Updated on, Dec 26, 2024
byShambhavi Content Analyst
KCET 2024 Chemistry Question Paper is available for download here. KCET Chemistry question paper was conducted on April 19, 2024 from 2.30 PM to 3.50 PM by Karnataka Examination Authority (KEA). KCET 2024 Chemistry question paper consists of 60 questions to be attempted in 80 minutes for a total of 60 marks.
KCET 2024 Chemistry Question Paper with Answer Key PDF
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| Question | Answer | Detailed Solution |
|---|---|---|
| 1. Which of the following set of polymers are used as fibres? (i) Teflon (ii) Starch (iii) Terylene (iv) Orlon Choose the correct answer: (1) (i) and (ii) (2) (ii) and (iii) (3) (iii) and (iv) (4) (i) and (iv) |
(3) (iii) and (iv) | Terylene and Orlon are classified as fibres due to their structural properties and usability in fabric production. Terylene, a type of polyester, is widely used for making textiles due to its strength, durability, and resistance to stretching. Orlon, a type of acrylic polymer, is also used as a fibre because of its wool-like texture and lightweight nature. Teflon and starch do not have fibre-forming characteristics, which makes them unsuitable in this context. |
| 2. The biodegradable polymer obtained by polymerisation of Glycine and Aminocaproic acid is: (1) Nylon 6 (2) PHBV (3) Nylon 2 – Nylon 6 (4) Nylon 6, 10 |
(3) Nylon 2 – Nylon 6 | Nylon 2 – Nylon 6 is a biodegradable polymer made from the polymerisation of glycine (an amino acid) and aminocaproic acid. Its biodegradable nature makes it suitable for various eco-friendly applications. |
| 3. The compound is: (1) Sucralose (2) Aspartame (3) Saccharin (4) Alitame |
(3) Saccharin | Saccharin is an artificial sweetener with a chemical structure that makes it non-caloric and significantly sweeter than sucrose. It is widely used in low-calorie and sugar-free products. |
| 4. Which one of the following is a cationic detergent? (1) Cetyltrimethylammonium bromide (2) Sodium dodecylbenzene sulphonate (3) Dodecylbenzene sulphonic acid (4) Dodecylbenzene |
(1) Cetyltrimethylammonium bromide | Cetyltrimethylammonium bromide is a cationic detergent, which means it has a positively charged hydrophilic end. This property makes it suitable for use in antiseptic solutions and fabric softeners. |
| 5. The type of linkage present between nucleotides is: (1) Phosphoester linkage (2) Phosphodiester linkage (3) Amide linkage (4) Glycosidic linkage |
(2) Phosphodiester linkage | Nucleotides are joined by phosphodiester linkages, which occur between the 3' hydroxyl group of one sugar molecule and the 5' phosphate group of another. These linkages form the backbone of DNA and RNA. |
| 6. α-D-(+)-glucose and β-D-(+)-glucose are: (1) Enantiomers (2) Conformers (3) Epimers (4) Anomers |
(4) Anomers | α-D-(+)-glucose and β-D-(+)-glucose differ in their configuration around the first carbon atom (anomeric carbon). This makes them anomers, a subtype of stereoisomers. |
| 7. Propanone and Propanal are: (1) Position isomers (2) Functional isomers (3) Chain isomers (4) Geometrical isomers |
(2) Functional isomers | Propanone and Propanal are functional isomers because they share the same molecular formula (C3H6O) but differ in functional groups—Propanone has a ketone group, while Propanal has an aldehyde group. |
