ICSE Class 10 Chemistry — 2025 Improvement

I & III

Reason — Copper is a main component of alloy bronze that is used to make medals and copper is also a main component of brass and duralumin whereas aluminium is not a part of stainless steel.

Sulphuric acid

Reason — An acid salt is formed when a dibasic or tribasic acid is only partially neutralised by a base.

• Nitric acid (HNO₃) and hydrochloric acid (HCl) are monobasic acids → they can form only normal salts, not acid salts.
• Acetic acid (CH₃COOH) is also monobasic → only forms sodium acetate.
• Sulphuric acid (H₂SO₄) is dibasic → with sodium hydroxide (NaOH), it can form an acid salt, sodium hydrogen sulphate (NaHSO₄), on partial neutralisation.

So, sodium hydroxide reacts with sulphuric acid to form an acid salt.

NaOH + H₂SO₄ ⟶ NaHSO₄ + H₂O

0.1 mole

Reason — Number of moles present in 10g of CaCO₃

Molar mass of CaCO₃ = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g

100 g of CaCO₃ = 1 mole

∴ 10 g of CaCO₃ = $\frac{10}{100}$

= 0.1 mole

AgNO₃

Reason — A white precipitate of AgCl is formed when dilute hydrochloric acid reacts with AgNO₃. Silver chloride is a white precipitate that is soluble in excess of Ammonium hydroxide solution and insoluble in dilute HNO₃.

AgNO₃ + HCl ⟶ AgCl + HNO₃

Both (A) and (R) are true, and (R) is the correct explanation of (A).

Reason — Cu²⁺ ions will be discharged in preference to Ca²⁺ ions because Cu²⁺ ions are present lower than Ca²⁺ ions in the electrochemical series.
Elements lower in the series get discharged more easily at the cathode during electrolysis because their cations can easily gain electrons. Hence, the assertion (A) is true.
The Cu²⁺ ions will be discharged in preference to Ca²⁺ ions because Ca²⁺ ions are placed above Cu²⁺ ions in the electrochemical series. Hence, the reason (R) is true and it correctly explains assertion (A).

(A) is true but (R) is false.

Reason — Hydraulic washing (gravity separation) is a method to separate impurities (gangue) from the ore. Hence, the assertion (A) is true.
During the hydraulic washing, denser particles settle at the bottom whereas lighter particles float on the surface. Hence, the reason (R) is false.

Amphoteric oxide

Reason — Amphoteric oxides can react with both acids and bases to form salt and water. For example: Al₂O₃

8 g of sulphur dioxide

Reason — One mole of any gas occupies 22.4 litres of volume at S.T.P.

The required moles of gas at S.T.P = $\frac{2.8}{22.4}$ = 0.125 mol

Substance	Molar mass(g)	Given mass (moles/g)	No. of moles	Volume at STP (L)
2 moles of carbon dioxide	44	2 mol	2	44.8
7.1 g of chlorine	71	7.1 g	0.10	2.24
8 g of sulphur dioxide	64	8g	0.125	2.8
56 g of carbon monoxide	28	56g	2.00	44.8

∴ 8 g of sulphur dioxide will occupy the volume 2.8 litres at S.T.P.

Fe₂(SO₄)₃

Reason —

Reddish-brown ppt with NaOH indicates Fe³⁺ ions.
$\text{Fe}^{3+} + 3\text{OH}^- \rightarrow \text{Fe(OH)}_3 \downarrow \space (\text{reddish-brown})$

White ppt with BaCl₂ indicates SO₄^2- ions (forms insoluble BaSO₄)
$\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \downarrow \space (\text{white})$

So the salt must contain Fe³⁺ and SO₄^2-, which is Fe₂(SO₄)₃.

