Q: Question 3(c)For the preparation of hydrochloric acid in the laboratory:(i) Why is direct absorption of hydrogen chloride gas in water not feasible?(ii) What arrangement is done to dissolve hydrogen chloride gas in water?

(i) Hydrogen chloride gas is sufficiently soluble, so, it is absorbed more quickly than it is being generated in the flask. This causes back-suction due to partial vacuum created in the tube. Hence, direct absorption of hydrogen chloride gas in water is not feasible.(ii) A special funnel arrangement is done in which an inverted funnel, connected to the hydrogen chloride gas supply is placed in the beaker in such a way that it just touches the water taken in the trough. This arrangement minimizes back suction.

Q: Question 4(a)The percentage composition of a gas is:Nitrogen 82.35%, Hydrogen 17.64%.Find the empirical formula of the gas. [N = 14, H = 1]

Element% compositionAt. wt.Relative no. of atomsSimplest ratioNitrogen82.351482.3514\dfrac{82.35 }{14}1482.35 = 5.885.885.88\dfrac{5.88}{5.88}5.885.88 = 1Hydrogen17.64117.641\dfrac{17.64}{1}117.64 = 17.6417.645.88\dfrac{17.64}{5.88}5.8817.64 = 3Simplest ratio of whole numbers = N : H = 1 : 3Hence, empirical formula is NH3

Question 1

Question 1(c)Write a balanced chemical equation for each of the following:(i) Action of concentrated sulphuric acid on carbon.(ii) Reaction of sodium hydroxide solution with iron (III) chloride solution.(iii) Action of heat on aluminum hydroxide.(iv) Reaction of zinc with potassium hydroxide solution.(v) Action of dilute hydrochloric acid on magnesium sulphite.

Accepted Answer

(i) C + 2H2SO4 [conc.] ⟶ CO2 + 2H2O + 2SO2(ii) FeCl3 + 3NaOH ⟶ 3NaCl + Fe(OH)3(iii) 2Al(OH)3 →1100°C\xrightarrow{1100\degree	ext{C}}1100°C​ Al2O3 + 3H2O(iv) Zn + 2KOH ⟶ K2ZnO2 + H2(v) MgSO3 + 2HCl ⟶ MgCl2 + H2O + SO2 [g]

Question 2

Question 1(f)Give a reason for each of the following:(i) Ionic compounds have a high melting point.(ii) Inert gases do not form ions.(iii) Ionisation potential increases across a period, from left to right.(iv) Alkali metals are good reducing agents.(v) Conductivity of dilute hydrochloric acid is greater than that of acetic acid.

Accepted Answer

(i) Ionic compounds have ionic bonds so there exists a strong force of attraction between the oppositely charged ions, so a large amount of energy is required to break the strong bonding between the ions. Hence, ionic compounds have a high melting point.(ii) Inert gases have completely filled octet which makes them extremely stable. They neither lose, nor gain electrons. Hence, they do not form ions.(iii) The ionisation potential of an element increases across a period because the atomic size decreases due to an increase in nuclear charge and electrons in the outermost shell are more strongly held because of which greater energy is required to remove the electron.(iv) Alkali metals have one electron in the outermost shell and hence during reactions they readily give the valence electron and get themselves oxidised. Hence, they are good reducing agents.(v) Dil. H2SO4 is a strong electrolyte and acetic acid is a weak electrolyte. Therefore, dil. H2SO4 allows large amount of electricity to flow through it and is a good conductor of electricity whereas acetic acid allows small amount of electricity to flow through it and is a poor conductor of electricity. Hence, electrical conductivity of dilute hydrochloric acid is greater than that of acetic acid.

Question 3

Question 1(g)Name the gas that is produced in each of the following cases:(i) Sulphur is oxidized by concentrated nitric acid.(ii) Action of dilute hydrochloric acid on sodium sulphide.(iii) Action of cold and dilute nitric acid on copper.(iv) At the anode during the electrolysis of acidified water.(v) Reaction of ethanol and sodium.