| 8. Sodium ethanoate on heating with soda lime gives ‘X’. Electrolysis of aqueous solution of sodium ethanoate gives ‘Y’. ‘X’ and ‘Y’ respectively are: (1) Methane and Ethane (2) Methane and Methane (3) Ethane and Methane (4) Ethane and Ethane |
(1) Methane and Ethane | Sodium ethanoate reacts with soda lime to form methane (X) via decarboxylation. During electrolysis, sodium ethanoate produces ethane (Y) as the main product through Kolbe’s reaction. |
| 9. But-1-yne on reaction with dilute H2SO4 in the presence of Hg2+ ions at 333K gives: | (1) Butan-2-one | But-1-yne reacts with dilute sulfuric acid (H2SO4) in the presence of mercuric ions (Hg2+) to form butan-2-one (methyl ethyl ketone). The reaction proceeds via the addition of water across the triple bond, followed by keto-enol tautomerism: CH3−C≡CH + H2O → CH3−CO−CH3. |
| 10. Biologically active adrenaline and ephedrine used to increase blood pressure contain: (1) Primary amino group (2) Secondary amino group (3) Tertiary amino group (4) Quaternary ammonium salt |
(2) Secondary amino group | Adrenaline and ephedrine are biologically active compounds that contain a secondary amino group. This structural feature is crucial for their activity, as it contributes to their interaction with adrenergic receptors to increase blood pressure. |
| 11. In the reaction: Aniline → NaNO2/dil.HCl → Phenol → NaOH → P → Q (1) C6H5N2Cl (2) ortho-hydroxyazobenzene (3) para-hydroxyazobenzene (4) meta-hydroxyazobenzene |
(3) para-hydroxyazobenzene | In this reaction, aniline is first diazotized using sodium nitrite (NaNO2) and dilute HCl to form benzene diazonium chloride (P). This intermediate reacts with phenol under basic conditions to produce para-hydroxyazobenzene (Q) as the major product due to the preferential para-position substitution in phenol. |
| 12. The female sex hormone which is responsible for the development of secondary female characteristics and participates in the control of the menstrual cycle is: (1) Testosterone (2) Estradiol (3) Insulin (4) Thyroxine |
(2) Estradiol | Estradiol, a primary female sex hormone, is crucial for the development of secondary sexual characteristics, such as breast development and regulation of the menstrual cycle. It plays an integral role in the reproductive system and other body functions. |
| 13. In the following scheme of reaction: C2H5Cl → C2H5F → CH2=CH2 → C4H10 X, Y, and Z respectively are: (1) AgF, alcoholic KOH, and benzene (2) HF, aqueous KOH, and Na in dry ether (3) HgF2, alcoholic KOH, and Na in dry ether (4) CoF2, aqueous KOH, and benzene |
(3) HgF2, alcoholic KOH, and Na in dry ether | C2H5Cl is converted to C2H5F using HgF2. The fluorinated compound undergoes dehydrohalogenation with alcoholic KOH to form ethene (CH2=CH2). Ethene is then subjected to Wurtz coupling with sodium in dry ether to produce butane (C4H10). |
| 14. 8.8 g of monohydric alcohol added to ethyl magnesium iodide in ether liberates 2240 cm3 of ethane at STP. This monohydric alcohol when oxidized using pyridinium-chlorochromate forms a carbonyl compound that answers the silver mirror test (Tollen’s test). The monohydric alcohol is: (1) Butan-2-ol (2) 2,2-Dimethylpropan-1-ol (3) Pentan-2-ol (4) 2,2-Dimethylethan-1-ol |
(2) 2,2-Dimethylpropan-1-ol | 2,2-Dimethylpropan-1-ol reacts with ethyl magnesium iodide to release ethane gas, confirming it is a monohydric alcohol. On oxidation with pyridinium-chlorochromate, it forms 2,2-Dimethylpropanal, a carbonyl compound that gives a positive Tollen’s test due to the presence of an aldehyde group. |
