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40 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers - Sample BITSAT Chemistry Test

Sample BITSAT Chemistry Test for JEE 2023 is part of BITSAT Mock Tests Series & Past Year Papers preparation. The Sample BITSAT Chemistry Test questions and answers have been prepared according to the JEE exam syllabus.The Sample BITSAT Chemistry Test MCQs are made for JEE 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Sample BITSAT Chemistry Test below.
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Sample BITSAT Chemistry Test - Question 1

Hydrogen bonding is maximum in

Detailed Solution for Sample BITSAT Chemistry Test - Question 1
ethyl chloride & triethyl doesnt have hydrogen bonding because of absence of N/O/F group.

di ethyl ether also doesnt have h-bonding because the h atom is not bonded with Oxygen.

ethanol has h-bonding because the H atom is joint with O atom & hence it has the maximum.
Sample BITSAT Chemistry Test - Question 2

Catalyst used in Rosenmund reduction is

Sample BITSAT Chemistry Test - Question 3

Which of the following does not have sp2-hybridised carbon?

Sample BITSAT Chemistry Test - Question 4

The total number of electrons present in all the s-orbitals, all the p-orbitals and all the d-orbitals of cesium ion are respedtively

Detailed Solution for Sample BITSAT Chemistry Test - Question 4
Cs+ = 2,8, 8,18,18 
No. of 's' electrons = 2 + 2 + 2 + 2 + 2 = 10 
No. of 'p' electrons = 6 + 6 + 6 + 6 = 24 
No. of 'd' electrons = 10 + 10 = 20 
Sample BITSAT Chemistry Test - Question 5

The effect of atoms or groups losing electrons towards a carbon atom is called

Sample BITSAT Chemistry Test - Question 6

The weight of one molecule of a compound C₆₀H₁₂₂ is

Detailed Solution for Sample BITSAT Chemistry Test - Question 6

The Molecule iS C₆₀H₁₂₂ 
Molecular mass of C₆₀H₁₂₂ --- [60x12+1x122]
=720+122
=842 Unified mass
Weight of one molecule of C₆₀H₁₂₂ :
=molecular weight of C₆₀H₁₂₂ / Avogadro's number
=842/6.023x10²³
=1.4x10⁻²¹ g

∴The weight of a molecule of a compound C₆₀H₁₂₂ is 1.4x10⁻²¹ g

Sample BITSAT Chemistry Test - Question 7

The number of significant figures in 40.00 is

Sample BITSAT Chemistry Test - Question 8

Which one of the following statements is true for proteins?

Detailed Solution for Sample BITSAT Chemistry Test - Question 8
Function of Proteins:
(i) Proteins occur as food reserves as glutelin, globulin casein in milk.  

(ii) They act as antibodies and as hormones.

(iii) Proteins are the most diverse molecule on the earth.

(iv) Proteins work as hormone as insulin and glucagon.

(v) Antibiotics as gramicidin, tyrocidin and penicillin are peptides.

(vi) They are structural component of cell.

(vii) They are biological buffers.

(viii) Proteins helps in defence, movement activity of muscles, visual pigments receptor molecules, etc.

Sample BITSAT Chemistry Test - Question 9

The no. of unpaired electrons in Mn2+ is

Detailed Solution for Sample BITSAT Chemistry Test - Question 9

Sample BITSAT Chemistry Test - Question 10

The alcohol obtained by the hydrolysis of oils and fats is

Sample BITSAT Chemistry Test - Question 11

Which of the following is paramagnetic species?

Detailed Solution for Sample BITSAT Chemistry Test - Question 11
Potassium superoxide is a yellow paramagnetic solid that is air and moisture sensitive. Its formula is KO2 and because potassium is +1, the superoxide has a charge of -1. ... The unpaired electron in the pi* orbital accounts for the paramagnetic behavior or potassium superoxide.
Sample BITSAT Chemistry Test - Question 12

The heat of neutralisation is maximum when

Sample BITSAT Chemistry Test - Question 13

Reaction of acids with alcohols is also known as

Sample BITSAT Chemistry Test - Question 14

In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant will

Sample BITSAT Chemistry Test - Question 15

Which one of the following is an incorrect statement?

Detailed Solution for Sample BITSAT Chemistry Test - Question 15

The electron affinity of fluorine is unexpectedly less than that of Cl. Because the size of F atom is quite small, as a result, the electron-electron repulsions in the relatively compact 2p-subshell are comparitively large and hence the incoming electron is not accepted with the same ease as in the case with Cl.

Sample BITSAT Chemistry Test - Question 16

The rate of a gaseous reaction is given by the expression K [A][B]. If the volume of the reaction vessel is suddenly reduced to 1/4th of the initial volume, the reaction rate relating to original rate will be

Detailed Solution for Sample BITSAT Chemistry Test - Question 16
Rate = kab. When volume is reduced to 1/4th, concentrations will become 4 times.
New rate=k(4a)(4b)=16kab=16 times. 
Sample BITSAT Chemistry Test - Question 17

The no. of unpaired electrons in the complex ion [CoF₆]3⁻ is (Atomic no. Co = 27)

Sample BITSAT Chemistry Test - Question 18

The law formulated by Nernst is

Detailed Solution for Sample BITSAT Chemistry Test - Question 18
The third law was developed by the chemist Walther Nernst during the years 1906-1912. It is often referred to as Nernst’s theorem or Nernst’s postulate. Nernst proposed that the entropy of a system at absolute zero would be a well-defined constant. Instead of being 0, entropy at absolute zero could be a nonzero constant, due to the fact that a system may have degeneracy (having several ground states at the same energy).
Sample BITSAT Chemistry Test - Question 19

2, 4, 6-Trinitrophenol is a/an

Detailed Solution for Sample BITSAT Chemistry Test - Question 19
An acid is a compound which can donate the protons easily.Carboxylic acids are termed as acids because they follow the definition of acid.So, the presence of -COOH group is not the criteria for being an acid.

