SOLUTION MODEL SPECIMEN PAPER–3 SECTION–I Answer 1. (a) (i) (D) Magnesium (ii) (C) Haematite (iii) (B) Lead sulphide (iv) (D) Acetic acid (v) (D) Addition Reaction (vi) (A) Ferrous Hydroxide (vii) (C) Ethanoic acid (viii) (D) Ammonium chloride (ix) (B) Aqua fortis (x) (B) Sulphur (b) (i) Hygroscopic substance (ii) Electronegativity (iii) Methanal (iv) Roasting (v) Mole (c) (i) When excess of ammonium hydroxide is added to copper sulphate solution, deep prussian blue solution of tetraamine copper hydroxide is formed. (ii) When caustic soda is added drop wise to zinc sulphate solution, a white precipitate of zinc hydroxide is formed which is soluble in excess of NaOH. (iii) When sulphur dioxide gas is passed through acidified potassium dichromate solution, it changes its colour from orange to green. (iv) When chlorine reacts with excess of ammonia, nitrogen trichloride is formed along with hydrochloric acid. (v) When dilute hydrochloric acid is added to sodium carbonate crystals, colourless and odourless carbon dioxide gas is released along with the formation of sodium chloride and water. (d) (i) Fe 2 O 3 (ii) an upward delivery of gas (iii) sulphuric acid (iv) substitution (v) 3 (e) 1. (e) 2. (c) 3. (b) 4. (a) 5. (d) (f) (i) Empirical formula weight = Vapour Density Molecular Weight = 2 x Vapour Density =2 x Empirical formula weight Empirical Formula = XY 2 So, Molecular Formula = 2 (XY 2 ) =X 2 Y 4 (ii) (a) Mass of Ca(NO 3 ) 2 taken = 16.4 g Moles of Ca(NO 3 ) 2 taken = 16.4 / 164 = 0.1 mole 2 moles of Ca(NO 3 ) 2 produce 4 volumes of NO 2 0.1 mole of Ca(NO 3 ) 2 will produce = 4 x 0.1 x 22.4 / 2 = 4.48 litres (b) 2 moles of Ca(NO 3 ) 2 produce 2 moles of CaO 0.1 mole of Ca(NO 3 ) 2 will produce = 0.1 moles of CaO = 0.1 x 56 = 5.6 g (g) (i) Ethyne (ii) Phosphoric acid (iii) Magnesium hydroxide (iv) Hydrogen sulphide (v) Nitrate
(b) (i) Hygroscopic substance (ii) Electronegativity(iii) Methanal (iv) Roasting(v) Mole
(c) (i) When excess of ammonium hydroxide is added to copper sulphate solution, deepprussian blue solution of tetraamine copper hydroxide is formed.
(ii) When caustic soda is added drop wise to zinc sulphate solution, a white precipitateof zinc hydroxide is formed which is soluble in excess of NaOH.
(iii) When sulphur dioxide gas is passed through acidified potassium dichromatesolution, it changes its colour from orange to green.
(iv) When chlorine reacts with excess of ammonia, nitrogen trichloride is formed alongwith hydrochloric acid.
(v) When dilute hydrochloric acid is added to sodium carbonate crystals, colourless andodourless carbon dioxide gas is released along with the formation of sodiumchloride and water.
(d) (i) Fe2O3 (ii) an upward delivery of gas(iii) sulphuric acid (iv) substitution(v) 3
(e) 1. (e) 2. (c) 3. (b) 4. (a) 5. (d)(f) (i) Empirical formula weight = Vapour Density
Molecular Weight = 2 x Vapour Density= 2 x Empirical formula weight
Empirical Formula = XY2
So, Molecular Formula = 2 (XY2)= X2Y4
(ii) (a) Mass of Ca(NO3)2 taken = 16.4 gMoles of Ca(NO3)2 taken = 16.4 / 164 = 0.1 mole
2 moles of Ca(NO3)2 produce 4 volumes of NO2 0.1 mole of Ca(NO3)2 will produce = 4 x 0.1 x 22.4 / 2
= 4.48 litres(b) 2 moles of Ca(NO3)2 produce 2 moles of CaO0.1 mole of Ca(NO3)2 will produce = 0.1 moles of CaO
Answer 2.(a) (i) Bauxite is mixed with concentrated sodium hydroxide solution, bauxite dissolves to
form sodium aluminate leaving behind the impurity ferric oxide unreacted. Thesolution is filtered where sodium is collected as filtrate and the impurities are leftbehind.
Al2O3.2H2O + 2NaOH → 2NaAlO2 + 3H2OSodium aluminate solution is diluted with water, white precipitate of aluminiumhydroxide is formed which is filtered, washed and dried.
NaAlO2 + 2H2O → NaOH + Al(OH)3
Aluminium hydroxide is heated to high temperatures to obtain Alumina.2Al(OH)3 → Al2O3 + 3H2O
(ii) At cathode :Al3+ + 3e– → Al
At anode :O2– + 2e– → [O][O] + [O] → O2
2C + O2 → CO2
2CO + O2 → 2CO2
(b) (i) Molecular formula – Z2
Electron dot diagram –
Type of bonding – Covalent(ii) Formula – X2Y, Type of bonding - Ionic(iii) Bond formed between Y and Z – Covalent. This is because neither Y nor Z is capable
of losing electrons. So, they share electrons with each other and form covalent bond.Answer 3.(a) (i) Element C (ii) Element F
(iii) Element C (iv) Element A(b) (i) An anion is formed by the gain of electron by the parent atom. The number of
protons remains the same in the anion as the parent atom and thus the nuclear forceof attraction for the outer electrons reduces and hence the distance between nuclearand outer electrons increases due to which anion is larger than its parent atom.
