Q: Question 1(x)The salt formed when concentrated sulphuric acid reacts with KNO3 above 200°C:K2SO4K2SO3KHSO4KHSO3

K2SO4Reason —KNO3+H2SO4[conc.]→>200°CK2SO4+HNO3\text{KNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\gt 200 \degree\text{C}} \text{K}_2\text{SO}_\text{4} + \text{HNO}_3KNO3+H2SO4[conc.]>200°CK2SO4+HNO3

Question 1

Question 1(iv)The vapour density of CH3OH is ............... (At. Wt. C=12, H=1, O=16)32181634

Accepted Answer

16Reason —Molecular mass of methyl alcohol [CH3OH] = 12 + 3(1) + 16 + 1 = 32Vapour density = molecular mass2\dfrac{	ext{molecular mass}}{2}2molecular mass​ = 322\dfrac{32}{2}232​ = 16

Question 2

Question 1(vi)The metallic hydroxide which forms a deep inky blue solution with excess ammonium hydroxide solution is:Fe(OH)2Cu(OH)2Ca(OH)2Fe(OH)3

Accepted Answer

Cu(OH)2Reason — Cu(OH)2 reacts with a solution of excess ammonium hydroxide to form a deep blue solution of tetrammine copper [II] sulphate.Cu(OH)2 + (NH4)2SO4 + 2NH4OH ⟶ [Cu(NH3)4]SO4 + 4H2O

Question 3

Question 1(ix)The nitrate which on thermal decomposition leaves behind a residue which is yellow when hot and white when cold:Lead nitrateAmmonium nitrateCopper nitrate.Zinc nitrate

Accepted Answer

Zinc nitrateReason —2Zn(NO3)2⟶2ZnO+4NO2+O2[Zinc[Zinc nitrate - white][Carboncarbonate][yellow - hot]dioxide][white][white - cold]\begin{matrix} 2	ext{Zn(NO}_3)_2 &\longrightarrow & 	ext{2ZnO} & + & 4	ext{NO}_2 &+ & 	ext{O}_2 \ 	ext{[Zinc} & & 	ext{[Zinc nitrate - white]} & & 	ext{[Carbon} \ 	ext{carbonate]} & & 	ext{[yellow - hot]} & & 	ext{dioxide]} \ 	ext{[white]} & & 	ext{[white - cold]} \end{matrix}2Zn(NO3​)2​[Zinccarbonate][white]​⟶​2ZnO[Zinc nitrate - white][yellow - hot][white - cold]​+​4NO2​[Carbondioxide]​+O2​Zinc nitrate decomposes to produces zinc oxide, which is yellow when hot and white when cold

Question 4

Question 1(x)The salt formed when concentrated sulphuric acid reacts with KNO3 above 200°C:K2SO4K2SO3KHSO4KHSO3

Accepted Answer

K2SO4Reason —KNO3+H2SO4[conc.]→>200°CK2SO4+HNO3	ext{KNO}_3 + 	ext{H}_2	ext{SO}_4 [	ext{conc.}]\xrightarrow{\gt 200 \degree	ext{C}} 	ext{K}_2	ext{SO}_	ext{4} + 	ext{HNO}_3KNO3​+H2​SO4​[conc.]>200°C​K2​SO4​+HNO3​

Question 5

Question 1(xii)The hydrocarbon formed when sodium propanoate and soda lime are heated together:MethaneEthaneEthenePropane

Accepted Answer

EthaneReason —CH3-CH2.COONasodium propionate+NaOHsodalime→ΔCaOC2H6ethane+Na2CO3\underset{	ext{sodium propionate}}{	ext{C}	ext{H}_3	ext{-C}	ext{H}_2	ext{.COONa}} + \underset{	ext{sodalime}}{	ext{NaOH}} \xrightarrow[\Delta]{	ext{CaO}} \underset{	ext{ethane}}{	ext{C}_2	ext{H}_6} + 	ext{Na}_2	ext{CO}_3sodium propionateCH3​-CH2​.COONa​+sodalimeNaOH​CaOΔ​ethaneC2​H6​​+Na2​CO3​

Question 6

Question 1(xv)The indicator which changes to pink colour in an alkaline solution is:Blue LitmusMethyl OrangeRed LitmusPhenolphthalein

Accepted Answer

PhenolphthaleinReason — Colourless phenolphthalein changes to pink colour in an alkaline solution.

Question 7

Question 2(i)Match the Column A with Column B:Column AColumn B(a) Sodium Chloride1. has two shared pair of electrons(b) Methane2. has high melting and boiling points(c) Hydrogen chloride gas3. a greenhouse gas(d) Oxidation reaction4. has low melting and boiling points(e) Water5. Zn - 2e- ⟶ Zn2+ 6. S + 2e- ⟶ S2-

Accepted Answer

Column AColumn B(a) Sodium Chloride2. has high melting and boiling points(b) Methane3. a greenhouse gas(c) Hydrogen chloride gas4. has low melting and boiling points(d) Oxidation reaction5. Zn - 2e- ⟶ Zn2+(e) Water1. has two shared pair of electrons

Question 8

Question 2(ii)The following sketch illustrates the process of conversion of Alumina to Aluminium:Study the diagram and answer the following:(a) Name the constituent of the electrolyte mixture which has a divalent metal in it.(b) Name the powdered substance 'X' sprinkled on the surface of the electrolyte mixture.(c) What is the name of the process?(d) Write the reactions taking place at the electrodes 'Y' (anode) and 'Z' (cathode) respectively.

