Q: Question 2(ii)Match Column A with Column B.Column AColumn B(a)Acid Salt(i)Black in colour(b)Manganese dioxide(ii)Brown ppt.(c)Lead hydroxide(iii)Hydrogen chloride(d)Ferric hydroxide(iv)Calcium hydrogen carbonate(e)Polar compound(v)Soluble in excess sodium hydroxide

Column AColumn B(a)Acid Salt(iv)Calcium hydrogen carbonate(b)Manganese dioxide(i)Black in colour(c)Lead hydroxide(v)Soluble in excess sodium hydroxide(d)Ferric hydroxide(ii)Brown ppt.(e)Polar compound(iii)Hydrogen chloride

Q: Question 2(iii)Complete the following by choosing the correct answers from the bracket:(a) HCl in the liquefied form is ............... [neutral / acidic](b) Organic compounds are generally soluble in ...............[Water / Organic solvents](c) An inert electrode used in electrolysis of copper sulphate solution is ............... [Copper/platinum](d) Hydrocarbons having triple bond is ............... [alkenes/alkynes](e) An acidic gas gives dense white fumes of ............... [NH4OH/NH4C1] with ammonia.

(a) HCl in the liquefied form is neutral(b) Organic compounds are generally soluble in Organic solvents.(c) An inert electrode used in electrolysis of copper sulphate solution is platinum.(d) Hydrocarbons having triple bond is alkynes.(e) An acidic gas gives dense white fumes of NH4C1 with ammonia.

Q: Question 2(v)(a) Draw the structure formula for the following :2-pentanal2-methyl butanol1-butyne(b) Name the following organic compound in IUPAC system :

(a)2-pentanal2-methyl butanol1-butyne(b)Butanoic acidBut-2-yne

Question 1

Question 1(i)The product formed at the cathode in electroplating of an article with Silver is:Hydrogen gasSilver ionsSilver atomsOxygen gas

Accepted Answer

Silver atomsReason — Ag+ ions from the electrolyte gain electrons (reduction) and are deposited as silver atoms on the article.Ag+ + e- ⟶ Ag

Question 2

Question 1(ii)Ionisation potential is maximum in:Alkaline earth metalsHalogensInert gasesAlkali metals

Accepted Answer

Inert gasesReason — Ionisation potential (or ionisation energy) is the energy required to remove the outermost electron from an atom in the gaseous state. It generally increases from left to right across a period (because effective nuclear charge increases) and decreases down a group (because atomic size and shielding increase).Noble gases have completely filled valence shells and very small atomic radii within their periods; therefore, much more energy is needed to remove an electron from them, giving them the maximum ionisation energy among the groups listed.

Question 3

Question 1(iii)If the RMM of carbon monoxide is 28, then its vapour density is:7561488

Accepted Answer

14Reason —Given,The relative molecular mass(RMM) of carbon monoxide = 28.The vapour density of carbon monoxide = ?Vapour density =  Relative molecular mass2\dfrac{	ext{ Relative molecular mass}}{2}2 Relative molecular mass​Vapour density = 282\dfrac{28}{2}228​ = 14Hence, the vapour density of carbon monoxide is 14.

Question 4

Question 1(viii)The drying agent used to dry Ammonia is:Concentrated Sulphuric acidCalcium oxideSulphurous acidCalcium hydroxide

Accepted Answer

Calcium oxideReason — The drying agent used to dry Ammonia is quicklime or calcium oxide (CaO). Other drying agents like conc. sulphuric acid and sulphurous acid are not used, as ammonia being basic, reacts with them. Calcium hydroxide (slaked lime) is slightly basic, but less effective than calcium oxide in drying. It may also partially react with ammonia.

