Section - I

Attempt all questions from this Section.

Answer 1. (a) (i) (D) Acetic acid (ii) (B) They can undergo addition as well as substitution reaction (iii) (C) Fluorine (iv) (B) 10 (v) (D) Ammonium chloride (b) (i) FeCl3 + 3NaOH ¾® Fe(OH)3 + 3NaCl (ii) 2Al + 2NaOH + 2H2O ¾® 2NaAlO2 + 3H2

(iii) Zn + 2NaOH ¾® Na2ZnO2 + H2 ↑ (iv) 2NaOH + SO2 ¾® Na2SO3 + H2O (v) CuSO4 + 2NaOH ¾® Na2SO4 + Cu(OH)2

(c) (i) Nitric oxide (ii) Ammonia (iii) Hydrogen chloride gas (iv) Oxygen (v) Nitrogen. (d) (i) conc. H2SO4 (ii) dil. HCl (iii) dil. H2SO4 (iv) conc. H2SO4

(v) conc. H2SO4. (e) (i) Molecular mass of urea (NH2CONH2) = (2 × 14) + (1 × 12) + (1 × 16) + (4 × 1) = 28 + 12 + 16 + 4 = 60 60 kg of urea contain N = 28 kg

5 kg of urea contain N = 2860

51

× = 146

= 73

= 2.33 kg \ Nitrogen supplied to the soil = 2.33 kg (ii) 1 mole of SO2 has mass = 64 g 1 mole of SO2 has volume at S.T.P. = 22.4 L 64 g of SO2 has volume at S.T.P. = 22.4

\ 8 g of SO2 has volume at S.T.P. = 22 464.

× 8

= 2.8 L (iii) 328 g of Ca (NO3)2 produces CaO = 112 g

\ 65.6 g of Ca (NO3)2 produce CaO = 112 65 6

328× .

produce CaO = 22.4 g

Model Specimen Paper Chemistry

3SOLVED

| 2 | ICSE Model Specimen Papers, X

(f) (i) (F) Calcination, (ii) (D) Hygroscopic, (iii) (C) Hydrocarbons, (iv) (E) Alkynes, (v) (A) Neutral oxide. (g) (i) Ethene, (ii) Methane, (iii) Propanal, (iv) Ethanoic acid, (v) Butanol. (h) (i) 1. 1- Methanal 2. Propanone 3. Ethoxy Ethane

(ii) 1.

H—C—C—C—C—H

—H —OH —H —H

—

H

—

H

—

H

—

H

2.

H C—C—CH3 3

—CH3

—

CH3

Section - IIAttempt any four questions from this Section.

Answer 2. (a) (i) The volume occupied by one mole of a gas at S.T.P. is 22.4 litres. (ii) Mass of 112 cm3 of phosphorus fluoride = 0.63 g

Mass of 22400 cm3 of phosphorus fluoride = 0 63

112 3. g

cm × 22400 cm3

= 126 g Relative molecular mass of the fluoride = 126 g Mass of fluorine in phosphorus fluoride = 126 – 31 = 95 g

Hence, number of fluorine in fluoride = 9519

= 5

Thus, the formula of phosphorus fluoride = PF5

(b) (i) NH3(g) + HCl(g) ¾® NH4Cl(g)

Ammonia Hydrogen Ammonium chloride chloride

(ii) 6FeSO4 + 3H2SO4 + 2HNO3 ∆ → 3Fe2(SO4)3 + 2NO + 4H2O

(c) (i) The white powder P is lead carbonate as it evolves CO2 on heating (property of carbonate salt) and metallic ion is lead as it leaves a yellow residue to lead monoxide.

PbCO3 ∆ → PbO + CO2.

(ii) The white crystalline compound (N) is ammonium chloride (NH4Cl).2NH4Cl + H2SO4 ¾® (NH4)2SO4 + 2HCl ↑

NH3 + HCl ¾® NH4Cl (dense white fumes) NH4Cl + NaOH ¾® NaCl + H2O + NH3 ↑ (Alkalies turn red litmus blue) \ The gases evolved are HCl and NH3. (iii) The green solid is ferrous sulphate (FeSO4). 1. The anion in F is sulphate [SO42–]. 2. The cation in F is Iron (II) Fe2+.Answer 3. (a) (i) In presence of catalyst like finely divided nickel, platinum, heating upto 473 K. (ii) Hot and concentrated alcoholic solution of potassium hydroxide. (iii) Platinum gauze catalyst at 800°C in presence of oxygen. (iv) Vanadium pentoxide is used as catalyst, for the conversion of sulphur dioxide to sulphur

trioxide.

Chemistry | 3 |

(b) (i) NaNO3 + H2SO4 ∆ < ° →200 C NaHSO4 + HNO3

Sodium Conc. Sodium Nitric nitrate hydrogen acid sulphate

(ii) C2H5Cl + NaOH ¾® C2H5OH + NaCl Monochloro aqueous Ethanol Sodium ethane chloride

(iii) Ca(OH)2 ∆ → CaO + H2O

Calcium Calcium hydroxide oxide

(iv) 1. Mercury 2. Calcium 3. AluminiumAnswer 4. (a) (i) 1 : 3 (ii) From air. (iii) Finely divided iron with little of molybdenum should be used as catalyst in this reaction.

(iv) N2(g) + 3H2(g) 400 450200

− °Catm. Fe

� ⇀������↽ ������� NH3(g), ∆H = – 92 kJ mol–1

(b) (i) Ethanol to acetic acid

C2H5OH dil. H SO OK Cr O

2 4

2 2 7

[ ] → CH3CHO [ ]OK Cr O2 2 7

→ CH3COOH

Ethanol Ethanoic acid (acetic acid)

(ii) Ethane to ethene

C2H6 5002 3

° →CAl O

C2H4 + H2 Ethane

Ethene

(iii) Methane to methyl alcohol CH4 + Cl2 ¾® CH3Cl + HCl Methane

CH3Cl + KOH ¾® CH3OH + KCl Methyl alcohol

(c) (i) Formation of NH4+ ion :

N×

××H

H

HAmmonia

molecule

+

Lone pair

H+

N×

×× HH

H

HAmmonium

ionHydrogen ion

(proton)

+

(ii)

OH

Water

+ H+

Hydronium ionHydrogen ion

×× OH ×× H

+

H

(iii)

ClH +× Cl×H H�+

Cl�–

Answer 5. (a) 2H2O(l) ¾® 2H2(g) + O2(g) 2 Vol. 1 Vol.

