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BITSAT Chemistry Test - 2 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Chemistry Test - 2

BITSAT Chemistry Test - 2 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Chemistry Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Chemistry Test - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Chemistry Test - 2 below.
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BITSAT Chemistry Test - 2 - Question 1

The correct match between items of List-I and List-II is:

Detailed Solution for BITSAT Chemistry Test - 2 - Question 1

→ Phenelzine contains hydrazine(2-phenylethylhydrazine)

→ Chloroxylenol contains phenol(4-Chloro-3,5-dimethylphenol)

→ Uracil is the pyrimidine base(2,4-Dihydroxypyrimidine)

→ Ranitidine contains furan ring

BITSAT Chemistry Test - 2 - Question 2

The binding of oxygen by haemoglobin (Hb) forming (HbO2), is partially regulated by the concentration of H3O + and dissolved CO2 in blood.

Release of O2 is favoured when there is

Detailed Solution for BITSAT Chemistry Test - 2 - Question 2

During exercise muscle need energy hence consume more oxygen and produce more CO2 and lactic acid gets accumulated as well. According to bohr's effect oxygen binding affinity is inversely proportional to CO2 concentration and acidity. Thus favours the dissociation​ or release of O2.

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BITSAT Chemistry Test - 2 - Question 3

Aerated water contains CO2 dissolved in water

CO2(g) + H2O(l) H2CO3(aq)


Variation of solubility (s) with pressure (p) is shown by 

Detailed Solution for BITSAT Chemistry Test - 2 - Question 3

Pressure increase solubility also increase in accordance with the Henry's law and option B is the closest to this. Hence B is correct.

BITSAT Chemistry Test - 2 - Question 4

If pKb for fluoride ion at 25°C is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :

Detailed Solution for BITSAT Chemistry Test - 2 - Question 4

H+  H+ + F
pKw = pKa + pKb
[For conjugate Acid-Base]
⇒ pKa = 14 – 10.87 = 3.17
Ka = 6.76 × 10–4

BITSAT Chemistry Test - 2 - Question 5

The pH of an aqueous solution of 1.0 M solution of a weak monoprotic acid which is 1% ionised is:

Detailed Solution for BITSAT Chemistry Test - 2 - Question 5

Concentration of solution =C=1M
Dissociation is α=0.01
After dissociation, the concentration of H+ ions will be Cα=1×0.01=10−2
pH=−log[H+]=−log(10−2) = +2

BITSAT Chemistry Test - 2 - Question 6

In which of the following set of compounds oxidation number of oxygen is not (- 2)?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 6


OF2 Fluorine is most electronegative atom with oxidation number = - 1
x - 2 = 0
x = + 2
H2O2   2 + 2x = 0 ⇒ x = - 1 

BITSAT Chemistry Test - 2 - Question 7

Which is chlorate (I) ion?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 7
  • ClO3: A very reactive inorganic anion.
  • The term chlorate can also be used to describe any compound containing the chlorate ion, normally chlorate salts. 
  • Example: Potassium chlorate, KClO3
BITSAT Chemistry Test - 2 - Question 8

Which pair is not in the correct order of lattice energy?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 8

Based on Fajans' rule,
Smaller the size of cation, larger the size of anion, larger the charge,
d-f-Aelectrons present then larger the polarising power, larger the covalent nature and larger the lattice energy.


BITSAT Chemistry Test - 2 - Question 9

Most stable compound is

Detailed Solution for BITSAT Chemistry Test - 2 - Question 9

By Fajans’ rule,
Smaller the size of cation, larger the size of anion, larger the charge, then larger the polarising power, larger the covalent nature, means low melting point.
Smaller the ionic nature, means low melting point 


Hence, most stable is LiF.

BITSAT Chemistry Test - 2 - Question 10

For the process, and 1 atmosphere pressure, the correct choice is

[JEE Advanced 2014]

Detailed Solution for BITSAT Chemistry Test - 2 - Question 10

At 100°C and 1 atmosphere pressure H2O (l) ⇋ H2O(g) is at equilibrium. For equilibrium

*Multiple options can be correct
BITSAT Chemistry Test - 2 - Question 11

Direction (Q. Nos. 11-14) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.

Q. For an ideal gas, consider only (p -V) work in going from initial state X to the final state Z. The final state Z can be reached either of the two paths shown in the figure. Which of the following choice (s) is (are) correct?

(Take ΔS as change in entropy and W as work done)

[IIT JEE 2012]

Detailed Solution for BITSAT Chemistry Test - 2 - Question 11

∆SX→Z = ∆SX→Y + ∆SY→Z(Entropy is a state function, so it is additive)
WX→Y→Z = WX→Y (work done in y→z is zero as the process is isochoric)

BITSAT Chemistry Test - 2 - Question 12

Comprehension Type

Direction (Q. Nos. 16 and 17) This section contains a paragraph, describing theory, experiments, data etc. Two questions related to the paragraph have been given. Each has only one correct answer among the four given options (a) ,(b), (c) and (d).

