JEE Exam  >  JEE Tests  >  VITEEE Chemistry Test - 3 - JEE MCQ

VITEEE Chemistry Test - 3 - JEE MCQ


Test Description

30 Questions MCQ Test - VITEEE Chemistry Test - 3

VITEEE Chemistry Test - 3 for JEE 2025 is part of JEE preparation. The VITEEE Chemistry Test - 3 questions and answers have been prepared according to the JEE exam syllabus.The VITEEE Chemistry Test - 3 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for VITEEE Chemistry Test - 3 below.
Solutions of VITEEE Chemistry Test - 3 questions in English are available as part of our course for JEE & VITEEE Chemistry Test - 3 solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt VITEEE Chemistry Test - 3 | 35 questions in 50 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
VITEEE Chemistry Test - 3 - Question 1

Primary and secondary alcohols on action of red hot copper give

Detailed Solution for VITEEE Chemistry Test - 3 - Question 1

Primary and secondary alcohols react with red hot copper to produce different products.

When red hot copper is applied:

  • Primary alcohols yield aldehydes.
  • Secondary alcohols produce ketones.

This reaction showcases the distinct behaviour of alcohols under heat in the presence of copper. Aldehydes and ketones are significant organic compounds with varied applications in chemistry and industry.

VITEEE Chemistry Test - 3 - Question 2

In the reaction, CH₃CHO + HCN → X, a chiral centre is introduced. The product X is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 2

In the reaction CH₃CHO + HCN, a chiral centre is introduced, resulting in product X.

The product X can possess specific properties based on its composition:

  • Chirality: The reaction introduces a chiral centre, leading to the formation of enantiomers.
  • Optical Activity: Depending on the nature of the enantiomers, X may be laevo-rotatory (left) or dextro-rotatory (right).
  • Racemic Mixture: If both enantiomers are present in equal amounts, X forms a racemic mixture, exhibiting no net optical rotation.
  • Meso Compound: If X has an internal plane of symmetry, it would be a meso compound, which is achiral despite having multiple chiral centres.

Thus, the precise nature of product X hinges on the distribution and configuration of the resulting enantiomers.

VITEEE Chemistry Test - 3 - Question 3

The cyanohydrin of a compound X on hydrolysis gives lactic acid; The X is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 3

To determine the cyanohydrin of compound X that hydrolyses to form lactic acid, consider the following:

  • The hydrolysis of a cyanohydrin typically yields an amino acid or a carboxylic acid and an alcohol.
  • Lactic acid has the formula C₃H₆O₃ and is a type of hydroxy acid.
  • To form lactic acid, the cyanohydrin must originate from a carbonyl compound that, upon hydrolysis, produces the correct structure.
  • Among the options provided, acetaldehyde (CH₃CHO), when reacted with hydrogen cyanide and subsequently hydrolysed, yields lactic acid.

Conclusion: The correct answer is acetaldehyde (CH₃CHO) as it leads to the formation of lactic acid upon hydrolysis of its cyanohydrin derivative.

VITEEE Chemistry Test - 3 - Question 4

In Cannizzaro's reaction

Detailed Solution for VITEEE Chemistry Test - 3 - Question 4

Cannizzaro's reaction involves the transformation of aldehydes under specific conditions.

  • Aldehyde is converted into both alcohol and sodium carboxylate.
  • This reaction occurs in the absence of any catalyst and in a basic medium.
  • It is particularly significant for non-enolizable aldehydes, which cannot undergo self-condensation.

In summary, Cannizzaro's reaction is a unique process that deals specifically with aldehydes, leading to the formation of two different products. It highlights the versatility and complexity of organic reactions.

