Question 1

Question 1(i)The element with highest ionization potential, is:HydrogenCaesiumRadonHelium

Accepted Answer

HeliumReason — As ionization potential decreases down the group and increases on moving from left to right in a periodic table, hence, helium, the element on the top right most corner of the periodic table has highest ionization potential.

Question 2

Question 1(v)The organic compound having a triple carbon-carbon covalent bond, is:C3H4C3H6C3H8C4H10

Accepted Answer

C3H4Reason — C3H4 is an Alkyne as it follows the general formula CnH2n-2 and Alkynes contain a triple carbon-carbon covalent bond.

Question 3

Question 1(b)State one relevant observation for each of the following reactions:(i) Action of concentrated nitric acid on copper.(ii) Addition of excess ammonium hydroxide into copper sulphate solution.(iii) A piece of sodium metal is put into ethanol at room temperature.(iv) Zinc carbonate is heated strongly.(v) Sulphide ore is added to a tank containing oil and water, and then stirred or agitated with air.

Accepted Answer

(i) Reddish brown fumes of nitrogen dioxide gas are liberated when copper reacts with conc. nitric acid.(ii) Deep blue or inky blue solution is formed when excess of ammonium hydroxide is added to copper sulphate solution.(iii) Effervescence of a gas is seen when a piece of sodium metal is put into ethanol at room temperature. The gas extinguishes a burning splint with a pop sound which confirms that the gas is hydrogen.(iv) A residue which is yellow when hot and white when cold is formed.(v) The ore is wetted by the oil and gets separated from the gangue in the form of froth. The impurities gets wetted by water and remain behind in the sink. Since the ore is lighter, it comes on the surface with the froth and impurities are left behind.

Question 4

Question 1(c)Write a balanced chemical equation for each of the following:(i) Reaction of carbon powder and concentrated nitric acid.(ii) Reaction of excess ammonia with chlorine.(iii) Reaction of lead nitrate solution with ammonium hydroxide.(iv) Producing ethane from bromo ethane using Zn/Cu couple in alcohol.(v) Complete combustion of ethane.

Accepted Answer

(i) C + 4HNO3 [conc.]⟶ CO2 + 2H2O + 4NO2(ii) 8NH3 [excess] + 3Cl2 ⟶ 6NH4Cl + N2(iii) Pb(NO3)2 + 2NH4OH ⟶ 2NH4NO3 + Pb(OH)2 ↓(iv) C2H5Brbromoethane+2[H]nascent hydrogen→alcoholZn/Cu coupleC2H6ethane+HBr\underset{	ext{bromoethane}}{	ext{C}_2	ext{H}_5	ext{Br}} + \underset{	ext{nascent hydrogen}}{2	ext{[H]}} \xrightarrow[	ext{alcohol}]{	ext{Zn/Cu couple}} \underset{	ext{ethane}}{	ext{C}_2	ext{H}_6} +	ext{HBr}bromoethaneC2​H5​Br​+nascent hydrogen2[H]​Zn/Cu couplealcohol​ethaneC2​H6​​+HBr(v) 2C2H6 + 7O2 [excess] ⟶ 4CO2 + 6H2O + Δ

Question 5

Question 1(f)Calculate:(i) The amount of each reactant required to produce 750 ml of carbon dioxide, when two volumes of carbon monoxide combine with one volume of oxygen to produce two volumes of carbon dioxide.2CO + O2 ⟶ 2CO2(ii) The volume occupied by 80 g of carbon dioxide at STP.(iii) Calculate the number of molecules in 4.4 gm of CO2.[Atomic mass of C= 12, O=16](iv) State the law associated in question no. (f)(i) above.