(i) From the above equation, it is clear that the volume of O2 evolved is half that of hydrogen.

| 4 | ICSE Model Specimen Papers, X

Hence, volume of O2 evolved = 2500

2

3cm = 1250 cm3

(ii) (1) When silver chloride is 143.5 g, KCl required = 74.5 g

When silver chloride is 2.87 g, KCl required = 74 5 2 87

143 5. .

.×

= 1.49 g

(2) When silver chloride is 143.5, AgNO3 required = 170 g

When silver chloride is 2.87 g, AgNO3 required = 170 2 87143 5× ..

= 3.40 g

(iii)

Element Percentage Weight

Atomic Weight

Relative No. of Atoms

Simplest Ratio

H 2.47 1 2 471.

= 2.472 471 23..

= 2

P 38.27 31 38 2731.

= 1.231 231 23..

= 1

O 59.26 16 59 2616.

= 3.703 701 23..

= 3

Simplest ratio, therefore is H : P : O :: 2 : 1 : 3 Therefore, the empirical formula is H2PO3 and molecular formula is (H2PO3)n Empirical formula weight = 2 × 1 + 31 + 3 × 16 = 81

n = Molecular weight

Empirical formula weight = 16281

= 2

Therefore, molecular formula is (H2PO3)2 , i.e., H4P2O6. (b) (i) 3rd period, 16th group. (ii) Z is a non-metal.Answer 6. (a) All ammonium salts on reaction with alkalies give metal salt, steam and ammonia :

Ca(OH)2 + 2NH4Cl ¾® CaCl2 + 2H2O + 2NH3 ↑

2NaOH + (NH4)2SO4 ¾® Na2SO4 + 2H2O + 2NH3↑

3KOH + (NH4)3PO4 ¾® K3PO4 + 3H2O + 3NH3↑

(b) (i) Iron + Chlorine :2Fe + 3Cl2 ¾® 2FeCl3

(ii) Copper oxide + dil. sulphuric acid :CuO + H2SO4 ¾® CuSO4 + H2O (Double decomposition)

(iii) Sodium hydroxide + dil. sulphuric acid :2NaOH + H2SO4 ¾® Na2SO4 + 2H2O (Neutralization)

(c) (i) Solution X (ii) Solution Z (iii) Solution Y.Answer 7. (a) (i) Mg + H2SO4 ¾® MgSO4 + H2 Magnesium dil. sulphuric acid Magnesium

Sulphate

Chemistry | 5 |

(ii) CuSO4. 5H2O ∆ → CuSO4 + 5H2O

Copper sulphate Coper pentahydrate sulphate (Blue) (White)

(iii) NaCl + H2SO4 ¾® NaHSO4 + HCl Sodium Sulphuric Sodium

chloride acid hydrogen

(conc.) sulphate

(iv) ZnCO3 ∆ → ZnO + CO2

Zinc Zinc oxide Carbon dioxide carbonate

(b) (i) A is cathode and B is anode. (ii) Molten fluorides of Al, Na and Ba. (iii) Graphite rods dipped in molten pure aluminium. (c) (i)

Use of Metal PropertyZinc in galvanisation. Not affected by air and moistureAluminium in thermite welding. Strong affinity for oxygen as

compared to iron. (ii) MgCl2

Mg ××

Cl

Cl

(2, 8, 7)

(2, 8, 2)

Mg2+ 2 Cl×

Ionic bond exists in MgCl2. NH3

NH

H

H××

×

Covalent bond exists in NH3. (iii) The elements of the second period shows resemblance in properties with the elements of

the next group of the third period, due to very less electronegativity difference.�

Section - I

Attempt all questions from this Section.

Answer 1. (a) (i) (B) A low melting point and low boiling point. (ii) (D) Ethyne (iii) (A) It is a strong reducing agent. (iv) (D) Atomic radius decreases and nuclear charge incrreases. (v) (C) A silver grey deposit at cathode and reddish brown fumes at anode.

(b) (i) Cu + 2H2SO4 ∆ → CuSO4 + SO2 + 2H2O

(ii) (NH4)2SO4 + 2NaOH ∆ → Na2SO4 + 2NH3 + 2H2O (iii) 2Al + 2NaOH + 2H2O ¾® 2NaAlO2 + 3H2

(iv) 4NH3 + 5O2 Pt C, 700° → 4NO + H2O

(v) 2NaNO3 ∆ → 2NaNO2 + O2

(c) (i) It forms a chalky white precipitate which dissolves in excess of ammonium hydroxide. (ii) The bromine vapour (reddish brown) decolourise. (iii) The blue colour of copper sulphate fades to form a colourless solution. (iv) Reddish brown gas (nitrogen dioxide) is evolved. The residue is yellow when hot, but

changes to white colour on cooling. (v) A reddish brown precipitate is formed, which does not redissolve in excess of sodium

hydroxide solution. (d) (i) Concentrated nitric acid. [When sulphur is heated with concentrated nitric acid, it

produces non-volatile sulphuric acid.] (ii) Concentrated sulphuric acid. (iii) Nitric acid. (iv) Dilute sulphuric acid. (v) Dilute hydrochloric acid. (e) (i) 44 g of CO2 = 6·023 × 1023 molecules 4·4 g CO2 = 6·023 × 1022 molecules (ii) 44 g CO2 = 1 mole = 22·4 L at STP.

4·4 g CO2 = 110

moles = 2·24 L at STP.

(iii) It will contain same numbers of oxygen molecules, i.e., 6·023 × 1023 molecules 6·023 × 1023 molecules = 32 g oxygen \ 6·023 × 1022 molecules = 3·2 g oxygen (e) (i) 2C2H6 + 7O2 ¾® 4CO2 + 6H2O 2 volume of ethane = 7 volume of oxygen

Model Specimen Paper Chemistry

6SOLVED

| 2 | ICSE Model Specimen Papers, X

2 volume of ethane = 100 litre \ 1 volume of ethane = 50 litre 7 volume of O2 = 50 × 7 = 350 litre \ 100 litre of ethane require = 350 litre of oxygen. (ii) According to equation, 4NH3 + 5O2 ¾® 4NO + 6H2O 5 vol. 4 vol. \ 5 volumes of reactants are consumed to give 4 volumes of nitrogen monoxide.

