Question 1

Question 1(b-i)An element with the atomic number 19 will most likely combine chemically with the element whose atomic number is:17111820

Accepted Answer

17Reason — The element with atomic number 19 has electronic configuration of (2, 8, 8, 1) and that with atomic number 17 has electronic configuration of (2, 8, 7). This element with atomic number 17 needs 1 electron to complete its octet whereas element with atomic number 19 has an extra electron. Hence, an element with atomic number 19 is most likely to combine with the element with atomic number 17.

Question 2

Question 1(c)Write balanced chemical equations for each of the following:(i) Action of warm water on AlN.(ii) Action of hot and concentrated Nitric acid on copper.(iii) Action of Hydrochloric acid on sodium bicarbonate.(iv) Action of dilute Sulphuric acid on Sodium Sulphite.(v) Preparation of ethanol from Ethyl Chloride.

Accepted Answer

(i) AlN + 3H2O ⟶ Al(OH)3 + NH3 [g](ii) Cu + 4HNO3 ⟶ Cu(NO3)2 + 2H2O + 2NO2(iii) NaHCO3 + HCl ⟶ NaCl + H2O + CO2 ↑(iv) Na2SO3 + H2SO4 (dil) ⟶ Na2SO4 + H2O + SO2(v) C2H5-Cl Chloroethane [ethyl chloride]+NaOH [aq.]→boilC2H5-OH Ethanol [ethyl alcohol]+NaCl\underset{	ext{ Chloroethane [ethyl chloride]}}{	ext{C}_2	ext{H}_5	ext{-Cl}} + 	ext{NaOH [aq.]} \xrightarrow{	ext{boil}} \underset{	ext{ Ethanol [ethyl alcohol]}}{	ext{C}_2	ext{H}_5	ext{-OH}} + 	ext{NaCl} Chloroethane [ethyl chloride]C2​H5​-Cl​+NaOH [aq.]boil​ Ethanol [ethyl alcohol]C2​H5​-OH​+NaCl

Question 3

Question 1(d)State your observations when:(i) Dilute Hydrochloric acid is added to Lead nitrate solution and the mixture is heated.(ii) Barium chloride solution is mixed with Sodium Sulphate Solution.(iii) Concentrated Sulphuric acid is added to Sugar Crystals.(iv) Dilute Hydrochloric acid is added to Copper carbonate.(v) Dilute Hydrochloric acid is added to Sodium thiosulphate.

Accepted Answer

(i) White precipitate of PbCl2 is formed which is soluble in hot water.Pb(NO3)2 + 2HCl ⟶ PbCl2 ↓ + 2HNO3(ii) White coloured precipitate of barium sulphate is formed when sodium sulphate is mixed with barium chloride.Na2SO4 (dil.) + BaCl2 (aq.) ⟶ 2NaCl + BaSO4 ↓ [white ppt. formed](iii) Black spongy charred mass of carbon is formed. Steam is seen to be evolved.C12H22O11→Conc. H2SO412C [sugar charcoal]+11H2O	ext{C}_{12}	ext{H}_{22}	ext{O}_{11}\xrightarrow{	ext{Conc. H}_2	ext{SO}_4} 	ext{12C} \small{\space[sugar\space charcoal]} + 	ext{11H}_2	ext{O}C12​H22​O11​Conc. H2​SO4​​12C [sugar charcoal]+11H2​O(iv) Effervescence of CO2 seen which turns lime water milky.CuCO3 + 2HCl ⟶ CuCl2 + H2O + CO2(v) Yellow particles of S precipitate.Na2S2O3 + 2HCl ⟶ 2NaCl + SO2 + S + H2O

Question 4

Question 1(f)(i) A gas of mass 32 gms has a volume of 20 litres at S.T.P. Calculate the gram molecular weight of the gas.(ii) How much Calcium oxide is formed when 82 g of calcium nitrate is heated? Also find the volume of nitrogen dioxide evolved:2Ca(NO3)2 ⟶ 2CaO + 4NO2 + O2[Ca = 40, N = 14, O = 16]

