(a) CuCO₃ + H₂SO₄ ⟶ CuSO₄ + H₂O + CO₂

(b) Zn + 2NaOH ⟶ Na₂ZnO₂ + H₂

(c) 2Fe + 3Cl₂ ⟶ 2FeCl₃

(a) Cryolite.

(b) Al³⁺ + 3e^- ⟶ Al

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

(a) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃

(b) C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

(c) C₂H₅Cl + aq. NaOH ⟶ C₂H₅OH + NaCl

(a) 800°C, Platinum

(b) 4NO₂ + 2H₂O + O₂ ⟶ 4HNO₂

(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↑

(a) Bronze

(b) Solder

(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.

A → Cathode
B → Anode
C → Electrolytic mixture

1. (c) Al

2. 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.

(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.

(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^-

(a) Y is the metallic element.

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

(c) Y is the reducing agent.

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.

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.

(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^-

(c) Y
Reason — The solution Y with pH 7 will not have any effect on litmus paper as it is neither acidic nor basic.

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

(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.

(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.

Electron dot diagram of ammonium ion is shown below:

(a) CaC₂ + 2H₂O ⟶ Ca(OH)₂ + C₂H₂

(b) CH₃COOH + C₂H₅OH ⟶ CH₃COOC₂H₅ + H₂O

(C) NaNO₃ + conc. H₂SO₄ ⟶ NaHSO₄ + 2HNO₃

(a) A and F

(b) C and E

(c) C₂H₅OH $\xrightarrow[170\degree\text{C}]{\text{Conc. H}_2\text{SO}_4}$ C₂H₄ ↑ + H₂O

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.

(c) Ethene (C₂H₄) can be prepared by the dehydration of ethyl alcohol (C₂H₅OH).