\ 27 litre of reactants are consumed to give = 27 45×

= 1085

= 21.6 litre. (f) (i) Aluminium (Al) (ii) Sodium (Na) (iii) Magnesium (Mg) (iv) Chlorine (Cl) (v) Chlorine (Cl) (vi) Phosphorus (P) (g) (i) Zinc ion is lower in electrochemical series. Thus, carbon monoxide provides enough

energy and hence reduces zinc oxide to zinc. Aluminium ion is higher in electrochemical series. The carbon monoxide does not produce enough energy and hence aluminium oxide is not reduced to aluminium.

(ii) Carbon tetrachloride is a covalent compound and hence does not have ions which are responsible for the conduction of current through a liquid.

(iii) Bromide ions discharge at anode to form bromine gas. This gas is highly corrosive and reacts with metallic anode. To avoid corrosion of anode, graphite is preferred as it does not react with bromine and is a good conductor of electricity.

(iv) Theaceticacidmoleculesdissociateverylittle(about4%)comparedtosulphuricacidmolecules (about 90%). Thus, electrical conductivity of dilute acetic acid is less thandilute sulphuric acid.

(v) The lead ions gain electrons at cathode to form lead atoms and hence are reduced. The bromide ions donate electrons at anode and hence are oxidised at the same time. Thus, on the whole decomposition of lead bromide is a redox reaction.

(h) (i) (a) Ethanoic acid (b) 1-chloro-2-methyl propane (c) Ethanedioic acid

(ii) (a)

H—C—C

—H

——O

—H

Ethanal

—

H

(b) H—C C — H�Ethyne

Section - IIAttempt any four questions from this Section

Answer 2. (a) (i) ZnS + 2HCl ¾® ZnCl2 + H2S↑ (ii) FeS + H2SO4 ¾® FeSO4 + H2S↑ (iii) HCl + NaOH ¾® NaCl + H2O (b) (i) Itiscolourlessandinflammableliquid. (ii) It has a pungent smell. (iii) It is lighter than air.

Chemistry | 3 |

(c) (i) Cryolite : It acts as a solvent for the electrolytic mixture. (ii) NaOH : It removes insoluble impurities from the ore.

(iii) Graphite : It is used as an electrode.Answer 3. (a) (i) Second period (ii) N (nitrogen) and it should be placed after carbon before oxygen. (iii) C (Carbon) (iv) The increasing order of electronegativity is Be < N < F.

(b) (i) 4NH3 + 5O2 Pt

C800° → 4NO + 6H2O

(ii) Platinum (iii) ItisthefirststepinmanufactureofnitricacidbyOstwald’s process. (iv) 4NH3 + 3O2 ¾® 2N2 + 6H2OAnswer 4. (a) (i) Vanadium pentaoxide or platinum. (ii) 450 to 500°C. (iii) 1. Excess of sulphur dioxide and oxygen. 2. High pressure to be used. (iv) SO3 is not directly dissolved in water because a highly exothermic reaction takes place

due to which sulphuric acid fog or mist is formed. This fog consists of sulphuric acid particles enclosing readily made steam as well as unreacted water and sulphur trioxide. The mist prevents the absorption of SO3 which passes out alongwith outgoing gases.

(b) Atomic no. 16 ® 2, 8, 6 K L M (i) 3 period (because it has three shells) (ii) 6 valence electrons (valence electrons are present in outermost orbit). (iii) Non-metal (elements having 5, 6 or 7 valence electrons are non-metals). (c) (i) Al4C3 + 12H2O ¾® 3CH4 + 4Al(OH)3

(ii) CH3OH + 2[O] K Cr OH SO2 2 7

2 4+ → HCOOH + H2O

(iii) C2H5OH + PCl5 ¾® C2H5Cl + POCl3 + HClAnswer 5. (a) (i) (NH4)2Cr2O7 = [(14 + 1 × 4) × 2] + (52 × 2) + (16 × 7) = 252 g

(ii) 252g (NH4)2Cr2O7 = 1 mole, 63 g (NH4)2Cr2O7 = 63252

= 0.25 mole

(iii) 1 mole of (NH4)2CrO7 liberates 0.25 moles of N2

\ 0.25 mole of (NH4)2CrO7 liberates 0.25 moles of N2 (iv) 0.25 mole of N2 at STP has volume = 22.4 × 0.25 = 5.6 litre (b) (i) To reduce the heat loss by radiation and to prevent burning of anode. (ii) To prevent them from rusting. (iii) On reacting with nitric acid a protective layer of Al2O3 is formed which prevents further

corrosion.Answer 6. (a) (i) Purple (ii) Green (iii) Pink (iv) Colourless

| 4 | ICSE Model Specimen Papers, X

(b) (i) Al2O3.2H2O + 2NaOH 140 150− ° →CDigestion 2NaAlO2 + 3H2O

(ii) NaAlO2 + 2H2O 50° →CPrecipitation

NaOH + Al(OH)3↓

(iii) 2Al(OH)3 1000° →CCalcination Al2O3 + 3H2O

(c) (i) Urea (ii) Methane (iii) Addition reactions.Answer 7. (a) (i) 3rd period, 16th group. (ii) Z is a non-metal. (iii) H2Z (iv) Covalent compound. (b) (i) Na2O (ii) Al2O3 (iii) SiO2

(c) (i) Ionic compound (ii) Y + 3e– ¾® Y3– (iii) 3X + Y2 ¾® X3Y2

�

Section - I

Attempt all questions from this Section.

Answer 1. (a) (i) (D) Salt (ii) (D) Lead (II) bromide (iii) (C) Ethyne (iv) (A) Formic acid (v) (A) Oxidation

(b) (i) CH3COONa + NaOH CaO → CH4 + Na2CO3

(ii) C2H5OH conc. H SOC2 4

170° → CH2 = CH2 + H2O

(iii) CH3COOH + NaOH ¾® CH3COONa + H2O

(iv) CH3CH2OH + CH3COOH conc. H SO2 4 → CH3COOC2H5 + H2O

(v) CH2 = CH2 + H2 Ni

C200° → CH3—CH3

(c) (i) Cu2+ and H+ ions move towards cathode and SO42 –, OH– ions move towards anode.

CuSO4 Cu2+ + SO42–

(cathode) (anode)

H2O H+ + OH–

(cathode) (anode)

(ii) When copper nitrate is heated at low temperature, it loses water of crystallisation and deep blue colour changes into green by anhydrous copper nitrate and copper nitrate when heated strongly it is decomposed to form copper oxide, nitrogen dioxide and oxygen.

