Question 1

Question 1(iii)Which of the following would weigh the least?(Atomic masses C=12, O=16, Na=23)2 gram atoms of oxygenone mole of sodium22.4 litres of carbon dioxide at STP6.023 x 1022 atoms of carbon

Accepted Answer

6.023 x 1022 atoms of carbonReason — 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 110\dfrac{1}{10}101​ of a mole of carbon.The atomic mass of carbon is 12 g/mol.Therefore, 110\dfrac{1}{10}101​ mole of carbon has a mass of 1210\dfrac{12}{10}1012​ = 1.2 grams.Comparing the masses:2 gram atoms of oxygen = 32 gramsOne mole of sodium = 23 grams22.4 litres of carbon dioxide at STP = 44 grams6.023 x 1022 atoms of carbon = 1.2 gramsHence, out of the given options, 6.023 x 1022 atoms of carbon will weigh the least.

Question 2

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?ElectronsOnly positive ionsOnly negative ionsBoth positive and negative ions

Accepted Answer

Both positive and negative ionsReason — Sulphuric acid will dissociate into H+ and SO42-. Also, in aqueous solution, the compound XSO4 would dissociate into X+ and SO42-. In both cases, both positive and negative ions are responsible for conducting electricity.

Question 3

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 ABoth A and R are true but R is not the correct explanation of AA is true but R is falseA is false but R is true

Accepted Answer

A is true but R is falseReason — Dry hydrogen chloride gas is collected by the upward displacement of air as it is heavier than air.

Question 4

Question 1(vii)Element 'P' has electronic configuration 2,8,8,1. The number of chlorine atoms present in the chloride of 'P' is:2134

Accepted Answer

1Reason — 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 5

Question 1(viii)1H2 is an isotope of hydrogen. In the modern Periodic Table it will:be placed before hydrogenbe placed after hydrogenbe placed at the same position as hydrogennot have any position in the Periodic Table

Accepted Answer

be placed at the same position as hydrogenReason — 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 6

Question 1(ix)A nitrate which forms a precipitate with ammonium hydroxide and is also soluble in excess of it:ferrous nitrateferric nitratelead nitratecopper nitrate

Accepted Answer

copper nitrateReason — 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 7

Question 1(x)Which of the following electronic configuration represents the most electropositive element?2, 12, 8, 12, 22, 8, 2

Accepted Answer

2, 8, 1Reason — 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 8

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 : 11 : 2.21 : 1010 : 1

Accepted Answer

1 : 10Reason —Molar mass of CO2 = 12 + 2 x 16 = 44 g/mole∴ Moles of 4.4g CO2 = 0.1Molar mass of H2 = 2 x 1 = 2 g/mole∴ Moles of 2g H2 = 1Under 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 = Vol. of CO2Vol. of H2\dfrac{	ext{Vol. of CO}_2}{	ext{Vol. of H}_2}Vol. of H2​Vol. of CO2​​ = 0.11\dfrac{0.1}{1}10.1​ = 1 : 10

Question 9

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 chlorideaqueous sodium sulphatedilute sulphuric acidsodium hydroxide solution

Accepted Answer

sodium hydroxide solutionReason — 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 10

Question 1(xiv)Which of the following about oxides is correct?A basic oxide is an oxide of a non-metalAcidic oxides contain ionic bondsAmphoteric oxides contain a metalBasic oxides are always gases

Accepted Answer

Amphoteric oxides contain a metalReason — 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 11

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.The chemical process taking place on the surface of the object isCu2+(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.(c) Write down the reaction taking place at the positive electrode during step 3.

Accepted 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.(c) Ag (s) ⟶ Ag+(aq) + e-

Question 12

Question 2(v)(a) Draw the structural diagram for the following compounds:propanoic acidpentan-2-ol2,2 dibromo butane(b) Give the IUPAC name of the following organic compounds:

Accepted Answer

(a) Structural Diagram:1. Propanoic acid :2. Pentan-2-ol3. 2,2 dibromo butane(b) IUPAC names are:1-propanolpent-1-ene

Question 13

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.

Accepted 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 + CO2ORCa(HCO3)2 + 2HNO3 ⟶ Ca(NO3)2 + 2H2O + CO2

Question 14

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.