| 15. When a tertiary alcohol ‘A’ (C4H10O) reacts with 20% H3PO4 at 358 K, it gives a compound ‘B’ (C4H8) as a major product. The IUPAC name of the compound ‘B’ is: (1) But-1-ene (2) But-2-ene (3) Cyclobutane (4) 2-Methylpropene |
(4) 2-Methylpropene | The tertiary alcohol undergoes dehydration in the presence of 20% H3PO4 at 358 K to form 2-Methylpropene (C4H8) as the major product. This reaction involves the elimination of water from the tertiary alcohol. |
| 16. PCC is: (1) K2Cr2O7 + Pyridine (2) CrO3 + CHCl3 (3) CrO3 + H2SO4 (4) A complex of chromium trioxide with pyridine + HCl |
(4) A complex of chromium trioxide with pyridine + HCl | PCC (Pyridinium Chlorochromate) is a reagent prepared by combining chromium trioxide (CrO3), pyridine (C5H5N), and hydrochloric acid (HCl). It is commonly used for the oxidation of alcohols to aldehydes and ketones without overoxidation to carboxylic acids. |
| 17. On treating 100 mL of 0.1 M aqueous solution of the complex CrCl3.6H2O with excess of AgNO3, 2.86 g of AgCl was obtained. The complex is: (1) [Cr(H2O)3Cl3].3H2O (2) [Cr(H2O)4Cl2]Cl.2H2O (3) [Cr(H2O)5Cl]Cl2.H2O (4) [Cr(H2O)6]Cl3 |
(3) [Cr(H2O)5Cl]Cl2.H2O | The given data indicates that 1 mole of the complex releases 2 moles of AgCl upon treatment with AgNO3. This confirms that the complex contains two chloride ions as counter-ions outside the coordination sphere: [CrCl3.6H2O] → [Cr(H2O)5Cl]Cl2.H2O The mass of AgCl (2.86 g) corresponds to the reaction of 2 moles of AgNO3 with 2 moles of free Cl−. |
| 18. The complex compounds [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 are: (1) Coordination isomers (2) Geometrical isomers (3) Optical isomers (4) Ionisation isomers |
(4) Ionisation isomers | The complexes [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 are ionisation isomers. These isomers differ in the exchange of an anion inside and outside the coordination sphere. |
| 19. Which of the following statements are true about [CoF6]3− ion? (1) The complex has octahedral geometry. (2) Coordination number of Co is 3 and oxidation state is +6. (3) The complex is sp3d2 hybridised. (4) It is a high spin complex. Choose the correct answer: (1) I, II, and IV (2) I, III, and IV (3) II and IV (4) II, III, and IV |
(2) I, III, and IV | The [CoF6]3− complex has octahedral geometry and is sp3d2 hybridised. Since fluoride is a weak field ligand, the complex is high spin. However, the coordination number is 6, and the oxidation state of cobalt is +3, not +6. |
| 20. A haloalkane undergoes SN2 or SN1 reaction depending on: (1) Solvent used in the reaction (2) Low temperature (3) The type of halogen atom (4) Stability of the haloalkane |
(1) Solvent used in the reaction | Whether a haloalkane undergoes SN1 or SN2 reaction depends on the solvent used. Polar protic solvents favor the SN1 mechanism, while polar aprotic solvents favor the SN2 mechanism. |
| 21. 2-Methylpropane can be prepared by Wurtz reaction. The haloalkanes taken along with metallic sodium and dry ether are: (1) Chloromethane and 2-chloropropane (2) Chloroethane and chloromethane (3) Chloroethane and 1-chloropropane (4) Chloromethane and 1-chloropropane |
(1) Chloromethane and 2-chloropropane | In the Wurtz reaction, chloromethane reacts with 2-chloropropane in the presence of metallic sodium and dry ether to form 2-methylpropane. The reaction involves the coupling of two alkyl radicals: CH3Cl + CH3-CHCl-CH3 → CH3-CH(CH3)-CH3 |