Cu(NO3)2·3H2O ∆ → Cu(NO3)2 + 3H2O Hydrated copper Anhydrous copper nitrate

nitrate (deep blue) (green mass)

2Cu(NO3)2 ∆ → 2CuO + 4NO2 + O2

(iii) The colourless hydrogen sulphide is obtained, which has a rotten egg smell.

ZnS + 2HCl ∆ → ZnCl2 + H2S Hydrogen sulphide

(iv) When copper is treated with conc. H2SO4, it gives sulphur dioxide and copper sulphate.

Cu + H2SO4 ∆ → CuSO4 + 2H2O + SO2

(v) When ammonium chloride is heated, it dissociates into ammonia and hydrogen chloride. It is a reversible reaction.

NH4Cl heatingcooling

� ⇀�����↽ ������ NH3 + HCl

Model Specimen Paper Chemistry

9SOLVED

| 2 | ICSE Model Specimen Papers, X

(d) (i) Nitrous oxide (N2O) (ii) Hydrogen chloride (HCl) (iii) Chlorine (Cl2) gas (iv) Chlorine gas (v) Hydrogen sulphide.

(e) 2NH3 + 212

O2 ¾® 2NO + 3H2O

212

× 22.4 dm3 2 moles 3 moles

56 dm3 at STP 60 g 5.4 g (i) When NO formed is 60 g the mass of steam formed = 54 g

\ NO formed is 1.5 g, the mass of steam formed = 54 1 5

60× .

= 1.35 g

(ii) For 5 moles of products oxygen required = 56 dm3 at STP

\ For 10 moles of products oxygen required = 56 10

60×

= 112 dm3 STP

(iii) 1 mole of N2O = 1 g molecule of N2O = (2 × 14) + 16 = 44 g 44 g of N2O = 1 mole

\ 3.3 g of N2O = 3 344.

mole

= 0.075 mole. (f) (i) Sodium (ii) Ammonia (iii) Nitrous oxide (iv) Chromium and nickel (v) Hydrogen chloride (g) (i) (D) Bleaching (ii) (C) Bessemerisation (iii) (F) Electrolysis (iv) (A) Annealing (v) (E) Atomic mass

(h) (i) (a)

H—C—C—C—H

—H—H—

H

—H

—

H

—

H—C—H—

H

(b)

H—C—C—C—H—

H

—H —OH —H

—

H

—

H2-Propanol

(2-methyl propane (an isomer of n-butane)

(c)

H—C—C—O—C—C—H—

H

—H —H

—

H

—

H

—H —H

—

H

Diethyle ether

(ii) (a) Propyne (b) 1, 2-dichloro ethane

Section - IIAttempt any four questions from this Section.

Answer 2. (a) (i) Covalent compounds exists as gases, liquids or soft solids because they are held by

relatively weaker forces. (ii) Anode is the oxidising electrode because anions lose electrons at anode.

Chemistry | 3 |

(b)

H—O H—

H

+

(H3O+) hydronium ion

(c) (i) 3rd period, 16th group (ii) Z is a non-metal. (iii) H2ZAnswer 3. (a) (i) Relative molecular mass of compound = 2 × Vapour density = 2 × 29 = 58 (ii) The empirical formula mass of C2H5 = (12 × 2) + (1 × 5) = 29

n = Molecular wt.Empirical wt.

= 5829

= 2

\ Molecular formula of the compound = (C2H5)2 = C4H10

(iii) Butane.

H—C—C—C—C—H—

H

—H —H

—H

—

H

—H —H

—

H

(iv) Molecular formula is C4H10 = (12 × 4) + (1 × 10) \ Molecular weight = 48 + 10 = 58 g (b) (i) The number of shells increases with increasing atomic number, therefore, size of atom

increases and hence, the atomic radius increases as we move from top to bottom in a group.

(ii) The lithium atom is smallest in its group, thus its nucleus attracts the outer electron strongly and hence, its first ionisation energy is very high.

(iii) (1) Beryllium, (2) Oxygen. (iv) CnH2n.Answer 4. (a) (i) Mg3N2 + 6H2O ¾® 3Mg(OH)2 + 2NH3

(ii) NH4Cl + NaOH ¾® NaCl + H2O + NH3

(iii) (NH4)2 SO4 ∆ → 2NH3 + H2SO4

(b) (i) Ammonia (ii) Acetic acid (iii) Hydrogen chloride (iv) Dilute sulphuric acid (v) Helium (c) (i) Covalent compounds are generally insoluble in water, whereas the electrovalent

compounds are generally soluble in water. (ii) Covalent compounds are generally gases, liquids or soft solids. However, electrovalent

compounds are crystalline in nature. (iii) Covalent compounds are volatile in nature and thus they have low melting and boiling

points whereas electrovalent compounds are non-volatile in nature and thus they have high melting and boiling point.

Answer 5. (a) (i) When dilute sulphuric acid is added to sodium nitrate no change is observed whereas

when dilute sulphuric acid is added to sodium sulphite colourless gas SO2 is evolved with the smell of burning sulphur which terms moist blue litmus paper red.

(ii) The starch iodide paper turns blue-black. (iii) Substance R is alkaline.

| 4 | ICSE Model Specimen Papers, X

(b) (i) Y is hydrochloride (HCl) gas. (ii) Gas Y is highly soluble in water. (iii) Ammonia gas.Answer 6. (a) (i) Ni2+ ion (ii) When anode is inert in nature oxygen gas is the product obtained. (b) (i) Electrolysis equation NaCl Na+ + Cl–

At cathode : Na+ + e– ® Na At anode : Cl– + e– ® Cl Cl + Cl ® Cl2 (ii) (a) CuSO4 Cu2+ + SO4

2–

(b) H2O H+ + OH–

(iii) 2Mg + O2 ¾® 2MgO

(c) (i) C2H4 + H2SO4 (conc.) 8030

° →Catms C2H5 — HSO4

C2H5 — HSO4 + H2O (boiling) ¾® C2H5OH + H2

(ii)

C

×

×

××

Cl

ClCl

Cl

×××

××

××× ××

××

××

××

××

××

××

××

××

××××

Structural formula of carbon tetrachloride is CCl4. The type of bond present in CCl4 is covalent bond.

