II PUC MODEL QUESTION PAPER 2026
CHEMISTRY (Subject Code: 34)
Maximum Marks: 70 Time: 3 Hours 15 Minutes
(15 minutes for reading the question paper)
1. The question paper has five parts. All parts are compulsory.
2. Write the serial number of the question before attempting it.
3. Figures in the margin indicate full marks.
4. Use log tables and simple calculator if necessary (use of scientific calculator is not allowed).
5. Part-A carries 20 marks.
PART - A
I. Select the correct option from the given choices. Each question carries 1 mark. (15 × 1 = 15)
II. Fill in the blanks by choosing the appropriate word from the brackets. (5 × 1 = 5)
PART - B
Answer any FIVE questions. Each question carries TWO marks. (5 × 2 = 10)
PART - C
Answer any FIVE questions. Each question carries THREE marks. (5 × 3 = 15)
(a) Ethanol → ethanal
(b) Propanone → propene
PART - D
Answer any THREE questions. Each question carries FIVE marks. (3 × 5 = 15)
(b) The resistance of a conductivity cell filled with 0.1 M KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 M KCl solution is 520 Ω, calculate the conductivity of 0.02 M KCl solution.
(Specific conductance of 0.1 M KCl = 1.29 × 10⁻² S cm⁻¹)
(b) The rate constant of a reaction is 2 × 10⁻² L mol⁻¹ s⁻¹ at 25°C. What will be the rate constant if the reaction is zero order?
(b) Write the IUPAC name of [Cr(en)₂Cl₂]⁺.
(a) HCN (b) NaBH₄ (c) Tollen's reagent.
PART - E
Answer any THREE questions. Each question carries THREE marks. (3 × 3 = 9)
(All workings must be shown)
Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu
Given E°Zn = –0.76 V, E°Cu = +0.34 V.
Note: This model paper is prepared strictly according to the latest Karnataka PUC blueprint and previous year question paper pattern (2025). For official updates, visit the Department of Pre-University Education website.
DETAILED ANSWER KEY WITH FULL EXPLANATIONS
Ea = 2 × 19.15 × (10/303×293) × 2.303 × 8.314 × 10³ ≈ 53.6 kJ/mol
Example: NaCl, KCl, CsCl
Method: Galvanization (Zn coating) or cathodic protection
Ea = Energy of activated complex – Energy of reactants
F F
\ /
Xe --- F --- F
/ \
F F
(lone pairs above & below)Example: Glycine ⁺H₃N–CH₂–COO⁻
Molarity (M) = moles of solute / volume in litres
Ex: 5.85 g NaCl in 500 mL → 0.2 M
Molality (m) = moles / kg of solvent
Ex: 5.85 g NaCl in 500 g water → 0.2 m
ln[A₀]/[A] = kt → k = 2.303/t log[A₀]/[A]
→ Forms complex with ZnS → prevents it from coming into froth
→ Allows PbS/Galena to float
(a) Lanthanoid contraction: Poor shielding of 4f electrons → steady decrease in size from La to Lu
(b) Actinoid contraction: Even stronger due to poorer 5f shielding
Step 1: Slow → R⁺ + X⁻ (carbocation)
Step 2: Fast → R⁺ + Nu⁻ → R–Nu
Example: (CH₃)₃C–Br + H₂O → (CH₃)₃C–OH + HBr → racemisation
(a) CH₃CH₂OH →[PCC] CH₃CHO
(b) CH₃COCH₃ →[N₂H₄/KOH] CH₃CH₂CH₃ (Wolf-Kishner)
Classification:
1. Pure protein (pepsin)
2. Conjugated (protein + non-protein cofactor)
(b) Cell constant = κ × R = 1.29 × 10⁻² × 100 = 1.29 cm⁻¹
For 0.02 M: κ = 1.29 / 520 = 2.48 × 10⁻³ S cm⁻¹
Zero order unit → mol L⁻¹ s⁻¹ → cannot be same
1. 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
2. 2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O
3. Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
(a) HCN → cyanohydrin
(b) NaBH₄ → 1°/2° alcohol
(c) Tollen’s → RCHO + [Ag(NH₃)₂]⁺ → RCOOH + Ag↓ (silver mirror)
Atoms per unit cell = 1
Volume of atoms = 4/3 π r³
Packing efficiency = (4/3 π r³ / (2r)³) × 100 = 52.4%
m = 10/60 / 0.2 = 0.833 m
ΔTf = 1 × 1.86 × 0.833 = 1.55 K
Freezing point = 0 – 1.55 = –1.55°C
New rate = k(2A)²(B/2) = k × 4A² × B/2 = 2 × original rate → increases 2 times
E = 1.10 – 0.059/2 log(0.1/0.01) = 1.10 – 0.0295 × 1 = 1.0705 V ≈ 1.07 V
12/6 = √(64/M) → 4 = 64/M → M = 16 g/mol → CH₄
ALL ANSWERS AS PER KARNATAKA PUC 2026 MARKING SCHEME
Use this for self-evaluation & board exam practice!