Accepted Answer

(a) Fluorspar [CaF2](b) Powdered coke(c) Hall-Heroult's process(d) Reactions taking place at the electrodes 'Y' (anode) and 'Z' (cathode):At anode:6O2- - 12e- ⟶ 6[O]3O + 3O ⟶ 3O2At cathode:4Al3+ + 12e- ⟶ 4Al

Question 9

Question 2(v)(a) Give the structural formula of the following organic compounds:2-chlorobutaneMethanalBut-2-yne(b) Give the IUPAC name of the following organic compounds:

Accepted Answer

(a) Structural formula1. 2-chlorobutane2. Methanal3. But-2-yne(b) IUPAC names are:Ethanoic acidButan-2-ol

Question 10

Question 3(i)Identify the cation in each of the following cases:(a) Ammonium hydroxide solution when added to Solution B gives a white precipitate which does not dissolve in excess of ammonium hydroxide solution.(b) Sodium hydroxide solution when added to Solution C gives a white precipitate which is insoluble in excess of sodium hydroxide solution.

Accepted Answer

(a) Pb2+Reason — When ammonium hydroxide solution is added to lead nitrate, it forms a chalky white precipitate of lead hydroxide [Pb(OH)2] which is insoluble in excess of ammonium hydroxide.Pb(NO3)2 + 2NH4OH ⟶ 2NH4NO3 + Pb(OH)2(b) Ca2+Reason — When sodium hydroxide solution is added to calcium nitrate, a white precipitate of calcium hydroxide is obtained which is insoluble in excess of sodium hydroxide solution.Ca(NO3)2 + 2NaOH ⟶ Ca(OH)2 ↓ + 2NaNO3

Question 11

Question 3(ii)Fill in the blanks by choosing the correct answer from the brackets:(a) During electrolysis, the compound ............... in its molten state liberates reddish brown fumes at the anode. [NaCl/PbBr2](b) The ion which could be discharged most readily during electrolysis is ............... [Fe2+/Cu2+]

Accepted Answer

(a) During electrolysis, the compound PbBr2 in its molten state liberates reddish brown fumes at the anode.(b) The ion which could be discharged most readily during electrolysis is Cu2+

Question 12

Question 3(iii)Arrange the following as per the instruction given in the brackets:(a) Al, K, Mg, Ca (decreasing order of its reactivity)(b) N, Be, O, C (increasing order of non-metallic character)(c) P, Si, F, Be (decreasing order of valence electrons)

Accepted Answer

(a) K > Ca > Mg > AlReason — According to the electrochemical series metals at the top of the series are most reactive and the reactivity decreases down the group.(b) Be < C < N < OReason — Across a period, left to right in a periodic table, the non-metallic character increases.(c) F > P > Si > BeReason — The valence electrons decreases from right to left in the periodic table.

Question 13

Question 4(i)State a relevant reason for the following:(a) Hydrogen chloride gas cannot be dried over quick lime.(b) Ammonia gas is not collected over water.

Accepted Answer

(a) Quicklime [CaO] is alkaline in nature and it reacts with HCl forming the respective chloride. Hence, it can't be used as a drying agent for Hydrogen chloride gas.CaO + 2HCl ⟶ CaCl2 + H2O(b) As ammonia gas is highly soluble in water, therefore, it is not collected over water.

Question 14

Question 4(iii)Solve the following numerical problem.Ethane burns in oxygen according to the chemical equation:2C2H6 + 7O2 ⟶ 4CO2 + 6H2OIf 80 ml of ethane is burnt in 300 ml of oxygen, find the composition of the resultant gaseous mixture when measured at room temperature.

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.80 ml:x\begin{matrix}	ext{C}_2	ext{H}_6 & : & 	ext{CO}_2 \ 2 	ext{ vol.} & : & 4 	ext{ vol.} \ 80 	ext{ ml} & : & 	ext{x} \end{matrix}C2​H6​2 vol.80 ml​:::​CO2​4 vol.x​∴x=42×80=160 ml	herefore x = \dfrac{4}{2} 	imes 80 = 160 	ext{ ml}∴x=24​×80=160 mlHence, volume of carbon dioxide formed = 160 ml(ii) To calculate the volume of unused O2 :C2H6:O22 vol.:7 vol.80 ml:x\begin{matrix}	ext{C}_2	ext{H}_6 & : & 	ext{O}_2 \ 2 	ext{ vol.} & : & 7 	ext{ vol.} \ 80 	ext{ ml} & : & 	ext{x} \end{matrix}C2​H6​2 vol.80 ml​:::​O2​7 vol.x​∴x=72×80=280 ml	herefore x = \dfrac{7}{2} 	imes 80 = 280 	ext{ ml}∴x=27​×80=280 mlUnused oxygen = 300 - 280 = 20 ml.Hence, volume of unused oxygen = 20 mlTherefore, the resultant gaseous mixture consists of 160 ml of carbon dioxide and 20 ml of unused oxygen.

