Solved 2024 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.

neither P nor Q

Reason —

Number of atoms = $\dfrac{\text{Mass}}{\text{Molecular Mass}}$ x N_A

where N_A is the Avogadro's number = 6.023 x 10²³

Number of atoms in 32 g of sulphur (s) = $\dfrac{32}{32}$ x N_A = N_A

Number of atoms in 16 g of oxygen (O) = $\dfrac{16}{16}$ x N_A = N_A

Number of atoms in 4 g of helium (He) = $\dfrac{4}{4}$ x N_A = N_A

Hence, all have same number of atoms which is = 6.023 x 10²³

W

Reason — Quick lime (CaO) is a drying agent and absorbs the moisture from ammonia gas without reacting with it, hence red litmus turns blue as ammonia gas is basic in nature.

dehydration

Reason — Concentrated sulphuric acid is a strong dehydrating agent due to its strong affinity for water, hence, the reaction taking place is a dehydration reaction.

C₆H₁₂O₆ $\xrightarrow {\text{Conc. H}_2\text{SO}_4}$ 6C + 6H₂O

Q

Reason — Anode should be the plate of silver and cathode should be the spoon to be electroplated. Electrolyte used should be silver nitrate.

1

Reason — Acetic acid has four hydrogen atoms in it but it ionises in aqueous solution to produce one hydrogen ion per molecule of the acid. Hence, basicity of acetic acid is 1.

3, 7

Reason — A has donated three electrons from its outer most shell and has +3 charge. B had 7 electrons and has taken one electrons and has -1 charge.

Colourless

Reason — According to the reactivity series of metals, magnesium is more reactive compared to copper. Hence, it replaces copper from CuSO₄ and the blue colour fades as colourless magnesium sulphate is formed and brown bits of copper metal form a precipitate:

Mg + CuSO₄ ⟶ MgSO₄ + Cu

17

Reason — Element with atomic number 17 has electronic configuration [2, 8, 7] and hence has 7 electrons in the valence shell. It is a non-metal and will form an acidic oxide.

It is a strong reducing agent

Reason — Nitric acid is a strong oxidizing agent and not reducing agent.

-OH is the functional group in methanol

Reason — The functional group present in methanol [CH₃OH] is alcohol [-OH].

Redox reaction

Reason — During electrolysis, the reaction at the cathode involves reduction of cations as they gain electrons to become neutral atoms while that at anode involves oxidation of anions as they lose electrons to become neutral. As redox reactions are reactions where oxidation and reduction takes place simultaneously. Hence, electrolysis is an example of redox reaction.

The catalyst used in Ostwald’s process is platinum.

Reason — Ostwald’s process :

$4\text{NH}_3 + 5\text{O}_2 \xrightarrow[700-800 \degree\text{C}]{\text{Pt}} 4\text{NO} + 6\text{H}_2\text{O} + 21.5 \text{K cals}$

6

Reason — The element in third period and sixteenth group will have atomic number 16 and electronic configuration 2, 8, 6. Hence, it has 6 valence electrons.

When sodium hydroxide is introduced into the flask, it undergoes a neutralization reaction with HCl to form salt and water:

HCl + NaOH ⟶ H₂O + NaCl

As HCl is used up in the reaction, this lowers the pressure inside the flask. The outside pressure being higher pushes the litmus solution inside, through the jet tube.

(b) The colour of the litmus solution will remain unchanged so the fountain will be blue in colour.
HCl and sodium hydroxide undergo neutralization reaction to form salt and water which are neutral products. Hence, colour of litmus solution will not change.

(c) No the observation will be the same. Ammonia gas and water produce ammonium hydroxide which is basic in nature, hence again a blue colour fountain will be seen.

(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 reddish brown coloured precipitate is formed when ammonium hydroxide is added to a solution of ferric chloride.

(d) Alkanes undergo substitution reactions.

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

(a) Co-ordinate bond

(b) Normal salt

(c) Halogenation

(d) Ionisation energy

(e) Alloy

(a) Structural diagrams are as follows:

1. 1- propanal

2. 1, 2 dichloro ethane

3. But-2-ene

(b) IUPAC name are as follows:

1. 1-propanol

2. 1-pentene

Calcium carbonate

CaCO₃ + 2HNO₃ [dil.] ⟶ Ca(NO₃)₂ + H₂O + CO₂↑

(a) Non-volatile nature of Sulphuric acid.

(b) Oxidising property of concentrated Sulphuric acid.

(a) Oxidising power of X > Y. Elements with high electron affinity accept electrons more easily hence have greater oxidising power.

(b) Electronegativity of X > Y. Element with high electron affinity has high electronegativity.

(c) X is to the right side of Y as electron affinity increases from left to right in a period.

(a)

1. False
   In an electrovalent compound, the cation attains the electronic configuration of the noble gas that comes before it in the periodic table.
2. True
   In the formation of a compound PQ₂, atom P gives one electron to each atom of Q. The compound PQ₂ is a good conductor of electricity in molten or aqueous solution state.

(b) Gram molecular mass of CO₂ = 12 + 2(16)
= 12 + 32 = 44 g

As,

44 g of CO₂ = 1 mole
∴ 22 g of CO₂ = $\dfrac{1}{44}$ x 22 = 0.5 moles

Hence, number of moles in 22 grams of carbon dioxide = 0.5

(a) The compound is cryolite [Na₃AlF₆].