(ii) In the preparation of hydrochloric acid, hydrogen chloride gas is dissolved in waterusing funnel arrangement to prevent back suction of water and this also allowsgreater surface area for the absorption of the gas.
Chemistry | 3
(iii) Oil of vitriol is sulphuric acid, which when reacts with sodium carbonate, briskeffervescence is given which is that of a colourless and odourless gas carbon dioxide.This gas can be confirmed by passing this gas through lime water which will turnmilky.
(iv) Iron sheets are cleaned with hydrochloric acid before dipping into molten zinc forgalvanising to remove any existing rust and prevent any further oxidation.
(c) (i) Manganese dioxide and copper oxide can be distinguished by observing the effecton addition of conc. HCl to each of them separately.When conc. HCl is added to MnO2, greenish coloured chlorine gas is evolved and asthe resulting solution is filtered, the filtrate is brown coloured MnCl2.
MnO2 + 4HCl (conc.) → MnCl2 + H2O + Cl2When conc. HCl is added to CuO, greenish coloured chlorine gas is evolved and asthe resulting solution is filtered, the filtrate is bluish coloured Cu2Cl2.
CuO+ 2HCl (conc.) → CuCl2 + H2O(ii) Potassium chloride and potassium nitrate can be distinguished by observing the
effect of addition of conc. H2SO4 to each of them separately.When conc. H2SO4 is added to solid KCl, a colourless pungent smelling gas HCl isevolved. When a glass rod dipped in NH4OH is brought near this gas, dense whitefumes of NH4Cl are observed, which confirms the evolution of HCl gas.
KCl + H2SO4 (conc.) → KHSO4 + HCl
NH4OH +HCl → NH4Cl + H2O
When conc. H2SO4 is added to solid KNO3, no change appears. When the contents ina test tube is heated, reddish brown coloured NO2 gas having pungent suffocatingsmell is evolved.
KNO3 + H2SO4 (conc.) → KHSO4 + HNO3
4HNO3 → 4NO2 + 2H2O + O2
Answer 4.
(a) (A) Anhydrous ferric chloride
Method of preparation – (i) Direct combination
Equation – 2Fe + 3Cl2 → 2FeCl3(B) Lead chloride
Method of preparation – (v) Reaction of two solutions of salts to form a precipitate.
Equation – Pb(NO3)2 + 2NaCl → 2 PbCl2 + 2NaNO3
(C) Sodium sulphate
Method of preparation – (iv) Titration of a dilute acid with a solution of soluble base.
Method of preparation – (iii) Reaction of a dilute acid with an insoluble base.Equation – Cu(OH)2 + H2SO4 (dil) → CuSO4 + 2H2O
(b) (i) Mass of the cathode increases while the mass of the anode decreases.
(ii) At cathode : Cu2+ +2e– → Cu
At anode : Cu – 2e– → Cu2+
(iii) Colour of the copper sulphate solution fades away i.e, it changes from blue tocolourless.
4 | ICSE Model Specimen Papers, X
(c) S. No. Electrolytic Dissociation Ionization1. The breaking of the molecule of a
substance by splitting into itsconstituent ions is called electrolyticdissociation.
The loss or gain of electrons in an atomby which it acquires a negative or apositive charge is called ionization.
2. It occurs under the effect of electriccurrent.
It occurs under the effect of heat, lightetc.
Answer 5.(a) (i) Sulphide (ii) Sulphate
(iii) Sulphite(b) (i) Solution R (ii) Solution R
(iii) Solution Q(c) (i) Isobutane
C
C
C
H
H H
H
H
H H
CH
H
H
IUPAC name: 2-methylpropane(ii) Neopentane
H—C|H
—H
H—C|H
|H
—–C
|||
|||
—–C|H
|H
—H
H—C|H
—H
IUPAC name: 2, 2-dimethylpropaneAnswer 6.(a) Molecular weight = 2 x Vapour Density
= 2 x 29= 58 g
Mass of hydrocarbon = 0.29 gMoles of hydrocarbon = 0.29 / 58
= 0.005 molesCxHy + (4x + y)/4O2 → xCO2 + y/2H2O
0.29 g 0.88 g0.005 moles of hydrocarbon produce 448 ml (448 / 22400 = 0.02 moles) of carbon dioxideNow, 0.005 moles of hydrocarbon completely burn to produce 0.02 moles of CO2So, 1 mole of hydrocarbon completely burn to produce x moles of CO2Hence, x = 0.02 / 0.005
= 4This shows that hydrocarbon CxHy is C4Hy
Chemistry | 5
And as the molecular weight of hydrocarbon C4Hy is 58g, this means that thehydrocarbon is butane (C4H10)(i) Mass of carbon dioxide = 0.02 x 44
= 0.88 g(ii) Mass of element carbon in 44 g carbon dioxide = 12 g
Mass of element carbon in 0.88 g carbon dioxide = 12 x 0.88 / 44 = 0.24 g(iii) Mass of hydrogen in 58 g hydrocarbon = 10 g
Mass of hydrogen in 0.29g hydrocarbon = 10 x 0.29 / 58 = 0.05 g(iv) Empirical formula of hydrocarbon = C2H5
(v) Molecular formula of hydrocarbon = C4H10
(b) The bond in which the sharing of electrons takes place on the part of entirely one atom iscalled as coordinate bond. or dative bond.
(c) (i) When ethylene is treated with alkaline solution of potassium permanganate, itdecolourizes the purple colour solution.
