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KCET 2018 Chemistry Paper with Solutions

KCET 2018 solved question paper is a perfect resource for the candidates who are appearing for KCET to have an idea of the difficulty level of the exam. The solutions are prepared in such a way that the candidates can have a crystal clear understanding of the KCET question paper. The solutions uploaded here will also enable the students to obtain a fair idea about the distribution of marks with respect to each topic. Practising these questions will help them understand the amount of preparation required to face the examination.

KCET 2018 - Chemistry

Question 1: 1.0 g of Mg is burnt with 0.28 g of O2 in a closed vessel. Which reactant is left in excess and how much?

  1. a. Mg, 5.8 g
  2. b. Mg, 0.58 g
  3. c. O2, 0.24 g
  4. d. O2, 2.4 g

Solution:

  1. Answer: (b)

    Given:

    Amount of Mg = 1.0g

    Amount of O2 = 0.25g

    Balanced chemical reaction

    2Mg + O2 → 2MgO

    Atomic weight of Mg = 24

    Atomic weight of O = 16

    Moles of Mg = given mass / atomic mass

    = [1g] / [24g / mol]

    = 0.0416 moles

    Moles of O2 = [0.28g] / [32g / mol]

    = 0.00875 moles

    In this chemical reaction, O2 acts as limiting reagent.

    1 mole O2 react with 2 mole Mg

    0.0875 Mole O2 react with = 2 × 0.00875 moles of mg

    = 0.0175 mole Mg

    Number of moles of Mg left after reaction with oxygen

    = 0.0416 - 0.0175

    = 0.0241 moles

    Weight of Mg left in excess = moles × atomic weight

    = 0.0241×24

    = 0.578

    ≈ 0.58 g

Question 2: The orbital nearest to the nucleus is

  1. a. 4f
  2. b. 5d
  3. c. 4s
  4. d. 7p

Solution:

  1. Answer: (c)

    Penetration power of the electron describes how much it is closer to the nucleus. Penetration power of an electron follows:

    1S > 2S > 2P > 3S > 3P > 4S > 3d > 4P > 5S > 4d > 5P > 6S

    Sub shell penetration power follows:

    S > P > D > F

    So, 4S orbital is closer to the nucleus among the other 4f, 5d, 7p orbital.

Question 3: Which of the following is the correct order of radius?

  1. a. H– > H > H+
  2. b. Na+ > F– > O2–
  3. c. F– > O2– > Na+
  4. d. Al3+ > Mg2+ > N3–

Solution:

  1. Answer: (a)

    The correct order of radius:

    H– > H > H+

    Size of an atom is directly proportional to the atomic radius if size increases, the atomic radius also increases. Addition of an extra electron in an atom increases the size of the atom as compared to neutral atom and removal of an electron (cation) makes the atom size shorter than the parent atom.

Question 4: The intramolecular hydrogen bond is present in

  1. a. Phenol
  2. b. o-Nitrophenol
  3. c. p-Nitrophenol
  4. d. p-Cresol

Solution:

  1. Answer: (b)

    Intramolecular hydrogen bonding:

    When H - atom of one molecule forms a bond with an atom of the electronegative element. Intramolecular H-bonding presents in O – nitrophenol.

    KCET 2018 Chemistry Paper With Solutions Q4

    Intramolecular hydrogen –bonding

Question 5: The state of hybrid orbitals of carbon in CO2, CH4 and CO32- respectively is

  1. a. sp3, sp2 and sp
  2. b. sp3, sp and sp2
  3. c. sp, sp3 and sp2
  4. d. sp2, sp3 and sp

Solution:

  1. Answer: (c)

    According to VSEPR theory, we can use the steric number to determine the hybridization of an atom. SN = number of lone pairs + number of atoms directly attached to the central atom.

    For
    KCET 2018 Chemistry Paper With Solutions Q5

    SN = 0 + 2 = 2 = S P hybridization

    That means one S and one P hybrid orbital participate in hybridization.

    For CH4:

    KCET 2018 Chemistry Paper Q5

    SN = 0 + 4 = SP3 hybridization

    Here, one S and three P hybrid orbitals participate in hybridization.

    For CO32-:

    Solved KCET 2018 Chemistry Paper

    SN = 0 + 3 = SP2 hybridization

    In CO32- ions

    That means one S and two P hybrid orbitals participate in hybridization.

Question 6: For an ideal gas, the compressibility factor is

  1. a. 0
  2. b. 1
  3. c. –1
  4. d. +2

Solution:

  1. Answer: (b)

    Compressibility factors of an ideal gas to describe the deviation of a real gas behaviour from an ideal gas, compressibility factor represented by Z.

