π Lessons & Marks Weightage (As per ICSE Board Pattern)
| π’ Lesson | π Marks Weightage | π Question Types |
|---|---|---|
| β‘ Electrolysis | 20β25 Marks | MCQ, 3β4 Mark Questions, Diagram-Based |
| π§ͺ Study of Compounds | 50β55 Marks | MCQ, 3β4 Mark Questions, Diagram-Based |
| π¬ Practical Chemistry | 10β15 Marks | MCQ, 3β4 Mark Questions, Diagram-Based |
| Total | 80 Marks |
π Detailed Breakdown of Study of Compounds:
| π§ͺ Compound | π Marks Weightage |
|---|---|
| π¨ Hydrogen Chloride | 10β12 Marks |
| π¨ Ammonia | 12β15 Marks |
| π¨ Nitric Acid | 12β15 Marks |
| π¨ Sulphuric Acid | 12β15 Marks |
π Subject: Chemistry
π Class: X (ICSE)
π Maximum Marks: 80
β° Time Allowed: 2 Hours
π General Instructions:
βοΈ Answers to this Paper must be written on the paper provided separately.
β³ You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper.
π The time given at the head of this Paper is the time allowed for writing the answers.
β Attempt all questions from Section A and any four questions from Section B.
π All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer.
β οΈ Omission of essential working will result in loss of marks.
π― The intended marks for questions or parts of questions are given in brackets
[ ].π Scientific calculators and log tables are not allowed.
π SECTION A (40 Marks)
(Attempt all questions from this Section.)
β Question 1 [15 Marks]
Choose the correct answers to the questions from the given options. (Do not copy the questions, write the correct answers only.)
(i) β‘ Which of the following is a non-electrolyte?
(a) NaCl solution
(b) Dilute HβSOβ
(c) Glucose solution
(d) CuSOβ solution
β
Answer: (c) Glucose solution
π Explanation: Glucose does not dissociate into ions in solution, hence it is a non-electrolyte.
(ii) β‘ During electrolysis of acidified water, the gas evolved at the cathode is:
(a) Oxygen
(b) Hydrogen
(c) Chlorine
(d) Nitrogen
β
Answer: (b) Hydrogen
π Explanation: At cathode (negative electrode), HβΊ ions gain electrons to form hydrogen gas.
(iii) β‘ The electrolyte used in the electroplating of silver is:
(a) Silver nitrate
(b) Silver chloride
(c) Silver sulphate
(d) Silver cyanide (potassium argentocyanide)
β
Answer: (d) Silver cyanide (potassium argentocyanide)
π Explanation: Potassium argentocyanide is used as electrolyte for silver plating.
(iv) π¨ Which gas is evolved when dilute HCl is added to sodium carbonate?
(a) Hydrogen
(b) Oxygen
(c) Carbon dioxide
(d) Chlorine
β
Answer: (c) Carbon dioxide
π Explanation: NaβCOβ + 2HCl β 2NaCl + HβO + COββ
(v) π¨ The gas which turns moist red litmus paper blue is:
(a) HCl
(b) NHβ
(c) COβ
(d) SOβ
β
Answer: (b) NHβ
π Explanation: Ammonia is basic in nature and turns red litmus blue.
(vi) π¨ Which of the following is used as a drying agent for ammonia gas?
(a) Conc. HβSOβ
(b) PβOβ
(c) Quick lime (CaO)
(d) Anhydrous CaClβ
β
Answer: (c) Quick lime (CaO)
π Explanation: Ammonia is basic, so it cannot be dried with acidic drying agents like HβSOβ or PβOβ
. CaO is basic and suitable.
(vii) π¨ The catalyst used in Ostwald’s process for manufacture of nitric acid is:
(a) Iron
(b) Platinum gauze
(c) Vanadium pentoxide
(d) Manganese dioxide
β
Answer: (b) Platinum gauze
π Explanation: Ostwald’s process uses platinum gauze as catalyst for oxidation of ammonia to nitric oxide.
(viii) π¨ In the contact process for manufacture of sulphuric acid, the catalyst used is:
(a) Platinum
(b) Vanadium pentoxide (VβOβ
)
(c) Iron
(d) Nickel
β
Answer: (b) Vanadium pentoxide (VβOβ
)
π Explanation: VβOβ
is used as catalyst in contact process.
(ix) π¨ Which of the following is a property of hydrogen chloride gas?
(a) It is coloured
(b) It is lighter than air
(c) It is heavier than air
(d) It has a sweet smell
β
Answer: (c) It is heavier than air
π Explanation: HCl gas is about 1.28 times heavier than air.