Question 5

Question 1(ix)The percentage of nitrogen present in urea (NH2)2CO is: [R.A.M. of N = 14, C= 12, O = 16, H = 1]23.3646.6719.3543.87

Accepted Answer

46.67Reason —Molar mass of urea (CON2H4) = 12 + 16 + 28 + 4 = 60 gMolar mass of nitrogen (N2) = 2 x 14 = 28 g60 g urea has mass of nitrogen = 28 g∴ 100 g urea will have mass= 28×10060\dfrac{28 	imes 100 }{60}6028×100​= 46.67%

Question 6

Question 1(xii)The organic compound prepared when calcium carbide reacts with water:MethaneEthaneAcetyleneEthene

Accepted Answer

AcetyleneReason — In the laboratory preparation of Acetylene, calcium carbide reacts with water to give acetylene and calcium hydroxide.CaC2calcium carbide+2H2Owater⟶C2H2Acetylene+Ca(OH)2calcium hydroxide\underset{	ext{calcium carbide}}{	ext{CaC}_2} + \underset{	ext{water}}{2	ext{H}_2	ext{O}} \longrightarrow \underset{	ext{Acetylene}}{	ext{C}_2	ext{H}_2} + \underset{	ext{calcium hydroxide}}{	ext{Ca(OH)}_2}calcium carbideCaC2​​+water2H2​O​⟶AcetyleneC2​H2​​+calcium hydroxideCa(OH)2​​

Question 7

Question 2(i)The figure given below illustrates the apparatus used in the laboratory preparation of nitric acid.(a) Name A (a liquid), B (a solid) and C (a liquid). (Do not give the formulae).(b) Write a balanced chemical equation for the above preparation.(c) Why is an all glass apparatus used ?(d) The acid prepared is yellow in colour. Why?(e) How is this colour removed?

Accepted Answer

(a)A — Concentrated sulphuric acid,B — Sodium nitrate or potassium nitrateC — Nitric acid.(b)KNO3[conc.]+H2SO4[conc.]→<200°CKHSO4[acid salt]+HNO3\underset{[	ext{conc.}]}{	ext{KNO}_3} + \underset{[	ext{conc.}]}{	ext{H}_2	ext{SO}_4} \xrightarrow{\lt 200 \degree	ext{C}} \underset{[	ext{acid salt}]}{	ext{KHSO}_4} + 	ext{HNO}_3[conc.]KNO3​​+[conc.]H2​SO4​​<200°C​[acid salt]KHSO4​​+HNO3​(c) All glass apparatus is used in the laboratory preparation of nitric acid since the vapours of nitric acid being highly corrosive attack rubber, cork, etc.(d) Pure nitric acid [HNO3] is colourless and unstable and decomposes slightly even at ordinary temperatures and in the presence of sunlight. The decomposition results in formation of reddish brown nitrogen dioxide [NO2] which remains dissolved in the acid thus imparting a slight yellowish brown colour.(e) If dry air or CO2 is bubbled through the yellow acid, the latter turns colourless because it drives out NO2 from warm acid which is further oxidised to nitric acid.

Question 8

Question 2(ii)Match Column A with Column B.Column AColumn B(a)Acid Salt(i)Black in colour(b)Manganese dioxide(ii)Brown ppt.(c)Lead hydroxide(iii)Hydrogen chloride(d)Ferric hydroxide(iv)Calcium hydrogen carbonate(e)Polar compound(v)Soluble in excess sodium hydroxide

Accepted Answer

Column AColumn B(a)Acid Salt(iv)Calcium hydrogen carbonate(b)Manganese dioxide(i)Black in colour(c)Lead hydroxide(v)Soluble in excess sodium hydroxide(d)Ferric hydroxide(ii)Brown ppt.(e)Polar compound(iii)Hydrogen chloride

Question 9

Question 2(iii)Complete the following by choosing the correct answers from the bracket:(a) HCl in the liquefied form is ............... [neutral / acidic](b) Organic compounds are generally soluble in ...............[Water / Organic solvents](c) An inert electrode used in electrolysis of copper sulphate solution is ............... [Copper/platinum](d) Hydrocarbons having triple bond is ............... [alkenes/alkynes](e) An acidic gas gives dense white fumes of ............... [NH4OH/NH4C1] with ammonia.

Accepted Answer

(a) HCl in the liquefied form is neutral(b) Organic compounds are generally soluble in Organic solvents.(c) An inert electrode used in electrolysis of copper sulphate solution is platinum.(d) Hydrocarbons having triple bond is alkynes.(e) An acidic gas gives dense white fumes of NH4C1 with ammonia.