    Z= PV / nRT

    For an ideal gas, the value of Z is always unity = 1.

Question 7: The relationship between Kp and Kc is Kp = Kc(RT)Δn. What would be the value of Δn for the reaction, NH4Cl(s) ⇌ NH3(g) + HCl(g)?

  1. a. 1
  2. b. 0.5
  3. c. 1.5
  4. d. 2

Solution:

  1. Answer: (d)

    For the reaction: NH4Cl(s) ⇌ NH3(g) + HCl(g)

    Δng = Number of moles of gaseous product – Number of moles of reactant

    Δng = 2 – 0

    Δng = 2

Question 8: Acidity of BF3 can be explained on which of the following concepts?

  1. a. Arrhenius concept
  2. b. Bronsted-Lowry concept
  3. c. Lewis concept
  4. d. Bronsted-Lowry as well as Lewis concept

Solution:

  1. Answer: (c)

    Lewis concept:

    According to Lewis acid-base theory bases are lone pair donors and acids are lone pair acceptors.

    BF3 molecules are electron deficient.

    Electronic configuration of boron = 1S2, 2S2, 2P1

    Excited state electronic configuration of boron = 1S2, 2S2, 2P1x, 2P1y, Pz

    After bond formation with the fluorine atom, boron has one vacant P-orbitals which makes the molecule electron deficient. BF3 very energetically reacts with water and ammonia that has available lone pairs.

Question 9: For the redox reaction

xMnO4– + yH2C2O4 + zH+ → mMn2+ + n CO2 + pH2O

  1. a. 10, 2, 5, 2
  2. b. 2, 5, 2, 10
  3. c. 6, 4, 2, 5
  4. d. 3, 5, 2, 10

Solution:

  1. Answer: (b)

    For redox reaction

    xMnO4– + yH2C2O4 + zH+ → mMn2+ + n CO2 + pH2O

    Skeletal equation

    (1) MnO4- + H2C2O4 → Mn+2 + CO2

    C oxidized, +3 → +4 ∴ H2C2O4 is the reducing agent

    Mn reduced, +7 → +2 ∴ MnO4– is the oxidizing agent

    (2) Balancing the equation other than H and O.

    Oxidized; H2C2O4 → 2CO2

    Reduction: MnO4– → Mn+2

    (3) Balance oxygen by addition of H2O

    Oxidation: H2C2O4 → 2CO2 + 2H+

    Reduction: MnO4– → Mn+2 + 4H2O

    (4) Balance hydrogen by adding H+

    Oxidation: H2C2O4 → 2CO2 + 2H+

    Reduction: 8H+ + MnO4– → Mn+2 + 4H2O

    (5) Balance charge by addition of electrons.

    Oxidation: H2C2O4 → 2CO2 + 2H+ + 2e–

    Reduction: 5e– + 8H+ + MnO4– → Mn+2 + 4H2O

    (6) 5× [Oxidation: H2C2O4 → 2CO2 + 2H+ + 2e–] 2 × [reduction: 5e– + 8H+ + MnO4– → Mn+2 + 4H2O]

    Add half reactions together, simplify (cancel species that are same on the both sides of the equation);

    5H2C2O4 + 10e– + 16H+ + 2MnO4– → 10CO2 + 10H+ + 10e– + 2Mn+2 + 8H2O

    So, the final balanced redox reaction:

    5H2C2O4(aq) + 2MnO4–(aq) + 6H+(aq) → 10CO2 (g) + 2Mn+2 (aq) + 8H2O (l)

Question 10: H2O2 is

  1. a. An oxidizing agent
  2. b. A reducing agent
  3. c. Both oxidizing and reducing agent
  4. d. Neither oxidizing nor reducing agent

Solution:

  1. Answer: (c)

    H2O2 acts as both oxidizing and reducing agent oxygen of H2O2 (-1 oxidation state) is reduced to H2O (–2 oxidation state) acts as an oxidizing agent. When the oxygen of H2O2 (-1 oxidation state) is oxidized to (0) oxidation state acts as a reducing agent.

Question 11: Dead burnt plaster is

  1. a. CaSO4
  2. b. CaSO4 · (1 / 2) H2O
  3. c. CaSO4 · H2O
  4. d. CaSO4 · 2H2O

Solution:

  1. Answer: (a)

    Plaster of Paris (CaSO4 . 1 / 2 H2O) is heated above 393K, it’s half water molecule last to crystallization and anhydrous calcium sulphate is left which is called as dead burnt plaster (CaSO4).

Question 12: Identify the following compound which exhibits geometrical isomerism:

  1. a. But-2-ene
  2. b. But-1-ene
  3. c. Butane
  4. d. Isobutane

Solution:

  1. Answer: (a)

    Geometrical isomerism (cis-trans isomerism):-

    This isomerism occurs where restricted rotation is found in a molecule. Usually, involve the carbon-carbon double bond containing compound.