Passage

Consider the following isomers of [Co(NH3)4Br2]+. The black sphere represents Co, grey sphere represents NH3 and unshaded sphere represents Br.

Q.

The oxidation state and coordination number of cobalt in the complex [Co(NH3)4Br2]+ are

Detailed Solution for BITSAT Chemistry Test - 2 - Question 12

For O.S:

X + 0 + 2(-1) = 1

X = 3

Coordination number is the number of molecules linked to the central atom.

Hence B is the correct answer.

BITSAT Chemistry Test - 2 - Question 13

Passage

Consider the following isomers of [Co(NH3)4Br2]+. The black sphere represents Co, grey sphere represents NH3 and unshaded sphere represents Br.

Q. 

Which of the structures is identical?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 13

Structure (a) = structure (c) and, structure (b) = structure (d)

BITSAT Chemistry Test - 2 - Question 14

What is the major product of the given reaction ?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 14

α-carbon is achiral, hence retention of configuration at β-carbon.

BITSAT Chemistry Test - 2 - Question 15

Which of the following are the examples of strong nucleophiles but weak base in protic solvents?

I. CH3S-
II. CH3O-
III. I-
IV. H2O
V. F-

Detailed Solution for BITSAT Chemistry Test - 2 - Question 15

Both CH3S- and I- are weak bases but strong nucleophiles.

BITSAT Chemistry Test - 2 - Question 16

The correct statement regarding 3-ethyl-3-hexanol is

Detailed Solution for BITSAT Chemistry Test - 2 - Question 16

All alcohols react (1°, 2° and 3°) with cerric nitrate and a colour change from yellow to red is observed.

BITSAT Chemistry Test - 2 - Question 17

An alcohol has molecular formula C6H12O X and it gives immediate turbidity with cold, concentrated HCI even in the absence of ZnCI2. X can also be obtained by treatment of an ether with excess of CH3MgBr followed by acid hydrolysis. Hence, the correct statement regarding X is

Detailed Solution for BITSAT Chemistry Test - 2 - Question 17

To determine the correct statement regarding the alcohol X with the molecular formula C66​H121​2O, we need to consider the clues given:

  1. Immediate turbidity with cold, concentrated HCl without ZnCl2: This suggests that the alcohol is a tertiary alcohol. Tertiary alcohols react immediately with Lucas reagent (conc. HCl + ZnCl2), but even in the absence of ZnCl2​, they can form a stable carbocation leading to turbidity.

  2. Obtained by treatment of an ether with CH3​MgBr followed by acid hydrolysis: This indicates that the ether must be symmetrical or such that it results in the formation of a tertiary alcohol upon Grignard reaction.

Given these conditions, let's evaluate the options:

  1. 3-methyl-3-pentanol:

    • Structure: CH3​-CH2​-C(CH3​)OH-CH2​-CH3
    • This is a tertiary alcohol (immediate turbidity with HCl).
    • This alcohol can be formed from the ether 3-methyl-3-pentyl ether by Grignard reaction with CH3​MgBr.
  2. 2-methyl-3-pentanol:

    • Structure: CH3​-CH2​-CH(OH)-CH2​-CH3
    • This is a secondary alcohol (slow reaction with HCl in the presence of ZnCl2).
  3. 2-methyl-2-pentanol:

    • Structure: CH3​-C(CH3​)OH-CH2​-CH3
    • This is a tertiary alcohol (immediate turbidity with HCl).
    • This alcohol can also be formed from the ether 2-methyl-2-pentyl ether by Grignard reaction with CH3​MgBr.

Since we are looking for a tertiary alcohol that gives immediate turbidity with HCl, and can be formed by treating an ether with CH3​MgBr followed by acid hydrolysis, the correct answers are: 3-methyl-3-pentanol and 2-methyl-2-pentanol

Therefore, the correct statement regarding X is:

Either (b) or (d) (2-methyl-3-pentanol is a mistake in the options; the correct pair is either 3-methyl-3-pentanol or 2-methyl-2-pentanol).

BITSAT Chemistry Test - 2 - Question 18

Diazonium salts are used in the preparation of

Detailed Solution for BITSAT Chemistry Test - 2 - Question 18

Diazonium reacts with phenols and napthols to form dyes.

BITSAT Chemistry Test - 2 - Question 19

When pyridine is treated with a mixture of nitric and sulfuric acids, the major product is:

Detailed Solution for BITSAT Chemistry Test - 2 - Question 19

In pyridine EAS take splace at 3 position.

BITSAT Chemistry Test - 2 - Question 20

Comprehension Type

Direction (Q. Nos. 13-15) This section contains a paragraph, describing theory, experiments, data, etc.
Three questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Passage

An alcohol (R — OH) can be converted into alkyl chloride by the treatm ent with HCI. Reaction involves protonation of alcohol followed by the formation of carbocation intermediate. Carbocation intermediate in the final step undergo nucleophilic attack by Cl- ion as :

Q. 