3C2H4 + 2KMnO4 + 4H2O → 3C2H6O2 + 2MnO2 + 2KOH(ii) When lead dioxide is heated with concentrated hydrochloric acid, HCl gets oxidized
to chlorine.PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O
(iii) When nitric acid is treated with hot coke dust, it oxidizes carbon to form carbondioxide.
C + 4HNO3 → CO2 + 4NO2 + 2H2O(iv) When a glass rod dipped in HCl is brought near ammonia gas, dense white fumes of
ammonium chloride are formed.NH3 + HCl → NH4Cl
(v) Sulphur dioxide is passed through hydrogen sulphidesolution,SO2 oxidizeshydrogen sulphide to sulphur.
SO2 + 2H2S → 3S + 2H2O(d) C + 2H2SO4 → CO2 + 2H2O + 2SO2
(i) 2 moles (2 x 98 = 196 g) of H2SO4 require 1 mole (12 g) of carbon19.6 g of H2SO4 will require = 12 x 19.6 / 196
= 1.2 g(ii) 2 moles of H2SO4 evolve 2 moles (2 x 22.4 = 44.8 l) of SO2
Moles of H2SO4 = 19.6 / 98 = 0.2 molesSo,0.2 moles of H2SO4 will evolve = 0.2 moles of SO2
= 0.2 x 22.4= 4.48 l
(iii) 2 moles of H2SO4 evolves1 mole (22.4 l) of CO2
Moles of H2SO4 = 19.6 / 98 = 0.2 molesSo, 0.2 moles of H2SO4 will evolv = (1 × 0.2 / 2) moles of CO2
= 0.1 × 22.4= 2.24 l
(e) (i) (d) (ii) (c) (iii) (a) (iv) (e) (v) (b)
(f) (i) Periodic properties (ii) Periodicity(iii) Dehydrogenation (iv) Denatured alcohol
(v) Ductility
2 | ICSE Model Specimen Papers, X
(g) (i) (5) Iron sulphide (ii) (4) Copper nitrate(iii) (2) Zinc nitrate (iv) (3) Sodium chloride(v) (1) Lead Carbonate
SECTION—II
Answer 2.(a) (i) Structure Molecular formula
H—C|H
|H
—C|H
|H
—C|H
|H
—C|H
|H
—C|H
|H
—H C5H12
Pentane
H—C|H
|H
—C|H
|H
—C|
CH3
|H
—C|H
|H
—H C5H12
2-Methylbutane
H—C|H
|H
—C|
CH3
|—C
|H
|H
—H
CH3 C5H122-Dimethylpropane
(ii) First two are position isomers while the third one is a chain isomer.
H—C|H
|H
—C|H
|H
—C|H
= C|H
|H
H—C|H
|H
—C|H
= C|H
—C|H
|H
—HH—C
|H
|H
—C = C|H
|H
H—C|H
—H
but-1-ene but-2-ene 2-methylprop-1-ene
(b) When ethanol is denatured by adding methanol, copper sulphate and pyridine to make itunfit for human consumption, it is called denatured alcohol.
(c) 1. It is used as a solvent.2. It is used in the manufacture of organic compounds like acetic acid, chloroform etc.
(d) (i) C2H4 + H2O → C2H5OH(ii) Oxidation of ethanol by acidified K2Cr2O7
C2H5OH + [O] → CH3CHOCH3CHO + [O] → CH3COOH
Answer 3.(a) (i) Chlorine (ii) 8
(iii) Chlorine (iv) Fluorine(v) Hydrogen
(b) (i) low (ii) 8, He, 2(iii) alkaline earth
Answer 4.(a) (i) Sodium sulphate–Neutralisation
(ii) Silver chloride – Precipitation(iii) Iron sulphide – Direct combination
Chemistry | 3
(b) (i) Copper sulphate and iron (II) sulphateWhen little NH4OH is added to CuSO4, pale blue or bluish white precipitate appearswhich dissolves in excess of NH4OH to give deep blue or inky blue solution.
When little NH4OH is added to Pb(NO3)2, a white precipitate appears which isinsoluble in excess of NH4OH.
Pb(NO3)2 + 2NH4OH → Pb(OH)2 + 2NH4NO3white ppt.
(iii) Iron (II) sulphate and iron (III) sulphate.When little NH4OH is added to FeSO4, dirty green precipitate appears which turnsreddish brown after sometime. It is insoluble in excess of NH4OH.
FeSO4 + 2NH4OH → Fe(OH)2 + (NH4)2SO4Dirty green
ppt.When little NH4OH is added to Fe2(SO4)3, reddish brown precipitate appears whichis insoluble in excess of NH4OH.
Fe2(SO4)3 + 6NH4OH → 2Fe(OH)3 + 3(NH4)2SO4(c) (i) Chlorine gas (ii) Zinc oxide
(iii) Chlorofluorocarbons (iv) CuSO4Answer 5.(a)
4 | ICSE Model Specimen Papers, X
(b) S. No. Name ofElectrolyte
Name ofCathode
Name ofAnode
Product atCathode
Product atAnode
1. CuSO4 (aq.) Copper Copper Copper Copper ions
2. PbBr2 (molten) Platinum Platinum Lead Bromine gas
(c) (i) Cu → Cu2+
It is an oxidation reaction.Cu – 2e– → Cu2+
(ii) Fe3+ → Fe2+
It is a reduction reaction.Fe3+ + e– → Fe2+
(iii) Cl– → ClIt is an oxidation reaction.
