JEE Main 2020 Jan 9h shift 2 Chemistry paper is provided here and consists of step by step solutions, prepared by our subject specialists. Practising these solutions will help students to improve the speed of solving problems for the JEE Main exam. These solutions are given in a way such that students can easily understand the problems. Students can access the solutions on our web page. It is also available in a PDF format which can be downloaded for free.
Question 1. 5 g of Zinc is treated separately with an excess of
I. dilute hydrochloric acid and
II. aqueous sodium hydroxide.
The ratio of the volumes of H_{2} evolved in these two reactions is:
Solution:
Zn + 2NaOH Na_{2}ZnO_{2 }+ H_{2}
Zn + 2HCl ZnCl_{2 }+ H_{2}
So, the ratio of volume of H2 released in both the cases is 1:1.
Answer:(c)
Question 2. The solubility product of Cr(OH)_{3} at 298 K is 6×10^{-31} . The concentration of hydroxide ions in a saturated solution Cr(OH)_{3} will be:
Solution:
Cr(OH)_{3 (s)} Cr^{3+}_{(aq)} + 3OH^{-}_{(aq)}
1-S S 3S
K_{sp} = 27S^{4}
6×10^{-31} = 27S^{4}
S = [(6/27)×10^{-31}]^{1/4}
[OH^{-}] = 3S = 3×[(6/27)×10^{-31}]^{1/4}
= (18×10^{-31})^{1/4} M
Answer: (a)
Question 3. Among the statements (a)-(d), the correct ones are:
a) Lithium has the highest hydration enthalpy among the alkali metals.
b) Lithium chloride is insoluble in pyridine.
c) Lithium cannot form ethynide upon its reaction with ethyne.
d) Both lithium and magnesium react slowly with H_{2}O.
Solution:
Only LiCl amongst the first group chlorides dissolve in pyridine because the solvation energy of lithium is higher than the other salts of the same group. Lithium does not react with ethyne to form ethynilide due to its small size and high polarizability. Lithium and Magnesium both have very small sizes and very high ionization potentials so, they react slowly with water. Amongst all the alkali metals, Li has the smallest size hence, the hydration energy for Li is maximum.
Answer: (a)
Question 4. The first and second ionization enthalpies of a metal are 496 and 4560 kJ mol^{-1} respectively. How many moles of HCl and H_{2}SO_{4}, respectively, will be needed to react completely with 1 mole of metal hydroxide?
Solution:
The given data for ionization energies clearly shows that IE_{2 }≫ IE_{1}. So, the element belongs to the first group. Therefore, we can say that this element will be monovalent and hence forms a monoacidic base of the type MOH.
MOH + HCl MCl + H_{2}O
2MOH + H_{2}SO_{4} M_{2}SO_{4 }+ 2H_{2}O
So, from the above equation we can say that, 1 mole of metal hydroxide requires 1 mole of HCl and 0.5 mole of H_{2}SO_{4}.
Answer: (b)
Question 5. In the figure shown below reactant A (represented by the square) is in equilibrium with product B (represented by circle). The equilibrium constant is:
Solution:
Let us assume the equation to be A ⇌ B,
Number of particles of A = 6
Number of particles of B = 11
K= 11/6 ≈ 2
Answer: (b)
Question 6. The correct order spin-only magnetic moments of the following complexes is:
I. [Cr(H_{2}O)_{6}]Br_{2}
II. Na_{4}[FeCN_{6}]
III. Na_{3}[Fe(C_{2}O_{4})_{3}] (_{0} > P)
IV. (Et_{4}N)_{2}[CoCl_{4}]
Solution:
Complex (I) has the central metal ion as Fe^{2+} with strong field ligands.
Configuration of Fe^{2+} = [Ar] 3d^{6}
Strong field ligands will pair up all the electrons and hence the magnetic moment will be zero.
Complex (II) has the central metal ion as Cr^{2+} with weak field ligands.
Configuration of Cr^{2+} = [Ar] 3d^{4}
As weak field ligands are present, pairing does not take place. There will be 4 unpaired electrons and hence the magnetic moment = √24 B.M.