Accepted Answer

(i) 20 lit. of gas weighs = 32 g.1 mole = 22.4 lit. of gas will weigh = 3220\dfrac{32}{20}2032​ x 22.4 = 35.84 g.Hence, gram mol. weight of the gas = 35.84 g.(ii)2Ca(NO3)2⟶2CaO+4NO2+O22[(40)+2(404[22.4]2(14+3(16))]+16) lit.=328 g112g\begin{matrix} 2	ext{Ca}(	ext{NO}_3)_2 & \longrightarrow & 2	ext{CaO} & + & 4	ext{NO}_2 & + & 	ext{O}_2 \ 2[(40) + & & 2(40 & & 4[22.4] \ 2(14 +3(16))] & & + 16) & & 	ext{ lit.} \ = 328 	ext{ g} & & 112 	ext{g} \ \end{matrix}2Ca(NO3​)2​2[(40)+2(14+3(16))]=328 g​⟶​2CaO2(40+16)112g​+​4NO2​4[22.4] lit.​+O2​(a) 328 g of Ca(NO3)2 produces 112 g of calcium oxide.∴ 82 g of Ca(NO3)2 will produce = 112328\dfrac{112}{328}328112​ x 82 = 27.99 g = 28 gHence, 28 g of calcium oxide is produced.(b) 328 g of Ca(NO3)2 produces 4(22.4) lit of nitrogen dioxide.∴ 82 g of Ca(NO3)2 will produce = 4×22.4328\dfrac{4 	imes 22.4}{328}3284×22.4​ x 82 = 22.4 lit.Hence, vol of nitrogen dioxide evolved = 22.4 lit.

Question 5

Question 1(g)Match the salts given in Column I with their method of preparation given in Column II:Column IColumn II(i) Pb(NO3)2 from PbO(A) Simple displacement(ii) MgCl2 from Mg(B) Titration(iii) FeCl3 from Fe(C) Neutralization(iv) NaNO3 from NaOH(D) Precipitation(v) ZnCO3 from ZnSO4(E) Combination

Accepted Answer

Column IColumn II(i) Pb(NO3)2 from PbO(C) Neutralization(ii) MgCl2 from Mg(A) Simple displacement (B) Titration(iii) FeCl3 from Fe(E) Combination(iv) NaNO3 from NaOH(B) Titration(v) ZnCO3 from ZnSO4(D) Precipitation

Question 6

Question 2(a)Use the letters only written in the Periodic Table given below to answer the questions that follow: IIIIIIIVVVIVIIO1       L2Q EGJZM 3R       4T       (i) State the number of valence electrons in atom J.(ii) Which element shown forms ions with a single negative charge?(iii) Which metallic element is more reactive than R?(iv) Which element has its electrons arranged in four shells?

Accepted Answer

(i) Atom J has 5 valence electrons as it belongs to the fifth group in the periodic table.(ii) M forms ions with a single negative charge as it has 7 electrons in the valence shell and obtaining one more electron completes it's octet.(iii) Metal T is more reactive than R because elements at the bottom of the group are more reactive.(iv) T has it's electrons arranged in 4 shells as it belongs to 4th period.

Question 7

Question 2(c)The following table shows the electronic configuration of the elements W, X, Y, Z:ElementWXYZElectronic configurations2,8,12,8,72,51(i) What type of bond is formed between:     1. W and X     2. Y and Z(ii) What is the formula of the compound formed between :     1. X and Z     2. W and X