Question 10

Question 2(v)(a) Draw the structure formula for the following :2-pentanal2-methyl butanol1-butyne(b) Name the following organic compound in IUPAC system :

Accepted Answer

(a)2-pentanal2-methyl butanol1-butyne(b)Butanoic acidBut-2-yne

Question 11

Question 3(iv)Give balanced chemical equations for the following:(a) Laboratory Preparation of Ethylene.(b) Preparation of Ethanol by hydrolysis of alkyl halide.(c) Ethene reacting with bromine.

Accepted Answer

(a) C2H5OH ethyl alcohol→170°CConc. H2SO4[excess]C2H4ethylene+H2O \underset{	ext{ ethyl alcohol}}{	ext{C}_2	ext{H}_5	ext{OH}} \xrightarrow[170\degree	ext{C}]{	ext{Conc. H}_2	ext{SO}_4	ext{[excess]}} \underset{ 	ext{ethylene}}{	ext{C}_2	ext{H}_4} + 	ext{H}_2	ext{O}\ ethyl alcoholC2​H5​OH​Conc. H2​SO4​[excess]170°C​ethyleneC2​H4​​+H2​O (b)C2H5Br Bromoethane [ethyl bromide]+NaOH aqueous→boilC2H5OHethanol+NaBr\underset{	ext{ Bromoethane [ethyl bromide]}}{	ext{C}_2	ext{H}_5	ext{Br}} + \underset{	ext{ aqueous}}{	ext{NaOH}} \xrightarrow{	ext{boil}} \underset{ 	ext{ethanol}} {	ext{C}_2	ext{H}_5	ext{OH}} + 	ext{NaBr} Bromoethane [ethyl bromide]C2​H5​Br​+ aqueousNaOH​boil​ethanolC2​H5​OH​+NaBr(c) 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​​

Question 12

Question 4(i)What is the role played by the following substances in the extraction of Aluminium ?(a) Caustic soda(b) Fluorspar

Accepted Answer

(a) Caustic soda is used in the Bayer purification of bauxite. Bauxite is heated under pressure with conc. caustic soda (NaOH) for 2 to 8 hours. Hot, concentrated NaOH dissolves the amphoteric alumina present in the ore, forming soluble sodium aluminate:Al2O3.2H2O [Impure bauxite]Amphoteric oxide+2NaOH [conc. soln]Base→150−200°C2NaAlO2[sodium aluminate]Salt+3H2OWater\underset{	ext{Amphoteric oxide}}{\underset{	ext{[Impure bauxite]}}{	ext{Al}_2	ext{O}_3.2	ext{H}_2	ext{O }}} + \underset{	ext{Base}}{\underset{	ext{[conc. soln]}}{2	ext{NaOH }}} \xrightarrow{150 - 200 \degree	ext{C}} \underset{	ext{Salt}}{\underset{	ext{[sodium aluminate]}}{2	ext{NaAlO}_2}} + \underset{	ext{Water}}{ 	ext{3H}_2	ext{O}}Amphoteric oxide[Impure bauxite]Al2​O3​.2H2​O ​​+Base[conc. soln]2NaOH ​​150−200°C​Salt[sodium aluminate]2NaAlO2​​​+Water3H2​O​The insoluble impurities mainly Fe2O3 and SiO2 are left behind and are referred to as Red mud.(b) Fluorspar is added to the electrolyte mixture as it lowers the fusion point of the mixture i.e., the mixture fuses around 950°C instead of 2050°C.

Question 13

Question 4(ii)Calculate:A gas cylinder is filled with hydrogen and it holds 50 g of gas. The same cylinder holds 200 g of gas X and 500 g of gas Y under same temperature and pressure conditions. Calculate the vapour density of gas X and molecular mass of gas Y.

Accepted Answer

Given,At same temperature and pressure Mass of hydrogen = 50 gMass of gas X = 200 gMass of gas Y = 500 gV.D. of gas X = mass of certain volume of gas X mass of equal volume of H2\dfrac{	ext{mass of certain volume of gas X }}{	ext{mass of equal volume of H}_2}mass of equal volume of H2​mass of certain volume of gas X ​ = 20050\dfrac{200}{50}50200​ = 4V.D. of gas Y = mass of certain volume of gas Ymass of equal volume of H2\dfrac{	ext{mass of certain volume of gas Y}}{	ext{mass of equal volume of H}_2}mass of equal volume of H2​mass of certain volume of gas Y​ = 50050\dfrac{500}{50}50500​ = 10Molecular weight = 2 x V.D. = 2 x 10 = 20 a.m.u.∴ The vapour density of gas X is 4 and Molecular mass of gas Y is 20 a.m.u.