Accepted Answer

(i) [By Lussac's law]2CO+O2⟶2CO22 vol.:1 vol.⟶2 vol.\begin{matrix} 2	ext{CO}& + & 	ext{O}_2 &\longrightarrow & 2	ext{CO}_2 \ 2 	ext{ vol.} & : & 1 	ext{ vol.} & \longrightarrow & 2	ext{ vol.} \end{matrix}2CO2 vol.​+:​O2​1 vol.​⟶⟶​2CO2​2 vol.​To calculate the amount of CO required :CO2:CO2 vol.:2 vol.750 ml:x\begin{matrix}	ext{CO}_2 & : & 	ext{CO} & \ 2 	ext{ vol.} & : & 2 	ext{ vol.} \ 750 	ext{ ml} & : & 	ext{x} \end{matrix}CO2​2 vol.750 ml​:::​CO2 vol.x​∴x=22×750=750 ml	herefore x = \dfrac{2}{2} 	imes 750 = 750 	ext{ ml}∴x=22​×750=750 mlTo calculate the amount of O2 required :CO2:O2 2 vol.:1 vol.750 cm.3:x\begin{matrix}	ext{CO}_2 & : & 	ext{O}_2 & \ 	ext{ 2 vol.} & : & 1 	ext{ vol.} \ 750 	ext{ cm.}^3 & : & 	ext{x} \end{matrix}CO2​ 2 vol.750 cm.3​:::​O2​1 vol.x​∴x=12×750=375 ml	herefore x = \dfrac{1}{2} 	imes 750 = 375 	ext{ ml}∴x=21​×750=375 mlHence, CO required is 750 ml and O2 required = 375 ml(ii) Gram molecular mass of CO2 = 12 + 32 = 44 g and occupies 22.4 lit. vol.∴ 80 g. will occupy = 22.444\dfrac{22.4}{44}4422.4​ x 80 = 40.73 lit.Hence, volume occupied = 40.73 lits.(iii) Gram molecular mass of CO2 = 12 + 32 = 44 g and contains 6.023 x 1023 molecules.∴ 4.4 gm. of CO2 will contain = 6.023×102344\dfrac{6.023 	imes 10^{23}}{44}446.023×1023​ x 4.4 = 6.023 x 1022 molecules.Hence, number of molecules = 6.023 x 1022 molecules.(iv) Gay Lussac's law is associated in the question.

Question 6

Question 2(b)Draw the electron dot diagram for the compounds given below. Represent the electrons by (.) and (x) in the diagram.[Atomic No.: Ca = 20, O = 8, Cl = 17, H = 1](i) Calcium oxide(ii) Chlorine molecule(iii) Water molecule

Accepted Answer

(i) Electron dot diagram of Calcium oxide is shown below:(ii) Electron dot diagram of Chlorine molecule is shown below:(iii) Electron dot diagram of Water molecule is shown below:

Question 7

Question 2(c)Choose the correct word which refers to the process of electrolysis from A to E, to match the description (i) to (iv):A: Oxidation B: Cathode C: Anode D: An electrolyte E: Reduction(i) Conducts electricity in aqueous or in molten state.(ii) Loss of electron takes place at anode.(iii) A reducing electrode.(iv) Electrode connected to the positive end or terminal of the battery.

Accepted Answer

(i) Conducts electricity in aqueous or in molten state — D: An electrolyte(ii) Loss of electrons takes place at anode — A: oxidation(iii) A reducing electrode — B: Cathode(iv) Electrode connected to positive end or terminal of the battery. — C: Anode

Question 8

Question 3(a)Baeyer's process is used to concentrate bauxite ore to alumina.Give balanced chemical equations for the reaction taking place for its conversion from bauxite to alumina.

Accepted Answer

Conversion of impure bauxite to 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​Conversion of sodium aluminate to aluminium hydroxide:NaAlO2[sodium aluminate]+2H2O→50−60°CNaOH[sodium hydroxide]+Al(OH)3[aluminium hydroxide]\underset{	ext{[sodium aluminate]}}{	ext{NaAlO}_2} + 2	ext{H}_2	ext{O} \xrightarrow{50 - 60 \degree	ext{C}} \underset{	ext{[sodium hydroxide]}}{	ext{NaOH}} + \underset{	ext{[aluminium hydroxide]}}{	ext{Al(OH)}_3}[sodium aluminate]NaAlO2​​+2H2​O50−60°C​[sodium hydroxide]NaOH​+[aluminium hydroxide]Al(OH)3​​Conversion of Al(OH)3 - to pure alumina - heat on Al(OH)3:2Al(OH)3[aluminium hydroxide]→1100°CAl2O3[pure alumina]+3H2O\underset{	ext{[aluminium hydroxide]}}{	ext{2Al(OH)}_3} \xrightarrow{1100\degree	ext{C}} \underset{	ext{[pure alumina]}}{	ext{Al}_2	ext{O}_3} + 	ext{3H}_2	ext{O}[aluminium hydroxide]2Al(OH)3​​1100°C​[pure alumina]Al2​O3​​+3H2​OPrecipitated Al(OH)3 is ignited to give pure alumina [Al2O3]

Question 9

Question 4(b)Draw the structures of isomers of pentane.

