JEE Advanced 2019 Chemistry Questions Paper 2

JEE Advanced Question Paper 2019 Chemistry Paper 2 is available here. Students can easily download these solutions from here in PDF format to practice offline. Practising these solutions will help students to improve their speed and accuracy, and they have an idea of what type of questions are asked from each chapter. Solving previous year question papers is an important aid for the students to analyse their performance. Students are advised to solve the last year papers to crack the JEE Advanced.

Paper 2 - Chemistry

1. With reference to aqua regia, choose the correct option(s):

  1. a) Aqua regia is prepared by mixing conc. HCl and conc. HNO3 in 3: 1 molar ratio.
  2. b) Reaction of gold with aqua regia produces an anion having Au in +3 oxidation state.
  3. c) Reaction of gold with aqua regia produces NO2 in the absence of air
  4. d) The yellow colour of aqua regia is due to the presence of NOCl & Cl2.

Solution:

  1. JEE Advanced Question Paper 2019 Chemistry Paper 2

    a) Aqua regia is HCl & HNO3 (conc.) in a 3 : 1

    b) Oxidation state of Au in [AuCl4]- is +3.

    c) NOCl/NO is formed

    d) NOCl is yellow in colour

    Answer: (a, b, d)


2. Choose the correct option that gives aromatic compound as major product:
JEE Advanced Question Paper 2019 Paper 2 Chemistry

    Solution:

    1. JEE Advanced Solved Question Paper 2019 Chemistry Paper 2

      Answer: a, b


    3. Which of the following reaction produce propane as major product? Solved JEE Advanced Question Paper 2019 Chemistry Paper 2

      Solution:

      1. Solved JEE Advanced Chemistry Paper 2 Question Paper 2019

        Answer: c, d


      4. Which of the following is/are correct

      1. a) Teflon is formed by polymerization of tetrafluoroethene.
      2. b) Natural rubber is the trans from of polyisoprene.
      3. c) Cellulose contains only α-D-glucose linkage
      4. d) Nylon-6 contains amide linkage.

      Solution:

      1. a)Fact

        b) Natural rubber is Cis form of polyisoprene

        c) Cellulose contains B – 1, 4 – glycosidic linkage

        d) Nylon 6 contains amide linkage.

        Solution: (a, d)


      5.JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 5

        Solution:

        1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solutions

          Answer: a, b


        6. Consider the following reaction(unbalanced)

        Zn + Hot conc.JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 6

        Zn + conc.JEE Advanced Question Paper 2019 Chemistry Paper 2 Solutions 6

        sssssssssssssssssJEE Advanced Question Paper 2019 Chemistry Paper 2 Answer 6(precipitate) X + Y

        Choose the correct option(s)

        1. a) R is a V-shaped molecule
        2. b) Z is dirty white in colour
        3. c) Bond order of Q is 1 in its ground state
        4. d) The oxidation state of Zn in T is +1

        Solution:

        1. Solution for JEE Advanced Question Paper 2019 Chemistry Paper 2 Question 6

          a) SO2 is v shaped.

          b) ZnS is dirty white in colour.

          c) Bond order of H2 is 1.

          d) Oxidation state of Zn in Na2ZnO2 is +2.

          Answer: a, b, c


        7. In the Mac. Arthur process of extraction JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 7

        1. a) R is [Au(CN4)](-)
        2. b) Z is [Zn(CN)4]2-
        3. c) Q is O2
        4. d) Y is Zn

        Solution:

        1. Answer: b, c, d


        8. For He+ the electron is in orbit with energy equal to 3.4eV. The azimuthal quantum number for that orbit is 2 and magnetic quantum number is 0. Then which of the following is/are correct.

        1. a) The subshell is 4d.
        2. b) The number of angular nodes in it is 2.
        3. c) The numbers of radial nodes in it is 3.
        4. d) The nuclear charge experienced in n = 4 is 2e less than that in n = 1, where e is electric charge.