Question 14

Question 4(iii)Explain the following :(a) HCl gas dissolved in toluene does not affect litmus.(b) An inverted funnel is used to dissolve Ammonia gas in water.(c) A bottle of liquor ammonia should be opened very carefully.

Accepted Answer

(a) HCl gas is covalent in nature and does not ionise in non-polar solvents like toluene. Since, there is no ionisation of HCl, so no hydronium (H3O+) ions are produced. As, acidity depends on the presence of these ions, neither blue nor red litmus shows any colour change; the solution is neutral to litmus.(b) The mouth of the inverted funnel provides a larger surface area over which ammonia gas can come into contact with water, so the gas dissolves rapidly. At the same time, keeping the funnel rim just below the water surface allows air to enter if the pressure inside the apparatus falls, preventing water from being sucked back into the flask that contains the gas.(c) Liquor ammonia (aqueous NH3) is stored under considerable internal pressure owing to the high volatility of ammonia. The bottle is first cooled in ice-cold water. The cooling lowers the vapour pressure, so when the stopper is removed the contents do not spurt out and there is no sudden escape of ammonia gas.

Question 15

Question 5(ii)Prepare lead sulphate from lead chloride.

Accepted Answer

Preparation of lead sulphate from lead chloride:Dissolve lead chloride (PbCl2) in hot water as the salt is only sparingly soluble in cold water but dissolves better when heated.While the solution is still hot, slowly add a dilute solution of sulphuric acid with constant stirring.PbCl2+H2SO4[aq]⟶PbSO4↓+2HCl	ext{PbCl}_2 + 	ext{H}_2	ext{SO}_4 [	ext{aq}] \longrightarrow	ext{PbSO}_4 \downarrow + 2	ext{HCl}PbCl2​+H2​SO4​[aq]⟶PbSO4​↓+2HClAllow the mixture to cool, then filter to collect the precipitate.Wash the precipitate with distilled water and finally dry it.

Question 16

Question 5(iii)What do you see in each of the following reactions:(a) When excess Ammonia gas is passed through an aqueous solution of Zinc Nitrate.(b) Copper Sulphate crystals are heated strongly.(c) Sodium hydroxide is added to Copper Sulphate solution in excess.

Accepted Answer

(a) When Ammonia gas is passed through an aqueous solution of Zinc Nitrate:A white, gelatinous precipitate of zinc hydroxide, Zn(OH)2, first appears.Zn(NO3)2 + 2NH4OH ⟶ 2NH4NO3 + Zn(OH)2 ↓On passing excess ammonia, the precipitate dissolves and the solution becomes clear and colourless owing to the formation of the soluble tetra-amminezinc(II) complex, [Zn(NH3)4]2+.(b) The blue crystals of hydrated copper sulphate (CuSO4·5H2O) lose their water of crystallisation, giving off steam and turning into a white powder of anhydrous CuSO4:CuSO4.5H2O(blue crystal)→HeatCuSO4(white powder)+5H2O\underset{	ext{(blue crystal)}}{	ext{CuSO}_4.5	ext{H}_2	ext{O}} \xrightarrow{	ext{Heat}} \underset{	ext{(white powder)}}{	ext{CuSO}_4} + 5	ext{H}_2	ext{O}(blue crystal)CuSO4​.5H2​O​Heat​(white powder)CuSO4​​+5H2​O(c) Copper(II) sulphate reacts with NaOH to form a blue precipitate of copper(II) hydroxide (Cu(OH)2), which is insoluble in excess NaOH.CuSO4 + 2NaOH ⟶ Cu(OH)2 ↓ + Na2SO4

Question 17

Question 6(i)Define(a) Electron affinity.(b) Catenation.

Accepted Answer

(a) The amount of energy released while converting a neutral gaseous isolated atom into a negatively charged gaseous ion (anion) by the addition of electron is called Electron Affinity (E.A.).(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.