    KCET 2018 Chemistry Paper With Solutions Q12

    In one case, the CH3 groups are on opposite sides of the double bond and in the other case, they are on the same side.

Question 13: During the fusion of organic compound with sodium metal, nitrogen present in the organic compound is converted into

  1. a. NaNO2
  2. b. NaNH2
  3. c. NaCN
  4. d. NaNC

Solution:

  1. Answer: (c)

    When an organic compound reacts with sodium (Na) metal, its nitrogen is converted in sodium cyanide.

    Chemical reaction:

    KCET 2018 Chemistry Paper With Solutions Q13

Question 14: The reagent ‘X’ used for the following reaction is

KCET 2018 Chemistry Paper With Solutions Q14

  1. a. Ni
  2. b. Pd/C
  3. c. LiAlH4
  4. d. Na/Liquid NH3

Solution:

  1. Answer: (b)

    Hydrogenation of an alkyne with Pd/C gives cis alkane.

    KCET 2018 Chemistry Papers Solutions

    Lithium aluminium hydride (LiAlH4) does not reduce alkynes, reducing only if an alcohol group is nearby.

Question 15: Which of the following ions will cause hardness in water?

  1. a. Ca2+
  2. b. Na+
  3. c. Cl–
  4. d. K+

Solution:

  1. Answer: (a)

    Hardness in water:

    Calcium and magnesium carbonates or bicarbonates dissolved in the water are the two most common minerals that make water "hard".

    Water which does not give lather with soap is hard water.

Question 16: Which of the following oxides shows electrical properties like metals?

  1. a. SiO2
  2. b. MgO
  3. c. SO2(s)
  4. d. CrO2

Solution:

  1. Answer: (d)

    Transition metal oxides show electrical properties like metals. CrO2 (chromium oxides) is a good conductor of electricity. It is ferromagnetic at room temperature. The Curie temperature is 118° C and the material readily is demagnetized with little energy input. The electrical conductivity of this black compound is quite high.

Question 17: Which of the following aqueous solutions should have the highest boiling point?

  1. a. 1.0 M NaOH
  2. b. 1.0 M Na2SO4
  3. c. 1.0 M NH4NO3
  4. d. 1.0 M KNO3

Solution:

  1. Answer: (b)

    Colligative properties of solutions are properties that depend upon the concentration of solute molecules or ions but not the identity of the solute. Concentrations of all the solutions are the same given one. So, colligative property (elevation in boiling point) depends only on the number of ions or molecules. The boiling point increases as the number of ions increases in the aqueous solutions.

    1M NaOH gives the number of ions = 2

    1M Na2SO4 gives the number of ions = 3

    1M NH4SO4 gives the number of ions = 2

    1M KNO3 gives the number of ions = 2

    So, 1M Na2SO4 has the maximum boiling point among them all.

Question 18: The charge required for the reduction of 1 mole of MnO4− to MnO2 is

  1. a. 1 F
  2. b. 3 F
  3. c. 5 F
  4. d. 7 F

Solution:

  1. Answer: (b)

    KCET 2018 Chemistry Paper With Solutions Q18

    Change is oxidation state = 7 – 4 = + 3

    The amount of electric charge carried by one mole of electrons (6.02 × 1023 electrons) is called the Faraday (F) and is equal to 96,500 coulombs.

    1-mole e– required 1F electric charge

    3 mole e– required 3F electric charge

    To reduce MnO4− to MnO2 we require 3 moles of the electron charge.

Question 19: For the reaction, 2SO2 + O2 ⇌ 2SO3, The rate of disappearance of O2 is 2 × 10–4 mol L–1 s–1. The rate of appearance of SO3 is

  1. a. 2 × 10–4 mol L–1 s–1
  2. b. 4 × 10–4 mol L–1 s–1
  3. c. 1 × 10–1 mol L–1 s–1
  4. d. 6 × 10–4 mol L–1 s–1

Solution:

  1. Answer: (b)

    For the reaction

    2SO2 + O2 ⇌ 2SO3

    Rate of reaction

    (-1 / 2) {d [SO2] / dt} = {-d [O2] / dt} = (1 / 2) {d [SO3] / dt}

    Rate of disappearance of O2 = 2 × 10–4 mol L–1 S–1

    Then,

    {-d [O2] / dt} = (1 / 2) {d [SO3] / dt}

    2 × 10–4 = (1 / 2) {d [SO3] / dt}

    {d [SO3] / dt} = 4 × 10–4 mol L–1 s–1

    Rate of appearance of SO3 = 4 × 10–4 mol L–1 s–1

Question 20: Which of the following electrolytes will have maximum coagulating value for AgI / Ag+ sol?