Complex (III) has the central metal ion as Co^{2+} with weak field ligands.
Configuration of Co^{2+} = [Ar] 3d^{7}
As weak field ligands are present no pairing can occur. There will be 3 unpaired electrons and hence the magnetic moment = √15 B.M.
Complex (IV) has the central metal ion as Fe^{3+} with strong field ligands.
Configuration of Fe^{3+} =[Ar] 3d^{5}
Strong field ligands will pair up the electrons but as we have a [Ar] 3d^{5} configuration, one electron will remain unpaired and hence the magnetic moment will be √3 B.M.
Answer: (c)
Question 7. The true statement amongst the following.
Solution:
Entropy is a function of temperature, at any temperature, the entropy can be given as:
Change in entropy is also a function of temperature, at any temperature, the entropy change can be given as:
Answer: (b)
Question 8. The reaction of H_{3}N_{3}B_{3}Cl_{3} (A) with LiBH_{4} in tetrahydrofuran gives inorganic benzene (B). Furthur, the reaction of (A) with (C) leads to H_{3}N_{3}B_{3}(Me)_{3}. Compounds (B) and (C) respectively, are:
Solution:
H_{3}N_{3}B_{3}Cl_{3 }+ LiBH_{4} B_{3}N_{3}H6_{+ LiCl + BCl3}
B_{3}N_{3}H_{3}Cl_{3} + 3CH_{3}MgBr B_{3}N_{3}H_{3}(CH_{3})_{3} + 3MgBrCl
So, we can say that,
B is B_{3}N_{3}H_{6} and C is CH_{3}MgBr
Answer: (d)
Question 9. A mixture of gases O_{2}, H_{2} and CO are taken in a closed vessel containing charcoal. The graph that represents the correct behaviour of pressure with time is:
Solution:
As H_{2},O_{2 }and CO gets adsorbed on the surface of charcoal, the pressure decreases. So, option (a) and (d) can be eliminated. After some time, as almost all the surface sites are occupied, the pressure becomes constant.
Answer: (c)
Question 10. The isomer(s) of [Co(NH_{3})_{4}Cl_{2}] that has/have a Cl-Co-Cl angle of 90^{0}, is/are:
Solution:
In cis-isomer, similar ligands are at an angle of 90^{0}.
Answer: (a)
Question 11.Amongst the following, the form of water with lowest ionic conductance at 298 K is:
Solution:
In distilled water there are no ions present except H^{+} and OH^{-} ions, both of which are immensely minute in concentration, that renders their collective conductivity negligible.
Answer: (a)
Question 12. The number of sp^{2} hybrid orbitals in molecule of benzene is:
Solution:
Benzene (C_{6}H_{6}) has 6 sp^{2} hybridized carbons. Each carbon has 3 -bonds and 1 -bond. 3 -bonds means that there are 3 sp^{2} hybrid orbitals for each carbon. Hence, the total number of sp^{2} hybrid orbitals is 18.
Answer: (a)
Question 13. Which of the following reactions will not produce a racemic product?
Solution:
Answer: (b)
Question 14. Which of the following has the shortest C-Cl bond?
Solution:
There is extended conjugation present in option (d), which will reduce the length of C-Cl bond to the greatest extent which can be represented as follows:
Answer: (d)
Question 15.Biochemical oxygen demand (BOD) is the amount of oxygen required (in ppm):
Solution:
Answer: (c)
Question 16. Which polymer has chiral, monomer(s)?
Solution:
Answer: (d)
Question 17. A, B and C are three biomolecules. The results of the tests performed on them are given below:
Molisch’s Test |
Barfoed Test |
Biuret Test | |
A |
Positive |
Negative |
Negative |
B |
Positive |
Positive |
Negative |
C |
Negative |
Negative |
Positive |
A, B and C are respectively
Solution:
Lactose, glucose and fructose gives positive Molisch’s test.
Glucose gives positive Barfoed’s test whereas sucrose gives a negative for Barfoed’s test.