(iii) C2H5OH [ ]OK Cr O2 2 7

→ CH3CHO + H2O [ ]OK Cr O2 2 7

→ CH3COOH

Answer 7. (a) (i) Haematite (Ferric (III) oxide), Fe2O3.

(ii) 4Fe + 3O2 + 3H2O H O2 → 2Fe2O3.3H2O (iii) 73% Fe, 18% Cr, 8% Ni, 1% C. Steel is widely used in the construction of roads. (iv)

S. No. Roasting Calcination1. Carbonates ores should be processed

with this procedure.Sulphide ores should be processed with this procedure.

2. Volatile impurities are given out. Organic and inorganic matter are oxidised.3. Carried out in presence of air. Carried out in absence of air.

(b) (i) The process is called galvanisation. (ii) It is done to prevent rusting of iron. (iii) Stainless steel is an alloy of iron, chromium, nickel and carbon. It has silvery white

shining surface, which is unaffected by dilute acid or alkali and resists corrosion. It is used in making household utensils, cutlery and food processing machinery.

(c) (i) The chemical equation representing the reaction is : 2C4H10 + 13O2 ¾® 8CO2 + 10H2O 2 × [12 × 4) + (1 × 10)] 13 moles 8 × 22.4 2 × [48 + 10] = 116 g = 179.2 litres

Chemistry | 5 |

116 g of butane for complete combustion requires 13 moles of oxygen.

(ii) \ 58 g of butane for complete combustion requires 13 58116× = 6.5 moles of O2.

116 g of butane forms 179.2 litres of CO2.

\ 58 g of butane forms 179 2 58116. × litres of CO2 = 89.6 litres.

(iii) (a) Empirical formula of C3H7COOH is C2H4O (b) Molecular weight = 2 × Vapour density = 2 × 32 = 64 64 g of compound Will occupy at S.T.P = 22.4 dm3

\ 1.6 g of compound

will occupy at S.T.P. = 22 4 1 6

64. .×

= 0.56 dm3.�

Section - I

Attempt all questions from this Section.

Answer 1. (a) (i) (A) Carbon (ii) (C) Three (iii) (C) Halogens (iv) (B) Magnesium (v) (B) Lead hydroxide (b) (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 + 8HC1 (cone.) → 3PbCl2 + Cl2 + 4H2O (c) (i) When acetylene is passed over bromine water, it decolourizes bromine solution in carbon

tetrachloride. (ii) When sulphur dioxide gas is passed iixto potassium dichromate solution, its colour

changes from orange to green and the chromium sulphate and potassium sulphate are formed.

(iii) When blue crystals of copper nitrate are heated, they lose their water of crystallisation. This reaction produces copper oxide, reddish brown nitrogen dioxide gas and oxygen gas.

(iv) When ammonium hydroxide solution is added to ferric nitrate solution, it forms gelatinous brown precipitate of ferric hydroxide and ammonium nitrate.

(v) When excess of sodium hydroxide is added to zinc chloride solution, white precipitate of zinc hydroxide is formed which dissolves in excess of sodium hydroxide and forms a colourless solution.

(d) (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 metals. (iv) Ammonia reduces metallic oxides to metals. (v) The brown ring of nitroso ferrous sulphate is formed at the junction of the two liquids. (e) (i) 1. 6 × 1023 molecules of N2 will weigh 28 g

24 × 1024 molecules of N2 will weigh = 28 24 10

6 10

24

23× ×

× = 1120 g Mass of N2 gas in cylinder is 1120 g. 2. 6 × 1023 molecules of N2 has a volume of 22.4 L

24 x 1024 molecules of N2 has a volume = 22 4 24 106 10

24

23. × ×

× = 896 L

So, volume of Nitrogen at S.T.P. = 896 L

Model Specimen Paper Chemistry

12SOLVED

| 2 | ICSE Model Specimen Papers, X

(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 dioxide 2 volumes of ethane require 7 volumes of oxygen

600 cc of ethane will require =

7 600

2×

cc of oxygen

= 2100 cc of oxygen

2. Unused volume of oxygen is (3000 - 2100) = 900 cc of oxygen. (f) (i) Ca2+ (ii) Solution of sodium argentocyanide (iii) Molecules (iv) Carboxylic acid (v) Ammonium chloride. (g) (i) Compound P – Ethanol Compound Q – Ethanoic acid Compound R – Ethyl ethanoate (ii) CH2COOH + C2H5OH → CH3COOC2H5 + H5O (iii) The above reaction is called as an Esterification reaction. (h) (a) (i) Methanol (ii) Methane (iii) Acetic acid.

(b) (i)

H—C—C—H

H

——

H

H

——

H

(ii)

H—C—C—C—OH

H

——

H

H

——

H O— —

Section - II

Attempt any four questions from this Section.Answer 2. (a)

METALS NON-METALSPhysical Properties1. Metals are generally solids at ordinary

temperature. (Mercury is liquid at room temperature.)

2. They volatilise at high temperature.

3. They are hard in nature. (Exceptions being sodium, potassium and

lead which are soft.)

Non-metals are usefully gases or liquids. (Carbon, sulphur, phosphorus are non-metallic)They volatilize at low temperature.(Exceptions being carbon, boron and silicon.)They are soft in nature.(Diamond is the hardest known naturally occurring substance.)

Chemical Properties1. They form basic oxides. (But ZnO, Al2O3, SnO2 and PbO are

amphoteric and Cr2O3 and Mn2O7 are acidic.)

2. They have 1, 2 or 3 valence electrons.3. They form cations by loss of electrons.

They form acidic oxides.(But CO, NO, H2O and N2O are neutral oxides.)

They 4, 5, 6, 7 or 8 valence electrons.They form anions by the gain of electrons.

Chemistry | 3 |

(b) (i) C2H5Br + 2H

Zn/HgC H OH2 5

C2H6 + HBr Ethyl bromide Ethane

(ii) C2H5COONa + NaOH

CaO300°C →

C2H6 + Na2CO3

(iii) CH3COONa + NaOH

CaO300°C →

CH4 + Na2CO3

(c) Characteristics of the electrochemical series. l This series indicates the ease with which atoms of the elements give up their electrons to

form ions. l Higher the metal in the series stronger is its tendency to exist as ions in solution. l The series provides a useful system of classification as the elements are placed in a

decreasing order of reactivity, i.e., the most reactive metals are placed on top and on going down the reactivity decreases.

l The series also shows which element will displace the other in a solution. Metals placed above in the metal reactivity series will displace metals placed below them.