Accepted Answer

(i) Type of bond formed(a) W (2,8,1) and X (2,8,7) — Electrovalent bond is formed as W donates 1 electron and X takes up that electron and both attain a stable octet.(b) Y (2, 5) and Z (1) — Covalent bond as electron sharing takes place between the two elements. The valency of Z element is 1 and that of Y is 3. Z needs one electrons to attain stable duplet structure of nearest noble gas - He [2] and Y needs three electrons to attain stable octet structure of nearest noble gas - Ne [2,8]One Y atom shares three electron pairs one with each of the three atoms of Z such that Z acquires a duplet configuration and Y attain a stable octet configuration resulting in the formation of three single covalent bonds.(ii) Formula of the compound formed(a) X (2,8,7) and Z (1) — XZ as X needs one 1 electron to complete it's octet and Z needs needs one electron to complete it's duplet hence, they both share one electron pair and forms ZX compound.(b) W (2,8,1) and X (2,8,7) — WX as W has the tendency to donate one electron and gain stability whereas X has the tendency to gain one electron to be stable. Hence, W donates one electron and X takes it and form WX compound.

Question 8

Question 3(a)Write a balanced chemical equation for each of the following:(i) Burning of ethane in plentiful supply of air.(ii) Action of water on Calcium carbide.(iii) Heating of Ethanol at 170°C in the presence of conc. Sulphuric acid.

Accepted Answer

(i) Carbon dioxide and water are produced:2C2H6+7O2[excess]⟶4CO2+6H2O+Δ2	ext{C}_2	ext{H}_6 + 7	ext{O}_2[	ext{excess}] \longrightarrow 4	ext{CO}_2 + 6	ext{H}_2	ext{O} + \Delta2C2​H6​+7O2​[excess]⟶4CO2​+6H2​O+Δ(ii) Calcium hydroxide and ethyne are producedCaC2calcium carbide+2H2Owater⟶C2H2ethyne [acetylene]+Ca(OH)2\underset{	ext{calcium carbide}}{	ext{CaC}_2} + \underset{	ext{water}}{2	ext{H}_2	ext{O}} \longrightarrow \underset{	ext{ethyne [acetylene]}}{	ext{C}_2	ext{H}_2} + 	ext{Ca(OH)}_2calcium carbideCaC2​​+water2H2​O​⟶ethyne [acetylene]C2​H2​​+Ca(OH)2​(iii) Ethene and water are producedC2H5OH 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 9

Question 3(b)Give the structural formulae of each of the following :(i) 2-methyl propane(ii) Ethanoic acid(iii) Butan-2-ol

Accepted Answer

(i) 2-methyl propane :(ii) Ethanoic acid :(iii) Butan – 2 – ol :

Question 10

Question 3(c)Equation for the reaction when compound A is bubbled through bromine dissolved in carbon tetrachloride is as follows:(i) Draw the structure of A.(ii) State your observation during this reaction.

Accepted Answer

(i) Compound A is Ethene. Its structure is shown below:(ii) Brown colour of bromine is discharged.

Question 11

Question 4(b)State your observations when ammonium hydroxide solution is added drop by drop and then in excess to each of the following solutions:(i) copper sulphate solution(ii) zinc sulphate solution.

Accepted Answer

(i) Pale blue precipitate of Cu(OH)2 is formed, which becomes deep blue when excess of ammonium hydroxide is added.CuSO4 + 2NH4OH ⟶ (NH4)2SO4 + Cu(OH)2 ↓Cu(OH)2 + (NH4)2SO4 + 2NH4OH [in excess] ⟶ 4H2O + [Cu(NH3)4]SO4(ii) Gelatinous white precipitate of zinc hydroxide is formed which becomes colourless solution on adding excess of ammonium hydroxide.ZnSO4 + 2NH4OH ⟶ (NH4)2SO4 + Zn(OH)2 ↓Zn(OH)2 +(NH4)2SO4 + 2NH4OH [in excess] ⟶ 4H2O + [Zn(NH3)4]SO4

Question 12

Question 5(a)A gas cylinder contains 12 x 1024 molecules of oxygen gas.If Avogadro's number is 6 x 1023; Calculate:(i) the mass of oxygen present in the cylinder.(ii) the volume of oxygen at S.T.P. present in the cylinder. [O=16]