2Cl– – 2e– → Cl2Answer 6.(a) (i) Mercury (ii) Graphite
(iii) Al2O3 (iv) Na2O(v) Iodine
(b) (i) Bauxite (ii) Baeyer’s process(iii) Alumina is added to cryolite in the electrolytic reduction of aluminium because
cryolite reduces the melting point of alumina from 2050ºC to 950ºC. It also increasesthe electrical conductivity of the electrolyte which helps in saving the electricity.
(iv) At cathode:Al3+ + 3e– → Al
At anode : O2– + 2e– → [O] [O] + [O] → O2
2C + O2 → CO2
2CO + O2 → 2CO2
(v) Froth floatationAnswer 7.
(a) Element Percentage composition bymass
Atomic mass Relative number of atoms Simple ratio
C 54.55 12 4.55 2
H 9.09 1 9.09 4
O 36.26 16 2.27 1Vapour Density = 44
Molecular Weight = 2 x Vapour Density= 2 x 44= 88 g
Empirical formula = C2H4O1Empirical Weight = (12 x 2) + (1 x 4) + (16 x 1)
= 44 gn = Molecular Weight / Empirical Weight
= 88 /44
Chemistry | 5
= 2Molecular formula = (Empirical formula)n
= (C2H4O1)2= C4H8O2
(b) (i) NH3
H—··N|H
—H
(ii) CH4
H H:C···· :H
H
(iii) H3O+
⎣⎢⎢⎢⎢⎡
⎦⎥⎥⎥⎥⎤H—O··
↓H
—H+
(c) Basis of comparison Covalent compounds Electrovalent compounds
(i) Solubility They are soluble in polar solvents. They are insoluble in organic solvents.
(ii) Structure They are generally solids. They are liquids or gases.
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SOLUTION
MODEL SPECIMEN PAPER–9SECTION–I
Answer 1.(a) (i) Electrovalent (ii) Cation
(iii) Copper oxide, Lead oxide (iv) Concentration(v) Galvanisation
(c) (i) A black coloured complex compound known as nitroso ferrous sulphate is formed.(ii) Reddish brown precipitate of ferric hydroxide appears which is insoluble in excess
of sodium hydroxide.(iii) Initially the sugar crystals become brown, steam is released which causes a lot of
frothing and finally a black porous mass of carbon is left behind.(d) (i) Name – Methoxy ethane
Functional group – ether
(ii) Name –Hexan-3-one
Functional group – ketone
(iii) 1. Name –2-methyl propanoic acid
Functional group – carboxylic acid
(iv) 1. Name –3-methyl butanal
Functional group – aldehyde(e) (i) F (ii) Li, Na, K
(ii) An inverted funnel is attached to the delivery tube to prevent the back suction ofwater and gives greater surface area for the absorption of the gas.
∴ Emprical formula of the compound is CH2BrMolecular formula = (Empirical formula)n
n =M. W.
Empirical formula weight
=2 ×V.D
Empirical formula weight
=2 × 94
(12 + 2 + 80) = 2 × 94
94 = 2
∴ Molecular formula = (CH2Br)2 = C2H4Br2(b) (i) Ostwald’s process
(ii) (1) Catalytic chamber
4NH3 + 5O2 Pt
⎯⎯→880°C
4NO + 6H2O
(2) Oxidation chamber
2NO + O2 50°C
⎯⎯→ 2NO2
(3) Absorption Tower
4NO2 + 2H2O + O2 → 4HNO3
(iii) Ratio –1:8
Answer 4
(a) (i) Calculated amount of water is used in dilution of Oleum to obtain concentratedsulphuric acid of desired concentration.
(ii) Vanadium pentoxide is preferred as a calalyst during catalytic oxidation of SO2because of the high oxidation state of vanadium.
(b) (i) Duralumin – 95%Al, 4%Cu, 0.5%Mg, 0.5%Mn
(ii) Brass – 60-70% Cu, 30-40% Zn
(iii) Stainless steel – 73% Fe, 18% Cr, 8%Ni, 1%C
(iv) Solder – 40 %Pb and 60 %Sn
(c) (i) Propanoic acid
H—C|H
|H
—C|H
|H
—C O
O—H
(ii) Diethyl ether
H—C|H
|H
—C|H
|H
—O—C|H
|H
—C|H
|H
—H
4 | ICSE Model Specimen Papers, X
(iii) Pentan-2-ol
H—C|H
|H
—C|H
|OH
— C|H
|H
—C|H
|H
—C|H
|H
—H
(iv) An isomer of n – butane
2-methyl propane
H—C|H
|H
—C|H
—C|H
|H
—H
H—C|H
—H
Answer 5.
(a) (i) Dehydrating agent (ii) Oxidising agent
(iii) Acidic
(b) (i) Solution C (ii) Salt and water
(iii) Solution D (iv) Solution A and B
(v) Solution D
(c) Positional Isomers of Butene
First two are position isomers while the third one is a chain isomer.
H—C|H
|H
—C|H
|H
—C|H
= C|H
|H
H—C|H
|H
—C|H
= C|H
—C|H
|H
—H H—C|H
|H
—C = C|H
|H
H—C|H
—H
but-1-ene but-2-ene 2 methylprop-1-ene
Answer 6.
(a) (i) Baeyer’s process (ii) Na3AlF6
(iii) Cryolite reduces the melting point or fusion temperature from 2050ºC to 950ºC ofpure alumina. It also increases the electrical conductivity of the electrolyte whichhelps in saving electricity.
