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JEE Main Chemistry Mock Test- 3 - JEE MCQ


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25 Questions MCQ Test Mock Tests for JEE Main and Advanced 2024 - JEE Main Chemistry Mock Test- 3

JEE Main Chemistry Mock Test- 3 for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2024 preparation. The JEE Main Chemistry Mock Test- 3 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Chemistry Mock Test- 3 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Chemistry Mock Test- 3 below.
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JEE Main Chemistry Mock Test- 3 - Question 1

The compound 'A' when treated with methyl alcohol and few drops of H₂SO₄ give wintergreen smell. The compound 'A' is

JEE Main Chemistry Mock Test- 3 - Question 2

1-Phenylethanol can be prepared by the reaction of benzaldehyde with

JEE Main Chemistry Mock Test- 3 - Question 3

The spectrum produced due to transition of an electron from M to L shell is

JEE Main Chemistry Mock Test- 3 - Question 4

The correct order of increasing energy of atomic orbitals is

JEE Main Chemistry Mock Test- 3 - Question 5

Combination of two AO's lead to the formation of

Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 5
According to Linear Combination Of Atomic Orbitals(LCAO), the combination of 2 AO', give 2 MO's one of which is bonding MO while the other is antibonding MO.
JEE Main Chemistry Mock Test- 3 - Question 6

Elimination of bromine from 2- bromobutane results in the formation of

JEE Main Chemistry Mock Test- 3 - Question 7

The name of  according to IUPAC nomenclature system is

JEE Main Chemistry Mock Test- 3 - Question 8

IUPAC name of CH₃-O-C₂H₅ is

JEE Main Chemistry Mock Test- 3 - Question 9

The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species?

Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 9

Higher the bond order, shorter will be the bond length. Thus, NO+ is having higher bond order than that of NO so NO+ has shorter bond length.

JEE Main Chemistry Mock Test- 3 - Question 10

The segment of DNA which acts as the instructional manual for the synthesis of the protein is

Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 10

A gene is the molecular unit of heredity of a living organism. It is widely accepted by the scientific community as a name given to some stretches ofdeoxyribonucleic acids (DNA) and ribonucleic acids (RNA) that code for apolypeptide or for an RNA chain that has a function in the organism, though there still are controversies about what plays the role of the genetic material

JEE Main Chemistry Mock Test- 3 - Question 11

K₄[Fe(CN)₆] is a

JEE Main Chemistry Mock Test- 3 - Question 12

What are the products formed in the reaction of xenon hexafluoride with silicon dioxide

JEE Main Chemistry Mock Test- 3 - Question 13

The hardness of transition elements is due to their

Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 13

Transition elements show metallic character as they have low ionization energies and have several vacant orbitals in their outermost shell. This property favors the formation of metallic bonds in the transition metals and so they exhibit typical metallic properties.These metals are hard which indicates the presence of covalent bonds. This happens because transition metals have unpaired d-electrons. The d-orbital which contains the unpaired electrons may overlap and form covalent bonds. Higher the number of unpaired electrons present in the transition metals, more is the number of covalent bonds formed by them. This further increases the hardness of the metal and its strength.

JEE Main Chemistry Mock Test- 3 - Question 14

Sodium nitroprusside reacts with sulphide ion to give a purple colour due to the formation of :

JEE Main Chemistry Mock Test- 3 - Question 15

To a solution containing equimolar mixture of sodium acetate and acetic acid, some more amount of sodium acetate solution is added. The pH of mixture solution

JEE Main Chemistry Mock Test- 3 - Question 16

The reason for geometrical isomerism by 2-butene is

JEE Main Chemistry Mock Test- 3 - Question 17

Ionisation potential of hydrogen is

Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 17

Hydrogen has higher ionisation potential than chlorine, because hydrogen has only one electron and small size. This electron experiences the direct attraction of the nucleus and there is no shielding effect. Whereas in Cl the combined effect of the increase in the atomic size and the screening effect more than compensates the effect of the increased nuclear charge. Consequently, the valence electrons become less firmly held by the nucleus and ionisation potential decreases.

JEE Main Chemistry Mock Test- 3 - Question 18

If chloroform is left open in air in presence of sun rays

JEE Main Chemistry Mock Test- 3 - Question 19

The bond dissociation energy of B–F in BF3 is 646 kJ mol-1 whereas that of C–F in CF4 is 515 kJ mol–1. The correct reason for higher B–F bond dissociation energy as compared to that of C–F is

Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 19

Because of pπ – pπ back bonding in BF3 molecule, all B-F bonds have partial double bond character.

JEE Main Chemistry Mock Test- 3 - Question 20

At 35°C, the vapor pressure of CS2 is 512 mm Hg and that of acetone is 344 mm Hg. A solution of CS2 in acetone has a total vapor pressure of 600 mm Hg. The false statement amongst the following is    (2020) (a) Raoult’s law is not obeyed by this system. (b) A mixture of 100 mL CS2 and 100 mL acetone has a volume < 200 mL. (c) CS2 and acetone are less attracted to each other than to themselves. (d) Heat must be absorbed in order to produce the solution at 35°C.
 

Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 20

Since, in the given mixture, the solvent – solvent and solute – solute interaction is more than that of solvent – solute interaction, thus, the above mixture shows positive deviation from Raoult’s law.
 

JEE Main Chemistry Mock Test- 3 - Question 21

A vessel at 1000 K contains COwith a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is

Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 21

Given total pressure = 0.8 atm

CO2(g) + C(s) ⇌ 2CO(g)

Total pressure = 0.5-P +2P = 0.8

P = 0.8-0.5 = 0.3

KP = P2CO/PCO2 = (2P)2/(0.5-P)

= (0.6)2/0.2

= 0.36/0.2

= 1.8 atm

*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 3 - Question 22

If  x = Group number of Europium (atomic number 63) 
    y = Period number of Americium (atomic number 95)
    z = Most stable oxidation state shown by Gadolinium (atomic number 64) 
then find the value of x + y + z?


*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 3 - Question 23

The decomposition of hydrogen peroxide follows first order kinetics.
2H2O2(aq) → 2H2O(ℓ) + O2(g)
If the volume of O2 gas liberated at STP in first 20 minutes of the start of decomposition is 25.00 ml, what should be the total volume of O2 gas (in ml) collected in time, t >> t1/2.
(t1/2 for the decomposition of H2O2 is 10 min)


Detailed Solution for JEE Main Chemistry Mock Test- 3 - Question 23

*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 3 - Question 24

How many H-atom can be exchanged by D-atom when the following compound is kept in  solution for long time?


*Answer can only contain numeric values
JEE Main Chemistry Mock Test- 3 - Question 25

Find out the number of following orders which are INCORRECT against the mentioned properties :
(i) Ortho hydrogen > Para hydrogen (Stability at low temperature)
(ii) Ethanol > Glycerol (Viscosity)
(iii) D2 < He (Boiling Point)
(iv) HF > H2O (Melting point)
(v)  H3BO3 > BF3 (Melting point)
(vi) NH3 < SbH3 (Boiling point)
(vii) H2O2 > H2O (Strength of hydrogen bond)
(viii) H2O < D2O (Freezing point)
(ix)  < HF (Strength of hydrogen bond)
(x) D2O < H2O (Bond energy)


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