C₃H₈

Reason — Let's check the molecular masses:

• CH₄ → 12 + 4×1 = 16
• C₂H₆ → 2×12 + 6×1 = 24 + 6 = 30
• C₃H₈ → 3×12 + 8×1 = 36 + 8 = 44
• C₄H₁₀ → 4×12 + 10×1 = 48 + 10 = 58

So, the alkane with molecular mass 44 is C₃H₈ (propane).

CO₂

Reason — The setup shows ethanoic acid reacting with sodium carbonate. This reaction produces carbon dioxide gas, which when passed through calcium hydroxide solution [lime water] turns it milky due to the formation of insoluble calcium carbonate:

$$\begin{aligned}\text{2CH₃COOH} + \text{Na₂CO₃} \rightarrow 2\text{CH₃COONa} + \text{H₂O} + \text{CO₂} \uparrow \\[1em] \text{CO₂} + \text{Ca(OH)₂} \rightarrow \text{CaCO₃} \downarrow + \text{H₂O}\end{aligned}$$

Hence, the gas evolved is carbon dioxide (CO₂).

N₂

Reason — When ammonia gas is passed over buff yellow PbO, ammonia reduces heated lead monoxide to greyish metallic lead with the liberation of nitrogen gas.

2NH₃ + 3PbO ⟶ 3Pb + 3H₂O + N₂ ↑

X & Y

Reason — X (NaCl solution) is an electrolyte, it contains free Na⁺ and Cl^- ions, so it conducts electricity well.

Y (acetic acid) is a weak electrolyte, it partially ionises to give H⁺ and CH₃COO^- , so it can conduct where the bulb may glow dimmer than with a strong electrolyte. Whereas, Z (sugar solution) contains molecular sucrose which does not dissociate into ions.

an ionic solid.

Reason — Electronic configuration 2, 2 means the atom has 2 electrons in its outer shell. Hence, it belongs to Group 2, it is a metal with valency 2 (it tends to lose 2 electrons to form X²⁺).
Oxygen is a non-metal with valency 2 (it tends to gain 2 electrons to form O^2-).
So, when X combines with oxygen, the compound formed will be of the type X²⁺ O^2-, i.e. metal + non-metal. This is ionic in nature and will exist as an ionic solid with high melting point, conducting electricity only in molten/solution state.

Metal

Reason — The amount of energy required to remove a loosely bound electron from the outermost shell of an atom is known as ionisation potential. Metals have loosely held outer electrons and tend to lose electrons easily. Hence, they have low ionisation potential.

(a) 4FeS₂ + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂

(b) Balanced chemical equation for formation of ‘B’

$2\text{SO}_2 + \text{O}_2 \xrightleftharpoons [450-500\text{^{\circ} C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + \text{45 K cals}$

(c) Sulphuric acid is not obtained by directly reacting SO₃ with water since sulphur trioxide does not dissolve in water satisfactorily and it gives a lot of heat and forms misty droplets of sulphuric acid. Hence, it is converted to oleum.

SO₃ + H₂SO₄ (conc.) ⟶ H₂S₂O₇ (oleum)

(d) C — Water (H₂O)

(e) H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

(a) The metal to be refined is kept at the anode during the process of electro-refining.

(b) Dilute HCl and dilute H₂SO₄ can be distinguished by adding BaCl₂ solution.

(c) Ammonia gas is collected by downward displacement of air.

(d) The gas formed when copper carbonate is heated is CO₂.

(e) Excess ammonia reacts with chlorine to form nitrogen.

(a) Acid

(b) Non-polar covalent bond

(c) Electroplating

(d) Substitution reaction

(e) Oxidising agent

(a) Structural diagrams

1.

2. Structural formula of methanal is shown below:

3. Structural formula of But-2-yne is shown below:

(b) IUPAC names

1. 2,2-dimethylpropane
2. Tetrachloromethane

(a) Zn undergoes oxidation.

Zn ⟶ Zn²⁺

(b) Pb²⁺ undergoes reduction.