Accepted Answer

(i) Nitrogen dioxide gas(ii) Hydrogen sulphide gas(iii) Nitric oxide gas(iv) Oxygen gas(v) Hydrogen gas

Question 4

Question 2(a)(i) What do you understand by a lone pair of electrons?(ii) Draw the electron dot diagram of Hydronium ion. (H=1; O=8)

Accepted Answer

(i) Lone pair of electrons are valence electrons that in a covalent bond are not shared with another atom.(ii) Electron dot diagram of Hydronium ion is shown below:

Question 5

Question 2(b)In Period 3 of the Periodic Table, element B is placed to the left of element A.On the basis of this information, choose the correct word from the brackets to complete the following statements:(i) The element B would have (lower / higher) metallic character than A.(ii) The element A would probably have (lesser / higher) electron affinity than B.(iii) The element A would have (greater / smaller) atomic size than B.

Accepted Answer

If the element are placed as B and then A in the 3rd period of the periodic table then(i) The element B would have higher metallic character than A as metallic character decreases across a period.(ii) The element A would probably have higher electron affinity than B as electron affinity increases across a period.(iii) The element A would have smaller atomic size than B as atomic size decreases across a period.

Question 6

Question 2(c)Copy and complete the following table which refers to the conversion of ions to neutral particles.ConversionIonic equationOxidation/ReductionChloride ion to chlorine moleculeLead [II] ion to lead

Accepted Answer

ConversionIonic equationOxidation/ReductionChloride ion to chlorine moleculeCl1- - 1e- ⟶ Cl [Cl2]Cl + Cl ⟶ Cl2OxidationLead [II] ion to leadPb2+ + 2e- ⟶ PbReduction

Question 7

Question 3(a)(i) Write the balanced chemical equation to prepare ammonia gas in the laboratory by using an alkali.(ii) State why concentrated sulphuric acid is not used for drying ammonia gas.(iii) Why is ammonia gas not collected over water?

Accepted Answer

(i) 2NH4Cl + Ca(OH)2 ⟶ CaCl2 + 2H2O + 2NH3 [g](ii) As sulphuric acid reacts chemically with ammonia to form ammonium sulphate hence, it is not used as a drying agent for drying ammonia.2NH3 + H2SO4 ⟶ (NH4)2SO4(iii) As ammonia gas is highly soluble in water hence, it is not collected over water.

Question 8

Question 3(b)(i) Name the acid used for the preparation of hydrogen chloride gas in the laboratory. Why is this particular acid preferred to other acids?(ii) Write the balanced chemical equation for the laboratory preparation of hydrogen chloride gas.

Accepted Answer

(i) Conc. H2SO4 is used for the preparation of hydrogen chloride gas in the laboratory.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.(ii) 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.]

Question 9

Question 3(c)For the preparation of hydrochloric acid in the laboratory:(i) Why is direct absorption of hydrogen chloride gas in water not feasible?(ii) What arrangement is done to dissolve hydrogen chloride gas in water?

Accepted Answer

(i) Hydrogen chloride gas is sufficiently soluble, so, it is absorbed more quickly than it is being generated in the flask. This causes back-suction due to partial vacuum created in the tube. Hence, direct absorption of hydrogen chloride gas in water is not feasible.(ii) A special funnel arrangement is done in which an inverted funnel, connected to the hydrogen chloride gas supply is placed in the beaker in such a way that it just touches the water taken in the trough. This arrangement minimizes back suction.

Question 10

Question 4(a)The percentage composition of a gas is:Nitrogen 82.35%, Hydrogen 17.64%.Find the empirical formula of the gas. [N = 14, H = 1]

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioNitrogen82.351482.3514\dfrac{82.35 }{14}1482.35​ = 5.885.885.88\dfrac{5.88}{5.88}5.885.88​ = 1Hydrogen17.64117.641\dfrac{17.64}{1}117.64​ = 17.6417.645.88\dfrac{17.64}{5.88}5.8817.64​ = 3Simplest ratio of whole numbers = N : H = 1 : 3Hence, empirical formula is NH3

Question 11

Question 4(b)Aluminum carbide reacts with water according to the following equation:Al4C3 + 12H2O ⟶ 4Al(OH)3 + 3CH4(i) What mass of aluminum hydroxide is formed from 12g of aluminum carbide?(ii) What volume of methane at s.t.p. is obtained from 12g of aluminum carbide?[Relative molecular weight of Al4C3 = 144; Al(OH)3 = 78]