VITEEE Chemistry Test - 3 - Question 5

Electrophile in the case of chlorination of benzene in presence of FeCl₃ is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 5

The electrophile involved in the chlorination of benzene, when using FeCl₃, is

  • The primary electrophile is Cl⁺, a positively charged chlorine ion.
  • FeCl₃ acts as a catalyst and helps generate the Cl⁺ from chlorine gas.
  • Cl⁻ is a negatively charged ion and does not participate as an electrophile.
  • The chlorine atom itself (Cl) is neutral and does not act as an electrophile.
VITEEE Chemistry Test - 3 - Question 6

Pure methane can be produced by

Detailed Solution for VITEEE Chemistry Test - 3 - Question 6

Pure methane can be produced through various methods:

  • Wurtz reaction: Generally used to form larger alkanes rather than methane.
  • Kolbe's electrolytic method: Produces carboxylic acids and is not efficient for methane.
  • Soda-lime decarboxylation: Effectively yields methane from carboxylic acids.
  • Reduction with H₂: Can produce methane from carbon compounds, but less common.

Among these, soda-lime decarboxylation is the most effective method for producing pure methane.

VITEEE Chemistry Test - 3 - Question 7

How many molecules are present in one gram of hydrogen gas?

Detailed Solution for VITEEE Chemistry Test - 3 - Question 7

To find how many molecules are present in 1 gram of hydrogen gas (H₂), follow these steps:

Step 1: Molar mass of hydrogen gas (H₂)

  • Each hydrogen atom (H) has a mass of 1 g/mol

  • Hydrogen gas is H₂, so:
    Molar mass = 2 g/mol

Step 2: Moles of H₂ in 1 gram

  • Moles = Mass / Molar Mass = 1 / 2 = 0.5 moles

Step 3: Number of molecules in 0.5 moles

  • 1 mole = 6.022 × 10²³ molecules (Avogadro’s number)

  • So,
    0.5 moles = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules

VITEEE Chemistry Test - 3 - Question 8

Vitamin B₁₂ contains

Detailed Solution for VITEEE Chemistry Test - 3 - Question 8

Vitamin B₁₂ contains a cobalt ion at its core, specifically in the form of Co(III). This is a key feature of the vitamin, making it essential for various bodily functions.

  • Cobalt is essential for the formation of red blood cells.
  • It plays a crucial role in the maintenance of the nervous system.
  • Vitamin B₁₂ is vital for DNA synthesis.

The presence of cobalt in Vitamin B₁₂ distinguishes it from other vitamins and minerals.

VITEEE Chemistry Test - 3 - Question 9

Which compound can exist in a dipolare (Zwitterion) structure?

Detailed Solution for VITEEE Chemistry Test - 3 - Question 9

A dipolar (Zwitterion) structure can be formed by certain compounds due to the presence of both positive and negative charges.

  • Key Characteristics:
    • A Zwitterion has both a positive and a negative charge within the same molecule.
    • This structure typically arises in amino acids and some other organic compounds.
  • Analysis of Options:
    • Option A: Contains an amino group and a carboxylic acid, suggesting potential for Zwitterion formation.
    • Option B: Also has an amino group and carboxylic acid, supporting dipolar characteristics.
    • Option C: Includes an amine and a carboxylic acid, indicating it can exist as a Zwitterion.
    • Option D: While it has carboxylic acid groups, it lacks an amino group, so it cannot form a Zwitterion.
  • Conclusion:
    • Options A, B, and C can exist in a dipolar structure.
    • Option D cannot form a Zwitterion.

VITEEE Chemistry Test - 3 - Question 10

Acidic hydrolysis of acetamide gives

Detailed Solution for VITEEE Chemistry Test - 3 - Question 10

Acidic hydrolysis of acetamide yields several products through the breakdown of the amide bond. The key outcomes are:

  • Acetic acid: This is a prominent product formed during the hydrolysis process.
  • Ammonia: Released due to the cleavage of the amide bond.
  • Water: A necessity for the hydrolysis reaction to occur.

While products like acetaldehyde and methyl amine can be associated with related reactions, they are not the primary products of acidic hydrolysis of acetamide. Thus, the main focus remains on acetic acid and ammonia as the primary outcomes.