Accepted Answer

(a) Non-volatile natureReason — 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 propertyReason — The oxidising property of concentrated sulphuric acid is due to the fact that on thermal decomposition, it yields nascent oxygen which oxidises metals.

Question 15

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.

Accepted Answer

Gram atom = Mass of elementAtomic mass\dfrac{	ext{Mass of element}}{	ext{Atomic mass}}Atomic massMass of element​∴ g atom of M = 11.256\dfrac{11.2}{56}5611.2​ = 0.2and, g atom of N = 4.816\dfrac{4.8}{16}164.8​ = 0.3simple ratio of gram-atoms of M to gram-atoms of N= 0.20.3\dfrac{0.2}{0.3}0.30.2​ = 23\dfrac{2}{3}32​ = M : NSo, the formula is M2N3

Question 16

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?(c) Identify the salt given to Abhishek for analysis.

Accepted Answer

(a) Nitrogen dioxideReason — 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 17

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.

Accepted 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 18

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 + 4H2O2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

Accepted Answer

Given, 20 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 12 litres and butane is 8 litresC3H8+5O2⟶3CO2+4H2O1 vol.:5 vol.⟶3 vol.2C4H10+13O2⟶8CO2+10H2O2 vol.:13 vol.⟶8 vol.\begin{matrix} 	ext{C}_3	ext{H}_8 & + & 5	ext{O}_2 & \longrightarrow & 3	ext{CO}_2 & + & 4	ext{H}_2	ext{O} \ 1 	ext{ vol.} & : & 5 	ext{ vol.} & \longrightarrow & 3	ext{ vol.} \ 2	ext{C}_4	ext{H}_{10} & + & 13	ext{O}_2 & \longrightarrow & 8	ext{CO}_2 & + & 10	ext{H}_2	ext{O} \ 2 	ext{ vol.} & : & 13 	ext{ vol.} & \longrightarrow & 8	ext{ vol.} \ \end{matrix}C3​H8​1 vol.2C4​H10​2 vol.​+:+:​5O2​5 vol.13O2​13 vol.​⟶⟶⟶⟶​3CO2​3 vol.8CO2​8 vol.​++​4H2​O10H2​O​1 Vol. C3H8 produces carbon dioxide = 3 VolSo, 12 litres C3H8 will produce carbon dioxide = 3 x 12 = 36 litres2 Vol. C4H10 produces carbon dioxide = 8 VolSo, 8 litres C4H10 will produce carbon dioxide = 82\dfrac{8}{2}28​ x 8 = 32 litresHence, 68 (i.e., 36 + 32) litres of CO2 is produced.

Question 19

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(c) will not have any effect on litmus paper?

Accepted Answer

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

Question 20

Question 7(iii)8.2 grams of calcium nitrate is decomposed by heating according to the equation2Ca (NO3)2 ⟶ 2CaO +4NO2 + O2Calculate the following:(a) Volume of nitrogen dioxide obtained at STP(b) Mass of CaO formed(Atomic weights : Ca- 40, N-14, O-16)

Accepted Answer

2Ca(NO3)2⟶2CaO+4NO2+O22[(40)+2(14)+6(16)]2(40+16)4(14)+8(16)1mole=328g=112g=184=32g\begin{matrix} 	ext{2Ca}(	ext{NO}_3)_2 & \longrightarrow & 2	ext{CaO}& +&4	ext{NO}_2 & +& 	ext{O}_2 \ 2[(40) + 2(14) +6(16)] & & 2(40 + 16) & & 4(14)+8(16) & & 1 	ext{mole} \ = 328 	ext{g} & & = 112	ext{g} & & =184 & & =32 	ext{g} \ \end{matrix}2Ca(NO3​)2​2[(40)+2(14)+6(16)]=328g​⟶​2CaO2(40+16)=112g​+​4NO2​4(14)+8(16)=184​+​O2​1mole=32g​(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 8.2328\dfrac{8.2}{328}3288.2​ = 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 8.2328\dfrac{8.2}{328}3288.2​ = 2.8 g of CaO.

Solved 2025 Specimen Paper ICSE Class 10 Chemistry

Section A 40 Marks

Section B 40 Marks