Albumin gives positive for Biuret test whereas alanine gives a negative Biuret test.
Answer (a)
Question 18. The decreasing order of basicity of the following amines is:
Solution:
The basicity of the compound depends on the availability of the lone pairs.
In compound IV, Nitrogen is sp^{3} hybridized.
In compound III, Nitrogen is sp^{2} hybridized and the lone pairs are not involved in resonance.
In compound I, Nitrogen is sp^{2} hybridized and the lone pairs are involved in resonance.
In compound II, Nitrogen is sp^{2} hybridized and the lone pairs are involved in resonance such that, they are contributing to the aromaticity of the ring. From the above points we can conclude that the basicity order should be IV > III > I > II.
Answer (b)
Question 19. The compound [P] is
p>
Solution:
Answer: (b)
Question 20. In the following reaction A is :
Solution:
Answer: (d)
Question 21. The sum of total number of bonds between chromium and oxygen atoms in chromate and dichromate ions is:
Solution:
Answer: (12)
Question 22.A sample of milk splits after 60 min. at 300K and after 40 min at 400K when the population of lactobacillus acidophilus in it doubles . The activation energy (in kJ/mol) for this process is closest to : (Given, R = 8.3 J mol^{-1}K^{-1}), ln(2/3) = 0.4, e^{-3} = 4.0)
Solution:
The generation time can be utilized to get an indication of the rate ratio. Let the amount generated be (x).
Rate = Amount generated/Time taken
Rate_{300 K} = (x)/60
Rate_{400 K} =(x)/40
Rate _{300K}/Rate _{400K} = 40/60
For the same concentration (which is applicable here), the rate ratio can also be equal to the ratio of rate constants.
E_{a }= 0.4 × 8.3 × 1200 = 3984 J/mol = 3.98 kJ/mol
Answer: (3.98)
Question 23. One litre of sea water (d =1.03g/cm^{3}) contains 10.3 mg of O_{2} gas. Determine the concentration of O_{2} in ppm:
Solution:
Ppm = (W_{Solute}/W _{Solution})×100
Using the density of the solution and its volume (1L = 1000 mL = 1000 cm^{3}), the weight of the solution can be calculated.
Wsolution = 1.03 × 1000 = 1030 g
Thus, ppm = ((10.3 x 10 ^{-3}g)/1030 g) × 100
Answer: (10.00)
Question 24. A cylinder containing an ideal gas (0.1 mol of 1.0 dm^{3} ) is in thermal equilibrium with a large volume of 0.5 molal aqueous solution of ethylene glycol at it freezing point. If the stoppers S_{1} and S_{2 }(as shown in the figure) suddenly withdrawn, the volume of the gas in liters after equilibrium is achieved will be: (Given, K_{f} (water) = 2.0 K kg mol^{-1} ,R = 0.08 dm^{3} atm K^{-1} mol^{-1})
Solution:
K_{f} = 2
Molality, ‘m’ = 0.5
T_{f} = K_{f} × m
= (0.5 × 2) = 1
So, the initial temperature now becomes 272 K. Further using the given value of moles and initial volume of the gas and the calculated initial temperature value, we can find out the initial pressure of the ideal gas contained inside the piston.
P_{gas} = nRT/V_{1}
= (0.1)(0.08)(272) = 2.176 atm
Now, on releasing the piston against an external pressure of 1 atm, the gas will expand until the final pressure of the gas, i.e. P_{2} becomes equal to 1 atm. During this expansion, since no reaction is happening and the temperature of the gas is not changing as well, the boyle’s law relation can be applied.
P_{1}V_{1} = P_{2}V_{2}
2.176 × 1 = 1 × V_{2}
V_{2} = 2.716
Answer: (2.18)
Question 25. Consider the following reactions;
The mass percentage of carbon in A is:
Solution:
Compound A is CH_{3}(CO)CH_{2}CH_{3} (C_{4}H_{8}O)
The percentage of carbon in compound A by weight is W_{Carbon}/W_{Compound }= 12×4/72 = 66.67
Answer (66.67)