Tag Archives: Contact Process

ICSE Class 10 Chemistry – Model Paper 1 (With Answers)

πŸ“Š Lessons & Marks Weightage (As per ICSE Board Pattern)

πŸ”’ LessonπŸ“ˆ Marks WeightageπŸ“ Question Types
⚑ Electrolysis20–25 MarksMCQ, 3–4 Mark Questions, Diagram-Based
πŸ§ͺ Study of Compounds50–55 MarksMCQ, 3–4 Mark Questions, Diagram-Based
πŸ”¬ Practical Chemistry10–15 MarksMCQ, 3–4 Mark Questions, Diagram-Based
Total80 Marks

πŸ“Œ Detailed Breakdown of Study of Compounds:

πŸ§ͺ CompoundπŸ“ˆ Marks Weightage
πŸ’¨ Hydrogen Chloride10–12 Marks
πŸ’¨ Ammonia12–15 Marks
πŸ’¨ Nitric Acid12–15 Marks
πŸ’¨ Sulphuric Acid12–15 Marks

πŸ“˜ Subject: Chemistry
πŸŽ“ Class: X (ICSE)
πŸ“Š Maximum Marks: 80
⏰ Time Allowed: 2 Hours


πŸ“‹ General Instructions:

  1. ✍️ Answers to this Paper must be written on the paper provided separately.

  2. ⏳ You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper.

  3. πŸ• The time given at the head of this Paper is the time allowed for writing the answers.

  4. βœ… Attempt all questions from Section A and any four questions from Section B.

  5. πŸ“ All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer.

  6. ⚠️ Omission of essential working will result in loss of marks.

  7. 🎯 The intended marks for questions or parts of questions are given in brackets [ ].

  8. πŸ“ Scientific calculators and log tables are not allowed.


πŸ“• SECTION A (40 Marks)

(Attempt all questions from this Section.)

❓ 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) ⚑ Which of the following is a non-electrolyte?
(a) NaCl solution
(b) Dilute Hβ‚‚SOβ‚„
(c) Glucose solution
(d) CuSOβ‚„ solution

βœ… Answer: (c) Glucose solution
πŸ“– Explanation: Glucose does not dissociate into ions in solution, hence it is a non-electrolyte.

(ii) ⚑ During electrolysis of acidified water, the gas evolved at the cathode is:
(a) Oxygen
(b) Hydrogen
(c) Chlorine
(d) Nitrogen

βœ… Answer: (b) Hydrogen
πŸ“– Explanation: At cathode (negative electrode), H⁺ ions gain electrons to form hydrogen gas.

(iii) ⚑ The electrolyte used in the electroplating of silver is:
(a) Silver nitrate
(b) Silver chloride
(c) Silver sulphate
(d) Silver cyanide (potassium argentocyanide)

βœ… Answer: (d) Silver cyanide (potassium argentocyanide)
πŸ“– Explanation: Potassium argentocyanide is used as electrolyte for silver plating.

(iv) πŸ’¨ Which gas is evolved when dilute HCl is added to sodium carbonate?
(a) Hydrogen
(b) Oxygen
(c) Carbon dioxide
(d) Chlorine

βœ… Answer: (c) Carbon dioxide
πŸ“– Explanation: Naβ‚‚CO₃ + 2HCl β†’ 2NaCl + Hβ‚‚O + CO₂↑

(v) πŸ’¨ The gas which turns moist red litmus paper blue is:
(a) HCl
(b) NH₃
(c) COβ‚‚
(d) SOβ‚‚

βœ… Answer: (b) NH₃
πŸ“– Explanation: Ammonia is basic in nature and turns red litmus blue.

(vi) πŸ’¨ Which of the following is used as a drying agent for ammonia gas?
(a) Conc. Hβ‚‚SOβ‚„
(b) Pβ‚‚Oβ‚…
(c) Quick lime (CaO)
(d) Anhydrous CaClβ‚‚

βœ… Answer: (c) Quick lime (CaO)
πŸ“– Explanation: Ammonia is basic, so it cannot be dried with acidic drying agents like Hβ‚‚SOβ‚„ or Pβ‚‚Oβ‚…. CaO is basic and suitable.

(vii) πŸ’¨ The catalyst used in Ostwald’s process for manufacture of nitric acid is:
(a) Iron
(b) Platinum gauze
(c) Vanadium pentoxide
(d) Manganese dioxide

βœ… Answer: (b) Platinum gauze
πŸ“– Explanation: Ostwald’s process uses platinum gauze as catalyst for oxidation of ammonia to nitric oxide.

