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BITSAT Chemistry Test - 1 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Chemistry Test - 1

BITSAT Chemistry Test - 1 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Chemistry Test - 1 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Chemistry Test - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Chemistry Test - 1 below.
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BITSAT Chemistry Test - 1 - Question 1

Which of the following will not be soluble in sodium carbonate solution?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 1


While 2,4,6-trinitrophenol, benzoic acid and benzene sulphonic acid are soluble in NaHCO3.This reaction is possible in the forward direction if acid is more acidic than H2CO3.o-nitrophenol is less acidic than H2CO3.Hence, it is not soluble in sodium hydrogen carbonate.

BITSAT Chemistry Test - 1 - Question 2

Which of the following expressions gives the de-Broglie relationship?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 2

De-Broglie Relationship

- The de-Broglie relationship relates the wavelength of a particle to its momentum.
- It is given by the equation: λ = h/p, where λ is the wavelength, h is the Planck constant, and p is the momentum of the particle.
- Therefore, the correct expression from the given options is: B) λ = h/mν.

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BITSAT Chemistry Test - 1 - Question 3

Which of the following reactions is used for detecting presence of carbonyl group ?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 3

Detection of Carbonyl Group

- Reaction with hydroxylamine: Carbonyl compounds react with hydroxylamine to form oximes. This reaction is used for detecting the presence of carbonyl group in organic compounds.

- Reaction with hydrazine: Carbonyl compounds react with hydrazine to form hydrazones. This reaction is also used for detecting the presence of carbonyl group in organic compounds.

- Reaction with phenyl hydrazine: Carbonyl compounds react with phenyl hydrazine to form phenylhydrazones. This reaction is commonly used for detecting the presence of carbonyl group, especially in sugars and other complex molecules.

- All of the above: The reactions with hydroxylamine, hydrazine, and phenyl hydrazine are all commonly used for detecting the presence of carbonyl group in organic compounds.

BITSAT Chemistry Test - 1 - Question 4

The total number of electrons present in all the ρ-orbitals of bromine is

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BITSAT Chemistry Test - 1 - Question 5

Point out incorrect statement about resonance

Detailed Solution for BITSAT Chemistry Test - 1 - Question 5

Incorrect Statement about Resonance:
 


  • Resonance structures should not have the same number of electron pairs: This statement is incorrect because resonance structures must have the same number of total electrons. The difference between resonance structures lies in the distribution of electrons rather than the total number of electrons. The total number of electrons must remain the same in all resonance structures.
BITSAT Chemistry Test - 1 - Question 6

A compound possesses 8% sulphur by mass. The least molecular mass is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 6

The minimum no. of sulphur atom which a molecule can possess is 1.

Mass of 1 sulphur atom =32g

As the compound has 8% sulphur by mass

∴ 8g of sulphur is present in 100g of compound

∴ 32g of sulphur is present in 100/8×32=400g of compound

Hence, the least molecular mass of the compound is 400 g.

BITSAT Chemistry Test - 1 - Question 7

The compound with carbon uses only its sp3 hybrid orbitals for bond formation is

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BITSAT Chemistry Test - 1 - Question 8

Which of the following gives maximum energy in metabolic process?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 8

Maximum energy in Metabolic Process


  • Lipids: Lipids provide the most energy in the metabolic process. When metabolized, lipids yield 9 calories per gram, which is more than double the amount of energy produced by carbohydrates and proteins.

  • Carbohydrates: Carbohydrates provide 4 calories per gram when metabolized. They are a quick source of energy but are not as efficient as lipids in providing sustained energy.

  • Protein: Proteins provide 4 calories per gram when metabolized. While they are essential for building and repairing tissues, they are not the most efficient source of energy.

  • Vitamins: Vitamins do not provide energy directly in the metabolic process. They are essential for various bodily functions but do not contribute to energy production.


Therefore, lipids give the maximum energy in the metabolic process compared to proteins, carbohydrates, and vitamins.

BITSAT Chemistry Test - 1 - Question 9

The main structure features of proteins is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 9

Main Structure Features of Proteins

Peptide Linkage: The main structure feature of proteins is the peptide linkage, which is formed between the carboxyl group of one amino acid and the amino group of another amino acid. This bond is known as a peptide bond and it is the basis for the formation of protein chains.
 

BITSAT Chemistry Test - 1 - Question 10

Which of the following is the strongest nucleophile?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 10

A negatively charged nucleophile is always stronger than a neutral one

BITSAT Chemistry Test - 1 - Question 11

Among the following the bond with highest bond dissociation energy is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 11

Comparison of Bonds in Question

- S-S bond has a higher bond dissociation energy compared to Se-Se, Te-Te, and O-O bonds.
- The reason for this is the size of the atoms and the bond length.
- Sulfur atoms are smaller than selenium and tellurium atoms, leading to stronger bonds.
- Additionally, the bond length of S-S is shorter than Se-Se and Te-Te bonds, making it stronger.
- Oxygen atoms in O-O bond have higher electronegativity, which leads to a weaker bond compared to S-S bond.