Answer 3. (a) (i) H2 + Cl2 → 2HCl (ii) CuCO3 → CuO + CO2

(iii) pHisdefinedaspotentialofhydrogen,thatisthenegativelogarithmtothebase10ofthehydrogen ion concentration in solution.

(iv) CaCl2 + (NH4)2CO3 → CaCO3 ↓ + 2NH4Cl (b) (i) All transition elements are metals. (ii) The atomic radius increases down the group. (iii) Non-metallic character increases across a period. (iv) Francium.Answer 4. (a) (i) Least reactive element of period 3 is x. (ii) The element with largest atomic radius in group 1 is B. (iii) The element with atomic number 11 is B. The formula for its oxide is B2O. (iv) Theelementwithhighestelectronaffinityinperiod3isu. (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– → OH 4OH → 2H2O + O2

(c) (i) On adding concentrated sulphuric acid to sugar, initially the crystals become brown, steamisreleasedwhichcausesalotoffrothingandfinallyablackporousmassisleftbehind.

(ii) Concentrated nitric acid is not used during preparation of HCI because it is volatile acid. (iii) Methane does not undergo addition reaction because it is bonded with four H-atom

while in ethene double bond breaks and provides site for addition.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

| 4 | ICSE Model Specimen Papers, X

(ii) 4NH3 + 3O2 → 2N2 + 6H2O According to the balanced equation, 4 volumes of ammonia require 3 volumes of oxygen to burn

800 cc of ammonia will require = 3 × 8004

= 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) (i) C + 2H2SO4 → CO2 + SO2 + 2H2O (ii) Mg + H2SO4 → MgSO4 + H2

(iii) C12H22O11 2 4conc.H SO→ 12C + 11H2OAnswer 6. (a) (i) The structural formulae for ethyl methyl ether

H—C—C—O—C—H

H

——

H

H

——

H

ethyl methyl ether

—H

H

—

Ethyl methyl ether (ii) The structural formulae for isobutanol

CH —3

CH—CH2—OH

CH3—

(iii) The structural formulae for 1-Chloro propyneCl—CH2—C≡CH

(b) (i)Common Name Chemical Name Formula

1. Cryolite2. Rust

Sodium hexafluoro aluminateHydrated ferric oxide

Na3AlF6

Fe2O3.xH2O (ii) 1. Slag :Thefusibleproductformedwhenfluxreactswithgangueduringtheextraction

of metals. 2. Calcination : It is the process of heating a concentrated ore in the absence of air or

oxygen. (c) Electron dot structure of ammonia gas.

H×

..H—N—H

H

—

H ×· N: + H+.

.×H

H×

H × ·N: H+→..

×H

Ammoniamolecule

Hydrogenion

H×

H ×· N: H+→.

.×H

coordinate bondlone pair ofelectrons

Ammonia molecule Hydrogen ion

Chemistry | 5 |

Answer 7. 7. (a) (i) C2H6 + [O] → C2H5OH (ii) 2C2H5OH + 2Na → 2C2H5ONa + H2

(iii) 2CH3COOH + 2H2 → 2CH3CH2OH + O2

(iv) CH4 → C + 2H2

(b) (i) Sulphur dioxide gas changes freshly made potassium dichromate (VI) paper from orange to green. When carbon dioxide gas is bubbled through lime water, it turns milky.

(ii) When zinc sulphate and ferrous sulphate are dissolved in NaOH in two separate test tubes, white precipitate of zinc hydroxide and dirty green precipate of ferrous hydroxide are formed respectively.

(iii) Iron (II) chlorides gives dirty green precipitate with NaOH solution while iron (III) chloride gives reddish brown precipitate with NaOH Solution.

(c) (i)

Element Percentage composition by mass

Atomic massRelative number of

atoms Simple ratio

C 41.37 12 3.45 3H 5.75 1 5.75 5N 16.09 14 1.15 1O 36.79 16 2.30 2

Empirical Formula = C3H5NO2

Empirical Formula Weight = (12 × 3) + (1 × 5) + (1 × 14) + (16 × 2) = 87 Molecular Weight = 2 × Vapour Density = 2 × 43.5 = 87

n =

Molecular WeightEmpirical Formula Weight

= 8787

n = 1 Molecular formula = C3H5NO2

(ii) C2H4 + 3O2 → 2CO2 + 2H2O 2 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.ll

Section - I

Attempt all questions from this Section.

Answer 1. (a) (i) (B) Tetraamine copper (II) sulphate (ii) (C) Aluminium (iii) (A) Lead chloride (iv) (D) Argon (v) (D) Ionizes when dissolved in water (b) (i) Pb(NO3)2(aq) + H2SO4(aq) → PbSO4(s) + 2HNO3(aq) (ii) Cu(s) + 2H2SO4(conc.) → CuSO4(aq) + 2H2O(l) + SO2(g) (iii) Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) (iv) 2NH3(g) + H2SO4(aq) → (NH4)2SO4(aq) (v) Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l) (c) (i) Initially the moist blue litmus paper turns red and then gets bleached. (ii) No change takes place in the colour of dry rose petals. (iii) The violet red colour of potassium permanganate disappears. (iv) Ammoniaburnswithalightblueflameformingnitrogenandsteam. (v) Dense white fumes of ammonium chloride are seen around the glass rod. (d) (i) Copper sulphate Substances used – Copper oxide and dilute sulphuric acid

CuO + H2SO4(dil.) → CuSO4 + H2O (ii) Iron sulphide Substances used – Iron and sulphur

Fe + S → FeS (iii) Zinc sulphate Substances used – Zinc and dilute sulphuric acid

Zn + H2SO4 (dil.) → ZnSO4 + H2

(iv) Magnesium carbonate Substances used – Magnesium sulphate and sodium carbonate MgSO4 + Na2CO3 → MgCO3 + Na2SO4

(v) Copper carbonate Substances used – Copper sulphate and sodium carbonate

CuSO4 + Na2CO3 → CuCO3 + Na2SO4

(e) (i) CaCO3 + 2HC1 (dil.) → CaCl2 + H2O + CO2 (ii) Molecular weight of calcium carbonate = 100 g Mass of 4.5 moles of calcium carbonate = 100 x 4.5 = 450 g