Pb²⁺ ⟶ Pb

(a) Reducing property means the tendency to lose electrons, On moving down a group atomic size increases. Outer electrons are farther from the nucleus and less strongly attracted. Hence, as we move down the group elements lose electrons more readily, making them stronger reducing agents.

(b) Aluminium oxide is highly stable and aluminium has a greater affinity for oxygen than carbon, so CO is not strong enough to reduce Al₂O₃.

(a) Mg > Si > P > S.

(b) I < Br < Cl < F

(c) Rb > K > Na > Li

(a) When Harsh added concentrated sulphuric acid to blue vitriol (CuSO₄.5H₂O), the blue colour crystals lose their blue colour and become white (anhydrous copper sulphate) this is because of loss of water molecules.

$\text{CuSO}_4.5\text{H}_2\text{O} \xrightarrow{\text{Conc. H}_2 \text{SO}_4} \text{CuSO}_4 + 5\text{H}_2\text{O}$

(b) When he passed ammonia gas over heated PbO, ammonia reduces heated yellow lead monoxide to greyish metallic lead.

2NH₃ + 3PbO ⟶ 3Pb + 3H₂O + N₂ [g]

(c) On adding sodium hydroxide solution to CuSO₄ solution, Harsh obtained a pale blue precipitate of Cu(OH)₂ which was insoluble in excess of sodium hydroxide.

CuSO₄ + 2NaOH ⟶ Cu(OH)₂ + Na₂SO₄

(a) O — Hydrogen chloride gas

Reason — When HCl is brought near ammonia gas, dense white fumes of ammonium chloride are formed.

HCl + NH₃ ⟶ NH₄Cl

(b) N — Hydrogen sulphide

Reason — Hydrogen sulphide(H₂S) gas turns the moist lead acetate paper black

(CH₃COO)₂Pb + H₂S ⟶ PbS + CH₃COOH

(c) P — Nitrogen dioxide

Reason — When lead nitrate is heated nitrogen dioxide gas is evolved.

2Pb(NO₃)₂ $\xrightarrow{\Delta}$ 2PbO + O₂ + 4NO₂

(a) Molten lead bromide conducts electricity.

Reason — Solid lead bromide is a non-conductor of electricity since it's ions are not free but held together by an electrostatic force of attraction. In molten state the ions become free and hence conduct electricity.

(b) Copper reacts with concentrated nitric acid to form nitrogen dioxide gas.

Reason — When copper reacts with dilute nitric acid it will form nitric oxide(NO) not nitrogen dioxide.

(c) Bromoethane reacts with aqueous sodium hydroxide to produce ethanol and sodium bromide.

Reason — Alcohol is prepared by the hydrolysis of alkyl halides (haloalkanes) on reaction with aqueous NaOH. Hydrolysis of bromoethane occurs with aqueous NaOH, not anhydrous/alcoholic NaOH.

(a) The student was unsuccessful in collecting any gas because calcium oxide (quick lime) react with hydrogen chloride.

CaO + 2HCl ⟶ CaCl₂ + H₂O

(b) The student can use conc. sulphuric acid as a drying agent. Conc. H₂SO₄ absorbs water vapour but does not react with HCl gas.

(a) Cu²⁺
Reason — During the electrolysis of CuSO₄ solution using copper electrodes. Both Cu²⁺ and H⁺ ions migrate to the cathode, Cu²⁺ ions discharged as neutral copper atoms.

(b) Ag
Reason — During electroplating of an article with Ag using sodium argentocyanide as an electrolyte, the anode is the plate of pure clean silver.

$$\begin{matrix} 2\text{C}_2\text{H}_6 & + & 7\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 6\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 7 \text{ vol.} & \longrightarrow & 4 \text{ vol.} & \\ \end{matrix}$$

[By Gay Lussac's law]

2 Vol. of C₂H₆ requires 7 Vol. of oxygen

∴ 60 cc C₂H₆ will require $\frac{7}{2}$ x 60

= 210 cc of Oxygen

Hence, unused oxygen = 250 - 210 = 40 cc

Similarly,

2 Vol. of C₂H₆ produces 4 Vol. of carbon dioxide

∴ 60 cc C₂H₆ produces $\frac{4}{2}$ x 60

= 120 cc of Carbon dioxide

Hence, carbon dioxide produced = 120 cc.