(iv) At cathode :
Al3+ + 3e– → Al
At anode :
O2– + 2e– → [O]
[O] + [O] → O2
2C + O2 → CO2
2CO + O2 → 2CO2
(v) Carbon anodes are periodically replaced because they get oxidized to carbondioxide.
(b) (i) CaC2 + 2H2O → Ca(OH)2 + C2H2
Chemistry | 5
(ii) C2H5Br + KOH(aq) → C2H5OH + KBr(iii) CH3I + 2[H] → CH4 + HI
(iv) (1) C2H5OH + H2SO4(conc) 110°C
⎯⎯→ C2H5HSO4+ H2O
(2) C2H5HSO4 160°C⎯⎯→ C2H4 + H2SO4
Answer 7.(a) (i) Iron (ii) Iron
(iii) Sulphur(b) (i) At cathode:
Al3+ + 3e– → Al(ii) Carbon anodes are periodically replaced because they get oxidised to carbon
dioxide.(iii) Hoope’s Process
(c) Pb3O4 + 8HCl → 3PbCl2 + 4H2O + Cl2(i) Molecular weight of red lead = (207 × 3) + (16 × 4)
= 621 + 64= 685 g
No. of moles of red lead taken = 6.85 / 685= 0.01 moles
1 mole of red lead forms 3 moles of lead chloride0.01 mole of red lead will form = (3 × 0.01) moles of lead chloride
= 0.03 molesMolecular weight of lead chloride = 207 + (35.5 x 2)
= 278 gLead chloride formed = 0.03 moles
= 0.03 × 278= 8.34 g
(ii) At STP,1 mole of red lead evolves 1 mole of chlorine
0.01 mole of red lead will form = 0.01 moles of chlorine= (0.01 × 22.4) L of chlorine= 0.224 L of chlorine
❏❏
SOLUTION
MODEL SPECIMEN PAPER–15SECTION–I
Answer 1.(a) (i) (C) Have same number of protons
(ii) (B) Liquid Ammonia(iii) (B) 6·0 × 6·023 × 1023
(c) (i) Carbon (ii) Covalent(iii) Hygroscopic substances (iv) Anode(v) Tungsten
(d) (i) Oxygen gas. (ii) 2KClO3 → 2KCl + 3O2
(iii) Molecular mass of KClO3 g = (39 + 35.5 + 16 × 3) = 122.5g67.2 L (3 × 22.4) of oxygen is evolved by 245 g (2 × 122.5) of KClO3
6.72 L of oxygen will be evolved by = 245 × 6.72 / 67.2= 24.50 g of KClO3
(iv) 22.4 L of oxygen contains 1 mole of oxygen6.72 L of oxygen contain = 1 × 6.72 / 22.4
= 0.3 molesNumber of molecules of oxygen in 6.72L of oxygen = 0.3 × 6 × 1023
= 1.8 × 1023 molecules(v) 1 mole of HCl gas occupy 22.4 L
0.01 mole of HCl gas will occupy (22.4 × 0.01) = 0.224 L(e) (i) Add conc. sulphuric acid to both the compounds separately.
Lead nitrate : No change appearsLead chloride : A colourless gas having pungent suffocating smell which fumes inmoist air evolves.
(ii) Add Tollen’s reagent (Ammoniacal silver nitrate solution)Ethyne – a yellow white precipitate of silver acetylide will be formed.Ethene – No reaction
(iii) Add dilute sulphuric acid to both the compounds separately.Sodium sulphite – A colourless gas having burning sulphur smell evolves.Sodium carbonate – A colourless and odourless gas evolves with brisk effervescence.
2 | ICSE Model Specimen Papers, X
(iv) Add sodium hydroxide to the salt solutions separatelyFerrous sulphate – Dirty green precipitate appears which changes to reddish brownafter some time and is insoluble in excess of sodium hydroxide.Copper sulphate – Bluish white precipitate appears which is insoluble in excess ofsodium hydroxide.
(v) Add NaOH solution to both the salts the ammonium salt will release the ammoniagas which can be smelled while sodium salt will not show any change.
Silver plating a spoon Pure block of silver Sodium argentocyanide Spoon
Purification of copper Impure copper Acidified copper sulphatesolution
Pure strip of copper
SECTION–II
Answer 2.(a) (i) Pure acetic acid is known as glacial acetic acid because on cooling, it forms
crystalline mass resembling ice.(ii) Esterification
(b) (i) Zinc in galvanisation Zinc is more reactive than iron
(ii) Aluminium in thermite welding Reducing agent
(c) (i) H–C ≡ C – H + 2AgNO3 + NH4OH → Ag+ C – ≡ C – Ag+ + 2NH4NO3 + 2H2O(ii) S + 2 H2SO4 (conc.) → 2H2O + 3SO2
(d) (i) When ammonium hydroxide is added to silver chloride, a complex diammine silverchloride is formed which is soluble in excess of ammonium hydroxide.
(ii) When manganese dioxide reacts with concentrated hydrochloric acid, a greenishyellow coloured gas with a pungent suffocating smell is evolved.
(iii) When dilute hydrochloric acid is added to sodium thiosulphate, the solution formedhas a yellow turbidity.
Answer 3
(a) Element Percentage compositionby mass
Atomic mass Relative number of atoms Simple ratio
C 4.8 12 0.4 1Br 95.2 80 1.19 3
(i) Empirical formula = CBr3
Empirical weight = 12 + (80 × 3)
= 12 + 240= 252
(ii) Vapour density = 252
Molecular weight = 2 × V.D.