  1. a. Na2S
  2. b. Na3PO4
  3. c. Na2SO4
  4. d. NaCl

Solution:

  1. Answer: (d)

    Different electrolytes have different coagulation values. Smaller the coagulation value of the electrolyte larger is its coagulating or precipitating power.

    Coagulating power ∝ (1 / Coagulation value or flocculation value)

    According to Hardy-Schulze law, the greater the charge on anion greater the coagulating power.

    Electrolyte

    Anion

    Charge

    Na2S

    S2-

    2

    Na3PO4

    PO43-

    3

    Na2SO4

    SO42-

    2

    NaCl

    Cl-

    1

    So, NaCl has lowest anionic charge and maximum coagulating power.

Question 21: Electrolytic refining is used to purify which of the following metals?

  1. a. Cu and Zn
  2. b. Ge and Si
  3. c. Zr and Ti
  4. d. Zn and Hg

Solution:

  1. Answer: (a)

    Electrolytic refining:

    It is a process of refining a metal (mainly copper and Zinc) by the process of electrolysis.

    During electrolysis, impure metal is used as anode with a thin strip of pure metal at the cathode. In this setup, an electrolyte (metal salt aqueous solution) depending on the metal is after use.

    The pure metal is obtained at the cathode when the electric current of a sufficient voltage is applied by dissolving metal to the anode. It is about 99.95% pure which makes it a good product.

Question 22: Dry ice is

  1. a. Solid CO
  2. b. Solid SO2
  3. c. Solid CO2
  4. d. Solid O2

Solution:

  1. Answer: (c)

    Dry ice: Dry ice is a common name for solid carbon dioxide (CO2). It gets this name because it does not melt into a liquid when heated; Instead, it changes directly into a gas (a process known as sublimation).

Question 23: Which of the following is an amphoteric oxide?

  1. a. V2O5, Cr2O3
  2. b. Mn2O7, Cr2O3
  3. c. CrO, V2O5
  4. d. V2O5, V2O4

Solution:

  1. Answer: (a)

    Amphoteric oxides can react with both acids and bases to form salt and water as the main products.

    The element that forms amphoteric oxides has some metals and some of the nonmetals characteristics.

    Amphoterism based on oxide's oxidation state.

    Both Cr2O3 and V2O5 are amphoteric in nature because of their high oxidation state.

    V2O5 reacts with alkalies as well as acids to give VO43– and VO4+ respectively.

    Cr2O3 (chromium (III) oxide) insoluble in water but dissolves in acid to produce hydrated chromium ions,[Cr(H2O)6]+3 which react with a base to give slat of [Cr(OH)6]3–

    It dissolved in concentrated alkali to yield chromite ions [CrO2]–.

Question 24: The IUPAC name of [Co(NH3)4Cl(NO2)] Cl is

  1. a. tetraamminechloridonitrito-N-cobalt(III) chloride
  2. b. tetraamminechloridonitrocobalt(II) chloride
  3. c. tetraamminechloridonitrocobalt(I) chloride
  4. d. tetraamminechloridodinitrocobalt(III) chloride

Solution:

  1. Answer: (a)

    First, write the name of the cationic part and then anionic part. After that ligands name in alphabetical order and then the name of the metal atom.

    For ambidentate ligands here we use –N– before the name of metal ion, because the nitrogen of – NO2 ligand is directly attached to the cobalt metal.

    After the name of metal ion with its oxidation state in roman number and then anionic ligand name [Co(NH3)4Cl(NO2)] Cl

    x+ (0) 4 + (–1) + (–1) = +1

    x = + 3

    IUPAC name of this complete is tetraamminechloridonitrito –N–cobalt (III) chloride.

Question 25: Which of the following statements is true in the case of alkyl halides?

  1. a. They are polar in nature
  2. b. They can form hydrogen bonds
  3. c. They are highly soluble in water
  4. d. They undergo addition reactions

Solution:

  1. Answer: (a)

    Polarity results from the uneven partial charge distribution between various atoms in a compound. More the electronegativity difference between two atoms more will be the polarity of that compound. In alkyl halide, the carbon-hydrogen bond is polar because all the halogen is more electronegative than carbon. Due to high electronegativity of halogens, they pull electrons of carbon towards them so, on carbon, a small positive charge (δ+) and on halogen partial negative charge (δ–) developed. As the Alkyl halides have a polar bond they are not soluble in water not form H-bond also.