(viii) πŸ’¨ In the contact process for manufacture of sulphuric acid, the catalyst used is:
(a) Platinum
(b) Vanadium pentoxide (Vβ‚‚Oβ‚…)
(c) Iron
(d) Nickel

βœ… Answer: (b) Vanadium pentoxide (Vβ‚‚Oβ‚…)
πŸ“– Explanation: Vβ‚‚Oβ‚… is used as catalyst in contact process.

(ix) πŸ’¨ 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

βœ… Answer: (c) It is heavier than air
πŸ“– Explanation: HCl gas is about 1.28 times heavier than air.

(x) πŸ’¨ The compound formed when ammonia reacts with excess chlorine is:
(a) Nβ‚‚
(b) NCl₃
(c) NHβ‚„Cl
(d) NO

βœ… Answer: (b) NCl₃
πŸ“– Explanation: 2NH₃ + 3Clβ‚‚ β†’ Nβ‚‚ + 6HCl (excess NH₃); NH₃ + 3Clβ‚‚ β†’ NCl₃ + 3HCl (excess Clβ‚‚)

(xi) πŸ”¬ In the laboratory preparation of HCl gas, the drying agent used is:
(a) Conc. Hβ‚‚SOβ‚„
(b) Quick lime
(c) NaOH
(d) KOH

βœ… Answer: (a) Conc. Hβ‚‚SOβ‚„
πŸ“– Explanation: Conc. Hβ‚‚SOβ‚„ is used to dry HCl gas as it does not react with it.

(xii) πŸ”¬ 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

βœ… Answer: (d) Blue
πŸ“– Explanation: HCl gives a blue flame colouration in the flame test.

(xiii) πŸ”¬ Which of the following is used to test for chloride ions?
(a) BaClβ‚‚ solution
(b) AgNO₃ solution
(c) NaOH solution
(d) FeSOβ‚„ solution

βœ… Answer: (b) AgNO₃ solution
πŸ“– Explanation: AgNO₃ gives a white precipitate of AgCl with chloride ions.

(xiv) πŸ”¬ The gas evolved when dilute HCl is added to iron(II) sulphide is:
(a) Hβ‚‚S
(b) SOβ‚‚
(c) Clβ‚‚
(d) Oβ‚‚

βœ… Answer: (a) Hβ‚‚S
πŸ“– Explanation: FeS + 2HCl β†’ FeClβ‚‚ + Hβ‚‚S↑ (rotten egg smell)

(xv) πŸ”¬ In the test for nitrate ions, the brown ring complex formed is:
(a) [Fe(Hβ‚‚O)β‚…NO]SOβ‚„
(b) [Fe(Hβ‚‚O)β‚…NO]²⁺
(c) [Fe(NO)]SOβ‚„
(d) [Fe(Hβ‚‚O)₆]²⁺

βœ… Answer: (b) [Fe(Hβ‚‚O)β‚…NO]²⁺
πŸ“– Explanation: Brown ring complex is nitroso ferrous sulphate, [Fe(Hβ‚‚O)β‚…NO]SOβ‚„.


❓ Question 2 [12 Marks]

(Solve the following)

(i) ⚑ State three differences between electrolytic conduction and metallic conduction. [4]

βœ… Answer:

FeatureElectrolytic ConductionMetallic Conduction
CarrierIonsElectrons
NatureChemical change occursNo chemical change
Temperature effectConductivity increases with temperatureConductivity decreases with temperature
TransferMatter is transferredOnly energy is transferred

(ii) πŸ’¨ Explain the laboratory preparation of ammonia gas with a balanced chemical equation and a labelled diagram. [4]

βœ… Answer:

πŸ“ Laboratory Preparation of Ammonia:

Chemicals: Ammonium chloride (NHβ‚„Cl) and slaked lime [Ca(OH)β‚‚]

Equation:
2NHβ‚„Cl + Ca(OH)β‚‚ β†’ CaClβ‚‚ + 2Hβ‚‚O + 2NH₃↑

Diagram:

        β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”
        β”‚  NHβ‚„Cl +    β”‚
        β”‚  Ca(OH)β‚‚    β”‚
        β””β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”˜
               β”‚
        β”Œβ”€β”€β”€β”€β”€β”€β–Όβ”€β”€β”€β”€β”€β”€β”
        β”‚   Hard      β”‚
        β”‚   Glass     β”‚
        β”‚   Test Tube β”‚
        β””β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”˜
               β”‚
        β”Œβ”€β”€β”€β”€β”€β”€β–Όβ”€β”€β”€β”€β”€β”€β”
        β”‚  Delivery   β”‚
        β”‚   Tube      β”‚
        β””β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”˜
               β”‚
        β”Œβ”€β”€β”€β”€β”€β”€β–Όβ”€β”€β”€β”€β”€β”€β”
        β”‚  NH₃ Gas    β”‚
        β”‚  Collected  β”‚
        β”‚  by Downwardβ”‚
        β”‚  Displacementβ”‚
        β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

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) πŸ’¨ 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]

βœ… Answer:

(a) NH₃ + 3Clβ‚‚ (excess) β†’ NCl₃ + 3HCl

(b) 3Cu + 8HNO₃ (dilute) β†’ 3Cu(NO₃)β‚‚ + 2NO + 4Hβ‚‚O

(c) Hβ‚‚SOβ‚„ + 2NaOH β†’ Naβ‚‚SOβ‚„ + 2Hβ‚‚O

(d) HCl + NH₃ β†’ NHβ‚„Cl (white fumes of ammonium chloride)


❓ Question 3 [12 Marks]

(Solve the following)

(i) ⚑ Explain the electrolysis of acidified water with a labelled diagram. [4]

βœ… Answer:

πŸ“ Electrolysis of Acidified Water:

Electrolyte: Acidified water (Hβ‚‚SOβ‚„ added to distilled water)

Electrodes: Platinum electrodes

Diagram:

        β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”
        β”‚     +           βˆ’       β”‚
        β”‚     β”‚           β”‚       β”‚
        β”‚  β”Œβ”€β”€β–Όβ”€β”€β”     β”Œβ”€β”€β–Όβ”€β”€β”    β”‚
        β”‚  β”‚Pt   β”‚     β”‚Pt   β”‚    β”‚
        β”‚  β”‚Anodeβ”‚     β”‚Cathodeβ”‚   β”‚
        β”‚  β””β”€β”€β”¬β”€β”€β”˜     β””β”€β”€β”¬β”€β”€β”˜    β”‚
        β”‚     β”‚           β”‚       β”‚
        β”‚  ═══β•ͺ═══════════β•ͺ═══    β”‚
        β”‚     β”‚  Acidified β”‚       β”‚
        β”‚     β”‚   Water    β”‚       β”‚
        β”‚     β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜       β”‚
        β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

Reactions:

  • At Cathode (βˆ’): 4H⁺ + 4e⁻ β†’ 2H₂↑ (Hydrogen gas)

  • At Anode (+): 2Hβ‚‚O β†’ Oβ‚‚ + 4H⁺ + 4e⁻ (Oxygen gas)

Overall: 2Hβ‚‚O β†’ 2Hβ‚‚ + Oβ‚‚

Ratio: Hydrogen : Oxygen = 2 : 1 (by volume)

(ii) πŸ’¨ Describe the contact process for manufacture of sulphuric acid with equations. [4]

βœ… Answer:

πŸ“ Contact Process for Hβ‚‚SOβ‚„:

Step 1: Burning of sulphur in air
S + Oβ‚‚ β†’ SOβ‚‚

Step 2: Catalytic oxidation of SOβ‚‚ to SO₃
2SOβ‚‚ + Oβ‚‚ β‡Œ 2SO₃ (Catalyst: Vβ‚‚Oβ‚…, Temperature: 450Β°C, Pressure: 2 atm)

Step 3: Absorption of SO₃ in conc. Hβ‚‚SOβ‚„
SO₃ + Hβ‚‚SOβ‚„ β†’ Hβ‚‚Sβ‚‚O₇ (Oleum)

Step 4: Dilution of oleum
Hβ‚‚Sβ‚‚O₇ + Hβ‚‚O β†’ 2Hβ‚‚SOβ‚„

Conditions:

  • Catalyst: Vanadium pentoxide (Vβ‚‚Oβ‚…)

  • Temperature: 450Β°C

  • Pressure: 2 atmospheres

(iii) πŸ”¬ Give one chemical test to distinguish between:
(a) HCl gas and NH₃ gas
(b) Dilute HCl and dilute HNO₃
(c) FeSOβ‚„ and FeCl₃
(d) SOβ‚‚ and COβ‚‚ [4]

βœ… Answer:

(a) HCl vs NH₃: HCl turns moist blue litmus red; NH₃ turns moist red litmus blue.