= 2 × 252
Chemistry | 3
= 504n = Molecular weight / Empirical weight
= 504 / 252= 2
Molecular formula = (Empirical formula)n= (CBr3)2
= C2Br6
(b) A : 4NH3 + 5O2 Pt
⎯⎯⎯⎯→800 °C
4NO + 6H2O
B : 2NO + O2 → 2NO2 (50°C)
C : 4NO2 + 2H2O + O2 Pt
⎯⎯⎯⎯→800 °C
4HNO3
D : Cu + 4HNO3 (conc.) →Cu(NO3)2 + 2H2O + 2NO2
(c) (i) During the manufacture of nitric acid by Ostwald’s process excess of oxygen ispreferred over air, because each and every step requires oxygen.
(ii) In the laboratory preparation of nitric acid the mixture of concentrated sulphuricacid and sodium nitrate should not be heated very strongly, because nitric aciddecomposes at high temperatures.
Answer 4.(a) Formula – CuSO4.5H2O
Molecular weight = 64 + 32 + (16 × 4) + 5 ( 2 × 1 + 16) – 250 gMass of water = 90 g
Percentage of water = 90 × 100 / 250 = 36%
(b) (i) During electroplating a direct current is used to maintain the same direction ofcurrent flow (ions) within the electrolyte. If alternating current is used, then thedirection of current will keep on changing and no plating would occur.
(ii) To ensure uniform plating, a small current should be used for a longer time.
(c) (i) Solution of H2SO4
(ii) Solution of CCl4(iii) Solution of CH3COOH
(d) (i) When dilute sulphuric acid is added to barium chloride, white precipitate of bariumsulphate is formed. When dilute sulphuric acid is added to lead nitrate, whiteprecipitate of lead sulphate and nitric acid are formed.
(ii) When dilute sulphuric acid is added to ferrous sulphide, green coloured solution ofFeCl2 is formed along with the evolution of a colourless gas and rotten smell.Whendilute sulphuric acid is added to copper sulphide, blue coloured solution of CuCl2 isformed along with the evolution of a colourless gas and rotten smell.
(iii) When sodium hydroxide is added to the solution of a copper salt (CuSO4), bluishwhite precipitate [Cu(OH)2] appears which is insoluble in excess of sodiumhydroxide.
CuSO4 + 2NaOH ⎯→Cu(OH)2 ↓ + Na2SO4
When sodium hydroxide is added to the solution of a zinc salt [(Zn(NO3)2], whiteprecipitate [Zn(OH)2] appears which is soluble in excess of sodium hydroxide.
Zn(NO3)2 + NaOH → Zn(OH)2↓ + NaNO3
Zn(OH)2 +NaOH → Na2ZnO2 + 2H2O
4 | ICSE Model Specimen Papers, X
Answer 5.(a) (i) (D) Sodium chloride
(ii) (B) Chromium sulphate(iii) (C) Lead (II) chloride(iv) (A) Iron (III) chloride
(b) (i) Period 2(ii) Nitrogen (N), It should be placed between C and O.(iii) Fluorine (F)(iv) Carbon (C)
(c) (i) Aluminium (Al)(ii) Iron (Fe)
Answer 6.(a) (i) Sodium argento cyanide is a complex salt.
(ii) Silver nitrate cannot be used as an electrolyte because it decomposes very rapidlyand leads to the formation of uneven layer.
(iii) A low current for a longer duration must be passed and direct current should beused instead of alternating current.
(iv) At cathodeAg+ + e– → Ag
At anodeAg – e– → Ag+
(b) (i) MgO + H2SO4 → MgSO4 + H2O
(ii) CH3COONa + NaOH CaO⎯→ CH4 + Na2CO3
(c) (i) Calcination is the process of heating of concentrated ore in the absence of air oroxygen whereas Roasting is the process of heating of concentrated ore in thesufficient supply of air or oxygen.
(ii) Ore of iron – Fe2O3 (Haemetite)Ore of aluminium – Al2O3.2H2O (Bauxite)
(b) (i) Lone pair of electrons are a pair of valence electrons that are not shared with anotheratom during the formation of a bond and is sometimes called a non-bonding pair.
(ii) H—O +
H
Lone pair ofelectrons H
··
(c) (i) He < Ne < Ar(ii) Li < N <Cl < F
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SOLUTION
MODEL SPECIMEN PAPER–18SECTION–I
Answer 1.(a) (i) (A) Al and Cu (ii) (A) Passing dry ammonia over heated copper oxide.
(iii) (C) Substitution (iv) (B) Oxygen is released at anode(v) (A) NaCl (vi) (A) He(vii) (A) CuSO4 (viii) (A) Avogadro’s Law(ix) (C) Consists of molecules (x) (D) FeCl3
(b) (i) Ferric chloride (ii) Brass(iii) Methane (iv) HCl(v) Aluminium
(c) (i) From the reaction,2 moles of decomposition of calcium nitrate produces 4 moles of nitrogen dioxide.1 mole of decomposition of calcium nitrate will produce 2 moles of nitrogen dioxide.
(ii) Molecular mass of Ca(NO3)2 = 164 g2 moles (164 × 2 = 328 g) of decomposition of calcium nitrate produces 1 molevolume (22.4L) of oxygen.65.6 g of decomposition of calcium nitrate will produce = (22.4 × 65.6 / 328)
= 4.48 L of oxygen(iii) 2 moles (164 × 2 = 328 g) of decomposition of calcium nitrate forms 2 moles (2 × 56 g
= 112 g) of calcium oxide.65.6 g of decomposition of calcium nitrate will form = (112 × 65.6 / 328) of calciumoxide
= 22.4 g of CaO
(iv) 2 moles (164 × 2 = 328 g) of decomposition of calcium nitrate forms 5 moles (4 molesof NO2 + 1 mole of O2) of gaseous product
Mass = 328 g
(v) 4 moles (22.4 × 4 = 89.6L) of nitrogen dioxide are produced by decomposition of 2moles (164 × 2 = 328g) of calcium nitrate.