Question 26: Phenol can be distinguished from ethanol by the reagent

  1. a. Bromine water
  2. b. Sodium metal
  3. c. Iron metal
  4. d. Chlorine water

Solution:

  1. Answer: (a)

    Bromine water test (Saturation test):- The bromine water test used to identify the alkene or alkane functional groups. phenols undergo substitution reaction in the presence of bromine water to give a brominated product. During the process, bromine water is decolourized and gives white precipitate.

    KCET 2018 Chemistry Paper With Solutions Q26

    Ethanol (C2H5OH) is a saturated compound; it does not react with a bromine water solution.

Question 27: Which of the following compounds undergoes haloform reaction?

  1. a. CH3COCH3
  2. b. HCHO
  3. c. CH3CH2Br
  4. d. CH3 – O– CH3

Solution:

  1. Answer: (a)

    Haloform reaction:

    Haloform reaction is a chemical reaction where haloform (R–X) is produced by the complete halogenations of acetone (CH3 – O– CH3). Acetophenone (PHCOCH3), or acetaldehyde (HCOCH3) in the presence of a base.

    The aldehyde or ketone must have a methyl group (–hydrogen) next to the carbonyl.

    Haloform reaction:

    KCET 2018 Chemistry Paper With Solutions Q27

Question 28: Which of the following will be the most stable diazonium salt (RN2+X-)?

  1. a. CH3 N2+ X–
  2. b. C6H5 N2+ X–
  3. c. CH3CH2 N2+ X–
  4. d. C6H5CH2 N2+ X–

Solution:

  1. Answer: (b)

    KCET 2018 Chemistry Paper With Solutions Q28

    Diazonium salts are usually prepared by the reaction (diazotization) of primary amines with nitrous acid; their most striking property is their stability.

    The aliphatic diazonium salts exist only as transient intermediates quickly decompose into a nitrogen molecular and a carbonium ion; Aromatic diazonium salts are stable enough to be isolated but react readily either by loss of nitrogen or by the formation of azo compounds.

    KCET 2018 Chemistry Paper Q28

    C6H5 N2+ X– is more stable in comparison to C6H5 N2+ X– because resonance stabilized due to direct bond with nitrogen.

Question 29: Which of the following bases is not present in DNA?

  1. a. Adenine
  2. b. Guanine
  3. c. Cytosine
  4. d. Uracil

Solution:

  1. Answer: (d)

    DNA is made up of nucleotides. A nucleotide has two compounds: a backbone, made from the sugar deoxyribose, phosphate groups, and nitrogenous bases, known as cytosine, thymine, adenine, and guanine.

    Adenine and Uracil are the main basic building block of RNA.

Question 30: Which one of the following is a polyamide polymer?

  1. a. Terylene
  2. b. Nylon-6, 6
  3. c. Buna-S
  4. d. Bakelite

Solution:

  1. Answer: (b)

    Polyamide polymers are the polycondensation product of a diacid chloride and a diamine.

    Nylon-6, 6 is a member of the polyamide group.

    The structural unit of polyamide is joined together by an amide

    KCET 2018 Chemistry Paper With Solutions Q30

Question 31: In F.C.C. the cell is shared equally by how many unit cells?

  1. a. 10
  2. b. 8
  3. c. 6
  4. d. 2

Solution:

  1. Answer: (c)

    In FCC arrangement, there are eight atoms at corners of the unit cell, and one atom centred and six in each of the faces. The atoms in the face are shared with the adjacent unit cell. So, an FCC unit cell is shared equally by six-unit cell.

    KCET 2018 Chemistry Paper With Solutions Q31

    Atoms present on the six faces of the face-centred unit cell.

Question 32: At a particular temperature, the ratio of molar conductance to specific conductance of 0.01 M NaCl solution is

  1. a. 105 cm3mol–1
  2. b. 103 cm3mol–1
  3. c. 10 cm3mol–1
  4. d. 105 cm2mol–1

Solution:

  1. Answer: (a)

    Given

    Concentration = 0.01 m

    KCET 2018 Chemistry Paper With Solutions Q32

Question 33: Isotonic solutions are solutions having the same

  1. a. Surface tension
  2. b. Vapour pressure
  3. c. Osmotic pressure
  4. d. Viscosity

Solution:

  1. Answer: (c)

    Isotonic solution

    Two solutions having the same osmotic pressure across a semipermeable membrane is referred to as an isotonic solution.

    It has the same osmolarity (solute concentration) as another solution.

Question 34: The temperature coefficient of a reaction is 2. When the temperature is increased from 30° C to 90°C, the rate of reaction is increased by

  1. a. 150 times
  2. b. 410 times
  3. c. 72 times
  4. d. 64 times

Solution:

  1. Answer: (d)

    Given that

    Temperature coefficient = 2

    T1 = 30°C

    T2 = 90° C

    Rate of the reaction (R) = (Temperature coefficient) ΔT/10

    =2{(90-30)/10}

    = 26

    R = 64

    So, the rate of reaction is increased by 64 times.