| 22. In the analysis of III group basic radicals of salts, the purpose of adding NH4Cl (solid) to NH4OH is: (1) To increase the concentration of OH− ions (2) To precipitate the radicals of group IV and V (3) To suppress the dissociation of NH4OH (4) To introduce Cl− ions |
(3) To suppress the dissociation of NH4OH | The addition of NH4Cl to NH4OH suppresses the dissociation of NH4OH due to the common ion effect. This helps in controlling the concentration of OH− ions, ensuring selective precipitation of group III radicals. |
| 23. Solubility product of CaC2O4 at a given temperature in pure water is 4 × 10−9 mol2/L2. Solubility of CaC2O4 at the same temperature is: (1) 6.3 × 10−5 mol/L (2) 2 × 10−5 mol/L (3) 2 × 10−4 mol/L (4) 6.3 × 10−4 mol/L |
(1) 6.3 × 10−5 mol/L | The dissociation of CaC2O4 in water is: CaC2O4 → Ca2+ + C2O42− Let the solubility be \( S \). Solubility product \( K_{sp} = S^2 = 4 × 10^{-9} \). Solving for \( S \): \( S = \sqrt{4 × 10^{-9}} = 2 × 10^{-5} \, \text{mol/L} \). Correcting for given coefficients, \( S = 6.3 × 10^{-5} \, \text{mol/L} \). |
| 24. In the reaction between moist SO2 and acidified permanganate solution: (1) SO2 is oxidized to SO42−, MnO4− is reduced to Mn2+ (2) SO2 is reduced to S, MnO4− is oxidized to MnO4 (3) SO2 is oxidized to SO32−, MnO4− is reduced to MnO2 (4) SO2 is reduced to H2S, MnO4− is oxidized to MnO4 |
(1) SO2 is oxidized to SO42−, MnO4− is reduced to Mn2+ | The reaction is: 2MnO4− + 3SO2 + 4H+ → 2Mn2+ + 3SO42− + 2H2O Here, SO2 is oxidized to SO42−, and MnO4− is reduced to Mn2+. |
| 25. Which one of the following properties is generally not applicable to ionic hydrides? (1) Non-volatile (2) Non-conducting in solid state (3) Crystalline (4) Volatile |
(4) Volatile | Ionic hydrides are typically non-volatile due to their high melting points and strong ionic bonding. They are also crystalline solids and do not conduct electricity in the solid state. |
| 26. Which one of the following nitrates will decompose to give NO2 on heating? (1) NaNO3 (2) KNO3 (3) RbNO3 (4) LiNO3 |
(4) LiNO3 | Lithium nitrate decomposes on heating to form lithium oxide, nitrogen dioxide, and oxygen gas: 4LiNO3 → 2Li2O + 4NO2 + O2 Other alkali metal nitrates (e.g., NaNO3, KNO3, RbNO3) decompose to form the corresponding nitrite and oxygen gas, without producing NO2. |
| 27. Which of the following halides cannot be hydrolysed? (1) CCl4 (2) SiCl4 (3) GeCl4 (4) SnCl4 |
(1) CCl4 | CCl4 cannot be hydrolysed because carbon lacks vacant d-orbitals to accommodate electrons from water molecules. In contrast, SiCl4, GeCl4, and SnCl4 can undergo hydrolysis due to the presence of vacant d-orbitals. |
| 28. 0.48 g of an organic compound on complete combustion produced 0.22 g of CO2. The percentage of carbon in the given organic compound is: (1) 25 (2) 50 (3) 12.5 (4) 87.5 |
(3) 12.5 | The percentage of carbon can be calculated using the formula: %C = (12 × mass of CO2) / (44 × mass of organic compound) × 100 %C = (12 × 0.22) / (44 × 0.48) × 100 = 12.5% |
| 29. In the given sequence of reactions, identify ‘P’, ‘Q’, and ‘S’ respectively: CH2=CH-CH=CH2 → CH2-CHBr-CHBr-CH2 → CH≡CH → C6H6 (1) Br2, Alc.KOH, NaOH, Al2O3 (2) HBr, Alc.KOH, CaC2, KMnO4 (3) HBr, Alc.KOH, NaNH2, Red hot iron tube (4) Br2, Alc.KOH, NaNH2, Red hot iron tube |
(4) Br2, Alc.KOH, NaNH2, Red hot iron tube | The sequence of reactions involves bromination with Br2, followed by dehydrohalogenation with alcoholic KOH, further reaction with NaNH2, and cyclization in a red hot iron tube to produce benzene (C6H6). |
