Q: Question 3(iv)Write one equation in each case to show the action of sulphuric acid on:(a) Calcium bicarbonate(b) Sugar(c) Sulphur

(a) Ca(HCO3)2 + H2SO4 ⟶ CaSO4 ↓ + 2H2O + 2CO2 ↑(b) C12H22O11Cane Sugar (s)→Conc. H2SO412C (s)Sugar Charcoal+11H2O\underset{\text{Cane Sugar}}{\text{C}{12}\text{H}{22}\text{O}_{11}} \text{ (s)} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \underset{\text{Sugar Charcoal}}{12\text{C (s)}} + 11\text{H}_2\text{O}Cane SugarC12H22O11 (s)Conc. H2SO4Sugar Charcoal12C (s)+11H2O(c) S + 2H2SO4 ⟶ 3SO2 + 2H2O

Question 1

Question 1(ii)The observation when ammonium salt reacts with caustic soda:A reddish brown gasA colourless gas which gives dense white fumes with conc. HCIA green coloured gas which turns moist blue litmus paper red.A colourless gas which turns lime water milky.

Accepted Answer

A colourless gas which gives dense white fumes with conc. HCIReason — When caustic soda (sodium hydroxide) is heated with an ammonium salt, ammonia gas (NH3) is evolved. Ammonia is colourless, has a pungent smell, turns damp red litmus blue, and produces dense white fumes of ammonium chloride (NH4Cl) when it comes in contact with concentrated hydrochloric acid.

Question 2

Question 1(iii)A compound with Empirical formula XY2, has vapour density equal to its empirical formula weight, its molecular formula is:XY2X3Y6X2Y4X6H12

Accepted Answer

X2Y4Reason — Vapour density = Empirical formula weight,∴ Molecular weight = 2 × Vapour density = 2 × (Empirical formula weight)Therefore,Molecular formula = 2 × (Empirical formula)So, the correct molecular formula should be X2Y4

Question 3

Question 1(v)Assertion (A): Addition reaction is a characteristic property of alkenesReason (R): Alkenes are double bond compounds.Both A and R are true and R is the correct explanation of A.Both A and R are true but R is not the correct explanation of A.A is true but R is false.A is false but R is true.

Accepted Answer

Both A and R are true and R is the correct explanation of A.Reason — Alkenes undergo addition reactions due to the presence of a reactive double bond (C=C). Hence, the assertion (A) is true. Alkenes are compounds with at least one carbon-carbon double bond. Addition reaction in alkenes is due to the presence of double bond. Hence, the reason (R) is true and it is the correct explanation of (A).

Question 4

Question 1(viii)The catalyst preferred in Haber's process is:Finely divided ironGraphiteVanadium pentoxidePlatinum

Accepted Answer

Finely divided ironReason — The speed of the reaction is improved by using a catalyst, which is finely divided iron, obtained by the reduction of iron oxide. It provides a surface for the reaction to occur and lowers the activation energy, allowing the reaction to proceed faster at given temperature and pressure

Question 5

Question 3(i)(a) State the type of bonding of the oxide of the element 'A' having electronic configuration 2, 8, 8. 1.(b) Write the formula of the oxide of A.

Accepted Answer

(a) Element A has electronic configuration of 2, 8, 8, 1, its atomic number is 19. Thus, element A will be potassium (K), and potassium is a metal, whereas oxygen is a non-metal, so the bonding in its oxide will be ionic bond.(b) A has valency of = 1Oxygen (O) has valency = 2So, two A atoms combine with one O atom to form A2O.

Question 6

Question 3(iv)Write one equation in each case to show the action of sulphuric acid on:(a) Calcium bicarbonate(b) Sugar(c) Sulphur

Accepted Answer

(a) Ca(HCO3)2 + H2SO4 ⟶ CaSO4 ↓ + 2H2O + 2CO2 ↑(b) C12H22O11Cane Sugar (s)→Conc. H2SO412C (s)Sugar Charcoal+11H2O\underset{ ext{Cane Sugar}}{ ext{C}{12} ext{H}{22} ext{O}_{11}} ext{ (s)} \xrightarrow{ ext{Conc. H}_2 ext{SO}_4} \underset{ ext{Sugar Charcoal}}{12 ext{C (s)}} + 11 ext{H}_2 ext{O}Cane SugarC12H22O11 (s)Conc. H2SO4Sugar Charcoal12C (s)+11H2O(c) S + 2H2SO4 ⟶ 3SO2 + 2H2O

Question 7

Question 4(ii)If 6.21 g of lead is combined with 4.26 g of chlorine, what is the empirical formula of the compound formed?(At. wt. Lead = 207, Chlorine = 35.5).