Question 35: Gold sol is not a

  1. a. Lyophobic sol
  2. b. Negatively charged sol
  3. c. Macromolecular sol
  4. d. Multimolecular colloid

Solution:

  1. Answer: (c)

    Gold sol is an example of multimolecular colloid. In multimolecular colloids many particles (atoms or small molecules) of the dispersed phase aggregate together to form species having the size of a colloidal particle (1–1000 nm).

Question 36: The common impurity present in bauxite is

  1. a. CuO
  2. b. ZnO
  3. c. Fe2O3
  4. d. Cr2O3

Solution:

  1. Answer: (c)

    Major impurities in bauxite are iron oxides [hematite (Fe2O3) & goethite], silicon dioxide, the clay mineral kaolinite as well as a small amount of anatase (TiO2).

Question 37: Very pure N2 can be obtained by

  1. a. Thermal decomposition of ammonium dichromate
  2. b. Treating aqueous solution of NH4Cl and NaNO2
  3. c. Liquefaction and fractional distillation of liquid airMidbrain
  4. d. Thermal decomposition of sodium azide

Solution:

  1. Answer: (d)

    Thermal decomposition of ammonium dichromate:

    KCET 2018 Chemistry Paper With Solutions Q37

    Treatment of an aqueous solution of NH4Cl and NaNO2

    NH4Cl + NaNO2 → NaCl + N2 ↑ + 2H2O

    Thermal decomposition of sodium azide:

    KCET 2018 Chemistry Paper Q37

    From the above decomposition of sodium azide gives maximum and very pure N2 gas.

Question 38: Which of the following oxidation states is common for all lanthanides?

  1. a. + 2
  2. b. + 3
  3. c. + 4
  4. d. + 5

Solution:

  1. Answer: (b)

    The lanthanides consist of the elements in the f-block of period six, in the periodic table. Lanthanides' most common and stable oxidation state is +3. It is obtained by removing three outermost electrons from 6S orbital and one electron from 4f orbital. Removal of one more electron from the 4f subshell is difficult because of high energy difference.

Question 39: The electronic configuration of transition element “X” is +3, the oxidation state is [Ar]3d5. What is its atomic number?

  1. a. 25
  2. b. 26
  3. c. 27
  4. d. 24

Solution:

  1. Answer: (b)

    The electronic configuration of X after removal of three electrons is X+3: [Ar] 3d5. So, before the removal of electrons.

    This compound contains X = 18 + 5 + 3 = 26e–.

    The atomic number of X is 26.

Question 40: n-Propyl chloride reacts with sodium metal in dry ether to give

  1. a. CH3-CH2-CH2–CH2–CH2–CH3
  2. b. CH3-CH2-CH3
  3. c. CH3-CH2-CH2–CH3
  4. d. CH3-CH2-CH2–CH2–CH2–CH2–CH3

Solution:

  1. Answer: (a)

    KCET 2018 Chemistry Paper With Solutions Q40

    n–propyl chloride reacts with sodium metal in presence of dry ether to give n-hexane, this reaction is known as Wurtz reaction.

Question 41: When the vapours of tertiary butyl alcohol are passed through heated copper at 573 K, the product formed is

  1. a. But-2-ene
  2. b. 2-Butanone
  3. c. 2-Methyl propene
  4. d. Butanal

Solution:

  1. Answer: (c)

    When the tertiary butyl alcohol vapours pass through heated copper, 2– methyl propene is formed with the removal of one water molecule.

    KCET 2018 Chemistry Paper With Solutions Q41

    According to the Saytzeff rule, more substituted alkene will be more stable.

Question 42: What is the increasing order of acidic strength among the following?

(i) p-methoxy phenol

(ii) p-methyl phenol

(iii) p-nitro phenol

  1. a. ii < iii <i
  2. b. iii < ii <i
  3. c. i< ii < iii
  4. d. i< iii < ii

Solution:

  1. Answer: (c)

    KCET 2018 Chemistry Paper With Solutions Q42

    P–nitrophenol is more acidic due to –I effect of –nitro group which makes the benzene ring e– deficient, so the H+ ion is removed easily. On the other hand, p-methoxy phenol is less acidic due to the high mesomeric effect of the –OCH3 groups in comparison to the weak hyperconjugation effect of the –CH3 group.

Question 43: Which of the following is more basic than aniline?

  1. a. Diphenylamine
  2. b. Triphenylamine
  3. c. p-nitroaniline
  4. d. Benzylamine

Solution:

  1. Answer: (d)

    Benzyl aniline is more basic than aniline.