Accepted Answer

Al4C3+12H2O⟶4Al(OH)3+3CH4144 g4(78)3(22.4)=312 g=67.2 lit.\begin{matrix} 	ext{Al}_4	ext{C}_3 & + & 12	ext{H}_2	ext{O} & \longrightarrow & 4	ext{Al(OH)}_3 & + & 3	ext{CH}_4 \ 144 	ext{ g} & & & & 4(78) & & 3(22.4) \ & & & & = 312 	ext{ g} & & = 67.2 	ext{ lit.} \ \end{matrix}Al4​C3​144 g​+​12H2​O​⟶​4Al(OH)3​4(78)=312 g​+​3CH4​3(22.4)=67.2 lit.​144 g of aluminium carbide forms 312 g of aluminium hydroxide.∴ 12 g of aluminium carbide will form 312144\dfrac{312}{144}144312​ x 12 = 26 g of aluminium hydroxideHence, 26 g of aluminium hydroxide is formed.(ii) 144 g of aluminium carbide forms 67.2 lit of methane.∴ 12 g of aluminium carbide will form 67.2144\dfrac{67.2}{144}14467.2​ x 12 = 5.6 lit.Hence, vol. of methane obtained at s.t.p. = 5.6 lit.

Question 12

Question 5(a)Complete the following table which relates to the homologous series of hydrocarbons.General formulaIUPAC name of the homologous seriesCharacteristic bond typeIUPAC name of the first member of the seriesCnH2n-2(A) ...............(B) ...............(C)...............CnH2n+2(D)...............(E) ...............(F)...............

Accepted Answer

General formulaIUPAC name of the homologous seriesCharacteristic bond typeIUPAC name of the first member of the seriesCnH2n-2(A) Alkynes(B) Triple covalent bond(C) EthyneCnH2n+2(D) Alkanes(E) Single covalent bond(F) Methane

Question 13

Question 5(b)(i) Name the most common ore of the metal aluminum from which the metalis extracted. Write the chemical formula of the ore.(ii) Name the process by which impure ore of aluminum gets purified by using concentrated solution of an alkali.(iii) Write the equation for the formation of aluminum at the cathode during the electrolysis of alumina.

Accepted Answer

(i) Bauxite [Al2O3.2H2O](ii) Bayer's Process(iii) 2Al3+ + 6e- ⟶ 2Al

Question 14

Question 6(a)A compound X (having vinegar like smell) when treated with ethanol in the presence of the acid Z, gives a compound Y which has a fruity smell.The reaction is:C2H5OH+X→ZY+H2O	ext{C}_2	ext{H}_5	ext{OH} + 	ext{X} \xrightarrow{	ext{Z}} 	ext{Y} + 	ext{H}_2	ext{O}C2​H5​OH+XZ​Y+H2​O(i) Identify Y and Z.(ii) Write the structural formula of X.(iii) Name the above reaction.

Accepted Answer

C2H5OHethyl alcohol+CH3COOHacetic acid→Conc. H2SO4CH3−COO−C2H5ethyl ethanoate+H2O\underset{	ext{ethyl alcohol}}{	ext{C}_2	ext{H}_5	ext{OH}} + \underset{	ext{acetic acid}}{	ext{CH}_3	ext{COOH}} \xrightarrow{	ext{Conc. H}_2	ext{SO}_4} \underset{	ext{ethyl ethanoate}}{	ext{CH}_3-	ext{COO}-	ext{C}_2	ext{H}_5} + 	ext{H}_2	ext{O}ethyl alcoholC2​H5​OH​+acetic acidCH3​COOH​Conc. H2​SO4​​ethyl ethanoateCH3​−COO−C2​H5​​+H2​O(i) Y is ethyl acetate and Z is conc. sulphuric acid(ii) Structural formula of X (ethanoic acid) is shown below:(iii) Esterification reaction

Question 15

Question 6(b)Ethane burns in oxygen to form CO2 and H2O according to the equation:2C2H6 + 7O2 ⟶ 4CO2 + 6H2O.If 1250 cc of oxygen is burnt with 300 cc of ethane.Calculate:(i) the volume of CO2 formed.(ii) the volume of unused O2.