Accepted Answer

Gram molecular mass of oxygen = 32 g = 1 mole and contains 6 x 1023 molecules i.e., 6 x 1023 molecules weigh 32 g∴ 12 x 1024 molecules will weigh = 326×1023\dfrac{32}{ 6	imes 10^{23}}6×102332​ x 12 x 1024 = 640 g(ii) 6 x 1023 molecules occupy 22.4 lit. of vol.∴ 12 x 1024 molecules will occupy = 22.46×1023\dfrac{22.4}{ 6	imes 10^{23}}6×102322.4​ x 12 x 1024 = 448 lit.Hence, mass of O2 present in the cylinder = 640 g and volume of O2 at S.T.P. present in the cylinder = 448 lit.

Question 13

Question 5(b)A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.[C=12, H=1]

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioCarbon82.761282.7612\dfrac{82.76}{12}1282.76​ = 6.896.896.89\dfrac{6.89}{6.89}6.896.89​ = 1Hydrogen17.24117.241\dfrac{17.24}{1}117.24​ = 17.2417.246.89\dfrac{17.24}{6.89}6.8917.24​ = 2.5Simplest ratio of whole numbers = C : H = 1 : 2.5 = 2 : 5Hence, empirical formula is C2H5Empirical formula weight = 2(12) + 5(1) = 29V.D. = 29Molecular weight = 2 x V.D. = 2 x 29n=Molecular weightEmpirical formula weight=2×2929=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{2 \times 29}{29} =2n=Empirical formula weightMolecular weight​=292×29​=2∴ Molecular formula = n[E.F.] = 2[C2H5] = C4H10

Question 14

Question 5(c)The equation :4NH3 + 5O2 ⟶ 4NO + 6H2O,represents the catalytic oxidation of ammonia. If 100 cm3 of ammonia is used calculate the volume of oxygen required to oxidize the ammonia completely.

Accepted Answer

[By Lussac's law]4NH3+5O2⟶4NO+6H2O4 vol.:5 vol.⟶4 vol.\begin{matrix} 4	ext{NH}_3 & + & 5	ext{O}_2 &\longrightarrow & 4	ext{NO} & + & 6	ext{H}_2	ext{O} \ 4 	ext{ vol.} & : & 5 	ext{ vol.} & \longrightarrow & 4	ext{ vol.} \end{matrix}4NH3​4 vol.​+:​5O2​5 vol.​⟶⟶​4NO4 vol.​+6H2​OTo calculate the volume of oxygen required :NH3:O24 vol.:5 vol.100 cm.3:x\begin{matrix}	ext{NH}_3 & : & 	ext{O}_2 \ 4 	ext{ vol.} & : & 5 	ext{ vol.} \ 100 	ext{ cm.}^3 & : & 	ext{x} \end{matrix}NH3​4 vol.100 cm.3​:::​O2​5 vol.x​∴x=54×100=125 cm.3	herefore x = \dfrac{5}{4} 	imes 100 = 125 	ext{ cm.}^3∴x=45​×100=125 cm.3Hence, volume of oxygen required = 125 cm3

Question 15

Question 5(d)By drawing an electron dot diagram show the formation of ammonium ion [N = 7, H = 1]

Accepted Answer

Electron dot diagram showing the formation of ammonium ion is given below:

Question 16

Question 6(a)Name the gas evolved when the following mixtures are heated:(i) Calcium hydroxide and Ammonium Chloride(ii) Sodium Nitrite and Ammonium Chloride

Accepted Answer

(i) Ammonia gas2NH4Cl + Ca(OH)2 ⟶ CaCl2 + 2H2O + 2NH3(ii) Nitrogen gasNH4Cl + NaNO2 ⟶ NaCl + NH4NO2NH4NO2 ⟶ 2H2O + N2

Question 17

Question 6(b)Write balanced chemical equations for each of the following:(i) When excess of ammonia is treated with chlorine.(ii) An equation to illustrate the reducing nature of ammonia.