    KCET 2018 Chemistry Paper With Solutions Q43

    The basic character of a compound depends on the availability of a lone pair of electrons.

    In aniline, lone pair of nitrogen atom delocalized to take part in resonance and not available to donate but in case of benzyl aniline nitrogen attached to the methyl group, not to the benzene ring and do not take part in resonance so, available to donate.

Question 44: The two forms of D-Glucopyranose are called

  1. a. Diastereomers
  2. b. Anomers
  3. c. Epimers
  4. d. Enantiomers

Solution:

  1. Answer: (b)

    D –glucopyranose: Anomers

    KCET 2018 Chemistry Paper With Solutions Q44

Question 45: Among the following, the branched-chain polymer is

  1. a. Polyvinyl chloride
  2. b. Bakelite
  3. c. Low-density polythene
  4. d. High-density polythene

Solution:

  1. Answer: (c)

    Branched-chain polymers are those polymers in which the monomers are joined to form a long chain with side chains or branches of different lengths.

    These branched-chain polymers are irregularly packed and therefore, they have low density, low boiling point and low melting point.

    So, the branched-chain polymer is a low-density polymer.

Question 46: Edge length of a cube is 300 pm. Its body diagonal would be

  1. a. 600 pm
  2. b. 423 pm
  3. c. 519.6 pm
  4. d. 450.5 pm

Solution:

  1. Answer: (c)

    Given

    Edge length of a cube (a) = 300 pm.

    Body = √3.a.

    Diagonal = √3 . 300

    = 1.732 × 300

    = 519.6 pm

    Body diagonal would be 519.6 pm.

Question 47: Which of the following is not a conductor of electricity?

  1. a. Solid NaCl
  2. b. Cu
  3. c. Fused NaCl
  4. d. Brine solution

Solution:

  1. Answer: (a)

    To conduct electricity compound must-have free electrons or ions. In solid-state NaCl, the compound has a fixed composition of ions. So these ions do not move from their place to conduct electricity. But in molten states (moist) ions are free to move to conduct electricity. So, in the molten state, they are a good conductor of electricity.

Question 48: For a cell reaction involving two-electron changes, E0cell = 0.3 V at 25° C. The equilibrium constant of the reaction is

  1. a. 10–10
  2. b. 3 × 10–2
  3. c. 10
  4. d. 1010

Solution:

  1. Answer: (d)

    Nernt’s equation

    For a cell reaction:

    F (faraday) = 96500

    R = gas constant (8.314 J mol–1 k–1)

    T = Temperature (k)

    n = charge

    E0cell = (2.303RT / nF) log k

    E0cell = {(2.303 * 8.314 * 298) / (2 * 96500)} log k

    = (0.059 / 2) log k

    Given

    E0cell = 0.3 V

    0.3 = 0.0295 log k or

    0.3 = 0.03 log k

    k = antilog 10

    k = 1010

    Equilibrium constant of the reaction is 1010.

Question 49: The value of rate constant of a pseudo-first-order reaction

  1. a. Depends only on temperature
  2. b. Depends on the concentration of reactants present in small amounts
  3. c. Depends on the concentration of reactants present in excess
  4. d. Is independent of the concentration

Solution:

  1. Answer: (c)

    A pseudo final order reaction can be defined as a second-order reaction that is made to behave like a first-order reaction.

    This reaction occurs when one reacting material is present in great excess.

    In pseudo-first-order reaction rate depends on the concentration of reaction present in excess.

Question 50: (CH3)3SiCl is used during polymerization of organosilicons because

  1. a. The chain length of organosilicon polymers can be controlled by adding (CH3)SiCl
  2. b. (CH3)3SiCl does not block the end terminal of silicone polymer
  3. c. (CH3)3SiCl does not block the end terminal of silicone polymer
  4. d. (CH3)3SiCl acts as a catalyst during polymerisation

Solution:

  1. Answer: (a)

    KCET 2018 Chemistry Paper With Solutions Q50

    During polymerization of organosilicons (CH3)3SiCl block the end terminal to control the chain length of organosilicon polymers.

Question 51: When PbO2 reacts with concentrated HNO3, the gas evolved is

  1. a. NO2
  2. b. O2
  3. c. N2
  4. d. N2O

Solution:

  1. Answer: (b)

    PbO2 + Conc. HNO3 → Pb (NO3)2 + H2O + O2

    When lead oxide reacts with concentrated nitric acids, removal of O2 occurs due to the reduction of PbO2.