Accepted Answer

ElementMass (g)Atomic massMolesSimplest ratioPb6.212076.21207\dfrac{6.21}{207}2076.21​ = 0.030.030.03\dfrac{0.03}{0.03}0.030.03​ = 1Cl4.2635.54.2635.5\dfrac{4.26}{35.5 }35.54.26​ = 0.120.120.03\dfrac{0.12}{0.03}0.030.12​ = 4Simplest ratio of whole numbers Pb : Cl = 1 : 4Hence, empirical formula is PbCl4

Question 8

Question 4(iii)Give a balanced equation for the following reactions:(a) Carbon with nitric acid.(b) Lead nitrate and hydrochloric acid.(c) Conversion of alcohol to ester.

Accepted Answer

(a) C + 4HNO3 ⟶ CO2 + 2H2O + 4NO2(b) Pb(NO3)2 + 2HCl (dil.) ⟶ PbCl2+ 2HNO3(c) C2H5OHethanol+CH3COOHacetic acid→ΔConc. H2SO4CH3−COO−C2H5ethyl acetate+H2O\underset{	ext{ethanol}}{	ext{C}_2	ext{H}_5	ext{OH}} + \underset{	ext{acetic acid}}{	ext{CH}_3	ext{COOH}} \xrightarrow[\Delta]{	ext{Conc. H}_2	ext{SO}_4} \underset{	ext{ethyl acetate}}{	ext{CH}_3-	ext{COO}-	ext{C}_2	ext{H}_5} + 	ext{H}_2	ext{O}ethanolC2​H5​OH​+acetic acidCH3​COOH​Conc. H2​SO4​Δ​ethyl acetateCH3​−COO−C2​H5​​+H2​O

Question 9

Question 5(i)An organic compound X has the following composition: O = 71.19% H= 2.22 %[At. mass C = 12 H = 1 O = 16 ](a) Find its empirical formula.(b) If in the gaseous state, its vapour density is 45. Find its molecular formula.

Accepted Answer

(a) Element% compositionAt. wt.Relative no. of atomsSimplest ratioHydrogen2.2212.221\dfrac{2.22}{1}12.22​ = 2.222.222.21\dfrac{2.22 }{2.21}2.212.22​ = 1Oxygen71.191671.1916\dfrac{71.19}{16}1671.19​ = 4.444.442.21\dfrac{4.44}{2.21 }2.214.44​ = 2Carbon26.591226.5912\dfrac{26.59}{12}1226.59​ = 2.212.212.21\dfrac{ 2.21 }{2.21}2.212.21​ = 1Simplest ratio of whole numbers = H : O : C = 1 : 2 : 1Hence, empirical formula is CHO2(b) Empirical formula weight = 12 + 1 + (2 x 16) = 13 + 32 = 45V.D. = 45Molecular weight = 2 x V.D. = 2 x 45 = 90n=Molecular weightEmpirical formula weight=9045=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{90}{45} = 2n=Empirical formula weightMolecular weight​=4590​=2So, molecular formula = 2[E.F.] = 2(CHO2) = C2H2O4