The gaseous mixture at room temperature (assuming water condenses) contains 120 cc CO₂ and 40 cc O₂.

(a) CH₃COOH

(b) SO₂

(c) NaHSO₄

(a) Electron dot diagram of hydronium ion is shown below:

(b) Electron dot diagram of oxygen molecule is shown below:

(c) Electron dot diagram of Calcium oxide is shown below:

(a) Conc. sulphuric acid

(b)

$\underset{[conc.]}{\text{KNO}_3} + \underset{[conc.]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 ^{\circ}\text{C}} \underset{[acid \space salt]}{\text{KHSO}_4} + \text{HNO}_3$

(c) The complete apparatus is made up of glass because nitric acid vapours attack rubber and cork.

(d) Pure acid is colourless but the acid obtained in the laboratory is slightly yellow. The yellow colour is due to dissolution of reddish brown coloured nitrogen dioxide gas in the acid. This gas is produced due to the thermal decomposition of a portion of nitric acid.
4HNO₃ ⟶ 2H₂O + 4NO₂ + O₂

(a) At anode : 6O^2- - 12e^- ⟶ 3O₂

(b) At cathode : 4Al³⁺ + 12e^- ⟶ 4Al

(a) Cryolite — Na₃AlF₆

(b) Haemetite — Fe₂O₃

(a) KHCO₃ ​+ HCl ⟶ KCl + H₂​O + CO₂ ​↑

(b)

$\underset{\text{sodium propionate}}{\text{C}_2\text{H}_5\text{COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \xrightarrow[300^{\circ}\text{C}]{\text{CaO}} \underset{\text{ethane}}{\text{C}_2\text{H}_6} \uparrow + \text{Na}_2\text{CO}_3$

(c) AlN + 3H₂​O ⟶ Al(OH)₃​ + NH₃​ ↑

(a) Catalyst — Finely divided iron [Fe]

(b) Temperature — 450-500°C [Optimum temperature]

(c) Pressure — 200 to 900 atmospheres [Optimum pressure]

Given:
1 mole of CO₂ occupies 24 dm³ (24 L) at room temperature and pressure

Gram molecular mass of CO₂ = 12 + 2 x 16 = 44 g

(a) 24 litres of CO₂ weighs = 44 g (at room temperature & pressure)

∴ Mass of 6 litres of CO₂ = $\frac{44}{24}$ x 6 = 11 g (at room temperature & pressure)

Hence, mass of 6 litres of CO₂ at room temperature & pressure is 11 g.

(b) Volume occupied by 44 g of CO₂ = 24 litres (at room temperature & pressure)

∴ Volume occupied by 60 g of CO₂ = $\frac{24}{44}$ x 60 ≈ 32.72 litres (at room temperature & pressure)

Hence, volume occupied by 60 g of CO₂ ≈ 32.72 litres (at room temperature & pressure)

(a) Electrolyte is a compound which either in aqueous solution or in molten state allow electric current to pass through it.

(b) The property of self linking of atoms of an element through covalent bonds in order to form straight chains, branched chains and cyclic chains of different sizes is known as catenation.

(a) Copper carbonate — Double Decomposition (Precipitation)

(b) Sodium sulphate — Neutralisation

(c) Ferric chloride — Direct combination

(a) C₂H₅Br + KOH (alcoholic) ⟶ C₂H₄ + KBr + H₂O

(b) 2C₂H₅OH $\xrightarrow[170^{\circ}\text{C}]{\text{H}_2\text{SO}_4}$ C₂H₅ — O — C₂H₅ + H₂O

(c) CH₄ + Cl₂ ⟶ CH₃Cl + HCl