In case of Picric acid, the phenolic -OH group readily give away proton as the resulting phenoxide ion is significantly stabilized by delocalization over  three nitro groups situated at 2-,4-,6- positions.So it is also an acid.
Sample BITSAT Chemistry Test - Question 20

The group of elements which are not transition elements but are discussed with them is

Sample BITSAT Chemistry Test - Question 21

Which of the following is formed when K₂Cr₂O₇ reacts with concentrated H₂SO₄ in cold?

Sample BITSAT Chemistry Test - Question 22

A depolariser used in dry cell is

Detailed Solution for Sample BITSAT Chemistry Test - Question 22
A common dry cell is the zinc-carbon cell, sometimes called the dry Leclanche cell, with a nominal voltage of 1.5 volts, the same as the alkaline cell (since both use the same zinc–manganese dioxide combination).
Sample BITSAT Chemistry Test - Question 23

Marsh gas contains which of the following

Sample BITSAT Chemistry Test - Question 24

Which one of the following is explosive and readily hydrolysed?

Sample BITSAT Chemistry Test - Question 25

The first product obtained during fractional distillation of petroleum is

Sample BITSAT Chemistry Test - Question 26

Which of the following is a strong electrolyte?

Sample BITSAT Chemistry Test - Question 27

In the keto-enol tautomerism of dicarbonyl compounds the enol form is preferred in contrast to the keto form, this is due to

Sample BITSAT Chemistry Test - Question 28

Chlorobenzene on heating with aq NH₃ under pressure in the presence of cuprous chloride gives

Detailed Solution for Sample BITSAT Chemistry Test - Question 28
Aniline is an organic compound with the formula C6H5NH2. Consisting of a phenyl group attached to an amino group, aniline is the prototypical aromatic amine. Its main use is in the manufacture of precursors to polyurethane and other industrial chemicals. Like most volatile amines, it has the odor of rotten fish.
Sample BITSAT Chemistry Test - Question 29

BF₃ acts as an acid according to the concept of

Detailed Solution for Sample BITSAT Chemistry Test - Question 29
It can accept a lone pair of electrons. So it acts as Lewis acid.
Sample BITSAT Chemistry Test - Question 30

Nitrogen is relatively inactive element because

Detailed Solution for Sample BITSAT Chemistry Test - Question 30
Nitrogen is relatively inactive element because bond dissociation energy of its molecule is fairly high.
N≡N bond energy is very high 945kJmol^−1.
Sample BITSAT Chemistry Test - Question 31

A metal X on heating in nitrogen gas gives Y, Y on treatment with H₂O gives a colourless gas which when passed through CuSO₄ soution gives a blue colour. Y is

Detailed Solution for Sample BITSAT Chemistry Test - Question 31

Sample BITSAT Chemistry Test - Question 32

In diborane, the two H - B - H angles are nearly

Sample BITSAT Chemistry Test - Question 33

One of the oxidants used with liquid propellants is

Sample BITSAT Chemistry Test - Question 34

What are the products formed when an aqueous solution of magnesium bicarbonate is boiled

Sample BITSAT Chemistry Test - Question 35

Sodium is made by the electrolysis of a molten mixture of about 40% NaCl and 60% CaCl₂ because

Detailed Solution for Sample BITSAT Chemistry Test - Question 35
Its melting point is so high and it can can melt under drastic condition. So this mixture is used to lower its melting point.
Sample BITSAT Chemistry Test - Question 36

Benzene and toluene form nearly ideal solutions. At 20oC, vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20oC for a solution containing 78 g of benzene and 46 g of toluene is ___ torr.

Detailed Solution for Sample BITSAT Chemistry Test - Question 36
Since benzene and toulene form nearly ideal solutions, we can apply raoult`s law
that is the partial pressure of benzene= mole fraction * vapour pressure of benzene in pure form
=2/3*75
=50
Sample BITSAT Chemistry Test - Question 37

The turbidity of a polymer solution measures

Sample BITSAT Chemistry Test - Question 38

Which of the following has Frenkel defects?

Sample BITSAT Chemistry Test - Question 39

X litres of CO is present at STP. It is completely oxidised to CO2. The volume ofCO2 formed is 11.2 litres at STP. What is the value of X.

Detailed Solution for Sample BITSAT Chemistry Test - Question 39
The balanced cheical rection is:
CO      +     1/2O2                -------------->            CO2
                                                                       moles = 11.207/22.4 = > 1.9735 
moles of CO = is also, 1.9735 from mole – 1 analysis.
Hence volume at STP for CO is:
1.9735  = X/22.4
that is same  as  = > 11.207 litre.
Hence the same Volume will be oxidized.
Sample BITSAT Chemistry Test - Question 40

The adsorption of solids, from a solution, is called

Detailed Solution for Sample BITSAT Chemistry Test - Question 40
When a solute(here solid ) is adsorbed from a solution is called Positive adsorption vice versa when a solvent is adsorbed it is called negative adsorption.
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