VITEEE Chemistry Test - 3 - Question 11

An organic acid having molecular formula C₂H₄O₂ is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 11

An organic acid with the molecular formula C₂H₄O₂ is known as:

  • Formic acid - HCOOH, the simplest carboxylic acid.
  • Acetic acid - CH₃COOH, commonly found in vinegar.
  • Oxalic acid - C₂HO₄, a dicarboxylic acid.
  • Propionic acid - C₃H₆O₂, a carboxylic acid used in food preservation.

Among these, both formic and acetic acids have the given molecular formula. However, acetic acid is more widely recognised in everyday use.

VITEEE Chemistry Test - 3 - Question 12

In a reaction involving ring substitution of C6H5Y, the major product is meta isomer. The group Y can be

Detailed Solution for VITEEE Chemistry Test - 3 - Question 12

Electron withdrawing groups favour meta substitution
- COOH group is metadirecting group

VITEEE Chemistry Test - 3 - Question 13

In a reversible reaction, the catalyst

Detailed Solution for VITEEE Chemistry Test - 3 - Question 13

In a reversible reaction, a catalyst:

  • Lowers the activation energy required for both the forward and reverse reactions.
  • This means that a catalyst facilitates the reaction without altering the overall energy change.
  • By decreasing the activation energy, it allows the reaction to proceed more quickly.
VITEEE Chemistry Test - 3 - Question 14

The reaction N₂ + O₂ → 2NO is endothermic . The forward reaction is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 14

The reaction N₂ + O₂ → 2NO is endothermic. The forward reaction is:

  • Favoured by a decrease in temperature. Lower temperatures shift the equilibrium towards the endothermic direction, promoting the formation of NO.
  • Unchanged by changes in pressure. Since the number of gas molecules is the same on both sides of the reaction (2 on each side), pressure alterations have no effect.
  • Not affected by adding a catalyst. While a catalyst speeds up the reaction rate, it does not influence the position of the equilibrium.
VITEEE Chemistry Test - 3 - Question 15

The enthalpy change of a reaction does not depend on

Detailed Solution for VITEEE Chemistry Test - 3 - Question 15

The enthalpy change of a reaction is influenced by various factors, but there are certain aspects it does not depend on:

  • State of reactants and products: The enthalpy change is not affected by whether the substances are solid, liquid, or gas.
  • Nature of reactants and products: The specific identity of the substances involved does not alter the enthalpy change.
  • Different intermediate reactions: The pathways or steps taken during the reaction do not impact the overall enthalpy change.
  • Initial and final enthalpy: What matters is the difference between the initial and final states, rather than the specific values at each stage.
VITEEE Chemistry Test - 3 - Question 16

An adiabatic expansion of an ideal gas always has

Detailed Solution for VITEEE Chemistry Test - 3 - Question 16

An adiabatic expansion of an ideal gas has specific characteristics:

  • Temperature decreases: During adiabatic expansion, the gas does work on its surroundings, leading to a reduction in temperature.
  • Heat transfer is zero: In an adiabatic process, there is no heat exchange with the environment, hence q = 0.
  • Work is done: The gas performs work on the surroundings, resulting in W ≠ 0.
  • Enthalpy change: The change in enthalpy ∆H can vary depending on the conditions.
VITEEE Chemistry Test - 3 - Question 17

If a gas at constant temperature and pressure expands, then its

Detailed Solution for VITEEE Chemistry Test - 3 - Question 17

When a gas expands at constant temperature and pressure, its internal energy behaves in the following way:

  • Internal energy remains constant.
  • For an ideal gas, internal energy is primarily a function of temperature.
  • Since the temperature is constant, there is no change in internal energy.
  • Expansion at constant temperature and pressure does not affect the internal energy directly.
VITEEE Chemistry Test - 3 - Question 18

Scurvy is caused due to the deficiency of vitamin

Detailed Solution for VITEEE Chemistry Test - 3 - Question 18

Scurvy is caused by a deficiency of vitamin C.