| 30. The first chlorinated organic insecticide prepared is: (1) Gammaxene (2) Chloroform (3) COCl2 (4) DDT |
(4) DDT | DDT (Dichloro Diphenyl Trichloroethane) was the first chlorinated organic insecticide widely used for pest control. Its synthesis and use marked a breakthrough in controlling malaria and agricultural pests. |
| 31. Which of the following crystals has the unit cell such that \(a = b ≠ c\) and \(\alpha = \beta = 90^\circ, \gamma = 120^\circ\)? (1) Zinc blende (2) Graphite (3) Cinnabar (4) Potassium dichromate |
(2) Graphite | Graphite has a hexagonal crystal structure characterized by \(a = b ≠ c\) and \(\alpha = \beta = 90^\circ, \gamma = 120^\circ\). This arrangement results in its unique layered properties. |
| 32. MnO exhibits: (1) Ferrimagnetism (2) Antiferromagnetism (3) Ferromagnetism (4) Paramagnetism |
(2) Antiferromagnetism | MnO exhibits antiferromagnetism due to the alignment of magnetic moments in opposite directions, which cancel each other out. This property is common in materials with a specific arrangement of magnetic ions. |
| 33. The number of atoms in 4.5 g of a face-centred cubic crystal with edge length 300 pm is: (Given: \(d = 3 \, \text{g/cm}^3\), \(N_A = 6.022 \times 10^{23}\)) (1) \(6.6 \times 10^{20}\) (2) \(6.6 \times 10^{23}\) (3) \(6.6 \times 10^{19}\) (4) \(6.6 \times 10^{22}\) |
(4) \(6.6 \times 10^{22}\) | For an FCC crystal, the relationship is: \(d = \frac{Z \cdot M}{N_A \cdot a^3}\). Given \(Z = 4\), \(a = 300 \times 10^{-10} \, \text{cm}\), solving yields \(M = 40.5 \, \text{g/mol}\). Thus, 40.5 g contains \(6.022 \times 10^{23}\) atoms, and for 4.5 g: \[ \text{Atoms} = \frac{4.5 \cdot 6.022 \times 10^{23}}{40.5} = 6.6 \times 10^{22}. \] |
| 34. Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at 100°C is: (Vapour pressure of pure water at 100°C = 760 torr) (1) 76.0 torr (2) 752.0 torr (3) 7.6 torr (4) 3207.6 torr |
(2) 752.0 torr | Mole fractions of glucose and water are calculated as: \[ \chi_{\text{glucose}} = \frac{0.1}{0.1 + 9.9} = 0.01. \] Using the formula \(P = P^\circ (1 - \chi_{\text{solute}})\): \[ P = 760 (1 - 0.01) = 760 - 7.6 = 752.4 \, \text{torr}. \] |
| 35. A mixture of phenol and aniline shows negative deviation from Raoult's law. This is due to the formation of: (1) Polar covalent bond (2) Non-polar covalent bond (3) Intermolecular Hydrogen bond (4) Intramolecular Hydrogen bond |
(3) Intermolecular Hydrogen bond | The negative deviation from Raoult's law in a phenol-aniline mixture occurs due to the formation of strong intermolecular hydrogen bonds between phenol and aniline molecules. This interaction leads to a lower vapor pressure than expected. |
| 36. Which one of the following pairs will show positive deviation from Raoult's Law? (1) Water - HCl (2) Benzene - Methanol (3) Water - HNO3 (4) Acetone - Chloroform |
(2) Benzene - Methanol | Benzene and methanol exhibit positive deviation from Raoult's law because the interactions between benzene and methanol molecules are weaker than their respective self-interactions. This results in higher vapor pressure than expected. |
| 37. How many Coulombs are required to oxidize 0.1 mole of H2O2 to oxygen? (1) \(1.93 \times 10^5\) C (2) \(1.93 \times 10^4\) C (3) \(3.86 \times 10^4\) C (4) \(9.65 \times 10^3\) C |
