Solved 2023 Question Paper ICSE Class 10 Chemistry

Argon

Reason — Argon belongs to period 3. It is an inert element and its octet is complete hence, it's electron affinity is zero.

Lithium

Reason — The given elements belong to the second period and Lithium is the first element of second period. Atomic size decreases from left to right in a period hence, Lithium will have the largest atomic radius.

Fluorspar

Reason — Fluorspar is calcium fluoride [CaF₂] whereas Bauxite [Al₂O₃.2H₂O], Corundum [Al₂O₃] and Cryolite [Na₃AlF₆] are ores of aluminium.

16

Reason —

Molecular mass of methyl alcohol [CH₃OH] = 12 + 3(1) + 16 + 1 = 32

Vapour density = $\dfrac{\text{molecular mass}}{2}$ = $\dfrac{32}{2}$ = 16

Ag - 1e^- ⟶ Ag¹⁺

Reason — During electroplating of an article with silver, the silver anode loses electrons to give Ag¹⁺ ions in the solution.

Cu(OH)₂

Reason — Cu(OH)₂ reacts with a solution of excess ammonium hydroxide to form a deep blue solution of tetrammine copper [II] sulphate.

Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH ⟶ [Cu(NH₃)₄]SO₄ + 4H₂O

Benzene

Reason — In benzene, six carbon atoms form a ring structure and there are three single and three double bonds between the carbon atoms. Hence, benzene is a cyclic compound.

concentrated sulphuric acid

Reason — Concentrated Sulphuric acid only removes the moisture content of HCl gas but does not react with it. Hence, it is used as a drying agent for HCl gas.

Zinc nitrate

Reason —

$\begin{matrix} 2\text{Zn(NO}_3)_2 &\longrightarrow & \text{2ZnO} & + & 4\text{NO}_2 &+ & \text{O}_2 \\ \text{[Zinc} & & \text{[Zinc nitrate - white]} & & \text{[Carbon} \\ \text{carbonate]} & & \text{[yellow - hot]} & & \text{dioxide]} \\ \text{[white]} & & \text{[white - cold]} \end{matrix}$

Zinc nitrate decomposes to produces zinc oxide, which is yellow when hot and white when cold

K₂SO₄

Reason —

$\text{KNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\gt 200 \degree\text{C}} \text{K}_2\text{SO}_\text{4} + \text{HNO}_3$

Non-volatile acid property

Reason — Concentrated sulphuric acid has a high boiling point, hence it is used to prepare hydrogen chloride gas from potassium chloride.

Ethane

Reason —

$\underset{\text{sodium propionate}}{\text{C}\text{H}_3\text{-C}\text{H}_2\text{.COONa}} + \underset{\text{sodalime}}{\text{NaOH}} \xrightarrow[\Delta]{\text{CaO}} \underset{\text{ethane}}{\text{C}_2\text{H}_6} + \text{Na}_2\text{CO}_3$

CH₃COOH

Reason — Acetic acid is a weak acid which does not dissociate completely in water, so it does not form acid salt but forms a normal salt.

C_nH_2n+2

Reason — The general formula of hydrocarbons with single covalent bonds is C_nH_2n+2

Phenolphthalein

Reason — Colourless phenolphthalein changes to pink colour in an alkaline solution.

(a) Fluorspar [CaF₂]

(b) Powdered coke

(c) Hall-Heroult's process

(d) Reactions taking place at the electrodes 'Y' (anode) and 'Z' (cathode):

At anode:
6O^2- - 12e^- ⟶ 6[O]
3O + 3O ⟶ 3O₂

At cathode:
4Al³⁺ + 12e^- ⟶ 4Al

(a) Metals are good reducing agents

(b) Non-polar covalent compounds are bad conductors of heat and electricity.

(c) Higher the pH value of a solution, the more alkaline it is.

(d) Silver chloride is a white precipitate that is soluble in excess of Ammonium hydroxide solution.

(e) Conversion of ethene to ethane is an example of hydrogenation.