44.8 l moles of nitrogen dioxide will be produced by decomposition of = 328 × 44.8
(ii) Carbon dioxide solution – pH less than 7Lime water – pH more than 7
(iii) Cu(OH)2
(iv) It is necessary to add acid to water before proceeding with the electrolysis of waterbecause distill water contains no ions due to which there is no passage of electriccurrent through it on adding few drops of acid (e.q. HCl) it acts as a catalyst and theionisation of water increases.
(v) In catalysis, the catalysts only alters the rate of reaction without taking part in thereaction. As water is feebly ionised, and on adding drop of acid (eg. HCl), which actsas catalyst, the ionisation of water increases. Hence electrolysis of acidulated water isan example of catalysis.
(c) (i) Fountain Experiment(ii) HCl gas is extremely soluble in water.(iii) Red
(ii) The paper is made up of cellulose and conc. H2SO4 acid being a powerfuldehydrating agent removes water from cellulose and black carbon is left behind, andFinally paper will turn black.Reaction of cellulose with conc. H2SO4
C6H10O5 → 6C + 5H2O(b) (i) Cathode – Lead metal
Anode – Bromine vapours(ii) Cathode – Potassium metal
Anode – Chlorine vapours
(iii) Cathode – Copper metal
Anode – Oxygen vapours
Answer 6.
(a) The chemical equation representing the reaction is
CO2 + C ⎯→ 2CO
1 vol. 2 vol.
Let, x ml of CO2 react with red hot carbon to form 2x ml of CO. Initial volume of carbondioxide = 500 ml.
Final volume of the reaction mixture = 700 ml
Thus, we have (500 – x) = (700 – 2x)or 2x – x = 700 – 500
x = 200Volume of carbon monoxide formed = 2 × 200 = 400 ml
Volume of carbon dioxide remains after the reaction= (500 – 200) = 300 ml.
(b) (i) A colourless gas having smell of burning sulphur evolves.2Al + 6H2SO4(conc.) → Al2(SO4)3 + 6H2O + 3SO2
4 | ICSE Model Specimen Papers, X
(ii) A soluble complex salt is formed with the liberation of colourless gas.2Al + NaOH + 2H2O → 2NaAlO2 + 3H2
(c) Ores of zinc – Zinc blende(ZnS) and Calamine(Fe2O4Zn)(d) Na2SO4 + Pb(NO3)2 → PbSO4 + 2NaNO3
1 mole of lead sulphate(At mass - 303 g) is formed by 1 mole of sodium sulphate (At mass- 142g)
30.30 g of lead sulphate will be formed by = 142 × 30.3 / 303= 14.2g
Mass of sodium sulphate present in the original solution = 14.2gAnswer 7.(a) (i) The apparatus is made of glass so as to prevent any damage of rubber and cork from
the vapours of nitric acid formed.(ii) KNO3 + H2SO4 (conc.) → KHSO4 + HNO3
(iii) Neutralisation reaction(iv) Oxygen gas
2KNO3 → 2KNO2 + O2
(b) (i) Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3.(ii) By downward displacement of air.(iii) Ammonia is not collected over water because it is highly soluble in water.(iv) Calcium oxide is normally used as a drying agent for ammonia.
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SOLUTION
MODEL SPECIMEN PAPER–21SECTION–I
Answer 1.(a) (i) (B) Number of outer electrons (ii) (A) Carbon
(b) (i) Ammonia is used as a cleansing agent.(ii) Elements of group three have three valence electrons.(iii) Metals of IA group are called alkali earth metals.(iv) Ammonia reduces metallic oxides to metals.(v) The brown ring of nitroso ferrous sulphate i s formed at the junction of the two
liquids.(c) (i) When acetylene is passed over bromine water, it decolourizes bromine solution in
carbon tetrachloride.(ii) When sulphur dioxide gas is passed into potassium dichromate solution, its colour
changes from orange to green and the chromium sulphate and potassium sulphateare formed.
(iii) When blue crystals of copper nitrate are heated, they lose their water ofcrystallisation. This reaction produces copper oxide, reddish brown nitrogen dioxidegas and oxygen gas.
(iv) When ammonium hydroxide solution is added to ferric nitrate solution, it formsdirty green precipitate of ferric hydroxide and ammonium nitrate.
(v) When excess of sodium hydroxide is added to zinc chloride solution, whiteprecipitate of zinc hydroxide is formed which dissolves in excess of sodiumhydroxide and forms a colourless solution.
(d) (i) (C) (ii) (A)(iii) (E) (iv) (B)(v) (D)
(e) (i) Compound P – EthanolCompound Q – Ethanoic acid
Compound R – Ethyl ethanoate
(ii) CH2COOH + C2H5OH → CH3COOC2H5 + H2O
(iii) The above reaction is called as an Esterification reaction.
(f) (i) S + 6HNO3 → H2SO4 + 6NO2 + 2H2O
(ii) 2C2H5OH + 2Na → 2C2H5ONa + H2
(iii) C2H5Br + NaOH(aq) → C2H5OH + NaBr
(iv) FeSO4 + 2NH4OH → Fe(OH)2 + (NH4)2SO4
(v) Pb3O4 + 8HCl (conc.) → 3PbCl2 + Cl2 + 4H2O(g) (i) 1. 6 × 1023 molecules of N2 will weigh 28 g
24 × 1024 molecules of N2 will weigh = 28 × 24 × 1024
6 × 1023
= 1120 gMass of N2 gas in cylinder is 1120 g.