(x) π¨ The compound formed when ammonia reacts with excess chlorine is:
(a) Nβ
(b) NClβ
(c) NHβCl
(d) NO
β
Answer: (b) NClβ
π Explanation: 2NHβ + 3Clβ β Nβ + 6HCl (excess NHβ); NHβ + 3Clβ β NClβ + 3HCl (excess Clβ)
(xi) π¬ In the laboratory preparation of HCl gas, the drying agent used is:
(a) Conc. HβSOβ
(b) Quick lime
(c) NaOH
(d) KOH
β
Answer: (a) Conc. HβSOβ
π Explanation: Conc. HβSOβ is used to dry HCl gas as it does not react with it.
(xii) π¬ The flame colour observed when a nichrome wire dipped in HCl is introduced into a non-luminous flame is:
(a) Golden yellow
(b) Brick red
(c) Apple green
(d) Blue
β
Answer: (d) Blue
π Explanation: HCl gives a blue flame colouration in the flame test.
(xiii) π¬ Which of the following is used to test for chloride ions?
(a) BaClβ solution
(b) AgNOβ solution
(c) NaOH solution
(d) FeSOβ solution
β
Answer: (b) AgNOβ solution
π Explanation: AgNOβ gives a white precipitate of AgCl with chloride ions.
(xiv) π¬ The gas evolved when dilute HCl is added to iron(II) sulphide is:
(a) HβS
(b) SOβ
(c) Clβ
(d) Oβ
β
Answer: (a) HβS
π Explanation: FeS + 2HCl β FeClβ + HβSβ (rotten egg smell)
(xv) π¬ In the test for nitrate ions, the brown ring complex formed is:
(a) [Fe(HβO)β
NO]SOβ
(b) [Fe(HβO)β
NO]Β²βΊ
(c) [Fe(NO)]SOβ
(d) [Fe(HβO)β]Β²βΊ
β
Answer: (b) [Fe(HβO)β
NO]Β²βΊ
π Explanation: Brown ring complex is nitroso ferrous sulphate, [Fe(HβO)β
NO]SOβ.
β Question 2 [12 Marks]
(Solve the following)
(i) β‘ State three differences between electrolytic conduction and metallic conduction. [4]
β Answer:
| Feature | Electrolytic Conduction | Metallic Conduction |
|---|---|---|
| Carrier | Ions | Electrons |
| Nature | Chemical change occurs | No chemical change |
| Temperature effect | Conductivity increases with temperature | Conductivity decreases with temperature |
| Transfer | Matter is transferred | Only energy is transferred |
(ii) π¨ Explain the laboratory preparation of ammonia gas with a balanced chemical equation and a labelled diagram. [4]
β Answer:
π Laboratory Preparation of Ammonia:
Chemicals: Ammonium chloride (NHβCl) and slaked lime [Ca(OH)β]
Equation:
2NHβCl + Ca(OH)β β CaClβ + 2HβO + 2NHββ
Diagram:
βββββββββββββββ β NHβCl + β β Ca(OH)β β ββββββββ¬βββββββ β ββββββββΌβββββββ β Hard β β Glass β β Test Tube β ββββββββ¬βββββββ β ββββββββΌβββββββ β Delivery β β Tube β ββββββββ¬βββββββ β ββββββββΌβββββββ β NHβ Gas β β Collected β β by Downwardβ β Displacementβ βββββββββββββββ
Method: The mixture is heated in a hard glass test tube. Ammonia gas is collected by downward displacement of air (since it is lighter than air).