(b) Dilute HCl vs Dilute HNO₃: Add AgNO₃ solution. HCl gives white precipitate of AgCl; HNO₃ gives no precipitate.

(c) FeSOβ‚„ vs FeCl₃: Add NaOH solution. FeSOβ‚„ gives dirty green precipitate of Fe(OH)β‚‚; FeCl₃ gives reddish-brown precipitate of Fe(OH)₃.

(d) SOβ‚‚ vs COβ‚‚: Pass through acidified Kβ‚‚Crβ‚‚O₇ solution. SOβ‚‚ turns it green; COβ‚‚ has no effect.


πŸ“— SECTION B (40 Marks)

(Attempt any four questions from this Section.)

❓ Question 4 [10 Marks]

(i) ⚑ Define the following terms:
(a) Electrolyte
(b) Non-electrolyte
(c) Electrolysis [3]

βœ… 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) πŸ’¨ Write short notes on:
(a) Aqua regia
(b) Brown ring test [3]

βœ… Answer:

(a) Aqua Regia: A mixture of 3 parts concentrated HCl and 1 part concentrated HNO₃. 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. Hβ‚‚SOβ‚„ along the walls of the test tube. A brown ring of [Fe(Hβ‚‚O)β‚…NO]SOβ‚„ forms at the junction.

(iii) πŸ”¬ 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]

βœ… Answer:

Gas X: Carbon dioxide (COβ‚‚)
Compound Y: Calcium carbonate (CaCO₃) or any carbonate/bicarbonate

Equation:
CaCO₃ + 2HCl β†’ CaClβ‚‚ + Hβ‚‚O + CO₂↑

Test: COβ‚‚ + Ca(OH)β‚‚ β†’ CaCO₃↓ + Hβ‚‚O (lime water turns milky)


❓ Question 5 [10 Marks]

(i) ⚑ 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]

βœ… 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 H⁺ and Cl⁻ ions which conduct electricity. HCl in toluene does not dissociate into ions, hence does not conduct.

(ii) πŸ’¨ How is nitric acid manufactured by Ostwald’s process? Write all equations. [3]

βœ… Answer:

πŸ“ Ostwald’s Process:

Step 1: Catalytic oxidation of ammonia
4NH₃ + 5Oβ‚‚ β†’ 4NO + 6Hβ‚‚O (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‚‚ + Hβ‚‚O β†’ 2HNO₃ + NO

(iii) πŸ”¬ Draw a labelled diagram for the laboratory preparation of hydrogen chloride gas and write the equation. [4]

βœ… Answer:

πŸ“ Laboratory Preparation of HCl Gas:

Chemicals: Sodium chloride (NaCl) and concentrated Hβ‚‚SOβ‚„

Equation:
NaCl + Hβ‚‚SOβ‚„ β†’ NaHSOβ‚„ + HCl↑ (at 150Β°C)
NaHSOβ‚„ + NaCl β†’ Naβ‚‚SOβ‚„ + HCl↑ (at 200Β°C)

Diagram:

        β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”
        β”‚  NaCl +     β”‚
        β”‚  Conc.      β”‚
        β”‚  Hβ‚‚SOβ‚„      β”‚
        β””β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”˜
               β”‚
        β”Œβ”€β”€β”€β”€β”€β”€β–Όβ”€β”€β”€β”€β”€β”€β”
        β”‚   Round     β”‚
        β”‚   Bottom    β”‚
        β”‚   Flask     β”‚
        β””β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”˜
               β”‚
        β”Œβ”€β”€β”€β”€β”€β”€β–Όβ”€β”€β”€β”€β”€β”€β”
        β”‚  Delivery   β”‚
        β”‚   Tube      β”‚
        β””β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”˜
               β”‚
        β”Œβ”€β”€β”€β”€β”€β”€β–Όβ”€β”€β”€β”€β”€β”€β”
        β”‚  Conc.      β”‚
        β”‚  Hβ‚‚SOβ‚„      β”‚
        β”‚  (Drying)   β”‚
        β””β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”˜
               β”‚
        β”Œβ”€β”€β”€β”€β”€β”€β–Όβ”€β”€β”€β”€β”€β”€β”
        β”‚  HCl Gas    β”‚
        β”‚  Collected  β”‚
        β”‚  by Upward  β”‚
        β”‚  Displacementβ”‚
        β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

Method: HCl gas is collected by upward displacement of air (since it is heavier than air).