Vitamin C plays a crucial role in maintaining health. Here are some key points about scurvy and vitamin C:

  • Deficiency of vitamin C leads to scurvy.
  • Symptoms include fatigue, swollen gums, and joint pain.
  • It is essential for collagen production, which helps in wound healing.
  • Sources of vitamin C include fruits like oranges, strawberries, and vegetables such as bell peppers.
VITEEE Chemistry Test - 3 - Question 19

Drug which helps to reduce anxiety and brings about calmness is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 19

The drug that reduces anxiety and promotes calmness is:

  • Tranquiliser: These medications are specifically designed to alleviate anxiety and induce relaxation.
  • Diuretic: This type of drug helps remove excess fluids from the body but does not impact anxiety.
  • Analgesic: Primarily used for pain relief and does not address anxiety issues.
  • Antihistamine: Mainly treats allergic reactions, with some causing drowsiness but not directly calming anxiety.
VITEEE Chemistry Test - 3 - Question 20

The number of isomers of C₄H₉ - is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 20

The number of isomers of C₄H₉ is:

  • The molecular formula C₄H₉ can represent various structural forms known as isomers.
  • There are several types of isomers, including:
    • Structural isomers: Different arrangements of atoms within the molecule.
    • Stereoisomers: Isomers that differ in spatial arrangement.
  • For C₄H₉, we specifically consider structural isomers.
  • The key isomers include:
    • Butane
    • Isobutane
    • Various branched forms
  • In total, there are 4 distinct isomers for C₄H₉.
VITEEE Chemistry Test - 3 - Question 21

The IUPAC name for the complex
[ Co(NO2)(NH3)5 ]Cl2 is :

Detailed Solution for VITEEE Chemistry Test - 3 - Question 21

The IUPAC name for the complex [ Co(NO2)(NH3)5 ]Cl2 is:

  • The complex contains a cobalt ion surrounded by five ammonia ligands and one nitrito ligand.
  • The nitrito ligand is attached through the nitrogen atom, which is indicated by the prefix "N" in the name.
  • Cobalt can exhibit different oxidation states; in this case, it is in the +2 oxidation state.
  • The correct IUPAC name is pentaamminenitrito-N-cobalt(II) chloride.
VITEEE Chemistry Test - 3 - Question 22

Which of the following complexes will be formed in the brown ring test for nitrates

Detailed Solution for VITEEE Chemistry Test - 3 - Question 22

The brown ring test for nitrates involves the formation of a specific complex.

  • The complex formed in this test is [Fe(H₂O)₅NO].
  • This complex typically exists in the 2+ oxidation state.
  • Thus, the correct answer is [Fe(H₂O)₅NO]²⁺.
VITEEE Chemistry Test - 3 - Question 23

In the diazotisation of aniline with sodium nitrite and hydrochloric acid, the excess of hydrochloric acid is used primarily to

Detailed Solution for VITEEE Chemistry Test - 3 - Question 23

The excess of hydrochloric acid in the diazotisation of aniline serves several important functions:

  • Suppresses the concentration of free aniline, which helps prevent undesired side reactions.
  • Reduces the hydrolysis of nitrous acid to phenol, ensuring more effective diazotisation.
  • Maintains a consistent stoichiometric amount of nitrous acid, which is crucial for the reaction.
  • Neutralises any base that may be liberated during the process, maintaining the desired acidic environment.
VITEEE Chemistry Test - 3 - Question 24

The oxidation no. of sulphur in H₂S is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 24

The oxidation number of sulphur in H₂S is

The oxidation state of sulphur in hydrogen sulphide (H₂S) can be determined using the following reasoning:

  • In H₂S, hydrogen has an oxidation state of +1.
  • Since there are two hydrogen atoms, their total contribution is +2.
  • Let the oxidation state of sulphur be represented as x.
  • According to the rule that the sum of oxidation states in a neutral molecule equals 0, we have:

x + 2 = 0

  • Solving this gives: x = -2.

Therefore, the oxidation number of sulphur in H₂S is -2.