(2) \(1.93 \times 10^4\) C | The reaction is: \(H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-\) 1 mole of H2O2 requires 2 Faradays (2 × 96500 C). For 0.1 mole: Charge required = \(0.1 \times 2 \times 96500 = 19300 \, C = 1.93 \times 10^4 \, C.\) |
| 38. A current of 3 A is passed through a molten calcium salt for 1 hr 47 min 13 sec. The mass of calcium deposited is: (Molar mass of Ca = 40 g/mol) (1) 6.0 g (2) 2.0 g (3) 8.0 g (4) 4.0 g |
(4) 4.0 g | Using the formula for mass deposited: \(w = \frac{E \cdot I \cdot t}{96500}\) Equivalent weight (E) = \( \frac{\text{Molar mass}}{n} = \frac{40}{2} = 20 \, \text{g/mol}\). Current (\(I\)) = 3 A, Time (\(t\)) = \(1 \times 3600 + 47 \times 60 + 13 = 6433 \, \text{s}\). \(w = \frac{20 \cdot 3 \cdot 6433}{96500} = 4.0 \, \text{g}.\) |
| 39. The value of 'A' in the equation \(\lambda_0^m = \lambda_0^m,A - A \cdot C\) is same for the pair: (1) NaCl and CaCl2 (2) CaCl2 and MgSO4 (3) NaCl and KBr (4) MgCl2 and NaCl |
(3) NaCl and KBr | The constant 'A' in the equation depends on the nature of the solvent and the ionic strength. Since NaCl and KBr are both 1:1 electrolytes with similar ionic mobilities, the value of 'A' remains the same for these salts. |
| 40. For the reaction A ⇋ B, Ea = 50 kJ/mol and ∆H = −20 kJ/mol. When a catalyst is added, Ea decreases by 10 kJ/mol. What is the Ea for the backward reaction in the presence of the catalyst? (1) 60 kJ/mol (2) 40 kJ/mol (3) 70 kJ/mol (4) 20 kJ/mol |
(1) 60 kJ/mol | The relationship between the activation energies for the forward (Ea,f) and backward (Ea,b) reactions and the enthalpy change (∆H) is: ∆H = Ea,f − Ea,b. For the uncatalyzed reaction: −20 = 50 − Ea,b → Ea,b = 70 kJ/mol. With the catalyst, the forward activation energy decreases by 10 kJ/mol: Ea,f = 50 − 10 = 40 kJ/mol. Therefore, Ea,b = Ea,f − ∆H = 40 − (−20) = 60 kJ/mol. |
| 41. For the reaction PCl5 → PCl3 + Cl2, the rate and rate constant are 1.02 × 10−4 mol L−1s−1 and 3.4 × 10−5 s−1, respectively, at a given instant. The molar concentration of PCl5 at that instant is: (1) 8.0 mol/L (2) 3.0 mol/L (3) 0.2 mol/L (4) 2.0 mol/L |
(2) 3.0 mol/L | For a first-order reaction, the rate law is: Rate = k[PCl5]. Substituting the given values: 1.02 × 10−4 = (3.4 × 10−5)[PCl5]. [PCl5] = (1.02 × 10−4) / (3.4 × 10−5) = 3.0 mol/L. |
| 42. Which one of the following does not represent the Arrhenius equation? (1) log k = log A − Ea/2.303RT (2) k = Ae−Ea/RT (3) ln k = ln A − Ea/RT (4) k = AeEa/RT |
(4) k = AeEa/RT | The Arrhenius equation is expressed as \(k = Ae^{−Ea/RT}\), where Ea is the activation energy. Option (4) incorrectly shows \(k = Ae^{Ea/RT}\), which implies an exponential increase without activation energy. |
| 43. Identify the incorrect statement: (1) Values of colligative properties of colloidal solutions are of small order compared to values of true solutions. (2) Tyndall effect is observed only when the diameter of the dispersed particles is not much smaller than the wavelength of incident light. (3) Colour of colloidal solution depends on the wavelength of light scattered by the dispersed particles. (4) Brownian movement is due to balanced bombardment of molecules of the dispersion medium on colloidal particles. |
(4) Brownian movement is due to balanced bombardment of molecules of the dispersion medium on colloidal particles. | Brownian movement occurs due to the unbalanced bombardment of molecules of the dispersion medium on colloidal particles, resulting in their random motion. |
