JEE Advanced 2019 Chemistry Questions Paper 1

JEE Advanced 2019 Chemistry Paper 1 solutions are available here. All questions are solved by subject experts and in a detailed manner. Students can also download the solved paper in a PDF format for offline practice to improve their speed and accuracy. Practising JEE Advanced 2019 Chemistry question Paper 1 will help the aspirants to understand the overall exam pattern and prepare accordingly for upcoming exam.

Paper 1 - Chemistry

1. Which of the following set represent correct formula for Malachite, Magnetite, Calamine & Cryolite?

  1. a) CuCO3, Fe2O3, ZnO, Al2O3
  2. b) CuCO3,Cu(OH), Fe3O4, ZnCO3, Na3 AlF6
  3. c) CuCO3, Fe3O4, ZnCO3, Al2O3
  4. d) CuCO3.Cu(OH), Fe2O3, ZnCO3, Na3AlF6

Solution:

  1. Malachite → CuCO3.Cu(OH)2

    Magnetite → Fe3O4

    Calamine → ZnCO3

    Cryollite → Na3AlF6

    Answer: (b)


2. Find the correct acidic strength order:
JEE Advanced Chemistry 2019 Paper

    Solution:

    1. Answer is (b)

    3. Sodium stearate is a strong electrolyte. Which of the following plot is correct regarding its conductance:
    JEE Advanced Chemistry 2019 Paper 1

      Solution:

      1. By definition, λmα (1/√C)

        Answer: (b)


      4. Which green coloured compound of chromium is formed in borax bead test?

      1. a) Cr(BO2)3
      2. b) Cr2O3
      3. c) CrB
      4. d) CrBO3

      Solution:

      1. JEE Advanced Chemistry 2019 Solved Paper 1

        Answer: (a)


      5. Choose the reaction, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation:
      JEE Advanced Chemistry 2019 Paper Solution 5

        Solution:

        1. By definition,

          Enthalpy of formation is defined as the Enthalpy change occurring when, a compound is formed from its constituent elements in standard state.

          Answer: (B, C)


        6. A Tin-chloride ‘P’ gives following reaction (unbalanced reaction)

        P + Cl- \rightarrow X [Monoanion pyramidal geometry]

        P + Me3N \rightarrow Y

        P + CuCl2 \rightarrow Z + CuCl

        Then which of the following is/are correct.

        1. a) Y contains co-ordinate bond
        2. b) X is sp3 hybridised.
        3. c) Oxidation state of Sn is X is +1.
        4. d) X contain lone pair on central atom.

        Solution:

        1. JEE Advanced Chemistry 2019 Paper Solution 6

          Answer: (a, b, d)


        7. 92238Ux190234Thx291234Pax3234Zx490230Th^{238}_{92}U \overset{x_1}{\rightarrow}^{234}_{90}Th \overset{x_2}{\rightarrow}^{234}_{91}Pa \overset{x_3}{\rightarrow}^{234}Z\overset{x_4}{\rightarrow}^{230}_{90}Th
        x1, x2, x3, x4 are either particles or radiation. Then

        1. a) x1 is deflected toward negatively charged plate.
        2. b) x2 is β-particle.
        3. c) x3 is γ-radiation.
        4. d) z is isotope of 238U

        Solution:

        1. x1 -> α - decay

          x2 -> β - decay

          x3 -> β - decay

          x4 -> α – decay

          Answer: (a, b, d)


        8. Fusion of MnO2 along with KOH and O2 forms X. Electrolytic oxidation of X yields Y. X undergoes disproportionation reaction in acidic medium to MnO2 and Y. The Manganese in X and Y is in the form W & Z respectively, then

        1. a) W & Z are coloured
        2. b) W is diamagnetic and Z is paramagnetic
        3. c) Both W & Z are tetrahedral in shape
        4. d) Both W & Z involve pπ-dπ bonding for π bond

        Solution:

        1. JEE Advanced Chemistry 2019 Paper Solution 8

          Answer: (a, c, d)


        9. JEE Advanced Chemistry 2019 Paper Solution 9

          Solution:

          1. JEE Advanced Solved Sample Paper 2019

            Answer: (c, d)


          10. Which of the following are true.

          1. a) Monosachharides can not be hydrolysed to give polyhydroxy aldehydes and ketones.
          2. b) Hydrolysis of sucrose gives dextrorotatory glucose and laevorotatory fructose
          3. c) Oxidation of glucose with bromine water gives glutamic acid.
          4. d) The two six membered hemiacetal form of D(+) glucose are anomers.