Question 10

Question 5(ii)Give the laboratory preparation of:(a) Ammonia(b) Nitric acid

Accepted Answer

(a)Ammonium chloride (NH4Cl) and calcium hydroxide (Ca(OH)2) are mixed together. (Excess of calcium hydroxide is mixed well with ammonium chloride).Reactants are finely grounded and taken in a round-bottom flask fitted in a slanting position, mouth downwards.On heating the mixture, ammonia gas is evolved.2NH4Cl + Ca(OH)2 ⟶ CaCl2 + 2H2O + 2NH3Ammonia gas is collected in inverted gas jars by the downward displacement of air because it is lighter than air and highly soluble in water and therefore, it cannot be collected over water.Quicklime (CaO) is used as a drying agent.(b)Nitric acid is obtained by distilling conc. sulphuric acid with potassium nitrate KNO3 (nitre) or sodium nitrate NaNO3 (Chile saltpetre).A mixture of equal parts, by weight, of potassium/sodium nitrate and concentrated sulphuric acid is heated gently to 180°C-200°C in a glass retort.Sulphuric acid is a non-volatile acid and produces volatile nitric acid on reacting with potassium nitrate or sodium nitrate.KNO3[conc.]+H2SO4[conc.]→<200°CKHSO4[acid salt]+HNO3\underset{[conc.]}{ ext{KNO}_3} + \underset{[conc.]}{ ext{H}_2 ext{SO}_4} \xrightarrow{\lt 200 \degree ext{C}} \underset{[acid \space salt]}{ ext{KHSO}_4} + ext{HNO}_3[conc.]KNO3+[conc.]H2SO4<200°C[acid salt]KHSO4+HNO3NaNO3[conc.]+H2SO4[conc.]→>200°CNaHSO4[acid salt]+HNO3\underset{[conc.]}{ ext{NaNO}_3} + \underset{[conc.]}{ ext{H}_2 ext{SO}_4} \xrightarrow{\gt 200 \degree ext{C}} \underset{[acid \space salt]}{ ext{NaHSO}_4} + ext{HNO}_3[conc.]NaNO3+[conc.]H2SO4>200°C[acid salt]NaHSO4+HNO3The vapours of nitric acid are condensed to a light yellow liquid by chilling the receiver with running cold water.

Question 11

Question 5(iii)Distinguish the following by a chemical test:(a) Ammonia solution and hydrogen chloride gas solution.(b) Saturated and unsaturated compounds.(c) Ferrous chloride and ferric chloride.

Accepted Answer

(a) Ammonia solution turns moist red litmus blue. Whereas, hydrogen chloride gas turns moist blue litmus red.(b) On adding bromine water, unsaturated compound decolourises bromine water due to addition reaction across the double or triple bond. However, saturated compound shows no reaction. So, there is no change in colour.(c) Add a few drops of aqueous NaOH:Ferrous chloride gives a dirty-green precipitate of ferrous hydroxide, Fe(OH)2, which soon turns brown on standing.Ferric chloride gives a reddish-brown precipitate of ferric hydroxide, Fe(OH)3.

Question 12

Question 5(iv)Select the correct property of sulphuric acid given in column A which relates to the respective conversions given in column B.Column AColumn(A)Dehydrating nature1. Na2SO3 + H2SO4 ⟶ Na2SO4 + H2O + SO2(B)Non- volatile acid2. S + 2H2SO4 ⟶ 3SO2 + 2H2O(C)Acidic nature3. CH3CH2OH + H2SO4 (Conc.) ⟶ H2C = CH2(D)Oxidising agent

Accepted Answer

Column AColumn(A)Dehydrating nature3. CH3CH2OH + H2SO4 (Conc.) ⟶ H2C = CH2(B)Non- volatile acid (C)Acidic nature1. Na2SO3 + H2SO4 ⟶ Na2SO4 + H2O + SO2(D)Oxidising agent2. S + 2H2SO4 ⟶ 3SO2 + 2H2O

Question 13

Question 6(i)Give balanced chemical equation for the following conversions :(a) Copper sulphate to inky blue solution(b) Methane to chloroform