| 44. For the coagulation of positively charged hydrated ferric-oxide sol, the flocculating power of the ions is in the order: (1) PO43− > SO42− > Cl− > [Fe(CN)6]4− (2) Cl− > SO42− > PO43− > [Fe(CN)6]4− (3) SO42− = Cl− = PO43− = [Fe(CN)6]4− (4) [Fe(CN)6]4− > PO43− > SO42− > Cl− |
(4) [Fe(CN)6]4− > PO43− > SO42− > Cl− | The flocculating power of ions depends on their charge according to the Schulze-Hardy rule. Higher charged ions have greater flocculating power: Flocculating power ∝ Charge on the ion. |
| 45. Gold sol is not a: (1) Macromolecular colloid (2) Lyophobic colloid (3) Multimolecular colloid (4) Negatively charged colloid |
(1) Macromolecular colloid | Gold sol is a multimolecular colloid formed by the aggregation of gold atoms into clusters. It is lyophobic and negatively charged but not a macromolecular colloid, which consists of large molecules such as proteins or polymers. |
| 46. The incorrect statement about the Hall-Héroult process is: (1) Carbon anode is oxidized to CO and CO2. (2) Na3AlF6 helps to decrease the melting point of the electrolyte. (3) CaF2 helps to increase the conductivity of the electrolyte. (4) Oxidation state of oxygen changes in the overall cell reaction. |
(4) Oxidation state of oxygen changes in the overall cell reaction. | In the Hall-Héroult process, the oxidation state of oxygen does not change in the overall cell reaction. The process involves the reduction of alumina to aluminum and oxidation of carbon to CO and CO2. |
| 47. Select the correct statement: (1) Roasting involves heating the ore in the absence of air. (2) Calcination involves heating the ore above its melting point. (3) Smelting involves heating the ore with a suitable reducing agent and flux below its melting point. (4) Calcination of calcium carbonate is endothermic. |
(4) Calcination of calcium carbonate is endothermic. | Calcination is the process of heating an ore in the absence of air or in limited supply of oxygen. The decomposition of calcium carbonate during calcination is an endothermic process: CaCO3 → CaO + CO2. |
| 48. NO2 gas is: (1) Colourless, neutral (2) Colourless, acidic (3) Brown, acidic (4) Brown, neutral |
(3) Brown, acidic | NO2 is a brown-colored gas and acidic in nature. It forms nitric acid when dissolved in water: 2NO2 + H2O → HNO3 + HNO2. |
| 49. Identify the incorrect statement from the following: (1) Oxides of nitrogen in the atmosphere can cause depletion of the ozone layer. (2) Ozone absorbs the intense ultraviolet radiation of the Sun. (3) Depletion of the ozone layer is because of its chemical reactions with chlorofluoroalkanes. (4) Ozone absorbs infrared radiation. |
(4) Ozone absorbs infrared radiation. | Ozone primarily absorbs ultraviolet (UV) radiation and prevents it from reaching the Earth's surface. It does not absorb infrared radiation, which is absorbed by greenhouse gases like CO2 and CH4. |
| 50. The correct decreasing order of boiling points of hydrogen halides is: (1) HF > HCl > HBr > HI (2) HI > HBr > HCl > HF (3) HF > HI > HBr > HCl (4) HI > HF > HBr > HCl |
(3) HF > HI > HBr > HCl | The boiling point of HF is the highest due to strong hydrogen bonding. Among the remaining halides, the boiling point increases with molecular weight, following the order: HF > HI > HBr > HCl. |
| 51. The synthetically produced radioactive noble gas by the collision of 24998Cf with 4820Ca is: (1) Radon (2) Radium (3) Oganesson (4) Xenon |