Accepted Answer

(i) 8NH3 [excess] + 3Cl2 ⟶ 6NH4Cl + N2(ii) 2NH3 + 3CuO ⟶ 3Cu + 3H2O + N2 [g]

Question 18

Question 6(c)A, B, C and D summarize the properties of sulphuric acid depending on whether it is dilute or concentrated.A: Typical acid propertyB: Non-volatile acidC: Oxidizing agentD: Dehydrating agentChoose the property [A, B, C or D] depending on which is relevant to each of the following:(i) Preparation of hydrogen chloride gas.(ii) Preparation of copper sulphate from copper oxide.(iii) Action of conc. sulphuric acid on sulphur.

Accepted Answer

(i) B: Non-volatile acidNaCl+H2SO4[conc.]→<200°CNaHSO4+HCl	ext{NaCl} + 	ext{H}_2	ext{SO}_4 [	ext{conc.}]\xrightarrow{\lt 200 \degree	ext{C}} 	ext{NaHSO}_	ext{4} + 	ext{HCl}NaCl+H2​SO4​[conc.]<200°C​NaHSO4​+HCl(ii) A: Typical acid propertyCuO + H2SO4 (conc.) ⟶ CuSO4 + H2O(iii) C: Oxidizing agentS + 2H2SO4 (conc.) ⟶ 3SO2 + 2H2O

Question 19

Question 6(d)Give reasons why:(i) Sodium Chloride will conduct electricity only in fused or aqueous solution state.(ii) In the electroplating of an article with silver, the electrolyte sodium argento-cyanide solution is preferred over silver nitrate solution.(iii) Although copper is a good conductor of electricity, it is a non-electrolyte.

Accepted Answer

(i) The ions Na+ and Cl– are not free but held together by strong electrostatic force of attraction. In fused or molten state the ions break free and move. Hence, NaCl will conduct electricity only in fused state or aq. soln. state.(ii) Migration of Ag1+ ions from sodium argentocyanide solution is slow compared to that from silver nitrate. Hence, an even deposition of metal silver is obtained on the article. Therefore, the electrolyte sodium argentocyanide soln. is preferred over silver nitrate solution.(iii) Copper does not undergo chemical decomposition due to flow of electric current through it. Hence, copper is a good conductor of electricity but it is a non-electrolyte.

Question 20

Question 7(a)(i) Name the solution used to react with Bauxite as a first step in obtaining pure aluminium oxide, in the Baeyer's process.(ii) Write the equation for the reaction where the aluminium oxide for the electrolytic extraction of aluminium is obtained by heating aluminium hydroxide.(iii) Name the compound added to pure alumina to lower the fusion temperature during the electrolytic reduction of alumina.(iv) Write the equation for the reaction that occurs at the cathode during the extraction of aluminium by electrolysis.(v) Explain why it is preferable to use a number of graphite electrodes as anode instead of a single electrode, during the above electrolysis.

Accepted Answer

(i) Sodium hydroxide(ii) Conversion of Al(OH)3 - to pure alumina - heat on Al(OH)32Al(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​O(iii) Cryolite (Na3AlF6)(iv) 2Al3+ + 6e- ⟶ 2Al(v) The oxygen evolved at the anode escapes as a gas or reacts with the carbon anode. The carbon anode is thus oxidized to carbon monoxide which either burns giving carbon dioxide or escapes out through an outlet. [2C + O2 ⟶ 2CO; 2CO + O2 ⟶ 2CO2] The carbon anode is hence consumed and renewed periodically after a certain period of usage. It is therefore preferable to use a number of graphite electrodes as anode instead of a single electrode.

Solved 2016 Question Paper ICSE Class 10 Chemistry

Section I 40 Marks

Section Ii 40 Marks