(iii) π¨ Write balanced chemical equations for:
(a) Ammonia reacts with excess chlorine
(b) Nitric acid reacts with copper
(c) Sulphuric acid reacts with sodium hydroxide
(d) Hydrogen chloride reacts with ammonia [4]
β Answer:
(a) NHβ + 3Clβ (excess) β NClβ + 3HCl
(b) 3Cu + 8HNOβ (dilute) β 3Cu(NOβ)β + 2NO + 4HβO
(c) HβSOβ + 2NaOH β NaβSOβ + 2HβO
(d) HCl + NHβ β NHβCl (white fumes of ammonium chloride)
β Question 3 [12 Marks]
(Solve the following)
(i) β‘ Explain the electrolysis of acidified water with a labelled diagram. [4]
β Answer:
π Electrolysis of Acidified Water:
Electrolyte: Acidified water (HβSOβ added to distilled water)
Electrodes: Platinum electrodes
Diagram:
βββββββββββββββββββββββββββ β + β β β β β β β ββββΌβββ ββββΌβββ β β βPt β βPt β β β βAnodeβ βCathodeβ β β ββββ¬βββ ββββ¬βββ β β β β β β ββββͺββββββββββββͺβββ β β β Acidified β β β β Water β β β ββββββββββββββ β βββββββββββββββββββββββββββ
Reactions:
At Cathode (β): 4HβΊ + 4eβ» β 2Hββ (Hydrogen gas)
At Anode (+): 2HβO β Oβ + 4HβΊ + 4eβ» (Oxygen gas)
Overall: 2HβO β 2Hβ + Oβ
Ratio: Hydrogen : Oxygen = 2 : 1 (by volume)
(ii) π¨ Describe the contact process for manufacture of sulphuric acid with equations. [4]
β Answer:
π Contact Process for HβSOβ:
Step 1: Burning of sulphur in air
S + Oβ β SOβ
Step 2: Catalytic oxidation of SOβ to SOβ
2SOβ + Oβ β 2SOβ (Catalyst: VβOβ
, Temperature: 450Β°C, Pressure: 2 atm)
Step 3: Absorption of SOβ in conc. HβSOβ
SOβ + HβSOβ β HβSβOβ (Oleum)
Step 4: Dilution of oleum
HβSβOβ + HβO β 2HβSOβ
Conditions:
Catalyst: Vanadium pentoxide (VβOβ )
Temperature: 450Β°C
Pressure: 2 atmospheres
(iii) π¬ Give one chemical test to distinguish between:
(a) HCl gas and NHβ gas
(b) Dilute HCl and dilute HNOβ
(c) FeSOβ and FeClβ
(d) SOβ and COβ [4]
β Answer:
(a) HCl vs NHβ: HCl turns moist blue litmus red; NHβ turns moist red litmus blue.
(b) Dilute HCl vs Dilute HNOβ: Add AgNOβ solution. HCl gives white precipitate of AgCl; HNOβ gives no precipitate.
(c) FeSOβ vs FeClβ: Add NaOH solution. FeSOβ gives dirty green precipitate of Fe(OH)β; FeClβ gives reddish-brown precipitate of Fe(OH)β.
(d) SOβ vs COβ: Pass through acidified KβCrβOβ solution. SOβ turns it green; COβ has no effect.
π SECTION B (40 Marks)
(Attempt any four questions from this Section.)
β Question 4 [10 Marks]
(i) β‘ Define the following terms:
(a) Electrolyte
(b) Non-electrolyte
(c) Electrolysis [3]
β Answer:
(a) Electrolyte: A substance which in aqueous solution or molten state conducts electricity and undergoes chemical decomposition.
(b) Non-electrolyte: A substance which does not conduct electricity in aqueous solution or molten state.
(c) Electrolysis: The process of decomposition of an electrolyte by the passage of electric current.
(ii) π¨ Write short notes on:
(a) Aqua regia
(b) Brown ring test [3]
β Answer:
(a) Aqua Regia: A mixture of 3 parts concentrated HCl and 1 part concentrated HNOβ. It is a strong oxidizing agent and dissolves noble metals like gold and platinum.
(b) Brown Ring Test: Test for nitrate ions. To the solution, add FeSOβ solution and then slowly add conc. HβSOβ along the walls of the test tube. A brown ring of [Fe(HβO)β NO]SOβ forms at the junction.
(iii) π¬ A gas ‘X’ is evolved when dilute HCl is added to a compound ‘Y’. The gas ‘X’ turns lime water milky. Identify X and Y. Write the balanced equation. [4]
β Answer:
Gas X: Carbon dioxide (COβ)
Compound Y: Calcium carbonate (CaCOβ) or any carbonate/bicarbonate
Equation:
CaCOβ + 2HCl β CaClβ + HβO + COββ
Test: COβ + Ca(OH)β β CaCOββ + HβO (lime water turns milky)
β Question 5 [10 Marks]
(i) β‘ Explain why:
(a) Copper cannot be used as anode in electrolysis of CuSOβ solution with platinum electrodes.
(b) A solution of HCl in water conducts electricity but a solution of HCl in toluene does not. [3]
β Answer:
(a) If copper is used as anode, it will dissolve into the solution instead of being deposited. Platinum is inert and does not react.
(b) HCl in water dissociates into HβΊ and Clβ» ions which conduct electricity. HCl in toluene does not dissociate into ions, hence does not conduct.