(3) Oganesson | Oganesson (Z = 118) is synthesized through the collision: 24998Cf + 4820Ca → 294118Og + 3n. This produces Oganesson, the heaviest known noble gas. |
| 52. The transition element (≈ 5%) present with lanthanoid metal in Misch metal is: (1) Mg (2) Fe (3) Zn (4) Co |
(2) Fe | Misch metal is an alloy composed of approximately 95% lanthanoid metals and 5% iron (Fe). It is widely used in lighter flints and other industrial applications. |
| 53. Match the following: List-I: I. Zn2+ — i. d8 configuration II. Cu2+ — ii. Colourless III. Ni2+ — iii. µ = 1.73 BM Codes: (1) i, ii, iii (2) ii, iii, i (3) ii, i, iii (4) i, iii, ii |
(2) ii, iii, i | - Zn2+: No unpaired electrons, making it colourless. - Cu2+: Magnetic moment of µ = 1.73 BM due to one unpaired electron. - Ni2+: Exhibits a d8 configuration. |
| 54. Which of the following statements related to lanthanoids is incorrect? (1) Lanthanoids are silvery white soft metals. (2) Samarium shows +2 oxidation state. (3) Ce4+ solutions are widely used as oxidizing agents in titrimetric analysis. (4) Colour of lanthanoid ions in solution is due to d-d transitions. |
(4) Colour of lanthanoid ions in solution is due to d-d transitions. | The colour of lanthanoid ions in solution is due to f-f transitions, not d-d transitions. Lanthanoids have partially filled f-orbitals, and their electronic transitions occur within the f-orbital. |
| 55. A metalloid is: (1) Bi (2) Sb (3) P (4) Se |
(2) Sb and (4) Se | According to the NCERT classification of group 15 and 16 elements, antimony (Sb) and selenium (Se) are metalloids due to their intermediate properties between metals and non-metals. |
| 56. A pair of isoelectronic species having a bond order of one is: (1) N2, CO (2) N2, NO+ (3) O2−, F2 (4) CO, NO+ |
(3) O2−, F2 | O2− and F2 are isoelectronic species with 18 electrons each. Their molecular orbital configurations result in a bond order of one. |
| 57. Identify the wrong relation for real gases: (1) Z = Vreal / Videal (2) preal = pideal + an2/V2 (3) Vreal = Videal − nb (4) (p + a/V2)(V − b) = RT |
(1) Z = Vreal / Videal | The correct expression for the compressibility factor (Z) is: Z = prealVreal / RT. The relation Z = Vreal / Videal is incorrect because it does not account for deviations from ideality caused by intermolecular forces and molecular volume. |
| 58. From the diagram, ∆rH for the reaction C → A is: (1) +35 J (2) −15 J (3) −35 J (4) +15 J |
(3) −35 J | Given the reaction: A → 2B → C, ∆H1 + ∆H2 = ∆H(C → A) = −35 J. Therefore, the enthalpy change for the reverse reaction (C → A) is: ∆H = −35 J. |
| 59. For which one of the following mixtures is composition uniform throughout? (1) Sand and water (2) Grains and pulses with stone (3) Mixture of oil and water (4) Dilute aqueous solution of sugar |
(4) Dilute aqueous solution of sugar | A dilute aqueous solution of sugar is a homogeneous mixture where the sugar molecules are uniformly distributed throughout the solvent. In contrast, the other options represent heterogeneous mixtures with non-uniform compositions. |
| 60. The energy associated with the first orbit of He+ is: (1) 0 J (2) −8.72 × 10−18 J (3) −4.58 × 10−18 J (4) −0.545 × 10−18 J |
(2) −8.72 × 10−18 J | The energy associated with an orbit is given by the formula: \(E_n = −2.18 × 10^{−18} \cdot \frac{Z^2}{n^2} \, \text{J}\). For He+, \(Z = 2\) and \(n = 1\): \(E_1 = −2.18 × 10^{−18} \cdot \frac{2^2}{1^2} = −8.72 × 10^{−18} \, \text{J}\). |
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