        Solution:

        1. E=E0z2n2E = E_0 \frac{z^2}{n^2}

          3.4 = 13.6 x 4/n2

          n = 2

          l = 2

          a) Subshell is 4 d

          b) Number of angular nodes is 2

          c) Number of radial nodes is 1.

          d) Nuclear charge would be the same.

          Answer: a, b


        9. Calculate the total number of cyclic ether (including stereo) having formula C4H8O.

          Solution:

          1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 9

            Total 10


          10. 1 mole of Rhombic sulphur is treated with conc. HNO3. Find the mass of H2O formed.

            Solution:

            1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 10

              Balancing

              JEE Advanced Question Paper 2019 Chemistry Paper 2 Solutions 10

              ∴ Mass of H2O = 288


            11. Mole fraction of urea in 900 gram water is 0.05. Density of Solution is 1.2 g/cm3. Find molarity of Solution.

              Solution:

              1. Number of moles of H2O = 900/18 = 50

                n1/(n1+50) = 0.05

                (n1 is No. of moles of urea)

                => n1 = 2.63

                Weight of urea = 2.63 × 60 = 157.8 g

                Total weight = 157.8 + 900 = 1057.8g

                Volume = 1057.8/1.2 = 881.5 cm3

                Molarity = 2.63/0.8 = 2.99


              12. Number of hydroxyl group in compound ‘Y’ is:

                Solution:

                1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 12

                  Total 6 – OH groups.


                13. In following reaction the value of K is 5 × 10-4 S-1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 13

                Initial pressure was 1 atm, while the final pressure was 1.45 atm at time y × 103 sec calculate ‘y’.

                  Solution:

                  1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solutions 13

                    t = 0

                    1

                    0

                    0

                    t = t

                    1 - P

                    P

                    P/2

                    t = ∞

                    0

                    1

                    0.5

                    P0 = 1 atm, Pt = 1.45 atm, P = 1 atm

                    t=12Kln(PP0PP1)t = \frac{1}{2K}ln(\frac{P_{\infty} - P_0}{P_{\infty} - P_1})

                    = 2.3 x 103

                    This implies, y = 2.3


                  14. Number of N-Mn-Cl bonds [N-Mn bonds is cis to Mn-Cl bond] in cis [Mn(en)2Cl2] are …….

                    Solution:

                    1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 14

                      This is the Cis form of [Mn(en)2Cl2]

                      Therefore, No. of N–Mn–Cl bonds = 6


                    15. Match the column
                    JEE Advanced Question Paper 2019 Chemistry Paper 2 Question 14

                    Which of the following is correct

                    1. a) P I
                    2. b) P II
                    3. c) P V
                    4. d) P III

                    Solution:

                    1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 15

                      Answer: (c)


                    16. Which of following is correct.

                    1. a) S IV
                    2. b) R I
                    3. c) R II
                    4. d) S III

                    Solution:

                    1. K.E. directly proportional to Z2/n2

                      Answer the question no. 17 & 18 on the basis of information given in Column – I & Column – II. Match the reactant in column – I with the possible intermediates and products of Column – II.

                      JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 16

                      Answer: (c)


                    17. Which of the following is correct?

                    1. a) P – II, III; S – II, III
                    2. b) P – II, IV; S – II, III
                    3. c) P – III, VI; S – II, III
                    4. d) P – I, III; S – IV, V

                    Solution:

                    1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 17

                      Answer: (a)


                    18. Which of the following is correct?

                    1. a) Q – I, IV, VI; R – II, III, V
                    2. b) Q – I, III, VI; R – II, IV, V
                    3. c) Q – I, II, VI; R – II, III, VI
                    4. d) Q – I, IV, V; R – III, I, V

                    Solution:

                    1. JEE Advanced Question Paper 2019 Chemistry Paper 2 Solution 18

                      Answer: (a)


                    Video Lessons - Paper 2 Chemistry

                    JEE Advanced 2019 Chemistry Paper 2 Solutions

                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions
                    JEE Advanced 2019 Chemistry Paper 2 Question and Solutions

                    BOOK

                    Free Class