Question 52: KMnO4 acts as an oxidizing agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidized to

  1. a. I2
  2. b. IO–
  3. c. IO3−
  4. d. IO4–

Solution:

  1. Answer: (c)

    In the alkaline medium KMnO4 acts as an oxidizing agent. When alkaline KMnO4 is treated with KI, Iodide ion is oxidized to iodate ion.

    KCET 2018 Chemistry Paper With Solutions Q52

Question 53: [Fe(NO2)3 Cl3] and [Fe(O–NO)3 Cl3] shows

  1. a. Linkage isomerism
  2. b. Geometrical isomerism
  3. c. Optical isomerism
  4. d. Hydrate isomerism

Solution:

  1. Answer: (a)

    Linkage isomerism: When two complexes differ only in the mode of attachment of an ambidentate ligand to the metal atom. [Fe(NO2)3Cl3] and [Fe(O–NPO)3Cl3 are pair of linkage isomers, in which ligand is joined to iron atom through nitrogen or oxygen, with the formulas NO2– (nitro) and (ONO–) (nitrito), respectively.

Question 54: Tertiary alkyl halide is practically inert to substitution by SN2 mechanism because of

  1. a. Insolubility
  2. b. Instability
  3. c. Inductive effect
  4. d. Steric hindrance

Solution:

  1. Answer: (d)

    Tertiary alkyl halides do not react by an SN2 mechanism because the substance blocks the approach of the nucleophile.

    The trigonal bipyramidal transition state cannot form because it is too statically crowded.

    KCET 2018 Chemistry Paper With Solutions Q54

Question 55: The products X and Z in the following reaction sequence are

KCET 2018 Chemistry Paper With Solutions Q55

  1. a. Isopropyl benzene and acetone
  2. b. Cumene peroxide and acetone
  3. c. Isopropyl benzene and isopropyl alcohol
  4. d. Phenol and acetone

Solution:

  1. Answer: (a)

    KCET 2018 Chemistry Paper Q55

Question 56: The appropriate reagent for the following transformation is

KCET 2018 Chemistry Paper With Solutions Q56

  1. a. Zn–Hg/HCl
  2. b. H2N–NH2, KOH/ethylene glycol
  3. c. Ni/H2
  4. d. NaBH4

Solution:

  1. Answer: (b)

    Ketones can be converted to a hydrazine derivative by reaction with hydrazine. These “Hydrazones” can be further converted to the corresponding alkane by reaction with base and heat.

    KCET 2018 Chemistry Paper Q56

Question 57: In the following reaction

KCET 2018 Chemistry Paper With Solutions Q57

The compound Z is

  1. a. Benzoic acid
  2. b. Benzaldehyde
  3. c. Acetophenone
  4. d. Benzene

Solution:

  1. Answer: (b)

    Toluene can be oxidized to Benzaldely using Etard reaction. The reaction begins with toluene with chromyl chloride, forming an Etard complex. Reducing conditions provided by saturated aqueous Sodium sulphide prevents further oxidation of the Etard Complex into a carboxylic acid.

    KCET 2018 Chemistry Paper Q57

Question 58: The reaction of Benzene diazonium chloride with aniline yields a yellow dye. The name of the yellow dye is

  1. a. p-Hydroxyazobenzene
  2. b. p-Aminoazobenzene
  3. c. p-Nitroazobenzene
  4. d. o-Nitroazobenzene

Solution:

  1. Answer: (b)

    Some liquid phenylamine (aniline) is added to a cold solution of benzene diazonium chloride, and the mixture is shaken vigorously. A yellow solid is produced.

    KCET 2018 Chemistry Paper With Solutions Q58

Question 59: The glycosidic linkage involved in linking the glucose units in amylose part of the starch is

  1. a. C1 – C4β-linkage
  2. b. C1 – C6α-linkage
  3. c. C1 – C6β-linkage
  4. d. C1 – C4α-linkage

Solution:

  1. Answer: (d)

    Amylase is a polysaccharide made of α–D-glucose bonded to each other by a (1, 4) glycoside bond.

    KCET 2018 Chemistry Paper With Solutions Q59

    Amylase is a linear polymer of glucose, arranged in three-dimensional helical structures.

Question 60: Ziegler-Natta catalyst is used to prepare

  1. a. Low-density polythene
  2. b. Teflon
  3. c. High-density polythene
  4. d. Nylon-6

Solution:

  1. Answer: (c)

    Ziegler –Natta catalyst is used for the production of high-density polyethylene. One of the common examples is a mixture of titanium tetrachloride (TiCl4) and trimethylaluminum (Al (C2H5)3).

Video Lessons - KCET 2018 - Chemistry









KCET 2018 Chemistry Question Paper with Solutions

KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions
KCET 2018 Chemistry Question Paper with Solutions