(ii) π¨ How is nitric acid manufactured by Ostwald’s process? Write all equations. [3]
β Answer:
π Ostwald’s Process:
Step 1: Catalytic oxidation of ammonia
4NHβ + 5Oβ β 4NO + 6HβO (Catalyst: Pt gauze, Temp: 800Β°C)
Step 2: Oxidation of NO to NOβ
2NO + Oβ β 2NOβ
Step 3: Absorption of NOβ in water
3NOβ + HβO β 2HNOβ + NO
(iii) π¬ Draw a labelled diagram for the laboratory preparation of hydrogen chloride gas and write the equation. [4]
β Answer:
π Laboratory Preparation of HCl Gas:
Chemicals: Sodium chloride (NaCl) and concentrated HβSOβ
Equation:
NaCl + HβSOβ β NaHSOβ + HClβ (at 150Β°C)
NaHSOβ + NaCl β NaβSOβ + HClβ (at 200Β°C)
Diagram:
βββββββββββββββ β NaCl + β β Conc. β β HβSOβ β ββββββββ¬βββββββ β ββββββββΌβββββββ β Round β β Bottom β β Flask β ββββββββ¬βββββββ β ββββββββΌβββββββ β Delivery β β Tube β ββββββββ¬βββββββ β ββββββββΌβββββββ β Conc. β β HβSOβ β β (Drying) β ββββββββ¬βββββββ β ββββββββΌβββββββ β HCl Gas β β Collected β β by Upward β β Displacementβ βββββββββββββββ
Method: HCl gas is collected by upward displacement of air (since it is heavier than air).
β Question 6 [10 Marks]
(i) β‘ What is electroplating? Explain the electroplating of iron with silver. [3]
β Answer:
Electroplating: The process of depositing a thin layer of one metal on another metal using electrolysis.
Silver Plating on Iron:
Cathode (β): Iron article
Anode (+): Pure silver rod
Electrolyte: Potassium argentocyanide [KAg(CN)β]
Reactions:
At Cathode: AgβΊ + eβ» β Ag (deposited on iron)
At Anode: Ag β AgβΊ + eβ» (dissolves from silver rod)
(ii) π¨ Give reasons:
(a) Ammonia is not dried by conc. HβSOβ.
(b) Nitric acid turns yellow on standing.
(c) HCl gas gives white fumes with ammonia. [3]
β Answer:
(a) Ammonia is basic and reacts with conc. HβSOβ to form ammonium sulphate. Hence, it cannot be dried.
(b) Nitric acid decomposes in presence of light to form NOβ which dissolves in it, giving a yellow colour.
(c) HCl reacts with NHβ to form white fumes of ammonium chloride (NHβCl).
(iii) π¬ Complete the following reactions:
(a) NHβCl + Ca(OH)β β
(b) Cu + 4HNOβ (conc.) β
(c) SOβ + HβO β
(d) AgNOβ + HCl β [4]
β Answer:
(a) 2NHβCl + Ca(OH)β β CaClβ + 2HβO + 2NHββ
(b) Cu + 4HNOβ (conc.) β Cu(NOβ)β + 2NOβ + 2HβO
(c) SOβ + HβO β HβSOβ
(d) AgNOβ + HCl β AgClβ + HNOβ (white precipitate)
β Question 7 [10 Marks]
(i) β‘ State Faraday’s laws of electrolysis. [3]
β Answer:
First Law: The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed.
m β Q = It
m = ZIt (where Z is electrochemical equivalent)
Second Law: When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited or liberated are proportional to their chemical equivalent weights.
mβ/mβ = Eβ/Eβ
(ii) π¨ What is the action of heat on:
(a) Ammonium chloride
(b) Copper nitrate
(c) Lead nitrate [3]
β Answer:
(a) NHβCl β NHβ + HCl (decomposes on heating)
(b) 2Cu(NOβ)β β 2CuO + 4NOβ + Oβ (copper nitrate decomposes)
(c) 2Pb(NOβ)β β 2PbO + 4NOβ + Oβ (lead nitrate decomposes)
(iii) π¬ A white precipitate is obtained when BaClβ solution is added to a solution of compound ‘X’. The precipitate is insoluble in dilute HCl. Identify ‘X’ and write the equation. [4]
β Answer:
Compound X: Sodium sulphate (NaβSOβ) or any sulphate
Precipitate: Barium sulphate (BaSOβ) β white, insoluble in dilute HCl
Equation:
NaβSOβ + BaClβ β BaSOββ + 2NaCl
Test: BaSOβ is insoluble in dilute HCl, confirming sulphate ions.