Model Specimen Paper Chemistry

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| 2 | ICSE Model Specimen Papers, X

(iii) At STP, 1 mole of CaCO3 liberates 1 molar volume of CO2. 4.5 moles of CaCO3 will liberate = 4.5 x 22.4 L of CO2. = 100.8 L of CO2

(iv) 1 mole of CaCO3 require 2 moles of HC1. 4.5 moles of CaCO3 will require = 9 moles of HCl. (f) (i) Copper hydroxide (ii) Magnesium hydroxide (iii) Electronegativity (iv) Ni (v) Thermal decomposition (g) (i) Hydrochloride gas and ammonia (ii) Hydrogen and oxygen (iii) Zinc chloride and copper chloride (iv) Nitrogen and oxygen (v) Hydrogen and nitrogen (h) (i) Isopentane (ii) Propan-2-ol

H—C—C—C—C—H

H—C—H

H

——

H

H

——

H

——

H

H

——

H

—

H

H—C—C — C—H

H—

—

H

H

—

H

——

H

—

OH

(iii) Methanoic acid (iv) But-2-ene

H—C—O—H— —O

CH

H—

—

H

—

—C

—

H

——

C

H

—

—C

—

H

—

—

H

H

(v) Neopentane

H—C—C—C—H

H—C—H

H

——

H

H

——

H

——

H

—

H

—

H—C—H

Section - IIAttempt any four questions from this Section.

Answer 2. (a) (i) Helium (ii) Fluorine (iii) Na > Mg > Al > Si > P > S > Cl (b) (i) Covalent (ii) QP4 (c) Hydronium ion formation

O O O H+ H+.. ..

.. ..

H H HH H H

..

Lone pair

.. ..

Coordinate bond

+ +

Donor

Acceptor

×× × × × ×

Chemistry | 3 |

Answer 3. (a) (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 (b) (i) Propanoic acid (ii) Diethyl ether

H—C—C—C

H

——

H

H

——

HO

O—H

H—C—C—O—C—C—H

H

——

H

H

——

H

H

——

H

H

——

H

(iii) Pentan-2-ol (iv) An isomer of n-butane

H—C—C—C—C—C—H

H

——

H

OH

——

H

H

——

H

H

——

H

H

——

H

H—C—C—C—H

H

——

H

H

—

H

——

H

—

H

—

H—C—H

Answer 4. (a) (i) Solution C (ii) Salt and water (iii) Solution D (iv) Solution A and B (b) (i) Ostwald’s process (ii) (1) Catalytic chamber

4NH3 + 5O2 Pt

C880° → 4NO + 6H2O

(2) Oxidation chamber

2NO + O2 50°C → 2NO2

(3) Absorption Tower4NO2 + 2H2O + O2―→ 4HNO3

(ii) Ratio – 1: 8 (c) (i) Calculated amount of water is used in dilution of oleum to obtain concentrated sulphuric

acid of desired concentration. (ii) Vanadium pentoxide is preferred as a calalyst during catalytic oxidation of SO2 because

of the high oxidation state of vanadium. (iii) SO3 + H2SO4―→ H2S2O7

(Oleum)

Answer 5. (a) (i) At cathode :

Al3+ + 3e– → Al (ii) Carbon anodes are periodically replaced because they get oxidised to carbon dioxide. (iii) Hoope’s Process

Pb3O4 + 8HC1 → 3PbCl2 + 4H2O + C12

(b) (i) Molecular weight of red lead = (207 × 3) + (16 × 4) = 621 + 64 = 685 g

| 4 | ICSE Model Specimen Papers, X

Number of moles of red lead taken = 6.85 / 685 = 0.01 moles 1 mole of red lead forms 3 moles of lead chloride 0.01 mole of red lead will form = (3 x 0.01) moles of lead chloride = 0.03 moles Molecular weight of lead chloride = 207 + (35.5 x 2) = 278 g Lead chloride formed = 0.03 moles = 0.03 x 278 = 8.34 g (ii) At S.T.P., 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 x 22.4) L of chlorine = 0.224 L of chlorineAnswer 6. (a)

Elements Percentage Ratio Atomic mass Relative No. of atoms Simplest ratioCHBr

12.672.1385.11

12180

12.67/12 = 1.0552.13/1 = 2.13

85.11/80 = 1.063

1.055/1.055 = 12.13/1.055 = 2 1.063/1.055 = 1

∴ Emprical formula of the compound is CH2Br Molecular formula = (Empirical fprmula)n

n =

Molecular WeightEmpirical formula weight

=

2×Vapour densityEmpirical formula weight

=

2 9412 2 80

2 9494

2×+ +

= × =( )

∴ Molecular formula = (CH2Br)2 = C2H4Br2

(b) (i) Flux : Afluxisasubstancethatisaddedtothechargeinafurnacetoremovethegangue. Slag : It is the fusibleproductwhenflux reactswith impurities (gangue)during the

extraction of metals. (ii) Minerals : The naturally occurring compounds of metals which are generally mixed

withothermattersuchassoil,sand,limestoneandrocksareknownasminerals. (c) (i) The factors on which the electronegativity values depends are : 1. Size of the atom. 2. Nuclear charge. (ii) Electronegativity increases from left to right in a period since the nuclear charge increases

due to an increase in atomic number. (iii) Mass number (A) = No. of protons (P) + No. of neutrons (n)Answer 7. (a) (i) Baeyer’s process (ii) Na3AlF6

Chemistry | 5 |

(iii) Cryolite reduces the melting point or fusion temperature from 2050°C to 950°C of pure alumina. It also increases the electrical conductivity of the electrolyte which helps in saving electricity.

(iv) At cathode : Al3+ 3e– → Al

At anode : O 2+ 2e– → [O] [O]+[O] → O2

2O + O2 → CO2

2CO + O2 → 2CO2

(b) (i) CaC2 + 2H2O → Ca(OH)2 + C2H2

(ii) C2H2 Br + KOH(aq) → C2H2OH + KBr (iii) CH3I + 2[H] → CH4 + HI (c) (i) Iron (ii) Iron (iii) Sulphur

Section - I

Attempt all questions from this Section.