Accepted Answer

Cu(OH)2+(NH4)2SO4+2NH4OH⟶[(Cu(NH3)4]SO4Tetraamminecopper (II) sulphate+4H2O	ext{Cu(OH)}_2 + (	ext{NH}_4)_2	ext{SO}_4 + 2	ext{NH}_4	ext{OH} \longrightarrow \underset{\underset{	ext{copper (II) sulphate}}{	ext{Tetraammine}}}{[(	ext{Cu(NH}_3)_4]	ext{SO}_4} + 4	ext{H}_2	ext{O}Cu(OH)2​+(NH4​)2​SO4​+2NH4​OH⟶copper (II) sulphateTetraammine​[(Cu(NH3​)4​]SO4​​+4H2​O(b)CH4 methane+Cl2→diffused sunlightΔCH3Clmonochloromethane+HCl\underset{	ext{ methane} }{	ext{CH}_4} + 	ext{Cl}_2 \underset{\Delta}{\xrightarrow{	ext{diffused sunlight}}} \underset{	ext{monochloromethane}}{	ext{CH}_3	ext{Cl}} + 	ext{HCl} methaneCH4​​+Cl2​Δdiffused sunlight​​monochloromethaneCH3​Cl​+HClCH3Clmonochloromethane+Cl2[excess]⟶CH2Cl2dichloromethane+HCl\underset{	ext{monochloromethane}}{	ext{CH}_3	ext{Cl}} + \underset{	ext{[excess]}}{	ext{Cl}_2} \longrightarrow \underset{	ext{dichloromethane}}{	ext{CH}_2	ext{Cl}_2} + 	ext{HCl}monochloromethaneCH3​Cl​+[excess]Cl2​​⟶dichloromethaneCH2​Cl2​​+HClCH2Cl2dichloromethane+Cl2⟶CHCl3trichloromethane or chloroform+HCl\underset{	ext{dichloromethane}}{	ext{CH}_2	ext{Cl}_2} + 	ext{Cl}_2 \longrightarrow \underset{	ext{trichloromethane or chloroform}}{	ext{CHCl}_3}+ 	ext{HCl}dichloromethaneCH2​Cl2​​+Cl2​⟶trichloromethane or chloroformCHCl3​​+HCl

Question 14

Question 6(ii)Look at the figure given alongside and answer the following questions ;(a) Explain why is a freshly prepared saturated solution of FeSO4 preferred in the above diagram.(b) Why is the ring formed at the junction of two liquids ?

Accepted Answer

(a) A freshly prepared ferrous sulphate solution is used, because on exposure to the atmosphere, it is oxidised to ferric sulphate, which will not give the brown ring test.(b) The brown ring of nitroso ferrous sulphate is formed at the junction of the two liquids because the conc. sulphuric acid being heavier settles down and the ferrous sulphate layer remains above it resulting in the formation of a brown ring at the junction.

Question 15

Question 6(iii)Give the structural formula of:(a) Acetic acid(b) Acetaldehyde(c) 1, 2 dichloroethane

Accepted Answer

(a) Acetic acid(b) Acetaldehyde(c)

Question 16

Question 6(iv)State your observations when:(a) Ethylene gas is passed through bromine water(b) Silver nitrate solution is added to hydrochloric acid.(c) Barium chloride is added to potassium sulphate solution.

Accepted Answer

(a) When ethylene gas is passed through bromine water, the reddish brown colour of bromine water disappears due to the formation of the colourless dibromoethane.C2H4 ethene+Br2→[inert solvent]CCl4C2H4Br21,2, dibromooethane [ethylene bromide]\underset{	ext{ ethene}}{	ext{C}_2	ext{H}_4} + 	ext{Br}_2 \xrightarrow[	ext{[inert solvent]}]{	ext{CCl}_4} \underset{	ext{1,2, dibromooethane [ethylene bromide]}}{	ext{C}_2	ext{H}_4	ext{Br}_2} etheneC2​H4​​+Br2​CCl4​[inert solvent]​1,2, dibromooethane [ethylene bromide]C2​H4​Br2​​(b) When silver nitrate reacts with Hydrochloric acid it gives a white precipitate of AgCl which is insoluble in nitric acid but soluble in ammonium hydroxide.AgNO3 + HCl ⟶ AgCl ↓ + HNO3(c) Barium chloride reacts with potassium sulphate to form barium sulphate (a white insoluble solid) and potassium chloride.BaCl2 + K2SO4 ⟶ BaSO4 + 2KCl

Question 17

Question 7(i)40 cm3 of methane reacts with chlorine. The reaction for the same isCH4 + 2Cl2 ⟶ CH2Cl2 + 2HCIFind(a) The volume of chlorine gas used.(b) The volume of hydrogen chloride gas formed.(c) How is hydrogen chloride gas dried?