 

BITSAT Chemistry Test - 1 - Question 12

The enthalpy change (∆H) for neutralisation of 1 M HCl by caustic soda in dilute solutions at 298 K is

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BITSAT Chemistry Test - 1 - Question 13


In the above reaction , product P is

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BITSAT Chemistry Test - 1 - Question 14

A and B are gaseous substances which react reversibly to give two gaseous substances C and D,accompanied by the liberation of heat.When the reaction reaches equilibrium,it is observed that Kp = Kc.The equilibrium cannot be disturbed by

BITSAT Chemistry Test - 1 - Question 15

The rate constant of first order reaction whose half life is 480 sec is

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BITSAT Chemistry Test - 1 - Question 16

Which of the following elements has the lowest ionisation potential

Detailed Solution for BITSAT Chemistry Test - 1 - Question 16

The ionisation potential increases from O to F to Ne, because of increasing nuclear charge. The ionisation potential is highest for Ne because of its stable (ns2 np6) electronic configuration. But the ionisation potential of O is lower than N although the nuclear charge of O is higher. This is due to the extra-stability of exactly half-filled 2p-orbitals of N.

BITSAT Chemistry Test - 1 - Question 17

A system is said to be in thermodynamic equilibrium when it is in

Detailed Solution for BITSAT Chemistry Test - 1 - Question 17

Thermodynamic Equilibrium


  • Chemical Equilibrium: In chemical equilibrium, the rates of the forward and reverse reactions are equal, and there is no net change in the concentrations of the reactants and products. This means that the system has reached a steady state in terms of chemical composition.

  • Mechanical Equilibrium: Mechanical equilibrium occurs when there is no net external force acting on the system. This means that the system is not accelerating and all the forces are balanced.

  • Thermal Equilibrium: Thermal equilibrium is achieved when there is no net heat transfer between the system and its surroundings. This means that the temperature throughout the system is uniform and there are no gradients.

  • All the Above Simultaneously: When a system is in thermodynamic equilibrium, it means that it is in a state where all the above conditions are met simultaneously. This indicates that the system is stable and not undergoing any changes with respect to chemical composition, mechanical forces, and thermal properties.

In conclusion, a system is said to be in thermodynamic equilibrium when it satisfies the conditions of chemical equilibrium, mechanical equilibrium, and thermal equilibrium simultaneously.

BITSAT Chemistry Test - 1 - Question 18

The detergents which are most suitable for cleansing synthetic fibres, are

Detailed Solution for BITSAT Chemistry Test - 1 - Question 18

  • Neutral Detergents: Neutral detergents are also suitable for cleansing synthetic fibres. These detergents are gentle on the fibres and help in maintaining the fabric's integrity while still effectively cleaning it.

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BITSAT Chemistry Test - 1 - Question 19

IUPAC name of Na₃[Co(NO₂)₆] is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 19

The IUPAC name of Na3[Co(N02)6] is sodium hexanitrocobaltate (III).

BITSAT Chemistry Test - 1 - Question 20

Four successive members of the first row transition elements are listed below with their atomic no. Which of them is expected to have the highest third ionisation enthalpy?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 20

Explanation:

• The third ionization enthalpy is the energy required to remove the third electron from an atom.

• The ionization enthalpy generally increases as we move across a period in the periodic table.

• Among the given elements, manganese (Z=25) is expected to have the highest third ionization enthalpy.

• This is because as we move from left to right across the transition metals in the first row, the effective nuclear charge increases due to an increase in the number of protons in the nucleus.

• Since manganese has the highest atomic number among the given elements, it will have the highest effective nuclear charge, making it more difficult to remove the third electron, resulting in the highest third ionization enthalpy.

BITSAT Chemistry Test - 1 - Question 21

Low spin complexes are generally formed by the elements of

Detailed Solution for BITSAT Chemistry Test - 1 - Question 21

Formation of Low Spin Complexes

Low spin complexes are generally formed by elements of the 3d-series and 4d-series.

Reasons for Low Spin Complex Formation

- Elements of the 3d-series and 4d-series have smaller atomic radii compared to those of the 5d-series.
- The smaller atomic radii result in stronger ligand field effects, which lead to the formation of low spin complexes.
- The stronger ligand field effects cause the d orbitals to split into higher energy t2g and lower energy eg sets, favoring the formation of low spin complexes.

Conclusion

Therefore, elements of the 3d-series and 4d-series are more likely to form low spin complexes due to their smaller atomic radii and stronger ligand field effects.