Question 15

Question 4(iv)The following questions are pertaining to the laboratory preparation of Ammonia gas from Magnesium nitride:(a) Write a balanced chemical equation for its preparation.(b) Why is this method seldom used?(c) How do you identify the gas formed?

Accepted Answer

(a) Mg3N2 + 6H2O ⟶ 3Mg(OH)2 + 2NH3(b) As nitrides are expensive, hence the method is seldom used.(c) Ways in which ammonia gas can be identified is:It has a sharp characteristic odour.It turns:moist red litmus blue,moist turmeric paper brown,phenolphthalein solution pink.When a glass rod dipped in conc. HCl is brought near the mouth of the jar containing NH3, dense white fumes of ammonium chloride are formed.NH3 + HCl ⟶ NH4Cl

Question 16

Question 5(iii)Give a balanced chemical equation for the following conversions with conditions:(a) Ethene from ethanol(b) Ethyne from calcium carbide(c) Monochloromethane from methane

Accepted Answer

(a) Ethene from ethanolC2H5OH 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 (b) Ethyne from calcium carbideCaC2calcium carbide+2H2Owater⟶C2H2ethyne+Ca(OH)2calcium hydroxide\underset{	ext{calcium carbide}}{	ext{CaC}_2} + \underset{	ext{water}}{2	ext{H}_2	ext{O}} \longrightarrow \underset{	ext{ethyne}}{	ext{C}_2	ext{H}_2} + \underset{	ext{calcium hydroxide}}{	ext{Ca(OH)}_2}calcium carbideCaC2​​+water2H2​O​⟶ethyneC2​H2​​+calcium hydroxideCa(OH)2​​(c) Monochloromethane from methaneCH4 methane+Cl2→Δdiffused sunlightCH3Clmonochloromethane+HCl\underset{	ext{ methane} }{	ext{CH}_4} + 	ext{Cl}_2 \xrightarrow[\Delta]{	ext{diffused sunlight}} \underset{	ext{monochloromethane}}{	ext{CH}_3	ext{Cl}} + 	ext{HCl} methaneCH4​​+Cl2​diffused sunlightΔ​monochloromethaneCH3​Cl​+HCl

Question 17

Question 6(i)Define / State:(a) Electronegativity(b) Gay-Lussac's Law of combining volumes

Accepted Answer

(a) Electronegativity — The tendency of an atom in a molecule to attract the shared pair of electrons towards itself is called its electronegativity.(b) Gay-Lussac's Law of combining volumes — When gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

Question 18

Question 6(ii)The Empirical formula of an organic compound is CHCl2.If its relative molecular mass is 168, what is its molecular formula?[At. Wt. C = 12, H = 1, Cl = 35.5]

Accepted Answer

Empirical formula is CHCl2Empirical formula weight = 12 + 1 + 2[35.5] = 12 + 1 + 71 = 84Molecular weight = 168n=Molecular weightEmpirical formula weight=16884=2\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{168}{84} = 2n=Empirical formula weightMolecular weight​=84168​=2So, molecular formula = 2(CHCl2) = C2H2Cl4

Question 19

Question 7(ii)Give reason for the following:(a) Ionisation potential decreases down a group.(b) Ionic compounds do not conduct electricity in solid state.

Accepted Answer

(a) On moving down the group, atomic size, as well as, nuclear charge increases. However, the effect of increase in atomic size dominates over the effect of increase in nuclear charge. Hence, ionization potential decreases down the group.(b) There is a strong electrostatic force of attraction between the oppositely charged ions of the ionic compounds that keeps ions in fixed position in solid state. Due to this immobility of ions, ionic compounds do not conduct electricity in solid state.

Question 20

Question 7(iii)Calculate:(a) The percentage of phosphorus in the fertilizer super phosphate Ca(H2PO4)2 correct to 1 decimal point.[At. Wt. H=1, P=31, O=16, Ca=40](b) Write the empirical formula of C8H18

Accepted Answer

(a) Molecular weight of Ca(H2PO4)2= 40 + 2[2(1) + 31 + 4(16)]= 40 + 2[2 + 31 + 64]= 40 + 2[97]= 40 + 194= 234 g234 g of Ca(H2PO4)2 contains 62 g of P∴ 100 g of Ca(H2PO4)2 will contain = 62234\dfrac{62}{234}23462​ x 100 = 26.49% = 26.5%Hence, 26.5% phosphorous is present in superphosphate Ca(H2PO4)2(b) Given,Molecular Formula = C8H18 = (C4H9)2Molecular formula = n[Empirical formula]∴ n = 2Hence, Empirical formula = C4H9

Solved 2023 Question Paper ICSE Class 10 Chemistry

Section A 40 Marks

Section B 40 Marks