(a) Electron affinity

(b) Catenation

(c) Haber’s process

(d) Basic salt

(e) Vapour Density

(a) Structural formula

1. 2-chlorobutane

2. Methanal

3. But-2-yne

(b) IUPAC names are:

1. Ethanoic acid
2. Butan-2-ol

(a) Pb²⁺

Reason — When ammonium hydroxide solution is added to lead nitrate, it forms a chalky white precipitate of lead hydroxide [Pb(OH)₂] which is insoluble in excess of ammonium hydroxide.

Pb(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Pb(OH)₂

(b) Ca²⁺

Reason — When sodium hydroxide solution is added to calcium nitrate, a white precipitate of calcium hydroxide is obtained which is insoluble in excess of sodium hydroxide solution.

Ca(NO₃)₂ + 2NaOH ⟶ Ca(OH)₂ ↓ + 2NaNO₃

(a) During electrolysis, the compound PbBr₂ in its molten state liberates reddish brown fumes at the anode.

(b) The ion which could be discharged most readily during electrolysis is Cu²⁺

(a) K > Ca > Mg > Al
Reason — According to the electrochemical series metals at the top of the series are most reactive and the reactivity decreases down the group.

(b) Be < C < N < O
Reason — Across a period, left to right in a periodic table, the non-metallic character increases.

(c) F > P > Si > Be
Reason — The valence electrons decreases from right to left in the periodic table.

(a) 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2NH₃ + 2H₂O

(b) CuSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Cu(OH)₂↓

(c) Cu + 4HNO₃ (conc.) ⟶ Cu(NO₃ )₂ + 2NO₂ + 2H₂O

(a) Quicklime [CaO] is alkaline in nature and it reacts with HCl forming the respective chloride. Hence, it can't be used as a drying agent for Hydrogen chloride gas.

CaO + 2HCl ⟶ CaCl₂ + H₂O

(b) As ammonia gas is highly soluble in water, therefore, it is not collected over water.

(a) Duralumin

(b) Stainless steel

[By Lussac's law]

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

(i) To calculate the volume of CO₂ formed :

$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{CO}_2 \\ 2 \text{ vol.} & : & 4 \text{ vol.} \\ 80 \text{ ml} & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{4}{2} \times 80 = 160 \text{ ml}$

Hence, volume of carbon dioxide formed = 160 ml

(ii) To calculate the volume of unused O₂ :

$\begin{matrix}\text{C}_2\text{H}_6 & : & \text{O}_2 \\ 2 \text{ vol.} & : & 7 \text{ vol.} \\ 80 \text{ ml} & : & \text{x} \end{matrix}$

$\therefore x = \dfrac{7}{2} \times 80 = 280 \text{ ml}$

Unused oxygen = 300 - 280 = 20 ml.

Hence, volume of unused oxygen = 20 ml

Therefore, the resultant gaseous mixture consists of 160 ml of carbon dioxide and 20 ml of unused oxygen.

(a) Mg₃N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃

(b) As nitrides are expensive, hence the method is seldom used.

(c) Ways in which ammonia gas can be identified is:

1. It has a sharp characteristic odour.
2. It turns:
   1. moist red litmus blue,
   2. moist turmeric paper brown,
   3. phenolphthalein solution pink.
3. When a glass rod dipped in conc. HCl is brought near the mouth of the jar containing NH₃, dense white fumes of ammonium chloride are formed.
   NH₃ + HCl ⟶ NH₄Cl

(a) Bronze is used in making statues, medals, coins

(b) Fuse metal is used for electrical fuse, soldering purposes.

(a) Ammonium ion

(b) A molecule of nitrogen

(a) Ethene from ethanol

$\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170\degree\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethene}}{\text{C}_2\text{H}_4} + \text{H}_2\text{O}\$

(b) Ethyne from calcium carbide

$\underset{\text{calcium carbide}}{\text{CaC}_2} + \underset{\text{water}}{2\text{H}_2\text{O}} \longrightarrow \underset{\text{ethyne}}{\text{C}_2\text{H}_2} + \underset{\text{calcium hydroxide}}{\text{Ca(OH)}_2}$

(c) Monochloromethane from methane

$\underset{\text{ methane} }{\text{CH}_4} + \text{Cl}_2 \xrightarrow[\Delta]{\text{diffused sunlight}} \underset{\text{monochloromethane}}{\text{CH}_3\text{Cl}} + \text{HCl}$

(a) Nitrate [NO₃^-]

(b) Sulphide [S₂^-]

(c) Sulphate [SO₄^2-]

(a) Electronegativity — The tendency of an atom in a molecule to attract the shared pair of electrons towards itself is called its electronegativity.