Answer 1. (a) (i) (C) Have same number of protons (ii) (B) Liquid ammonia (iii) (B) 6.0 × 6.023 × 1023 (iv) (C) Methanol (v) (C) Fluorospar (b) (i) Pb3O4 + 8HCl(conc.) → 3PbCl2 + Cl2 + 4H2O (ii) NH3 + 3C12 → NCl3 + 3HCl (iii) 3Cu + 8HNO3 (dil.) → 3Cu(NO3)2 + 4H2O + 2NO (iv) 2P + 5H2SO4 → 5SO2 + 2H3PO4 + 2H2O

(v) CH4 + 2O2

Cu tubeatm120

→ CO2 + 2H2O + Energy

(c) (i) It decomposes sulphur dioxide into sulphur and oxygen. (ii) Purple colour of the solution fades away. (iii) It forms auric chloride. (iv) It gives a plastic like product rasin. (v) The copper of the anode dissolves in the solution producing copper ions. (d)

Process Anode Electrolyte CathodeSilver plating a spoon Pure block of silver Sodium argentocyanide SpoonPurification of copper Impure copper Acidified copper

sulphate solutionPure strip of cop-per

(e) (i) Oxygen gas. (ii) 2KClO3 → 2KCl + 3O2 (iii) Molecular mass of KClO3 g = (39 + 35.5 + 16 × 3) = 122.5 g 67.2 L of oxygen will be evolved by 245 g = 245 × 6.72 / 67.2 = 24.50 g of KClO3 (f) (i) Carbon (ii) Covalent (iii) Hygroscopic substance (iv) Anode (v) Tungsten (g) (i) Add conc. sulphuric acid to both the compounds separately. Lead nitrate : No change appears. Lead chloride : A colourless gas having pungent suffocating smell which fumes in moist

air evolves. (ii) Add Tollen’s reagent (Ammonical 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.

Model Specimen Paper Chemistry

19SOLVED

| 2 | ICSE Model Specimen Papers, X

(iv) Add sodium hydroxide to the salt solutions separately Ferrous sulphate – Dirty green precipitate appears which changes to reddish brown,

after some time and is insoluble in excess of sodium hydroxide. Copper sulphate – Bluish white precipitate appears which is insoluble in excess of

sodium hydroxide. (v) Add NaOH solution to both the salts the ammonium salt will release the ammonia gas

which can be smelled while sodium salt will not show any change.

(h) (i)

H—C—OH

— —O

(Methanoic acid)

(ii)

H—C—C—H

— —O

(Ethanal)

—H

—

H

(iii)

H—C C—H≡(Ethyne)

(iv)

H—C—C—C—H

— —O

(Acetone)

—H

—

H

—H

—

H

(v)

H—C—C—C—H

(2-Methyl propane)—H

—

H—H

—H

—H

——

—H—C—H

——

H

Section - IIAttempt any four questions from this Section

Answer 2. (a)

Element Percentage compo-sition by 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 × vapour density = 2 × 252 = 504 n = Molecular weight / Empirical weight = 504 / 252 = 2 Molecular formula = (Empirical formula)n

= (CBr3)2

= C2Br6

(b) A : 4NH3 + 5O2

Pt

C800 ° → 4NO + 6H2O

B : 2NO + O2 → 2NO2 (50 °C)

Chemistry | 3 |

C : 4NO2 + 2H2O + O2

Pt

C800 ° → 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 is preferred over air, because each and every step requires oxygen.

(ii) In the laboratory preparation of nitric acid the mixture of concentrated sulphuric acid and sodium nitrate should not be heated very strongly, because nitric acid decomposes at high temperatures.

Answer 3. (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)Answer 4. (a) (i) Sodium argento cyanide is a complex salt. (ii) Silver nitrate cannot be used as an electrolyte because it decomposes very rapidly and

leads to the formation of uneven layer. (iii) A low current for a longer duration must be passed and direct current should be used

instead of alternating current. (iv) At cathode

Ag+ + e– → Ag At anode

Ag – e– → Ag+

(b) (i) Solution of H2SO4

(ii) Solution of CCl4 (iii) Solution of CH3COOH (c) (i) When dilute sulphuric acid is added to barium chloride, white precipitate of barium

sulphate is formed. When dilute sulphuric acid is added to lead nitrate, white precipitate of lead sulphate and nitric acid are formed.

(ii) When dilute sulphuric acid is added to ferrous sulphide, green coloured solution of FeCl2 is formed along with the evolution of a colourless gas and rotten smell. When dilute sulphuric acid is added to copper sulphide, blue coloured solution of CuO2 is formed 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), bluish white precipitate [Cu(OH)2] appears which is insoluble in excess of sodium hydroxide.

CuSO4 + 2NaOH → Cu(OH)2 ↓ + Na2SO4

When sodium hydroxide is added to the solution of a zinc salt [(Zn(NO3)2], white precipitate [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 + 2H2OAnswer 5. (a) (i) When ammonium hydroxide is added to silver chloride, a complex diammine silver

chloride is formed which is soluble in excess of ammonium hydroxide. (ii) When manganese dioxide reacts with concentrated hydrochloric acid, a greenish yellow

coloured gas with a pungent suffocating smell is evolved.

| 4 | ICSE Model Specimen Papers, X

(iii) When dilute hydrochloric acid is added to sodium thiosulphate, the solution formed has a yellow turbidity.

(b) Formula of hydrated CuSO4 — CuSO4.5H2O Molecular weight = 64 + 32 + (16 × 4) + 5(2 × 1 + 16) = 250 g Mass of water = 90 g

Percentage of water =

90 100

250×

= 36%

Answer 6. (a) (i) Pure acetic acid is known as glacial acetic acid because on cooling, it forms crystalline

mass resembling ice. (ii) This reaction is known as esterification.

CH3COOH + C2H5OH + H2SO4 → CH3COOC2H5 + HSO4– + H3O+

(b) (i) Lone pair of electrons are a pair of valence electrons that are not shared with another atom during the formation of a bond and is sometimes called a non-bonding pair.

(ii)

H—O+

H

H

Lone pair of

electrons

(c) (i) He < Ne < Ar (ii) Li < N < Cl < F (iii) F < I < AtAnswer 7.

(a) (i) 2CuO + C ∆ → 2Cu + CO2 (ii) FeO + H2 ∆ → Fe + H2O (iii) 2Al + 3H2O → Al2O3 + 3H2↑

(b) (i) CuSO4 → Cu2+ + SO42– (ii) NaCl → Na+ + Cl–

(iii) Cu2+ + 2e– → Cu (c) (i) Na > Mg > Al > Cl (ii) Al < Mg < Na (iii) H > Li > Cs