Which of the following alcohol reacts most easily?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 20

As mentioned in mechanism, reaction proceed via carbocation intermediate. Hence, alcohol forming most stable carbocation reacts most easily. Alcohol (c) forms aromatic (highly stable) carbocation, hence most reactive.

BITSAT Chemistry Test - 2 - Question 21

An alcohol (R — OH) can be converted into alkyl chloride by the treatm ent with HCI. Reaction involves protonation of alcohol followed by the formation of carbocation intermediate. Carbocation intermediate in the final step undergo nucleophilic attack by Cl- ion as :

Q. 

Which of the following can catalyse the above reaction? 

Detailed Solution for BITSAT Chemistry Test - 2 - Question 21

ZnCI2 is a Lewis acid, helps in the form ation of carbocation intermediate, hence catalyse the reaction.

BITSAT Chemistry Test - 2 - Question 22

Which of the following is not a bactericidal Antibiotic:

Detailed Solution for BITSAT Chemistry Test - 2 - Question 22

Tetracycline is a bacteriostatic antibiotic. Bactericidal antibiotics are antibiotics which can kill bacteria and Bacteria-static are which can inhibit the growth of bacteria. Ofloxacin is bactericidal antibiotic. Ofloxacin is a fluoroquinolone carboxylic acid antibacterial showing bactericidal effects by inhibition of DNA gyrase.

BITSAT Chemistry Test - 2 - Question 23

Assuming that both, amide and ester bonds of Aspartame are hydrolysed in the stomach, the amino acids obtained is:

Detailed Solution for BITSAT Chemistry Test - 2 - Question 23

BITSAT Chemistry Test - 2 - Question 24

Following are the structures of three drugs A, B and C , respectively,
(a) 
(b) 
(c) 

Detailed Solution for BITSAT Chemistry Test - 2 - Question 24

Paracetamol is N-(4-hydroxyphenyl) ethanamide, which is a hydroxy group containing compound.

Phenacetin is N-(4-ethoxyphenyl) ethanamide, which is an amide and ethoxy group containing compound.

Aspirin is 2-acetoxybenzoic acid, which is an ester.

BITSAT Chemistry Test - 2 - Question 25

Which of the following drug is capable of destroying microorganisms?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 25

Antibiotics can be classified as bactericidal and bacteriostatic on the basis of the effect, it has on microbes. Bactericidal drugs have killing/destroying effect on microbes while, bacteriostatic have inhibiting effect on microbes. A few examples of the two types of antibiotics are as follows: Bactericidals are Penicillin, Aminoglycosides, Ofloxacin. Bacteriostatics are Erythromycin, Tetracycline, Chloramphenicol.

BITSAT Chemistry Test - 2 - Question 26

Which of the following is employed as Antihistamine?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 26

Chloramphenicol is a broad spectrum antibiotic. Norethindrone is an antifertility drug. Omeprazol is an antacid and Diphenhydramine is an antihistamine mainly used to treat allergies.

BITSAT Chemistry Test - 2 - Question 27

Which of the following is an artificial edible colour?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 27

Tartrazine impart colour, it is artificial edible colour. It is used generally for yellow colour in food. It is stable compound because it contains 6 membered aromatic ring in which four atoms are nitrogen atoms, and it's chemical formula is C2H2N4.

BITSAT Chemistry Test - 2 - Question 28

What are lysergic acid diethylamide (LSD) and mescaline?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 28

LSD and mescaline are psychedelic drugs.
These produce visual and auditory hallucinations.

BITSAT Chemistry Test - 2 - Question 29

The substances which do not act as antiseptic are:
(a) Iodoform
(b) H2O2
(c) Diphenhydramine
(d) Soframycin
(e) Amoxycillin
(f) Ranitidine

Detailed Solution for BITSAT Chemistry Test - 2 - Question 29

Ranitidine is antacid; amoxicillin is an antibiotic while diphenhydramine is an antihistamine.
Antiseptics and disinfectants are also the chemicals which either kill or prevent the growth of microorganisms. Antiseptics are applied to the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.  Hydrogen peroxide, iodoform and soframycin act as antiseptics.

BITSAT Chemistry Test - 2 - Question 30

A chemist has 4 samples of artificial sweetener A , B , C and D . To identify these samples, he performed certain experiments and noted the following observations:
(i) A and D both form blue-violet colour with ninhydrin.
(ii) Lassaigne extract of C gives positive AgNO3 test and negative Fe4[Fe(CN)6]3 test.
(iii) Lassaigne extract of B and D gives positive sodium nitroprusside test. Based on these observations which option is correct?

Detailed Solution for BITSAT Chemistry Test - 2 - Question 30

A− Aspartame

B− Saccharine

C− Sucralose

D− Alitame

(i) A & D give positive test with ninhydrin because both have free carboxylic and amine groups.
(ii) C form precipitate with AgNO3 in the Lassaigne extract of the sugar because it has chlorine atoms.
(iii) B & D give positive test with sodium nitroprusside because both have sulphur atoms

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