Accepted Answer

The structures of isomers of pentane are shown below:Pentane [n-pentane]2-Methyl butane [iso-pentane]2,2 Dimethyl Propane [neo-pentane]

Question 10

Question 4(c)Hydrogen chloride gas is prepared in the laboratory using concentrated sulphuric acid and sodium chloride. Answer the questions that follow based on this reaction:(i) Give the balanced chemical equation for the reaction with suitable condition(s) if any.(ii) Why is concentrated sulphuric acid used instead of concentrated nitric acid?(iii) How is the gas collected?(iv) Name the drying agent not used for drying the gas.

Accepted Answer

(i) Balanced chemical equation for the reaction are :NaCl+H2SO4→<200°CNaHSO4+HCl [g.]	ext{NaCl} + 	ext{H}_2	ext{SO}_4 \xrightarrow{\lt 200 \degree	ext{C}} 	ext{NaHSO}_4 + 	ext{HCl [g.]}NaCl+H2​SO4​<200°C​NaHSO4​+HCl [g.](ii) As conc. H2SO4 is non-volatile and has a high boiling point, therefore, it displaces the volatile hydrogen chloride from the salt sodium chloride. Hence, conc. H2SO4 is used as a reactant in the laboratory preparation of HCl from sodium chloride.Conc. nitric acid cannot be used in place of conc. H2SO4 since it is volatile and may volatize out along with the hydrogen chloride.(iii) The gas is collected by upward displacement of air method.(iv) Quicklime (CaO) and phosphorous pentoxide [P2O5] are not used as drying agent, since they react with HCl forming their respective chlorides.

Question 11

Question 5(b)Calculate the percentage of:(i) Fluorine(ii) Sodium and(iii) Aluminium in sodium aluminium fluoride [Na3AlF6], to the nearest whole number.[Atomic Mass: Na = 23, Al= 27, F= 19]

Accepted Answer

(i) Molecular weight of sodium aluminium fluoride (Na3AlF6)= 3(23) + 27 + 6(19)= 69 + 27 + 114= 210 g210 g of sodium aluminium fluoride contains 114 g of fluorine∴ 100 g of sodium aluminium fluoride will contain = 114210\dfrac{114}{210}210114​ x 100 = 54.28% = 54%(ii) 210 g of sodium aluminium fluoride contains 69 g of sodium∴ 100 g of sodium aluminium fluoride will contain = 69210\dfrac{69}{210}21069​ x 100 = 32.85% = 33%(iii) 210 g of sodium aluminium fluoride contains 27 g of aluminium∴ 100 g of sodium aluminium fluoride will contain = 27210\dfrac{27}{210}21027​ x 100 = 12.85% = 13%Hence, percentage of F = 54%, Na = 33%, Al = 13%

Question 12

Question 5(c)(i) State the volume occupied by 40 gm of methane at STP, if its vapour density (V.D.) is 8.(ii) Calculate the number of moles present in 160 gm of NaOH.[Atomic Mass: Na = 23, H= 1, O= 16]

Accepted Answer

(i) Gram molecular mass of methane = V.D. x 2 = 8 x 2 = 16 g16 g of methane occupies 22.4 lit.∴ 40 g of methane will occupy = 22.416\dfrac{22.4}{16}1622.4​ x 40 = 56 lits.(ii) Gram molecular mass of NaOH = 23 + 16 + 1 = 40 g40 g of NaOH = 1 mole∴ 160 g of NaOH = 16040\dfrac{160}{40}40160​ = 4 moles.