Accepted Answer

CH4+2Cl2⟶CH2Cl2+2HCl1 vol.:2 vol.⟶1 vol.:2 vol.\begin{matrix} 	ext{CH}_4 & + & 2	ext{Cl}_2 & \longrightarrow & 	ext{CH}_2	ext{Cl}_2 & + & 2	ext{HCl} \ 1 	ext{ vol.} & : & 2 	ext{ vol.} & \longrightarrow & 1 	ext{ vol.} & : & 2 	ext{ vol.}\ \end{matrix}CH4​1 vol.​+:​2Cl2​2 vol.​⟶⟶​CH2​Cl2​1 vol.​+:​2HCl2 vol.​(a) Volume of chlorine gas used = ?For 1 Vol of methane = 2V of Cl2 required∴ For 40 cm3 of methane = 40 x 2 = 80 cm3 of Cl2 is required.(b) Volume of HCl gas formed = ?[By Gay Lussac's law]1 Vol of methane produces = 2 Vol. HCl∴ 40 cm3 of methane produces = 80 cm3 HClHence, volume of HCl gas formed = 80 cm3 and chlorine gas required = 80 cm3(c) Hydrogen chloride gas is dried by passing through a washer bottle containing concentrated sulphuric acid.

Question 18

Question 7(ii)Give reasons for the following:(a) Electronegativity increases across a Period from left to right in the Modern periodic table(b) Aluminium and zinc powder cannot be distinguished by strong alkalis.(c) Water is not added to sulphuric acid to dilute it.

Accepted Answer

(a) Across a period the nuclear (effective) charge on the atoms increases while the number of electron shells remains the same. As a result, atomic radius decreases and the attraction of the nucleus for the bonding electrons becomes stronger, so electronegativity increases from left to right.(b) When treated with a strong alkali (like sodium hydroxide, NaOH), both aluminium and zinc powder dissolve and form soluble complex salts with the evolution of hydrogen gas. As both metals behave in the same way, the alkali test cannot distinguish between them.(c) Water is never poured on acid to dilute it as large amount of heat is evolved which changes poured water to steam. The steam so formed causes spurting of acid which can cause burn injuries.

Question 19

Question 7(iii)(a) Name a drying agent for ammonia.(b) How is hydrochloric acid prepared from HCl gas?(c) Give the equations occurring at anode in the extraction of Aluminium.(d) In the laboratory preparation, what must be added to sodium chloride to obtain HCI gas ?

Accepted Answer

(a) Quicklime (CaO)(b) Hydrochloric acid is prepared by dissolving hydrogen chloride gas in water using a special funnel arrangement in which an inverted funnel, connected to the hydrogen chloride gas supply is placed in the beaker in such a way that it just touches the water taken in the trough. This arrangement minimizes back suction.(c) Reaction at anode:6O2- - 12e- ⟶ 6[O]3[O] + 3[O] ⟶ 3O2Anode is oxidised to carbon monoxide which further forms carbon dioxide2C + O2 ⟶ 2CO2CO + O2 ⟶ 2CO2(d) In the laboratory preparation of HCl, concentrated sulphuric acid (H2SO4) must be added to sodium chloride (NaCl) to obtain HCl gas.NaCl+H2SO4→<200°CNaHSO4+HCl [g] ext{NaCl} + ext{H}_2 ext{SO}_4 \xrightarrow{\lt 200 \degree ext{C}} ext{NaHSO}_4 + ext{HCl [g]}NaCl+H2SO4<200°CNaHSO4+HCl [g] ↑2NaCl+H2SO4→>200°CNa2SO4+2HCl [g] ext{2NaCl} + ext{H}_2 ext{SO}_4 \xrightarrow{\gt 200 \degree ext{C}} ext{Na}_2 ext{SO}_4 + ext{2HCl [g}]2NaCl+H2SO4>200°CNa2SO4+2HCl [g] ↑

Question 20

Question 8(i)Define Isomerism'. Give an example (with IUPAC. name) showing Position isomerism.

Accepted Answer

Isomerism is the phenomenon due to which two or more compounds have the same molecular formula but differ in molecular arrangement or in structural formula.Position isomers — When two or more compounds with the same molecular formula differ in the position of substituent atom or functional group on the carbon atom, they are called position isomers.Example : But-1-yne and But-2-yneBut-1-yneBut-2-yne

Model Paper 3

Section A 40 Marks

Section B 40 Marks