2 | ICSE Model Specimen Papers, X
2. 6 × 1023 molecules of N2 has a volume of 22.4 L
24 × 1024 molecules of N2 has a volume = 22·4 × 24 × 1024
6 × 1023 = 896 L
So, Volume of Nitrogen at STP = 896 L(ii) 1. 2C2H6 + 7O2 → 4CO2 + 6H2O
2 volumes of ethane form 4 volumes of carbon dioxide
600 cc of ethane will form =4 × 600
2 cc of carbon dioxide
= 1200 cc of carbon dioxide2 volumes of ethane require 7 volumes of oxygen
600 cc of ethane will require =7 × 600
2 cc of oxygen
= 2100 cc of oxygen2. Unused volume of oxygen is (3000 – 2100) = 900 cc of oxygen.
SECTION—II
Answer 2.(a) (i) The structural formulae for ethyl methyl ether
H — C — C — O|H
|H
|H
|H
HC
H
Hethyl methyl ether
(ii) The structural formulae for isobutanolH3C
H3CCH — CH2 — OH
(iii) The structural formulae for 1-Chloro propyneCl–CH2–C ≡ CH
(b) (i) Common Name Chemical Name Formula
1. Cryolite Sodium hexafluoro aluminate Na3AlF6
2. Rust Hydrated ferric oxide Fe2O3xH2O(ii) 1. Slag : The fusible product formed when flux reacts with gangue during the
extraction of metals.2. Calcination : It is the process of heatinga concentrated ore in the absence of air
or oxygen.(c) Electron dot structure of ammonia gas.
H — ˙N — H|H
Hydrogen
ion
Ammonia
moleculeAmmonium
ion
Chemistry | 3
Answer 3.
(a) (i) A can form covalent bond.
(ii) Least reactive element of period 3 is x.
(iii) The element with largest atomic radius in group 1 is F.
(iv) The element with atomic number 11 is B. The formula for its oxide is B2O.
(v) The element with highest electron affinity in period 3 is u.(b) (i) Colour of the copper sulphate solution fades away i.e. it changes from blue to
colourless.(ii) Mass of the cathode increases while the mass of the anode decreases.
(iii) At anode : OH– – e– → OH4OH → 2H2O + O2
(c) (i) On adding concentrated sulphuric acid to sugar, initially the crystals become brown,steam is released which causes a lot of frothing and finally a black porous mass isleft behind.
(ii) Concentrated nitric acid is not used during preparation of HCl because it is volatileacid.
Answer 4.
(a)Name of the
process Input CatalystEquations for thecatalysed reaction Output
Contact Process Sulphur dioxide+
Oxygen
Vanadiumpentoxide
(V2O5)
2SO2 + O2 ↔2SO3 Sulphurtrioxide (SO3)
(b) (i) Element X is Nitrogen (N)(ii) Electronic configuration – 2, 8, 5
Number of valence electrons – 5(iii) Electron dot structure
Type of bond– Covalent bond (Triple bond)(c)
Ions present in the Ions discharged atProcess Nature of anode electrolyte Cathode Anode
Electroplating ofan iron rod with
silver
Pure silver Electrolyte shouldcontain the ions of
the silver
Silver Silver
Answer 5.
(a) (i) Mass of 1 mole of water = 18 g
Mass of 0.2 moles of water = 0.2 × 18 g
= 3.6 g
(ii) 4NH3 + 3O2 → 2N2 + 6H2O
According to the balanced equation,
4 volumes of ammonia require 3 volumes of oxygen to burn
4 | ICSE Model Specimen Papers, X
800 cc of ammonia will require = 3 × 800/4= 600 cc of oxygen
So, 600 cc of oxygen will be required to burn 800 cc of ammonia.4 volumes of ammonia produces 2 volumes of nitrogen
800 cc of ammonia will produces = 2 × 800 /4= 400 cc of nitrogen
So, 400 cc of nitrogen will be produced when 800 cc of ammonia burns.(b) Whenever gases combine chemically they do so in volume which bear a simple ratio to
each other as well as the product of gases under the same conditions of temperature andpressure.
C2H4 + 3O2 ⎯→ 2CO2 + 2H2O2 volumes of CO2 is produced by 1 volume ethane
∴ 200 cm3 of CO2 produced by 12 × 200 = 100 cm3 ethane
∴ 2 volumes of CO2 = 3 volumes O2
200 cm3 of CO2 =32 × 200 = 300 cm3 of oxygen
∴ Volume of ethane is 100 cm3 and volume of Oxygen is 300 cm3.Answer 6.(a) (i) Number of valence electrons of the elements of Group 1[A] is 1 and those in
elements of Group 17[A] are 7.(ii) Elements of Group 1[A] are good conductors of electricity while those of Group
17[A] are poor conductors of electricity.(iii) Elements of Group 1 [A] are good reducing agents while those of Group 17[A] are
good oxidizing agents.(iv) Elements of Group 1[A] are metals while those of Group 17[A] are non-metals.
(c) (i) Sulphur dioxide gas changes freshly made potassium dichromate (VI) paper fromorange to green. When carbon dioxide gas is bubbled through lime water, it turnsmilky.
(ii) When zinc sulphate and ferrous sulphate are dissolved in NaOH in two separate testtubes, white precipitate of zinc hydroxide and dark green precipate of ferroushydroxide are formed respectively.