BITSAT Chemistry Test - 1 - Question 22

The units of electrochemical equivalent are

Detailed Solution for BITSAT Chemistry Test - 1 - Question 22

Explanation:

- Electrochemical equivalent is defined as the mass of a substance liberated at an electrode during electrolysis when a current of 1 ampere flows for 1 second.
- The unit of current is ampere (A), the unit of time is second (s), and the unit of mass is gram (g).
- Therefore, the unit of electrochemical equivalent is gram amp-1 sec-1.
- This unit represents the amount of substance (in grams) deposited or liberated at an electrode per ampere of current flowing for one second during electrolysis.

BITSAT Chemistry Test - 1 - Question 23

Which of the following isomers will have the highest boiling point ?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 23


- Among isomers with the same molecular formula, the linear isomer (in this case, Hexane) will have the highest boiling point due to its maximized surface area for van der Waals interactions.
- Therefore, Hexane, being a straight-chain alkane with no branching, will exhibit the highest boiling point among its isomers.

BITSAT Chemistry Test - 1 - Question 24

When n-propyl iodide is heated with alcoholic KOH,one of the products is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 24

Explanation:

Reaction Overview:
- When n-propyl iodide is heated with alcoholic KOH, it undergoes an elimination reaction to form propene as one of the products.

Mechanism of Reaction:
- The reaction proceeds via an E2 (Elimination Bimolecular) mechanism.
- Alcoholic KOH acts as a base and abstracts a proton from the beta-carbon (carbon adjacent to the iodine atom) of n-propyl iodide.
- This results in the formation of a carbanion intermediate.
- The leaving group (iodide ion) leaves, leading to the formation of a double bond between the two adjacent carbons.
- This gives rise to propene as one of the products.

Product Formation:
- Propene (C₃H₆) is formed as a result of the elimination of a proton and iodide ion from n-propyl iodide.
- The molecular formula of propene is C₃H₆, which matches the product formed in this reaction.

Therefore, the correct answer is A: Propene (C₃H₆).

BITSAT Chemistry Test - 1 - Question 25

The process of separation of racemic mixture into its constitutes is called

Detailed Solution for BITSAT Chemistry Test - 1 - Question 25

Resolution of Racemic Mixture

- Definition: Resolution is the process of separating a racemic mixture, which contains equal amounts of both enantiomers, into its individual enantiomers.

- Principle: The separation is based on the differences in the physical or chemical properties of the enantiomers.

BITSAT Chemistry Test - 1 - Question 26

The volume strength of 1.5 N H₂O₂ solution is

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BITSAT Chemistry Test - 1 - Question 27

The solubility of AgI in NaI solution is less than that in pure water because

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Explanation:


  • Common Ion Effect: When AgI is introduced into a solution containing NaI, the presence of I- ions from NaI will shift the equilibrium of AgI dissolution reaction to the left. This is due to the common ion effect, where the presence of a common ion (I-) reduces the solubility of AgI in the solution. Therefore, the solubility of AgI in NaI solution is less than that in pure water.


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BITSAT Chemistry Test - 1 - Question 28

How many monochlorobutanes will be obtained on chlorination of n-butane?

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BITSAT Chemistry Test - 1 - Question 29

Of the following which is paramagnetic and has three electron bond in its structure

Detailed Solution for BITSAT Chemistry Test - 1 - Question 29

Paramagnetic Molecule with Three Electron Bond


  • Paramagnetic Molecules: Paramagnetic molecules have unpaired electrons, which are attracted to a magnetic field.

  • Three Electron Bond: A three-electron bond is a type of chemical bond in which three electrons participate in the bonding between two atoms.

  • Identifying the Molecule:

    • N₂O (Nitrous Oxide): This molecule does not have unpaired electrons and does not exhibit paramagnetic properties.

    • NO (Nitric Oxide): This molecule has an odd number of electrons, one unpaired electron, and exhibits paramagnetic properties. It also has a three-electron bond between the nitrogen and oxygen atoms.

    • N₂O₃ (Dinitrogen Trioxide): This molecule does not have unpaired electrons and does not exhibit paramagnetic properties.

    • N₂O₅ (Dinitrogen Pentoxide): This molecule does not have unpaired electrons and does not exhibit paramagnetic properties.

Therefore, the molecule that is paramagnetic and has a three-electron bond in its structure is NO (Nitric Oxide).

BITSAT Chemistry Test - 1 - Question 30

Which of the following is the strongest base?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 30

Strongest Base

- NH₃ (Ammonia) is the strongest base among the given options.
- The strength of a base is determined by its ability to accept a proton (H⁺ ion) and form a stable conjugate acid.
- NH₃ has a lone pair of electrons on the nitrogen atom, which can readily accept a proton to form NH₄⁺.
- The other options, PH₃, AsH₃, and SbH₃, also have lone pairs of electrons, but nitrogen in NH₃ is more electronegative than phosphorus, arsenic, and antimony in the other compounds, making NH₃ a stronger base.

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