VITEEE Chemistry Test - 3 - Question 25

Hydrolysis of benzonitrile gives

Detailed Solution for VITEEE Chemistry Test - 3 - Question 25

Hydrolysis of benzonitrile results in the formation of:

  • Benzylamine
  • Aniline
  • Benzoic acid
  • Benzene

During hydrolysis, benzonitrile reacts with water, leading to the conversion of the nitrile group. This process typically yields:

  • Benzoic acid through hydrolysis.
  • Benzylamine as a potential product if ammonia is present.

However, the primary product from this reaction is generally benzoic acid.

VITEEE Chemistry Test - 3 - Question 26

The quanitity of electricity required to librate 112 cm3 of hydrogen at STP from acidified water is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 26

2H+ + 2e- → H2
2e- → 2F
H2 → 22.4 L at STP
To liberate 112 cm3 at STP electron required
= (2 x 96500 x 112)/ 22400 = 965 C

VITEEE Chemistry Test - 3 - Question 27

An electrolytic cell contains a solution of Ag₂SO₄ and has Pt electrodes.A current is passed till 1.6 g of O₂ has been liberated at anode. The amount of silver deposited at cathode will be

Detailed Solution for VITEEE Chemistry Test - 3 - Question 27

To find the amount of silver deposited at the cathode, follow these steps:

  • Calculate the moles of oxygen liberated:
    • Using the molar mass of O₂ (32 g/mol), the moles of 1.6 g is:
    • 1.6 g / 32 g/mol = 0.05 moles of O₂.
  • Determine the charge involved:
    • Each mole of O₂ requires 4 moles of electrons for its formation (4e- + 2H₂O → O₂ + 4H+).
    • Thus, for 0.05 moles of O₂, the total moles of electrons transferred is:
    • 0.05 moles × 4 = 0.2 moles of electrons.
  • Calculate the amount of silver deposited:
    • The reduction of silver ions (Ag+ + e- → Ag) shows that 1 mole of silver is deposited per mole of electrons.
    • Therefore, 0.2 moles of electrons will deposit 0.2 moles of silver.
  • Find the mass of silver:
    • The molar mass of silver (Ag) is approximately 107.88 g/mol.
    • Thus, the mass of silver deposited is:
    • 0.2 moles × 107.88 g/mol = 21.576 g.

The amount of silver deposited at the cathode is approximately 21.6 g.

VITEEE Chemistry Test - 3 - Question 28

The reagent used for the preparation of higher ethers from halogenated ethers is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 28

Grignard reagant is used to prepare higher ethers from halogenated ethers. The reaction is as follows

VITEEE Chemistry Test - 3 - Question 29

For which ore of the metal, froth floatation method is used for concentration?

Detailed Solution for VITEEE Chemistry Test - 3 - Question 29

The froth flotation method is used primarily for the concentration of ores that contain specific types of minerals.

This technique is particularly effective for:

  • Sulphide ores such as those containing copper, lead, and zinc.
  • Ores with a significant amount of gangue materials that need to be separated from valuable minerals.
  • Ores where the desired minerals are hydrophobic and can be separated through flotation.

Common examples include:

  • Galena (lead ore)
  • Chalcopyrite (copper ore)
  • Sphalerite (zinc ore)

In summary, froth flotation is a highly effective method for concentrating ores that are compatible with this process.

VITEEE Chemistry Test - 3 - Question 30

A pseudo solid is

Detailed Solution for VITEEE Chemistry Test - 3 - Question 30

A pseudo solid is a material that exhibits properties of both solids and liquids. Here are key points regarding pseudo solids:

  • Definition: Pseudo solids behave like solids under certain conditions but can flow like liquids over time.
  • Examples: Common examples include materials such as glass, which appears solid but lacks a crystalline structure.
  • Characteristics: They may have a fixed shape yet can undergo deformation when force is applied for extended periods.
  • Applications: Pseudo solids are often used in various industries, including construction and manufacturing, due to their unique properties.
View more questions
Information about VITEEE Chemistry Test - 3 Page
In this test you can find the Exam questions for VITEEE Chemistry Test - 3 solved & explained in the simplest way possible. Besides giving Questions and answers for VITEEE Chemistry Test - 3, EduRev gives you an ample number of Online tests for practice
Download as PDF