❓ Question 6 [10 Marks]

(i) ⚑ What is electroplating? Explain the electroplating of iron with silver. [3]

βœ… Answer:

Electroplating: The process of depositing a thin layer of one metal on another metal using electrolysis.

Silver Plating on Iron:

  • Cathode (βˆ’): Iron article

  • Anode (+): Pure silver rod

  • Electrolyte: Potassium argentocyanide [KAg(CN)β‚‚]

Reactions:

  • At Cathode: Ag⁺ + e⁻ β†’ Ag (deposited on iron)

  • At Anode: Ag β†’ Ag⁺ + e⁻ (dissolves from silver rod)

(ii) πŸ’¨ Give reasons:
(a) Ammonia is not dried by conc. Hβ‚‚SOβ‚„.
(b) Nitric acid turns yellow on standing.
(c) HCl gas gives white fumes with ammonia. [3]

βœ… Answer:

(a) Ammonia is basic and reacts with conc. Hβ‚‚SOβ‚„ 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 NH₃ to form white fumes of ammonium chloride (NHβ‚„Cl).

(iii) πŸ”¬ Complete the following reactions:
(a) NHβ‚„Cl + Ca(OH)β‚‚ β†’
(b) Cu + 4HNO₃ (conc.) β†’
(c) SOβ‚‚ + Hβ‚‚O β†’
(d) AgNO₃ + HCl β†’ [4]

βœ… Answer:

(a) 2NHβ‚„Cl + Ca(OH)β‚‚ β†’ CaClβ‚‚ + 2Hβ‚‚O + 2NH₃↑

(b) Cu + 4HNO₃ (conc.) β†’ Cu(NO₃)β‚‚ + 2NOβ‚‚ + 2Hβ‚‚O

(c) SOβ‚‚ + Hβ‚‚O β†’ Hβ‚‚SO₃

(d) AgNO₃ + HCl β†’ AgCl↓ + HNO₃ (white precipitate)


❓ Question 7 [10 Marks]

(i) ⚑ State Faraday’s laws of electrolysis. [3]

βœ… Answer:

First Law: The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed.
m ∝ 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.
m₁/mβ‚‚ = E₁/Eβ‚‚

(ii) πŸ’¨ What is the action of heat on:
(a) Ammonium chloride
(b) Copper nitrate
(c) Lead nitrate [3]

βœ… Answer:

(a) NHβ‚„Cl β†’ NH₃ + HCl (decomposes on heating)

(b) 2Cu(NO₃)β‚‚ β†’ 2CuO + 4NOβ‚‚ + Oβ‚‚ (copper nitrate decomposes)

(c) 2Pb(NO₃)β‚‚ β†’ 2PbO + 4NOβ‚‚ + Oβ‚‚ (lead nitrate decomposes)

(iii) πŸ”¬ 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]

βœ… Answer:

Compound X: Sodium sulphate (Naβ‚‚SOβ‚„) or any sulphate

Precipitate: Barium sulphate (BaSOβ‚„) β€” white, insoluble in dilute HCl

Equation:
Naβ‚‚SOβ‚„ + BaClβ‚‚ β†’ BaSO₄↓ + 2NaCl

Test: BaSOβ‚„ is insoluble in dilute HCl, confirming sulphate ions.


❓ Question 8 [10 Marks]

(i) ⚑ Explain the terms:
(a) Electrochemical equivalent
(b) Current efficiency [3]

βœ… 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) πŸ’¨ 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]

βœ… Answer:

(a) 4NH₃ + 5Oβ‚‚ β†’ 4NO + 6Hβ‚‚O (Catalyst: Pt, Temp: 800Β°C)

(b) S + 6HNO₃ (conc.) β†’ Hβ‚‚SOβ‚„ + 6NOβ‚‚ + 2Hβ‚‚O

(c) Cu + 2Hβ‚‚SOβ‚„ (conc.) β†’ CuSOβ‚„ + SOβ‚‚ + 2Hβ‚‚O

(iii) πŸ”¬ Draw a labelled diagram of the apparatus used to test for chloride ions and explain the test. [4]

βœ… Answer:

πŸ“ Test for Chloride Ions:

Diagram:

        β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”
        β”‚   Test Tube             β”‚
        β”‚   β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”     β”‚
        β”‚   β”‚  Solution +   β”‚     β”‚
        β”‚   β”‚  AgNO₃        β”‚     β”‚
        β”‚   β”‚  (White ppt)  β”‚     β”‚
        β”‚   β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜     β”‚
        β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜

Method:

  1. Take the given solution in a test tube.

  2. Add a few drops of dilute HNO₃.

  3. Add AgNO₃ solution.

  4. A white precipitate of AgCl is formed.

  5. The precipitate is soluble in ammonium hydroxide but insoluble in dilute HNO₃.

Equation:
NaCl + AgNO₃ β†’ AgCl↓ + NaNO₃


❓ Question 9 [10 Marks]

(i) ⚑ 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]

βœ… 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

βœ”οΈ Mass of copper deposited = 1.1844 g

(ii) πŸ’¨ Explain the term “oleum” and write its formula. [3]

βœ… Answer:

Oleum: It is pyrosulphuric acid (Hβ‚‚Sβ‚‚O₇), formed when SO₃ is dissolved in concentrated Hβ‚‚SOβ‚„.

Formula: Hβ‚‚Sβ‚‚O₇

Formation: SO₃ + Hβ‚‚SOβ‚„ β†’ Hβ‚‚Sβ‚‚O₇

(iii) πŸ”¬ 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]

βœ… Answer:

Compound X: Ammonium salt (e.g., NHβ‚„Cl)
Gas Y: Ammonia (NH₃)

Equation:
NHβ‚„Cl + NaOH β†’ NaCl + Hβ‚‚O + NH₃↑

Test: Ammonia turns red litmus blue. It also gives white fumes with HCl.


❓ Question 10 [10 Marks]

(i) ⚑ What is the role of cryolite in the electrolytic extraction of aluminium? [3]

βœ… Answer:

Role of Cryolite (Na₃AlF₆):

  • It lowers the melting point of alumina (Alβ‚‚O₃) 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) πŸ’¨ Describe the test for nitrate ions with equations. [3]

βœ… Answer:

πŸ“ Brown Ring Test for Nitrates:

  1. Take the solution in a test tube.

  2. Add freshly prepared FeSOβ‚„ solution.

  3. Slowly add conc. Hβ‚‚SOβ‚„ along the walls of the test tube.

  4. A brown ring forms at the junction of the two liquids.

Equation:
2NO₃⁻ + 4Hβ‚‚SOβ‚„ + 6FeSOβ‚„ β†’ 3Feβ‚‚(SOβ‚„)₃ + 2NO + 4Hβ‚‚O + 2SO₄²⁻
[Fe(Hβ‚‚O)₆]SOβ‚„ + NO β†’ [Fe(Hβ‚‚O)β‚…NO]SOβ‚„ + Hβ‚‚O (Brown ring)

(iii) πŸ”¬ A gas ‘X’ has a pungent smell and turns acidified Kβ‚‚Crβ‚‚O₇ green. Identify ‘X’ and write two tests to confirm it. [4]

βœ… Answer:

Gas X: Sulphur dioxide (SOβ‚‚)

Tests:

  1. Acidified Kβ‚‚Crβ‚‚O₇ test: SOβ‚‚ turns orange Kβ‚‚Crβ‚‚O₇ green due to reduction.
    Kβ‚‚Crβ‚‚O₇ + Hβ‚‚SOβ‚„ + SOβ‚‚ β†’ Kβ‚‚SOβ‚„ + Crβ‚‚(SOβ‚„)₃ + Hβ‚‚O
    (Orange β†’ Green)

  2. Lime water test: SOβ‚‚ turns lime water milky due to formation of CaSO₃.
    SOβ‚‚ + Ca(OH)β‚‚ β†’ CaSO₃↓ + Hβ‚‚O


πŸŽ“ END OF QUESTION PAPER


πŸ“ž Contact Dhingra Classes Today!

πŸ“ Address: Plot No 3, XQ3G+H3M Deacon Homes, 301C, opp. Metro Zone, near Guru Govind Sing, Samarth Nagar, Dnyaneshwar Nagar, Pathardi Phata, Nashik, Maharashtra 422009

πŸ“± Phone: 98230 62106

🌐 Website: https://www.dhingraclassesnashik.com/

πŸ“§ Email: dhingraclassesnsk@gmail.com