(b) Gay-Lussac's Law of combining volumes — When gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

Empirical formula is CHCl₂

Empirical formula weight = 12 + 1 + 2[35.5] = 12 + 1 + 71 = 84

Molecular weight = 168

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{168}{84} = 2$

So, molecular formula = 2(CHCl₂) = C₂H₂Cl₄

(a) Copper

(b) Sodium Sulphite

(c) Ferrous Sulphate

(a) When a few drops of sodium hydroxide solution are added to the copper nitrate solution, light blue coloured ppt. of Cu(OH)₂ is obtained which is insoluble with excess NaOH.

(b) Ammonia burns in the atmosphere of excess oxygen with a greenish yellow flame, forming nitrogen and water vapour.

(c) Buff yellow PbO is reduced to greyish metallic lead.

(a) Isomers

(b) Homologous series

(a) On moving down the group, atomic size, as well as, nuclear charge increases. However, the effect of increase in atomic size dominates over the effect of increase in nuclear charge. Hence, ionization potential decreases down the group.

(b) There is a strong electrostatic force of attraction between the oppositely charged ions of the ionic compounds that keeps ions in fixed position in solid state. Due to this immobility of ions, ionic compounds do not conduct electricity in solid state.

(a) Molecular weight of Ca(H₂PO₄)₂
= 40 + 2[2(1) + 31 + 4(16)]
= 40 + 2[2 + 31 + 64]
= 40 + 2[97]
= 40 + 194
= 234 g

234 g of Ca(H₂PO₄)₂ contains 62 g of P

∴ 100 g of Ca(H₂PO₄)₂ will contain = $\dfrac{62}{234}$ x 100 = 26.49% = 26.5%

Hence, 26.5% phosphorous is present in superphosphate Ca(H₂PO₄)₂

(b) Given,

Molecular Formula = C₈H₁₈ = (C₄H₉)2

Molecular formula = n[Empirical formula]

∴ n = 2

Hence, Empirical formula = C₄H₉

(a) Impure block of copper.

(b) Pure copper is deposited on the thin sheet of pure copper placed at the cathode.

(c) Cu²⁺ + 2e^- ⟶ Cu

(a) Molecular weight of C₂H₂ = 2[12] + 2[1] = 26

Molecular weight of C₃H₆ = 3[12] + 6[1] = 42

Molecular weight of CH₄ = 12 + 4[1] = 16

Molecular weight of C₂H₄ = 2[12] + 4[1] = 28

Hence, increasing order of molecular weights is :

CH₄ < C₂H₂ < C₂H₄ < C₃H₆

(b) Ag⁺, Cu²⁺, Zn²⁺, Na⁺

(a) Conc. H₂SO₄ dehydrates sucrose to a black spongy charged mass of carbon sugar charcoal.

$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{sugar charcoal}}{12\text{C}} + \text{11H}_2\text{O}$

The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

$\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{[White Anhydrous]}}{\text{CuSO}_4} + 5\text{H}_2\text{O}$

(b) Sulphuric acid forms sulphate salts whereas hydrochloric acid forms chloride salts.

(a) Ammonia to nitric oxide using oxygen and platinum catalyst.

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[800 \degree \text{C}]{\text{Pt}.} 4\text{NO} + 6\text{H}_2\text{O} + \Delta$

(b) Sodium hydroxide to sodium sulphate using sulphuric acid.

2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

(c) Ferrous sulphide to hydrogen sulphide using hydrochloric acid

FeS + 2HCl ⟶ FeCl₂ + H₂S

(a) NH₄Cl

(b) PbO

(c) CCl₄