Accepted Answer

[By Lussac's law]2C2H6+7O2⟶4CO2+6H2O2 vol.:7 vol.⟶4 vol.\begin{matrix} 2	ext{C}_2	ext{H}_6 & + & 7	ext{O}_2 &\longrightarrow & 4	ext{CO}_2 & + & 6	ext{H}_2	ext{O} \ 2	ext{ vol.} & : & 7 	ext{ vol.} & \longrightarrow & 4	ext{ vol.} \end{matrix}2C2​H6​2 vol.​+:​7O2​7 vol.​⟶⟶​4CO2​4 vol.​+6H2​O(i) To calculate the volume of CO2 formed :C2H6:CO22 vol.:4 vol.300 cc:x\begin{matrix}	ext{C}_2	ext{H}_6 & : & 	ext{CO}_2 \ 2 	ext{ vol.} & : & 4 	ext{ vol.} \ 300 	ext{ cc} & : & 	ext{x} \end{matrix}C2​H6​2 vol.300 cc​:::​CO2​4 vol.x​∴x=42×300=600 ccx = \dfrac{4}{2} 	imes 300 = 600 	ext{ cc}x=24​×300=600 ccHence, volume of carbon dioxide formed = 600 cc(ii) To calculate the volume of unused O2 :C2H6:O22 vol.:7 vol.300 cc:x\begin{matrix}	ext{C}_2	ext{H}_6 & : & 	ext{O}_2 \ 2 	ext{ vol.} & : & 7 	ext{ vol.} \ 300 	ext{ cc} & : & 	ext{x} \end{matrix}C2​H6​2 vol.300 cc​:::​O2​7 vol.x​∴x=72×300=1050 ccx = \dfrac{7}{2} 	imes 300 = 1050 	ext{ cc}x=27​×300=1050 ccUnused oxygen = 1250 - 1050 = 200 cc.Hence, volume of unused oxygen = 200 cc

Question 16

Question 6(c)Three solutions P, Q and R have pH value of 3.5, 5.2 and 12.2 respectively.Which one of these is a:(i) Weak acid?(ii) Strong alkali?

Accepted Answer

(i) QReason — On a pH scale, acids have pH less than 7 whereas weak acids have pH towards 7. Hence, Q will be a weak acid with pH 5.2 .(ii) RReason — On a pH scale, alkali have pH more than 7 and alkalinity increases as the pH value moves away from 7. Hence, R will be a strong alkali with pH 12.2 .

Question 17

Question 7(a)Give a chemical test to distinguish between the following pairs of chemicals:(i) Lead nitrate solution and Zinc nitrate solution(ii) Sodium chloride solution and Sodium nitrate solution

Accepted Answer

(i) When NaOH is added to each of the compounds, lead nitrate forms a chalky white precipitate of lead hydroxide [Pb(OH)2] whereas a gelatinous white precipitate of zinc hydroxide [Zn(OH)2] is formed in case of zinc nitrate.(ii) Add silver nitrate soln. to the given solns., sodium chloride reacts to form a white ppt. which is soluble in NH4OH and insoluble in dil. HNO3. The other soln. is sodium nitrate.NaCl + AgNO3 ⟶ AgCl + NaNO3NaNO3 + AgNO3 ⟶ no white ppt.

Question 18

Question 7(b)Write a balanced equation for the preparation of each of the following salts:(i) Copper sulphate from Copper carbonate.(ii) Zinc carbonate from Zinc sulphate.

Accepted Answer

(i) CuCO3 + H2SO4 ⟶ CuSO4 + H2O + CO2(ii) ZnSO4 + (NH4)2CO3 ⟶ (NH4)2SO4 + ZnCO3

Question 19

Question 7(c)(i) What is the type of salt formed when the reactants are heated at a suitable temperature for the preparation of Nitric acid?(ii) State why for the preparation of Nitric acid, the complete apparatus is made up of glass.

Accepted Answer

(i) Acid salt(ii) The vapours of nitric acid are highly corrosive and attack rubber, cork. etc. For this reason, all glass apparatus is used in the laboratory preparation of nitric acid.

Question 20

Question 7(d)Which property of sulphuric acid is shown by the reaction of concentrated sulphuric acid with:(i) Ethanol?(ii) Carbon?

Accepted Answer

(i) Dehydrating propertyC2H5OH→Conc. H2SO4C2H4 [ethene−ethylene]+H2O	ext{C}_{2}	ext{H}_{5}	ext{OH}\xrightarrow{	ext{Conc. H}_2	ext{SO}_4} 	ext{C}_{2}	ext{H}_{4} \small{\space[ethene-ethylene]} + 	ext{H}_2	ext{O}C2​H5​OHConc. H2​SO4​​C2​H4​ [ethene−ethylene]+H2​O(ii) Oxidizing propertyC + 2H2SO4 (conc.) ⟶ CO2 + 2SO2 + 2H2O

Solved 2018 Question Paper ICSE Class 10 Chemistry

Section I 40 Marks

Section Ii 40 Marks