Question 13

Question 6(a)Identify the salts P, Q, R from the following observations:(i) Salt P has light bluish green colour. On heating, it produces a black coloured residue. Salt P produces brisk effervescence with dil. HCl and the gas evolved turns lime water milky, but no action with acidified potassium dichromate solution.(ii) Salt Q is white in colour. On strong heating, it produces buff yellow residue and liberates reddish brown gas. Solution of salt Q produces chalky white insoluble precipitate with excess of ammonium hydroxide.(iii) Salt R is black in colour. On reacting with concentrated HCl, it liberates a pungent greenish yellow gas which turns moist starch iodide paper blue black.

Accepted Answer

(i) Copper carbonate(ii) Lead nitrate(iii) Manganese dioxide

Question 14

Question 6(b)Identify the substance underlined in each of the following:(i) The electrode that increases in mass during the electro-refining of silver.(ii) The acid that is a dehydrating as well as a drying agent.(iii) The catalyst used to oxidize ammonia into nitric oxide.

Accepted Answer

(i) CathodeReason — Pure copper is deposited on the thin sheet of pure copper placed at the cathode(ii) Sulphuric acidReason — Being hygroscopic sulphuric acid absorbs moisture but not enough to form a solution. Hence, it is a drying agent.Sulphuric acid is a dehydrating agent as it can remove chemically combined water molecules from blue vitriol.CuSO4.5H2O + H2SO4 ⇌ CuSO4 + 5H2O(iii) Platinum4NH3+5O2→800°CPt.4NO+6H2O+Δ4	ext{NH}_3 + 5	ext{O}_2 \xrightarrow[800 \degree 	ext{C}]{	ext{Pt.}} 	ext{4NO} + 	ext{6H}_	ext{2}	ext{O} + \Delta4NH3​+5O2​Pt.800°C​4NO+6H2​O+Δ

Question 15

Question 6(c)Copy and complete the following paragraph using the options given inbrackets:Alkenes are a homologous series of (i) ............... (saturated / unsaturated) hydrocarbons characterized by the general formula (ii) ............... (CnH2n/CnH2n). Alkenes undergo (iii)............... (addition / substitution) reactions and also undergo (iv) ............... (hydrogenation / dehydrogenation) to form alkanes.

Accepted Answer

Alkenes are a homologous series of unsaturated hydrocarbons characterized by the general formula CnH2n. Alkenes undergo addition reactions and also undergo hydrogentaion to form alkanes.

Question 16

Question 7(a)Write balanced chemical equations, for the preparation of the given salts.(i) to (iii) by using the methods A to C respectively:A: Neutralization B: Precipitation C: Titration(i) Copper sulphate(ii) Zinc carbonate(iii) Ammonium sulphate

Accepted Answer

Preparation of copper sulphate by neutralizationCuO + H2SO4 ⟶ CuSO4 + H2OPreparation of zinc carbonate by precipitationZn(NO3)2 + Na2CO3 ⟶ 2NaNO3 + ZnCO3Preparation of ammonium sulphate by titration2NH4OH + H2SO4 ⟶ (NH4)2SO4 + 2H2O

Question 17

Question 7(b)Name the following elements:(i) An alkaline earth metal present in group 2 and period 3.(ii) A trivalent metal used to make light tools.(iii) A monovalent non-metal present in fluorspar.

Accepted Answer

(i) Magnesium(ii) Aluminium(iii) Fluorine

Question 18

Question 7(c)An aqueous solution of nickel (II) sulphate was electrolyzed using nickel electrodes.Observe the diagram and answer the questions that follow:(i) What do you observe at the cathode and anode respectively?(ii) Name the cation that remains as a spectator ion in the solution.(iii) Which equation for the reaction at the anode is correct?Ni ⟶ Ni2+ + 2e-+Ni + 2e- ⟶ Ni2+Ni2+ ⟶ Ni + 2e-Ni2+ + 2e- ⟶ Ni

Accepted Answer

(i) The observations at cathode and anode are:At cathode: Ni2+ ions move towards cathode and are discharged as neutral Ni atoms by gain of electrons. Thus, Ni is deposited at cathode.At anode: Diminishes in mass. Ni2+ ions are formed.(ii) H+ ions.(iii) Ni ⟶ Ni2+ + 2e-

Solved 2020 Question Paper ICSE Class 10 Chemistry

Section I 40 Marks

Section Ii 40 Marks