(b) 2Al³⁺ + 6e^- ⟶ 2Al

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{20}{2} \\[0.5em] = 10 \text{ g}$

Molecular weight = 2 x Vapour density
= 2 x 10 = 20 g

Hence, relative molecular mass of gas is 20 g

(a) AlN + 3H₂O ⟶ Al(OH)₃ + NH₃ [g]

(b) C + 4HNO₃ ⟶ CO₂ + 2H₂O + 4NO₂

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

(a) Temperature : 450-500°C [Optimum temperature]

(b) Catalyst : Finely divided iron [Fe]

(c) N₂ + 3H₂ ⇌ 2NH₃ + Δ

(a) Black Copper [II] oxide is reduced to reddish brown copper by Ammonia.

(b) 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂↑

(a) Bronze

(b) Solder

(a) P is blue vitriol [CuSO₄.5H₂O]

(b) Blue colour slowly fades as ZnSO₄ is formed.

(c) White ppt. of BaSO₄ is obtained.
CuSO₄ + BaCl₂ ⟶ BaSO₄ ↓ + CuCl₂

(a) In case of ethene, the valencies of atleast 2 carbon atoms are not fully satisfied by hydrogen atoms. The availability of electrons in the double bond makes them more reactive and hence they undergo addition reactions only.

(b) Hydrocarbons are highly combustible and they produce good amount of heat hence they are used as a fuel.

(c) Hydrogen chloride gas is not collected over water since it is highly soluble in water.

(a) Zinc blende [ZnS]

(b) Bauxite [Al₂O₃]

(a) When sodium chloride is added to a solution of lead nitrate, an insoluble white precipitate of lead chloride is formed.

2NaCl + Pb(NO₃)₂ ⟶ 2NaNO₃ + PbCl₂ ↓

(b) When barium chloride solution is added to a solution of Zinc sulphate, white ppt. of barium sulphate is obtained which is insoluble in dil. HCl or nitric acid.

ZnSO₄ + BaCl₂ ⟶ BaSO₄ ↓ [white ppt.] + ZnCl₂

(a) Anode is the oxidizing electrode. At anode, anions lose electrons and are oxidized.

(b) Cu - 2e^- ⟶ Cu²⁺

(a) Y is the metallic element.

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

(c) Y is the reducing agent.

Empirical formula is C₃H₄N

Molecular weight = 108

Amount of carbon in 1 mole of the compound = molar mass of carbon x number of carbon atoms in the empirical formula.

= 12 x 3 = 36 g/mol

Hence, in 1 mole of the compound C₃H₄N, there is approximately 36 grams of carbon.

(a) 1. On adding the acid, pH value will decrease.

2. On adding the solution of a base, pH value will increase.

(b) As the atomic number is 15, so the electronic configuration will be 2, 8, 5. Hence, it will belong to VA group.

$\begin{matrix} 2\text{Ca}(\text{NO}_3)_2 & \longrightarrow & 2\text{CaO} & + & 4\text{NO}_2 & + & \text{O}_2 \\ 2[(40) + & & 2(40 & & 4[22.4] \\ 2(14 +3(16))] & & + 16) & & \text{ lit.} \\ = 328 \text{ g} & & = 112 \text{g} \\ \end{matrix}$

(i) 328 g of Ca(NO₃)₂ produces 4(22.4) lit of nitrogen dioxide.

∴ 8.2 g of Ca(NO₃)₂ will produce $\dfrac{4 \times 22.4}{328}$ x 8.2 = 2.24 lit of nitrogen dioxide.

Hence, vol of nitrogen dioxide evolved = 2.24 lit.

(ii) 328 g of Ca(NO₃)₂ produces 112 g of calcium oxide.

∴ 8.2 g of Ca(NO₃)₂ will produce $\dfrac{112}{328}$ x 8.2 = 2.8 g of calcium oxide.

Hence, 2.8 g of calcium oxide is produced.

(a) Zinc and aluminium cannot be distinguished by heating the metal powder with concentrated sodium hydroxide solution as they both react with conc. alkalis to form soluble sodium salts and hydrogen gas.

Zinc reacts to form sodium zincate

Zn + 2NaOH ⟶ H₂ + Na₂ZnO₂ [sodium zincate]

Aluminium reacts to form sodium aluminate

2Al + 2NaOH + 2H₂O ⟶ 3H₂ + 2NaAlO₂ [sodium aluminate]

(b) Yes, calcium nitrate and lead nitrate can be distinguished using ammonium hydroxide solution. Ammonium hydroxide on reaction with lead nitrate gives chalky white precipitate of Pb(OH)₂. No precipitation occurs on adding Ammonium hydroxide to calcium nitrate even when it is added in excess.

Pb(NO₃)₂ + 2NH₄OH ⟶ Pb(OH)₂ + 2NH₄NO₃

Electron dot diagram showing the structure of Hydronium ion is given below:

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

(b) $\underset{\text{ethanol}}{\text{C}_2\text{H}_5\text{OH}} + \underset{\text{acetic acid}}{\text{CH}_3\text{COOH}} \xrightarrow[\Delta]{\text{Conc. H}_2\text{SO}_4} \underset{\text{ethyl acetate}}{\text{CH}_3-\text{COO}-\text{C}_2\text{H}_5} + \text{H}_2\text{O}$

(c) $\underset{[\text{conc.}]}{\text{KNO}_3} + \underset{[\text{conc.}]}{\text{H}_2\text{SO}_4} \xrightarrow{\lt 200 \degree\text{C}} \underset{[\text{acid salt}]}{\text{KHSO}_4} + \text{HNO}_3$

(a) Ammonia

(b) Sulphide [S^2-]

(c) Mainly ions.