          Solution:

          1. Answer is: (a, b, d)


          11. Identify the option where all four molecules possess permanent dipole moment at room temperature.

          1. a) BF, O3, SF6, XeF6
          2. b) BeCl2, CO2, BCl3, CHCl3
          3. c) SO2, C6H5Cl, H2Se, BrF5
          4. d) NO2, NH3, POCl3, CH3Cl

          Solution:

          1. For Option c


            JEE Advanced Solved Chemistry 2019 Paper 1

            Answer: (c, d)


          12. Which of the following is/are correct regarding root mean square speed (Urms) & average translation K.E. (Eav) of molecule in a gas at equilibrium.

          1. a) Eav is doubled when its temperature is increased 4 times
          2. b) Urms is inversely proportional to the square root of its molecular mass
          3. c) Eav at a given temperature doesn’t depend on its molecular mass
          4. d) Urms is doubled when its temperature is increased 4 times

          Solution:

          1. Eav = (3/2)RT [independent of mass)

            urms = 3RTM\sqrt{\frac{3RT}{M}}

            Answer: (b, c, d)


          13. XeF4 + O2F2 \rightarrow product. The total number of lone pairs on the xenon containing product is: (l)

            Solution:

            1. 19
              JEE Advanced Chemistry Paper 1 in 2019

              Distorted octahedral shape.


            14. For the following reaction, equilibrium constant Kc at 298 K is 1.6 × 1017.

            Fe(aq)2++S(aq)2FeS (s)Fe^{2+}_{(aq)} + S^{2-}_{(aq)} \rightleftharpoons FeS \ (s)

            When equal volume of 0.06 M Fe+2 and 0.2 M S-2 solution are mixed, then equilibrium concentration of Fe+2 is found to be Y × 10-17 M. Y is:

              Solution:

              1. Fe2++S2FeSFe^{2+} + S^{2-} \rightarrow FeS

                Ini

                0.06

                0.2

                0

                After mix

                0.03

                0.1

                At 0 q/m

                x

                0.07

                JEE Advanced Chemistry 2019 Paper Solution 14

              15.Solved JEE Advanced Chemistry 2019 Paper 1

              Number of atoms of Br in compound ‘T’

                Solution:

                1. Solved Paper 1 for JEE Advanced Chemistry 2019JEE Advanced Chemistry 2019 Paper Answer 15

                  Total Br atoms = 4


                16. Which of the following compounds contain bond between same type of atoms.

                N2O4, B3 N3H6, H2S2O3, N2O, H2S2O8, B2H6

                  Solution:

                  1. JEE Advanced Chemistry 2019 Paper Solution 16

                    Answer: N2O4, H2S2O3, N2O, H2S2O8


                  17. A + B + C -> Product
                  JEE Advanced Chemistry 2019 Paper Solution 17

                  When [A] = 0.15

                  [B] = 0.25

                  [C] = 0.15

                  Rate of reaction is Y × 10-5 m/s. Find Y.

                    Solution:

                    1. Let r=K[A]x[B]y[C]zr = K[A]^x[B]^y[C]^z

                      From (1), (2) -> y = 0

                      (1), (3) -> z = 1

                      (1), (4) -> x = 1

                      Therefore, rate law becomes

                      r=K[A]1[B]0[C]1r = K[A]^1[B]^0[C]^1

                      From (2)

                      K = 3 x 10-3

                      When [A] = 0.15, [B] = 0.25, [C] = 0.15

                      r = 3 x 10-3 [0.15]1 [0.15]1

                      = 6.75 x 10-5 moll-1 s-1

                      So, y x 10-6

                      Hence, y = 6.75

                      Answer: (6.75)


                    18. On dissolving 0.5 g of non-volatile, non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm of Hg to 640 mm of Hg. The depression of freezing point of benzene (in K) upon addition of the solute is __________________.

                    [Given data: Molar mass & molar freezing point depression of benzene is 78 g mol-1 & 5.12 K Kg mol-1]

                      Solution:

                      1. P0PSPS=nsolutensolvant\frac{P^0 - P_S}{P_S} = \frac{n_{solute}}{n_{solvant}}

                        6506406401×0.58×78M×39\frac{650 - 640}{640} - \frac{1 \times 0.58 \times 78}{M \times 39}

                        Therefore, M = 64 g

                        Solved Paper 1 for JEE Advanced Chemistry 2019

                      Video Lessons - Paper 1 Chemistry

                      JEE Advanced 2019 Chemistry Paper 1 Solutions

                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions
                      JEE Advanced 2019 Chemistry Paper 1 Question and Solutions