β Question 8 [10 Marks]
(i) β‘ Explain the terms:
(a) Electrochemical equivalent
(b) Current efficiency [3]
β Answer:
(a) Electrochemical Equivalent (Z): The mass of substance deposited or liberated by 1 coulomb of electricity. Z = E/F (where E is equivalent weight, F is Faraday constant).
(b) Current Efficiency: The ratio of actual mass deposited to the theoretical mass deposited, expressed as percentage.
(ii) π¨ Write the chemical equations for:
(a) Catalytic oxidation of ammonia
(b) Reaction of nitric acid with sulphur
(c) Reaction of sulphuric acid with copper [3]
β Answer:
(a) 4NHβ + 5Oβ β 4NO + 6HβO (Catalyst: Pt, Temp: 800Β°C)
(b) S + 6HNOβ (conc.) β HβSOβ + 6NOβ + 2HβO
(c) Cu + 2HβSOβ (conc.) β CuSOβ + SOβ + 2HβO
(iii) π¬ Draw a labelled diagram of the apparatus used to test for chloride ions and explain the test. [4]
β Answer:
π Test for Chloride Ions:
Diagram:
βββββββββββββββββββββββββββ β Test Tube β β βββββββββββββββββ β β β Solution + β β β β AgNOβ β β β β (White ppt) β β β βββββββββββββββββ β βββββββββββββββββββββββββββ
Method:
Take the given solution in a test tube.
Add a few drops of dilute HNOβ.
Add AgNOβ solution.
A white precipitate of AgCl is formed.
The precipitate is soluble in ammonium hydroxide but insoluble in dilute HNOβ.
Equation:
NaCl + AgNOβ β AgClβ + NaNOβ
β Question 9 [10 Marks]
(i) β‘ Calculate the mass of copper deposited when a current of 2 A is passed through CuSOβ solution for 30 minutes. (E.C.E. of Cu = 0.000329 g/C) [3]
β Answer:
Given: I = 2 A, t = 30 min = 1800 s, Z = 0.000329 g/C
m = ZIt
m = 0.000329 Γ 2 Γ 1800
m = 0.000329 Γ 3600
m = 1.1844 g
βοΈ Mass of copper deposited = 1.1844 g
(ii) π¨ Explain the term “oleum” and write its formula. [3]
β Answer:
Oleum: It is pyrosulphuric acid (HβSβOβ), formed when SOβ is dissolved in concentrated HβSOβ.
Formula: HβSβOβ
Formation: SOβ + HβSOβ β HβSβOβ
(iii) π¬ A compound ‘X’ on heating with NaOH gives a gas ‘Y’ which turns red litmus blue. Identify X and Y. Write the equation and name the test. [4]
β Answer:
Compound X: Ammonium salt (e.g., NHβCl)
Gas Y: Ammonia (NHβ)
Equation:
NHβCl + NaOH β NaCl + HβO + NHββ
Test: Ammonia turns red litmus blue. It also gives white fumes with HCl.
β Question 10 [10 Marks]
(i) β‘ What is the role of cryolite in the electrolytic extraction of aluminium? [3]
β Answer:
Role of Cryolite (NaβAlFβ):
It lowers the melting point of alumina (AlβOβ) from about 2050Β°C to about 950Β°C.
It increases the conductivity of the electrolyte.
It dissolves alumina and makes it a good conductor.
(ii) π¨ Describe the test for nitrate ions with equations. [3]
β Answer:
π Brown Ring Test for Nitrates:
Take the solution in a test tube.
Add freshly prepared FeSOβ solution.
Slowly add conc. HβSOβ along the walls of the test tube.
A brown ring forms at the junction of the two liquids.
Equation:
2NOββ» + 4HβSOβ + 6FeSOβ β 3Feβ(SOβ)β + 2NO + 4HβO + 2SOβΒ²β»
[Fe(HβO)β]SOβ + NO β [Fe(HβO)β
NO]SOβ + HβO (Brown ring)
(iii) π¬ A gas ‘X’ has a pungent smell and turns acidified KβCrβOβ green. Identify ‘X’ and write two tests to confirm it. [4]
β Answer:
Gas X: Sulphur dioxide (SOβ)
Tests:
Acidified KβCrβOβ test: SOβ turns orange KβCrβOβ green due to reduction.
KβCrβOβ + HβSOβ + SOβ β KβSOβ + Crβ(SOβ)β + HβO
(Orange β Green)Lime water test: SOβ turns lime water milky due to formation of CaSOβ.
SOβ + Ca(OH)β β CaSOββ + HβO
π END OF QUESTION PAPER
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