Question 18

Question 6(ii)1250 cc of oxygen was burnt with 300 cc of ethane (C2H6). Calculate the volume of the unused oxygen and the volume of the carbon dioxide formed:2C2H6 + 7O2 ⟶ 4CO2 + 6H2O

Accepted Answer

2C2H6+7O2⟶4CO2+6H2O2 vol.:7 vol.⟶4 vol.\begin{matrix} 2	ext{C}_2	ext{H}_6 & + & 7	ext{O}_2 & \longrightarrow & 4	ext{CO}_2 & + & 6	ext{H}_2	ext{O} \ 2 	ext{ vol.} & : & 7 	ext{ vol.} & \longrightarrow & 4 	ext{ vol.} & \ \end{matrix}2C2​H6​2 vol.​+:​7O2​7 vol.​⟶⟶​4CO2​4 vol.​+​6H2​O[By Gay Lussac's law]2 Vol. of C2H6 requires 7 Vol. of oxygen∴ 300 cc C2H6 will require 72\dfrac{7}{2}27​ x 300= 1050 cc of OxygenHence, unused oxygen = 1250 - 1050 = 200 ccSimilarly,2 Vol. of C2H6 produces 4 Vol. of carbon dioxide∴ 300 cc C2H6 produces 42\dfrac{4}{2}24​ x 300= 600 cc of Carbon dioxideHence, carbon dioxide produced = 600 cc.

Question 19

Question 6(iii)Write a balanced reaction where sulphuric acid shows the following properties:(a) Oxidising agent(b) Non Volatile Acid(c) Dehydrating agent

Accepted Answer

(a) C + 2H2SO4 ⟶ CO2 + 2H2O + 2SO2 ↑(b) NaCl + H2SO4 ⟶ NaHSO4 + HCl(c) C2H5OH ethyl alcohol→170°CConc. H2SO4[excess]C2H4ethene↑+H2O\underset{	ext{ ethyl alcohol}}{	ext{C}_2	ext{H}_5	ext{OH}} \xrightarrow[170\degree	ext{C}]{	ext{Conc. H}_2	ext{SO}_4	ext{[excess]}} \underset{ 	ext{ethene}}{	ext{C}_2	ext{H}_4} ↑ + 	ext{H}_2	ext{O} ethyl alcoholC2​H5​OH​Conc. H2​SO4​[excess]170°C​etheneC2​H4​​↑+H2​O

Question 20

Question 7(iii)An organic compound contains: H = 6.32 %, N = 17.76%. In the vapour state, this compound is 39.5 times as heavy as the same volume of hydrogen.(a) Find the molecular formula of the compound. (At wt: H = 1 N = 14 )(b) Calculate the number of hydrogen atoms in one mole of this compound.

Accepted Answer

Given:Percentage of H = 6.32%Percentage of N = 17.76%Percentage of carbon = 100 - (6.32 + 17.76) = 75.92 %Atomic weights: H = 1, N = 14(a)Element% compositionAt. wt.Relative no. of atomsSimplest ratioHydrogen6.3216.321\dfrac{6.32 }{1}16.32​ = 6.326.321.27\dfrac{6.32 }{1.27}1.276.32​ = 5Nitrogen17.761417.7614\dfrac{ 17.76}{14}1417.76​ = 1.271.271.27\dfrac{1.27}{1.27 }1.271.27​ = 1Carbon75.921275.9212\dfrac{75.92 }{12}1275.92​ = 6.326.321.27\dfrac{ 6.32 }{1.27}1.276.32​ = 5Simplest ratio of whole numbers = H : N : C = 5 : 1 : 5Hence, empirical formula is C5H5NEmpirical formula weight = (5 × 12) + (5 × 1) + (1 × 14) = 60 + 5 + 14 = 79 g/molV.D. = 39.5Molecular weight = 2 x V.D. = 2 x 39.5 = 79n=Molecular weightEmpirical formula weight=7979=1\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{79}{79} = 1n=Empirical formula weightMolecular weight​=7979​=1∴ Molecular Formula = n[E.F.] = 1[C5H5N] = C5H5N(b) From the formula C5H5N, there are 5 hydrogen atoms per molecule.In 1 mole of the compound the number of molecules = 6.022 × 1023∴ Total hydrogen atoms = 5 × (6.022 × 1023) = 3.011 × 1024 hydrogen atoms.

Model Paper 2

Section A 40 Marks

Section B 40 Marks